Class 11 Commerce Applied Mathematics - Differentiation

Find $\displaystyle \frac{dy}{dx}$ of $ax + by^2 = \cos y$

$ax + by^2 = \cos y$
Differentiating both sides with respect to

$x$ , we obtain

$\displaystyle \frac{d}{dx} (ax) + \frac{d}{dx} (by^2) = \frac{d}{dx} (\cos y)$

$\displaystyle\Rightarrow a + b \times 2y \frac{dy}{dx} = -\sin y \frac{dy}{dx}$

$\Rightarrow \displaystyle (2by + \sin y) \frac{dy}{dx} = -a$

$\therefore \displaystyle \frac{dy}{dx} = \frac{-a}{2by + \sin y}$

Find the derivative of the following functions: $\displaystyle \sin x \cos x$

Let

$\displaystyle f\left ( x \right )= \sin x \cos x$
Thus using product rule,

$f'(x) = \sin x (-\sin x)+\cos x(\cos x)$

$=\cos^2x-\sin^2x$

$=\cos 2x$

Find $\displaystyle \frac{dy}{dx}$ of $2x + 3y = \sin y$

Given,

$2x + 3y = \sin y$

Differentiating both sides respect to x, we get,

$\Rightarrow \displaystyle \dfrac{d}{dx} (2x) + \dfrac{d}{dx} (3y) = \dfrac{d}{dx} (\sin y)$

$\Rightarrow \displaystyle 2 + 3 \dfrac{dy}{dx} = \cos y \dfrac{dy}{dx}$

$\Rightarrow \displaystyle 2 = (\cos y - 3) \dfrac{dy}{dx}$

$\therefore \displaystyle \dfrac{dy}{dx} = \dfrac{2}{\cos y - 3}$

Find $\displaystyle \frac{dy}{dx}$ of $xy + y^2 = \tan x + y$

$xy + y^2 = \tan x + y$
Differentiating both sides w.r.t

$x$ we get,

$\displaystyle\Rightarrow y+x\frac{dy}{dx}+2y\frac{dy}{dx}=\sec^2x +\frac{dy}{dx}$

$\Rightarrow \displaystyle \frac{dy}{dx}=\frac{\sec^2x-y}{x+2y-1}$

Find the derivative of the function given by $f(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)$ and hence find $f'(1).$

The given equation is

$f(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)$
Taking logarithm on both the sides, we obtain

$\log f(x) = \log (1 + x) + \log (1 + x^2) + \log (1 + x^4) + \log (1 + x^8)$
Differentiating both sides with respect to

$x$ , we obtain

$\displaystyle \frac{1}{f(x)} . \frac{d}{dx} [ f(x) ] = \frac{d}{dx} \log (1 + x) + \frac{d}{dx} \log (1 + x^2) + \frac{d}{dx} \log (1 + x^4) + \frac{d}{dx} \log (1 + x^8)$

$\Rightarrow \displaystyle \frac{1}{f(x)} . f'(x) = \frac{1}{1 + x} . \frac{d}{dx} (1 + x) + \frac{1}{1 + x^2} . \frac{d}{dx} (1 + x^2) + \frac{1}{1 + x^4} . \frac{d}{dx} (1 + x^4) + \frac{1}{1 + x^8} . \frac{d}{dx} (1 + x^8)$

$\Rightarrow \displaystyle f'(x) = f(x) \left[ \frac{1}{1 + x} + \frac{1}{1 + x^2} . 2x + \frac{1}{1 + x^4} . 4x^3 + \frac{1}{1 + x^8} . 8x^7 \right]$

$\therefore \displaystyle f'(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8) \left[ \frac{1}{1 + x} + \frac{2x}{1 + x^2} + \frac{4x^3}{1 + x^4} + \frac{8x^7}{1 + x^8} \right]$
Hence,

$\displaystyle f'(1) = (1 + 1) (1 + 1^2) (1 + 1^4) (1 + 1^8) \left[ \frac{1}{1 + 1} + \frac{2 \times 1}{1 + 1^2} + \frac{4 \times 1^3}{1 + 1^4} + \frac{8 \times 1^7}{1 + 1^8} \right]$

$\displaystyle = 2 \times 2 \times 2 \times 2 \left[ \frac{1}{2} + \frac{2 }{2} + \frac{4}{2 } + \frac{8}{2} \right]$

$\displaystyle = 16 \times \left( \frac{1 + 2 + 4 + 8}{2} \right)$

$\displaystyle = 16 \times \frac{15}{2} = 120$

Find derivative of the following functions (it is to be understood that $a, b, c, d, p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers) : $\displaystyle \left ( x^{2}+ 1 \right )\cos \, x$

Let

$f (x) = \displaystyle \left ( x^{2}+1 \right )\cos x$
Thus by product rule

$\displaystyle f'(x)=\left ( x^{2}+1 \right )\frac{d}{dx}\left ( \cos x \right )+\cos x\frac{d}{dx}\left ( x^{2}+1 \right )$

$\displaystyle = \left ( x^{2} +1\right )\left (-\sin x \right )+\cos x\left ( 2x \right )$

$=\displaystyle -x^{2}\sin x-\sin x+2x\cos x$

Differentiate the given function w.r.t. $x$ :
$\sin^3x + \cos^6 x$

Let

$y = \sin^3 x + \cos^6 x$

$\therefore \displaystyle{\frac{dy}{dx} = \frac{d}{dx} (\sin^3 x) + \frac{d}{dx} (\cos ^6 x)}$

$=\displaystyle{3 \sin^2 x.\frac{d}{dx} (\sin x) + 6\cos^5 x.\frac{d}{dx}(\cos x)}$

$= 3\sin^2 x.\cos x + 6 \cos^5x.(-\sin x)$

$= 3\sin x \cos x(\sin x - 2\cos^4 x)$

If $y = x^{x}$ , then find $\dfrac {dy}{dx}$

We have,

$y = x^{x}$

Taking

$\log$ on both the sides, we get

$\log y = x\log x$
On differentiating w.r.t.

$x$ , we get

$\dfrac {1}{y}\dfrac {dy}{dx} = \dfrac {x}{x} + \log x$

$\Rightarrow \dfrac {dy}{dx} = y + y \log x$

$\Rightarrow \dfrac {dy}{dx} = x^{x}(1 + \log x)$ ....

$(\because y = x^{x})$

Find the derivative of the following functions (it is to be understood that $a, b, c, d, p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers) : $(ax + b)\displaystyle ^{n} (cx + d)\displaystyle ^{m}$

Let

$f (x) = (ax + b)\displaystyle ^{n} (cx + d)\displaystyle ^{m}$
Thus using product rule,

$\displaystyle f'\left ( x \right )=\left ( ax+b \right )^{2}\left \{ mc\left ( cx+d \right )^{m-1} \right \}+\left ( cx+d \right )^{m}\left \{ na\left ( ax+b \right )^{n-1} \right \}$

$\displaystyle \quad = \left ( ax+b \right )^{n-1}\left ( cx+d \right )^{m-1}\left [ mc\left ( ax+b \right )+na\left ( cx+d \right ) \right ]$

If $f(x)=\tan { 2x } \tan { 3x } \tan { 5x }$ , then prove that $f'(x)=5\sec ^{ 2 }{ 5x } -3\sec ^{ 2 }{ 3x } -2\sec ^{ 2 }{ 2x }$

$\tan { 5x } =\tan { \left( 2x+3x \right) } =\cfrac { \tan { 2x } +\tan { 3x } }{ 1-\tan { 2x } \tan { 3x } }$

$\therefore f(x)=\tan { 5x } -\tan { 2x } -\tan { 3x }$

$\therefore f'(x)=5\sec ^{ 2 }{ 5x } -3\sec ^{ 2 }{ 3x } -2\sec ^{ 2 }{ 2x }$

Find the $\dfrac{dy}{dx}$ by implicit differentiation. $x^2 - 8xy + y^2 = 8$

$x^2 - 8xy + y^2 = 8$

Now differentiating both sides with respect to

$x$ we get,

$2x-8y+\left(-8x+2y\right)\dfrac{dy}{dx}=0$

or,

$\dfrac{dy}{dx}=\dfrac{8y-2x}{2y-8x}$

Find $\cfrac { dy }{ dx }$ where ${ x }^{ 3 }+{ y }^{ 3 }+3xy=7$ .

${ x }^{ 3 }+{ y }^{ 3 }+3xy=7\\$

Differentiate on both sides

$\Longrightarrow 3{ x }^{ 2 }dx+3{ y }^{ 2 }dy+3xdy+3ydx=0\\ \Longrightarrow (3{ x }^{ 2 }+3y)dx+(3{ y }^{ 2 }+3x)dy=0$

$\Longrightarrow \dfrac { dy }{ dx } =-\dfrac { { x }^{ 2 }+y }{ { y }^{ 2 }+x }$

Differentiable $\log_7(log x)$ with respect to $x$ .

$\gamma =\log _{ 7 }{ (\log { x } ) } =\cfrac { \log _{ 10 }{ (\log { x } ) } }{ \log _{ 10 }{ 7 } } \quad \quad (\log _{ m }{ a } =\cfrac { \log _{ n }{ a } }{ \log _{ n }{ m } } )\\ \cfrac { dy }{ dx } =\cfrac { 1 }{ \log _{ 10 }{ 7 } } \cfrac { d }{ dx } (\log { (\log { x } ) } )\\ =\cfrac { 1 }{ \log _{ 10 }{ 7 } } \cfrac { 1 }{ \log _{ 10 }{ x } } .\cfrac { 1 }{ x } \\ \cfrac { dy }{ dx } =\cfrac { 1 }{ x.\log _{ 10 }{ 7 } .\log _{ 10 }{ x } }$

Find the value of $f'(e)$ where $f(x)=\log_e(\log_ex)$ .

Given,

$f(x)=\log_e(\log_ex)$ .

Now,

$f'(x)=\dfrac{1}{x\log_ex}$ .

Then

$f'(e)=\dfrac{1}{e}$ .

$y = \sqrt{log \,x+\sqrt{log \,x + \sqrt{log \,x}}} +$ _____ $\infty$ show that $\dfrac{dy}{dx} = \dfrac{1}{x(2y - 1)}$ .

Given,

$y = \sqrt{log \,x+\sqrt{log \,x + \sqrt{log \,x}}} +$ _____

$\infty$ .......(1)

Squaring both sides we get,

$y^2=\log x+y$ [ Using (1)]

Differentiating both sides we get,

$(2y-1)\dfrac{dy}{dx}=\dfrac{1}{x}$

or,

$\dfrac{dy}{dx}=\dfrac{1}{x(2y-1)}$

If $\sqrt{x}+\sqrt{y}=4$ then find $\dfrac{dy}{dx}$ at $x=1$ .

Given,

$\sqrt{x}+\sqrt{y}=4$ .......(1).

Now differentiating both sides with respect to

$x$ we get,

$\dfrac{1}{2\sqrt{x}}+\dfrac{1}{2\sqrt{y}}\dfrac{dy}{dx}=0$

or,

$\dfrac{dy}{dx}=-\dfrac{\sqrt{y}}{\sqrt{x}}$

or,

$\left.\dfrac{dy}{dx}\right|_{x=1}=-\dfrac{3}{1}=-3$ [ When

$x=1$ then form (1) we get

$\sqrt{y}=3$ .]

Find $\dfrac{dy}{dx}$ :

$x+y^2=\log y + x^2$

$x+y^2=\log y +x^2$

Differentiating w.r.t. x, we get,

$1+2y\dfrac{dy}{dx}=\dfrac{1}{y}.\dfrac{dy}{dx}+2x$

$\dfrac{dy}{dx}(2y-\dfrac{1}{y})=2x-1$

$\dfrac{dy}{dx}(\dfrac{2y^2-1}{y})=2x-1$

$\dfrac{dy}{dx}=\dfrac{y(2x-1)}{2y^2-1}$

If $f(x)= \left| x-2 \right| +\left| x-5 \right|$ then find $f'(4)$ .

$\begin{array}{l} f'\left( x \right) \, =\, \left| { x-2 } \right| +\left| { x-5 } \right| \\ \, \, \, \, \, \, \, \, \, \, \, \, =\left\{ \begin{array}{l} -\left( { x-2 } \right) -\left( { x-5 } \right) \, ,x<2 \\ x-2-\left( { x-5 } \right) \, ,2<x<5 \\ x-2+x-5\, ,\, x<5 \end{array} \right. \\ \, \, \, \, \, \, \, \, \, \, \, =\left\{ \begin{array}{l} -x+2-x+5\, ,\, \, x<2 \\ x-2-x+5\, ,\, \, \, \, \, 2<x<5 \\ 2x-7\, ,\, \, \, \, \, x>5 \end{array} \right. \\ \, \, \, \, \, \, \, \, \, \, \, =\left\{ \begin{array}{l} -2x+7\, \, \, ,\, \, \, \, \, x<2 \\ 3\, ,\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, 2<x<5 \\ 2x-7\, \, \, ,\, \, \, \, \, \, \, \, x>5 \end{array} \right. \\ f'\left( x \right) \, =\left\{ \begin{array}{l} -2\, \, \, ,\, \, \, \, \, x<2 \\ 0\, \, \, \, \, \, ,\, \, \, \, \, 2<x<5 \\ 2\, \, \, \, \, \, ,\, \, \, \, \, \, \, \, x>5 \end{array} \right. \\ So,\, \, f'\left( 4 \right) =0 \end{array}$

If $y=3x^{2}$ ,Find $\dfrac {dy}{dx}=?$

Consider given the equation,

Differentiate both sides with respect to $x$

$\dfrac{dy}{dx}=6x$

Hence, this is the answer.

If $e^{y}+xy=e$ , find $y''(0)$

Given

$e^{y}+x{y}=e$ ............

$(1)$

when

$x=0$ ,equation

$(1)$ becomes

$\\e^y=e\implies y=1$

Differentiating (

$1)$ on both sides

$\implies e^{y}{\dfrac{dy}{dx}}+y+x{\dfrac{dy}{dx}}=0\\\implies y'=\dfrac{-y}{e^{y}+x}\\\implies y'(0)=\dfrac{-1}{e^{1}+0}=\dfrac{-1}{e}$

$y''=-\dfrac{(e^{y}+x)\dfrac{dy}{dx}-y(e^{y}\dfrac{dy}{dx}+1)}{(e^{y}+x)^{2}}\\\implies y''(0)=\dfrac{(e^{1}+0)(\dfrac{-1}{e})-(1)({e^{1}}\times \dfrac{-1}{e}+1)}{(e^{1}+0)^{2}}=\dfrac{1}{e^{2}}$

Find $\dfrac{dy}{dx}$ when $x^2 + y^2 = c^2$ .

Given,

$x^2 + y^2 = c^2$

Now, differentiating both sides with respect to

$x$ we get,

$2x+2y\dfrac{dy}{dx}=0$

or,

$\dfrac{dy}{dx}=-\dfrac{x}{y}$

$\dfrac{{dy}}{{dx}}\, + \,\dfrac{y}{x}\,\,{\rm I}n\,y\, = \,\dfrac{y}{{{x^2}}}{\left( {{\rm I}n\,y} \right)^2}$

${{dy} \over {dx}} + {y \over x}{\mathop{\rm l}\nolimits} ny = {y \over {{x^2}}}{\left( {lny} \right)^2}$

${1 \over {y{{\left( {\ln y} \right)}^2}}}{{dy} \over {dx}} + {1 \over {x\left( {lny} \right)}} = {1 \over {{x^2}}}$

$put\,{1 \over {lny}} = t$

$- {1 \over {{{\left( {lny} \right)}^2}}} \times {1 \over y}{{dy} \over {dx}} = {{dt} \over {dt}}$

$- {{dt} \over {dx}} + {t \over x} = {1 \over {{x^2}}}$

${{dt} \over {dx}} - {t \over x} = - {1 \over {{x^2}}}$

$I.F. = {e^{\smallint - {1 \over x}dn}} = {e^{ - \log x}} = {1 \over x}$

$so{l^n}\,is\,t \times {1 \over x} = \smallint - {1 \over {{x^2}}} \times {1 \over x}dn + c$

$t \times {1 \over x} = {1 \over {2{x^2}}} + c$

${1 \over {x{\mathop{\rm l}\nolimits} ny}} = {1 \over {2{x^2}}} + c$

Differentiate: $\sqrt { \dfrac { \left( x-3 \right) \left( { x }^{ 2 }+4 \right) }{ { 3 }x^{ 2 }+4x+5 } }$

$y = \,\sqrt {\cfrac{{\left( {x - 3} \right)\left( {{x^2} + 4} \right)}}{{3{x^2} + 4x + 5}}}$

Taking

${log}$ on both sides

$\log y = \cfrac{1}{2}\log \left[ {\cfrac{{\left( {x - 3} \right)\left( {{x^2} + 4} \right)}}{{3{x^2} + 4x + 5}}} \right]$

$\Rightarrow \log y = \cfrac{1}{2}\left[ {\log \left( {x - 3} \right) + \log \left( {{x^2} + 4} \right) - \log \left( {3{x^2} + 4x + 5} \right)} \right]$

Differentiating w.r.t.x

$\cfrac{1}{y}\cfrac{{dy}}{{dx}} = \cfrac{1}{2}\left[ {\cfrac{1}{{x - 3}} + \cfrac{1}{{{x^2} + 4}} \times 2x-\cfrac{{ 1\left( {6x + 4} \right)}}{{3{x^2} + 4x + 5}}} \right]$

$\Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{1}{2}\sqrt {\cfrac{{\left( {x - 3} \right)\left( {{x^2} + 4} \right)}}{{3{x^2} + 4x + 5}}} \left[ {\cfrac{1}{{x - 3}} + \cfrac{{2x}}{{{x^2} + 4}} - \cfrac{{6x + 4}}{{3{x^2} + 4x + 5}}} \right]$

If $\sin y = x\sin (a + y)$ , then show
$\dfrac{dy}{dx}=\dfrac{\sin \left( a+y \right)}{\cos y-x.\cos \left( a+y \right)}$

Given ,

$\sin y=x.\sin \left( a+y \right)$

$Differentiate\,\,with\,respect\,\,to\,y\,\,we\,get$

$\cos y=x\cos \left( a+y \right)+\sin \left( a+y \right).\dfrac{dx}{dy}$

$\dfrac{dx}{dy}=\dfrac{\cos y-x\cos \left( a+y \right)}{\sin \left( a+y \right)}$

$\dfrac{dy}{dx}=\dfrac{\sin \left( a+y \right)}{\cos y-x.\cos \left( a+y \right)}$

Find $\dfrac{{dy}}{{dx}},x = a\left( {\cos t + \log \tan \frac{t}{2}} \right),y = a\sin t$

$x = a(\cos t + \log \tan \frac{t}{2})$ ; $$y = a\sin t$

$\frac{{dx}}{{dt}} = a\left( { - \sin t + \frac{1}{{\tan t/2}} \times {{\sec }^2}t/2 \times \frac{1}{2}} \right)$

$= a( - \sin t + \frac{1}{2}\frac{{\cos t/2}}{{\sin t/2{{\cos }^{2t/2}}}})$

$= a( - \sin t + \frac{1}{{\sin t}})$

$= a\left( {\frac{{{{\cos }^2}t}}{{\sin t}}} \right)$

$\frac{{dy}}{{dx}} = a\cos t$

$\therefore \frac{{dy}}{{dx}} = \frac{{a\cos t}}{{a(\frac{{\cos 2t}}{{\sin t}})}}$

$= \frac{{\sin t}}{{\cos t}} = \tan t$

If $e^x + e^y = e^{x+y}$ , find $\dfrac{dy}{dx}$ .

Given,

$e^x + e^y = e^{x+y}$

Now differentiating both sides with respect to

$x$ we get,

$e^x+e^y\dfrac{dy}{dx}=e^{x+y}\left(1+\dfrac{dy}{dx}\right)$

or,

$e^x-e^{x+y}=(e^{x+y}-e^y)\dfrac{dy}{dx}$

or,

$\dfrac{dy}{dx}=e^{x-y}\dfrac{1-e^y}{e^x-1}$

or,

$\dfrac{dy}{dx}=-e^{x-y}\dfrac{1-e^y}{1-e^x}$

Find $\dfrac{dy}{dx}$ for $y=e^{\sqrt{x}}$ .

Given,

$y=e^{\sqrt{x}}$ .

Now differentiating both sides with respect to

$x$ we get,

$\dfrac{dy}{dx}=e^{\sqrt{x}}.\dfrac{1}{2\sqrt{x}}$

Differentiate ${\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + {x^2}} - 1}}{x}} \right)$ with respect to ${\sin ^{ - 1}}x$

$\cfrac{dtan^{-1}(\cfrac{\sqrt{1+x^2} -1}{x})}{dsin^{-1}x}$

$\cfrac{dtan^{-1}(\cfrac{\sqrt{1+x^2} -1}{x})}{dsin^{-1}x} \cfrac{dx}{dx}$

$\cfrac{dtan^{-1}(\cfrac{\sqrt{1+x^2} -1}{x})}{dx} \cfrac{dx}{dsin^{-1}x}$

$\cfrac{dtan^{-1}(\cfrac{\sqrt{1+x^2} -1}{x})}{dx} = \cfrac{1}{(\cfrac{\sqrt{1+x^2}-1}{x})^2+1}*\cfrac{\cfrac{x^2}{\sqrt{1+x^2}}-\sqrt{1+x^2}-1}{x^2}$

$\implies \cfrac{dtan^{-1}(\cfrac{\sqrt{1+x^2} -1}{x})}{dx} = \cfrac{\sqrt{x^2+1}-1}{(2((x^2+1)^\cfrac{3}{2} )-x^2-1)}$

$\cfrac{dsin^{-1}x}{dx} = \cfrac{1}{\sqrt{1-x^2}}$

Hence we get,

$\cfrac{dtan^{-1}(\cfrac{\sqrt{1+x^2} -1}{x})}{dx} \cfrac{dx}{dsin^{-1}x} =\cfrac{(\sqrt{x^2+1}-1)(\sqrt{1-x^2})}{\big(2((x^2+1)^\cfrac{3}{2}\big) -x^2-1)}$

Differentiate with respect to $x$ .
$y = x\tan x$

$\begin{array}{l}y = x\tan x\\\frac{{dy}}{{dx}} = x{\sec ^2}x + \tan x\end{array}$

Differentiate the following w.r.t.x:
$5^x\cdot \sec^{-1}2x$

$y=5^x\sec^{-1} 2x$

$\dfrac{dy}{dx} = 5x \dfrac{d}{dx} \sec6{-1} 2x + \sec^{-1}2x \dfrac{d}{dx} (5^x)$

$=5^x\cdot \dfrac{1}{|2x|\sqrt{4x^2-1}} (2) + sec^{-1} 2x\cdot 5^x ln 5$

$=5^x \left(\dfrac{1}{|x|\sqrt{4x^2-1}} + \sec^{-1}2x ln\, 5\right)$

Differentiate $x = \tan (e^{-y})$

$x=\tan (e^{-y})$

$\dfrac{dx}{dx}=-e^{-y}\sec ^2(e^{-y})\dfrac{dy}{dx}$

$1=-e^{-y}\sec ^2(e^{-y})\dfrac{dy}{dx}$

$\dfrac{dy}{dx}=\dfrac{-1}{e^{-y}\sec ^2(e^{-y})}$

Find $\dfrac {dy}{dx}$
$\dfrac{{dx}}{{dt}} = ap\cos pt,\dfrac{{dy}}{{dt}} = - bp\sin pt$

We have,

$\dfrac{dx}{dt}=ap\cos pt$ ……. (1)

$\dfrac{dy}{dt}=-bp\sin pt$ …….. (2)

On dividing equation (2) and (1), we get

$\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{-bp\sin pt}{ap\cos pt}$

$\dfrac{dy}{dx}=-\dfrac{b\tan pt}{a}$

Hence, this is the answer.

Differentiate the following w.r.t.x:
$\sin^{-1}x+\cos^{-1}x$

Let

$y=\sin^{-1}x+\cos^{-1}x$

We know,

$\sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2}$

Therefore,

$y=\dfrac{\pi}{2}$

Hence,

$\dfrac{dy}{dx}=0$

If $y = {3^{2\log x + 7}}$ then $\dfrac{{dy}}{{dx}}$

$y=3^{2\log x+7}$

$=3^{7}.3^{2\log x}$

$=3^{7}.3^{\log x^{2}}$

$\dfrac{dy}{dx}=3^{7}.3^{\log x^{2}}.\dfrac{d}{dx}(\log x^{2}_.\log 3$

$=3^{7}.3^{\log x^{2}}.\dfrac{2x}{x^{2}}.\log 3$

$=\dfrac{2}{x}.3^{7}.3^{\log x^{2}}.\log 3$

$\dfrac{dy}{dx}=\dfrac{2\log 3}{x}3^{7+\log x^{2}}$

Differentiate the following function:
${x}^{\log{x}}+{(\log{x})}^{x}$

$y=x^{\log x}+(\log x)^x$

$u=x^{\log x}\Rightarrow \log u+=(\log x)^2$

$\Rightarrow \dfrac{1}{u}\dfrac{du}{dx}=\dfrac{2\log (x)}{x}$

$\Rightarrow \dfrac{du}{dx}=\dfrac{2(\log x)x}{x}$

$v=(\log x)^x\Rightarrow \log v=x \log (\log x)$

$\Rightarrow \dfrac{1}{v}\dfrac{dv}{dx}=\dfrac{x}{x\log x}+\log (\log x)$

$\Rightarrow \dfrac{dy}{dx}=\left(\log x)^x\left(\dfrac{1}{\log x}+\log(\log x\right)\right)$

$\therefore \dfrac{dy}{dx}=\dfrac{dv}{dx}+\dfrac{du}{dx}$

$=\dfrac{2(\log x)^{\log x}}{x}+(\log x)^x\left(\dfrac{1}{\log x}+\log (\log x)\right)$

1062341_1090909_ans_ae33013ab1dd482793a0db207a01113b.png

If $y=\sin {\sqrt {1-x^{2}}}$ , then find $\dfrac{dy}{dx}$ .

We have,

$y=\sin \sqrt{1-{{x}^{2}}}$

On differentiating w.r.t $x$ , we get

$\dfrac{dy}{dx}=\cos \sqrt{1-{{x}^{2}}}\left( \dfrac{1}{2\sqrt{1-{{x}^{2}}}}\times \left( 0-2x \right) \right)$

$\dfrac{dy}{dx}=\dfrac{-x\cos \sqrt{1-{{x}^{2}}}}{\sqrt{1-{{x}^{2}}}}$

Hence, this is the answer.

Find:
$\dfrac{d}{{dx}}\left\{ {\dfrac{{5 + 4x}}{{x + 3}}} \right\}$

$\cfrac{d}{{dx}}\left\{ {\cfrac{{5 + 4x}}{{x + 3}}} \right\}$

$= \cfrac{{\left( {x + 3} \right)\cfrac{d}{{dx}}\left( {5 + 4x} \right) - \left( {5 + 4x} \right)\cfrac{d}{{dx}}\left( {x + 3} \right)}}{{{{\left( {x + 3} \right)}^2}}}$

$= \cfrac{{\left( {x + 3} \right).4 - \left( {5 + 4x} \right).1}}{{{{\left( {x + 3} \right)}^2}}} = \cfrac{{4x + 12 - 5 - 4x}}{{{{\left( {x + 3} \right)}^2}}}$

$= \cfrac{7}{{{{\left( {x + 3} \right)}^2}}}$

If $y = {a^{\dfrac{1}{2}{{\log }_a}\cos x}}.$ Find $\dfrac{{dy}}{{dx}}$

$\displaystyle y = {a^{\dfrac{1}{2}{{\log }_a}\cos x}}$

$\displaystyle = a^{log}a\sqrt{cosx}$

$\displaystyle = \sqrt{cosx}$

$\displaystyle \therefore \frac{dy}{dx} = \frac{d}{dx}(\sqrt{cosx}) = \frac{1}{2\sqrt{cosx}} (-sinx)$

$\displaystyle \frac{dy}{dx} = \frac{-1}{2}\frac{sinx}{\sqrt{cosx}}$

Find the derivative of $f\left ( x \right )=1+x+x^{2}+--+x^{50}atx=1$ .

$f{\left( x \right)} = 1 + x + {x}^{2} + ....... + {x}^{50}$

For

$f'{\left( x \right)}$ at

$x = 1$

$f'{\left( x \right)} = 0 + 1 + 2x + 3{x}^{2} + ..... + 50 {x}^{49}$

$x = 1$ ,

$f'{\left( 1 \right)} = 1 + 2 \left( 1 \right) + 3 {\left( 1 \right)}^{2} + ..... + 50 {\left( 1 \right)}^{49}$

$\Rightarrow f'{\left( 1 \right)} = 1 + 2 + 3 + ..... + 50$

As we know that sum of

$n$ natural numbers is given as-

$S = \dfrac{n \left( n + 1 \right)}{2}$

$\therefore f'{\left( 1 \right)} = \dfrac{50 \left( 50 + 1 \right)}{2} = 1275$

Hence the value of

$f'{\left( x \right)}$ at

$x = 1$ is

$1275$ .

Find the derivate of $\ {e^{\sqrt {2x + 1} }}$ with respect to $x$ at $x=12$

$\dfrac{d}{dx}\left({e}^{\sqrt{2x+1}}\right)$ at

$x=12$

$\dfrac{d}{dx}\left({e}^{\sqrt{2x+1}}\right)\quad =\left({e}^{\sqrt{2x+1}}\right).\dfrac{1}{2\sqrt{2x+1}}.2$

$=\dfrac { { e }^{ \sqrt { 2x+1 } } }{ \sqrt { 2x+1 } }$

$x=12,\quad \dfrac { { e }^{ \sqrt { 2x+1 } } }{ \sqrt { 2x+1 } }=\dfrac{{e}^{\sqrt{25}}}{\sqrt{25}}$

$={\dfrac{1}{5}}{e}^{5}$

If $y = \sqrt x + \dfrac{1}{{\sqrt x }},$ then show that $2x\dfrac{{dy}}{{dx}} + y = 2\sqrt x$

$y = \sqrt{x}+\dfrac{1}{\sqrt{x}}$

$\dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x}}+ \dfrac{\dfrac{-1}{2\sqrt{x}}}{x}$

$= \dfrac{1}{2\sqrt{x}}-\dfrac{1}{2x\sqrt{x}}$

$2x\dfrac{dy}{dx} = 2x(\dfrac{1}{2\sqrt{x}}-\dfrac{1}{2x\sqrt{x}})$

$= 2(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}) = \sqrt{x} -\dfrac{1}{\sqrt{x}}$

$\therefore 2x\dfrac{dy}{dx}+y = \sqrt{x}-\dfrac{1}{\sqrt{x}}+\sqrt{x}+\dfrac{1}{\sqrt{x}}$

$= 2\sqrt{x}$

$\therefore \boxed{2x\dfrac{dy}{dx}+y = 2\sqrt{x}.}$

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Find the derivative of $sin^{2}x$ with respect to x using product rule.

$\dfrac{d}{dx} \left( \sin^{2}{x} \right)$

$= \dfrac{d}{dx} \left( \sin{x} . \sin{x} \right)$

Applying product rule, we have

$\dfrac{d}{dx} \left( \sin{x} . \sin{x} \right)$

$= \sin{x} \; \cfrac{d}{dx} \left( \sin{x} \right) + \dfrac{d}{dx} \left( \sin{x} \right) \; \sin{x}$

$= \sin \; \cos{x} + \cos{x} \; \sin{x} \; \left( \because \dfrac{d}{dx} \left( \sin{x} \right) = \cos{x} \right)$

$= 2 \sin{x} \cos{x}$

$= \sin{2x} \; \left( \because 2 \sin{x} \cos{x} = \sin{2x} \right)$

Therefore,

$\dfrac{d}{dx} \left( \sin^{2}{x} \right) = \sin{2x}$

Let $y=\log(\sin x)$ find $\dfrac{d^2y}{dx^2}$ .

Given,

$y=\log(\sin x)$

Now differentiating both sides with respect to

$x$ we get,

$\dfrac{dy}{dx}=\dfrac{\cos x}{\sin x}$

or,

$\dfrac{dy}{dx}=\cot x$

Now again differentiating both sides with respect to

$x$ we get,

$\dfrac{d^2y}{dx^2}=-\ cosec^2 x$ .

Find $\dfrac{dy}{dx}$ , if $x\ \sin\ y+y\ \sin\ x=0$ .

$x \sin{y} + y \sin{x} = 0$

Differentiating above equation w.r.t.

$x$ , we have

$\left( 1 . \sin{y} + \cos{y} \cfrac{dy}{dx}.x \right) + \left( \cfrac{dy}{dx} \sin{x} + \cos{x} . y \right) = 0$

$\Rightarrow \cfrac{dy}{dx} \left( x \cos{y} + \sin{x} \right) = - \left( y \cos{x} + \sin{y} \right)$

$\Rightarrow \cfrac{dy}{dx} = - \cfrac{\left( y \cos{x} + \sin{y} \right)}{\left( x \cos{y} + \sin{x} \right)}$

$y=x^3-x^2-x+7$ find x =? when $\dfrac{dy}{dx}=0$

Given

$y=x^3-x^2-x+7$

Differentiating it w.r.t x

$\dfrac{dy}{dx}=3x^2-2x-1$

$\dfrac{dy}{dx}=0$ [Given]

$3x^2-2x-1=0$

$3x^2-3x+x-1=0$

$3x(x-1)+1(x-1)=0$

$(3x+1)(x-1)=0$

$x=1,\ \dfrac{-1}{3}$

Differentiate $\log(sin\ x)(log\ x)$

Let

$y = \log{\left( \sin{x} \right)} \log{x}$

Differentiating above equation by Lebinitz rule, we have

$\cfrac{dy}{dx} = \left( \cfrac{1}{\sin{x}} . \cos{x} \right) \log{x} + \cfrac{1}{x} \log{\left( \sin{x} \right)}$

$\Rightarrow \cfrac{dy}{dx} = \cfrac{\cos{x}}{\sin{x}} \log{x} + \cfrac{\log{\left( \sin{x} \right)}}{x}$

$\Rightarrow \cfrac{dy}{dx} = \tan{x} \; \log{x} + \cfrac{\log{\left( \sin{x} \right)}}{x}$

$\Rightarrow \cfrac{dy}{dx} = \log{\left( {x}^{\tan{x}} \right)} + \log{\left( {\sin{x}}^{\frac{1}{x}} \right)}$

$\Rightarrow \cfrac{dy}{dx} = \log{\left( \left( {x}^{\tan{x}} \right) \left( {\sin{x}}^{\frac{1}{x}} \right) \right)}$

Hence

$y' = \log{\left( \left( {x}^{\tan{x}} \right) \left( {\sin{x}}^{\frac{1}{x}} \right) \right)}$

Differentiate: ${\sin}^{2}{3x}.{\tan}^{3}{2x}$

Let

$y = sin^2 3x \, tan^3 2x$

$\dfrac{dy}{dx} = 2sin 3x . cos 3x . 3 \, tan^3 2x + sin^2 3x . 3tan^2 2x .sec^2 2x.2$

$\Rightarrow \dfrac{dy}{dx} = 3sin 6x \, tan^3 2x + 6 sin^3 3x \, tan^2 2x sec^2 2x$

Find $\dfrac {dy}{dx}$ for
$y= \left( \sin 20x+{{a}^{2x}}+10 \right) \\$

Consider given the differentiation,

$\Rightarrow \dfrac{d}{dx}\left( \sin 20x+{{a}^{2x}}+10 \right) \\$

$\Rightarrow 20.\cos 20x+2.{{a}^{2x}}.{{\log }_{e}}a+0 \\$

$\Rightarrow 20.\cos 20x+2.{{a}^{2x}}.{{\log }_{e}}a \\$

Hence,this is the answer,

If $x\sqrt { 1-{ y }^{ 2 } } +y\sqrt { 1-{ x }^{ 2 } } =1$ , prove that $\dfrac { dy }{ dx } =-\sqrt { \frac { { 1-y }^{ 2 } }{ { 1-x }^{ 2 } } . }$

$x\sqrt{1-y^2}+y\sqrt{1-x^2}=1$

Differentiating throughout with respect to x we get

$x\dfrac{1}{2\sqrt{1-y^2}}\cdot (-2y)\dfrac{dy}{dx}+\sqrt{1-y^2}+y\dfrac{1}{2\sqrt{1-x^2}}\cdot (-2x)+\sqrt{1-x^2}\dfrac{dy}{dx}=0$

$\Rightarrow \dfrac{-xy}{\sqrt{1-y^2}}\dfrac{dy}{dx}+\sqrt{1-y^2}+\dfrac{(-xy)}{\sqrt{1-x^2}}+\sqrt{1-x^2}\dfrac{dy}{dx}=0$

$\Rightarrow \left(\dfrac{-xy}{\sqrt{1-y^2}}+\sqrt{1-x^2}\right)\dfrac{dy}{dx}=\dfrac{dy}{\sqrt{1-x^2}}-\sqrt{1-y^2}$

$\Rightarrow \left(\dfrac{-xy+\sqrt{(1-x^2)(1-y^2)}}{\sqrt{1-y^2}}\right)\dfrac{dy}{dx}=\left(\dfrac{xy-\sqrt{(1-x^2)(1-y^2)}}{\sqrt{1-x^2}}\right)$

$\Rightarrow \dfrac{-(xy-\sqrt{(1-x^2)(1-y^2)})}{\sqrt{1-y^2}}$

$\dfrac{dy}{dx}$

$=\dfrac{(xy-\sqrt{(1-x^2)(1-y^2)})}{\sqrt{1-x^2}}$

$\Rightarrow \dfrac{dy}{dx}=-\dfrac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$

Hence

$\dfrac{dy}{dx}=-\sqrt{\dfrac{1-y^2}{1-x^2}}$ .

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$y=x^{3}-3x+2$
Find $\frac{dy}{dx}$ if the given function is continuous.

$y={ x }^{ 3 }-3x+2$

$\dfrac { dy }{ dx } ={ 3x }^{ 2 }-3$

If ${x^y} + {y^x} = {a^b}$ then show that $\dfrac{{dy}}{{dx}} = - \left[ {\dfrac{{y{x^{y - 1}} + {y^x}\log y}}{{{x^y}\log x + x{y^{x - 1}}}}} \right]$

$x^y+y^x=a^b$

$e^{y ln x}+e^{x ln y}=a^b$

Differentiating w.r.t

$x$

$x^y\left[\dfrac{dy}{dx}ln x+\dfrac{y}{x}\right]+y^x\left[ln y+\dfrac{x}{y}\dfrac{dy}{dx}\right]=0$

$\dfrac{dy}{dx}\left[x^yln x+y^{x-1}x\right]=-\left[y^xln y+yx^{y-1}\right]$

$\dfrac{dy}{dx}=-\dfrac{-y^xln y+yx^{y-1}}{x^yln x+xy^{x-1}}$ .

Find $\dfrac{dy}{dx}$ if $xy=e^{x-y}$

$xy=e^{x-y}$ (Differentiating both sides)

$y+xyy^\backprime =e^{x-y}(1-y^\backprime) \\y+xyy^\backprime =e^{x-y}-e^{x-y}\times y^\backprime \\y^\backprime (xy+e^{x-y})=e^{x-y}-y\\y^\backprime =\cfrac{e^{x-y}-y}{xy+e^{x-y}}$

${\left( {{x^2} + {y^2}} \right)^2} = xy,$ then $\dfrac{dy}{dx}$

We have,

$(x^2+y^2)^2=xy$

On differentiating w.r.t

$x$ , we get

$\dfrac{d}{dx}(x^2+y^2)^2=\dfrac{d}{dx}(xy)$

$2(x^2+y^2)\dfrac{d}{dx}(x^2+y^2)=x\dfrac{d}{dx}(y)+y\dfrac{d}{dx}(x)$

$2(x^2+y^2)(2x+2y\dfrac{dy}{dx})=x\dfrac{dy}{dx}+y\times 1$

$4(x^2+y^2)(x+y\dfrac{dy}{dx})=x\dfrac{dy}{dx}+y$

$4x(x^2+y^2)+4xy(x^2+y^2)\dfrac{dy}{dx}=x\dfrac{dy}{dx}+y$

$4xy(x^2+y^2)\dfrac{dy}{dx}-x\dfrac{dy}{dx}=y-4x(x^2+y^2)$

$\left[4xy(x^2+y^2)-x\right]\dfrac{dy}{dx}=y-4x(x^2+y^2)$

$\dfrac{dy}{dx}=\dfrac{y-4x(x^2+y^2)}{[4xy(x^2+y^2)-x]}$

Hence, this is the answer.

Find derivative of $f(x)=x \cos x$

We have,

$\begin{array}{l} f\left( x \right) =x\cos x \\ f'\left( x \right) =\cos x-x\sin x \end{array}$

Hence, this is the answer.

If $y\sqrt { { x }^{ 2 }+1 } =log\{ \sqrt { { x }^{ 2 }+1 } -x\} ,$ show that ${ (x }^{ 2 }+1)\dfrac { dy }{ dx } +xy+1=0.$

$\sqrt{x^{2}+1}= log \left \{ \sqrt{x^{2}+1}-x \right \}$

Differentiating with reset to be x we get

$\sqrt{x^{2}+1}\frac{dy}{dx}+\frac{1}{2\sqrt{x^{2}+1}}.2x.y=\frac{1}{\sqrt{x^{2}+1}-x}\left \{ \frac{1}{2\sqrt{x^{2}+1}}2x-1 \right \}$

$\sqrt{x^{2}}\frac{dy}{dx}+\frac{xy}{\sqrt{x^{2}+1}}=\frac{1}{\sqrt{x^{2}+1}-x}\left \{ \frac{x}{\sqrt{x^{2}+1}-1} \right \}$

$\Rightarrow \sqrt{x^{2}+1}\frac{dy}{dx}+\frac{xy}{\sqrt{x^{2}+1}}=\frac{1}{\sqrt{x^{2}+1}-x}\left \{ \frac{x-\sqrt{x^{2}+1}}{\sqrt{x^{2}+1}} \right \}$

$\Rightarrow \sqrt{x^{2}+1}\frac{dy}{dx}+\frac{xy}{\sqrt{x^{2}+1}}=\frac{-1}{\sqrt{x^{2}+1}}$

$\Rightarrow (x^{2}+1)\frac{dy}{dx}+xy=-1$

$\Rightarrow (x^{2}+1)\frac{dy}{dx}+xy+1=0.$

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Find f'(-3) using definition if $f\left( x \right) =\frac { 2x }{ x-3 }$

$\begin{aligned} f ( -3 ) = & \frac { 2 \cdot ( - 3 ) } { ( - 3 ) - 3 } \\ & = \frac { - 6 } { - 6 } \\ & = 1 \end{aligned}$

Find $\dfrac { d }{ dx } { \sin }^{ 3 }4x{ \cos }^{ 8 }5x$

$\dfrac{d}{dx}(\sin^34x\cos^85x)$

Let

$u=\sin^34x$ and

$v=\cos^85x$

$\therefore \dfrac{d}{dx}(\sin^34x\cos^85x)=\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$

$=\sin^34x\cdot \dfrac{d}{dx}(\cos^85x)+\cos^85x\dfrac{d}{dx}(\sin^34x)$ …..

$(1)$

Here

$\dfrac{d}{dx}(\cos^85x)=\dfrac{d}{d(\cos 5x)}\{(\cos 5x)^8\}\cdot \dfrac{d(\cos 5x)}{dx}$

$=8(\cos 5x)^7\cdot \dfrac{d}{d(5x)}(\cos (5x))=\dfrac{d(5x)}{dx}$

$=8\cos^{-1}5x\cdot (-\sin 5x)\cdot 5\dfrac{dx}{dx}$

$[\because \dfrac{d(\cos x)}{dx}=-\sin x]$

$=-40\sin 5x\cos^75x$ .

and

$\dfrac{d}{dx}(\sin^34x)=\dfrac{d}{d(\sin 4x)}\{(\sin 4x)^3\}.\dfrac{d(\sin 4x)}{dx}$

$=3\sin^24x\cdot \dfrac{d(\sin 4x)}{d(4x)}\cdot \dfrac{d(4x)}{dx}$

$=3\sin^24x\cos 4x 4\dfrac{dx}{dx}$

$[\because \dfrac{d}{dx}(\sin x)=\cos x]$

$=12\sin^24x\cos 4x$

From

$(1)$ ,

$\dfrac{d}{dx}\sin^34x\cos^85x=\sin^34x.(-40\sin 5x\cos^75x)+\cos^85x(12\sin^24x\cos 4x)$

$=-40\sin^34x\sin 5x\cos^75x+12\cos^85x\sin^24x\cos 4x$

$=4(-10\sin^34x\sin 5x\cos^75x+3\sin^24x\cos 4x\cos^85x)$

$=4\cos^75x\cdot \sin^24x(3\cos 4x\cos 5x-10\sin 4x\sin 5x)$

$=4\cos^75x\sin^24x\left(\dfrac{3\cos (4x+5x)+\cos(4x-5x)}{2}-10\times \dfrac{\cos (4x-5x)-\cos (4x+5x)}{2}\right)$

$=4\cos^75x\sin^24x\left(\dfrac{3\cos 9x+3\cos (-x)-10\cos (-x)+\cos 9x}{2}\right)$

$=2\cos^75x\sin^24x(4\cos 9x-7\cos x)$

$[\because \cos (-x)=\cos x]$

$\therefore \dfrac{d}{dx}(\sin^34x\cos^85x)=2\cos^75x\sin^24x(4\cos 9x-7\cos x)$ .

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Differentiate $xe^x$ from first principles.

1345622_1344955_ans_618594c79b574cc0bce903f863e6bce3.jpg

If $\cos (x+y)=y \sin x$ , then find $\dfrac { dy }{ dx } .$

Given

$\cos(x+y)=y\sin x$

Differentiating both sides with respect to x, we get

$-\sin(x+y)\left\{1+\dfrac{dy}{dx}\right\}=y\cos x+\sin x\dfrac{dy}{dx}$

$-\sin(x+y)-\sin(x+y)\dfrac{dy}{dx}=y\cos x+\sin x\dfrac{dy}{dx}$

$\Rightarrow -\sin(x+y)\dfrac{dy}{dx}-\sin x\dfrac{dy}{dx}=y\cos x+\sin(x+y)$

$\Rightarrow \left\{-\sin (x+y)-\sin x\right\}\dfrac{dy}{dx}=y\cos x+\sin(x+y)$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{y\cos x+\sin(x+y)}{\sin(x+y)-\sin x}$ .

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Differentiate the following $w.r.t.x$ .
$\sqrt{x}.cotx$

$\begin{array}{l} y=\sqrt { x } \cot x \\ \dfrac { { dy } }{ { dx } } =\dfrac { 1 }{ { 2\sqrt { x } } } \cot x+\sqrt { x } \left( { -\cos e{ c^{ 2 } }x } \right) \\ =\dfrac { { \cot x } }{ { 2\sqrt { x } } } -\sqrt { x } \cos e{ c^{ 2 } }x. \end{array}$

Find $\dfrac{dy}{dx}$ :

$\sin x +\cos y = 5x+4$

$\sin x + \cos y = 5x+4$

$\cos x -sin y \dfrac{dy}{dx}=5$

$\sin y \dfrac{dy}{dx}=\cos x-5$

$\dfrac{dy}{dx}=\dfrac{\cos x-5}{\sin y}$

Differentiate of the function $x^{3}+y^{3}+3x^{2}y=a^{3}$ with respect to $x$ .

Given function is

${x}^{3} + {y}^{3} + 3 {x}^{2} y = {a}^{3}$

Differentiating above function w.r.t.

$x$ , we have

$3 {x}^{2} + 3 {y}^{2} \cfrac{dy}{dx} + 6xy + 3 {x}^{2} \cfrac{dy}{dx} = 0$

$\cfrac{dy}{dx} \left( 3{x}^{2} + 3 {y}^{2} \right) = -3 {x}^{2} - 6xy$

$\Rightarrow \cfrac{dy}{dx} = - \cfrac{{x}^{2} + 2xy}{{x}^{2} + {y}^{2}}$

Find $\dfrac{dy}{dx}$ :

$\sin x - 3x = 5y$

$\sin x - 3x=5y$

$\cos x - 3=5\dfrac{dy}{dx}$

$\dfrac{dy}{dx}=\dfrac{\cos x -3}{5}$

If $x^py^q=(x+y)^{p+q}$ then prove that $\dfrac{dy}{dx}=\dfrac{y}{x}$ .

Given,

$x^py^q=(x+y)^{p+q}$

Now taking logarithm both sides we get,

or,

$p\log x+q\log y=(p+q)\log(x+y)$

Now differentiating both sides with respect to

$x$ we get,

$\dfrac{p}{x}+\dfrac{q}{y}\dfrac{dy}{dx}=\dfrac{(p+q)}{x+y}\left(1+\dfrac{dy}{dx}\right)$

or,

$\dfrac{p}{x}-\dfrac{p+q}{x+y}=\left(\dfrac{p+q}{x+y}-\dfrac{q}{y}\right)\dfrac{dy}{dx}$

or,

$\dfrac{py-qx}{x(x+y)}=\dfrac{py-qx}{y(x+y)}\dfrac{dy}{dx}$

or,

$\dfrac{dy}{dx}=\dfrac{y}{x}$ .

If $f(x)=x+\log x$ find $f^{\prime}(x)$

$f(x)=x+\log x\\f^{\prime}(x)=\dfrac{d}{dx}x+\dfrac{d}{dx}\log x\\f^{\prime}(x)=1+\dfrac1x$

Find drivative of $y=(2-\sin x)(e^{x}+x^{3}+2)$ with respect to $x$ .

From rule

$\dfrac{d}{dx} (u.x) = u'v + uv'$

$\Rightarrow \dfrac{dy}{dx} = (-\cos x)(e^x + x^3 + 2) + (2-\sin x)(e^x + 3x^2)$

$=e^x(2-\sin x - \cos x) - \cos x.x^3 + x^2 (6 - 3 \sin x) - 2\cos x$ .

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Find $\dfrac{dy}{dx}$ if $3x+4y=9$

$3x+4y=9\\\dfrac{d}{dx}3x+\dfrac{d}{dx}4y=0\\3+4\dfrac{dy}{dx}=0\\\dfrac{dy}{dx}=\dfrac{-3}{4}$

Find the derivative of $\dfrac{x+\cos x}{\tan x}$

Let

$y=\dfrac{x+\cos x}{\tan x}$ ………….

$(1)$

differentiating equation

$(1)$ wrt x we get

$\dfrac{dy}{dx}=\dfrac{d}{dx}\left(\dfrac{x+\cos x}{\tan x}\right)$

$\dfrac{dy}{dx}=\dfrac{\tan x(1-\sin x)-\sec^2x(x+\cos x)}{(\tan x)^2}$

$\dfrac{dy}{dx}=\dfrac{1-\sin x}{\tan x}-\left(\dfrac{\sec x}{\tan x}\right)^2(x+\cos x)$ .

1204134_1399836_ans_a16e2a5ee92d41078005c2f8c7e75481.JPG

Differentiate $\sin^{2} x$ wrt $x$

$\dfrac{d(\sin^2 x)}{d x}=2\sin x \times \dfrac{d (\sin x)}{d x}=2\sin x \cos x=\sin 2 x$

Find $\dfrac{dy}{dx}$
$y=\sin x.\ \cos x$

$y=\sin x \cos x\\\dfrac{dy}{dx}=\dfrac{d}{dx}(\sin x\cos x)\\\dfrac{dy}{dx}=\cos x\cos x +\sin x(-\sin x)\\\cos ^2x-\sin ^2x\\\cos 2x$

$f((x)=\log x+\sin x$ find $f^{\prime }(x)$

$f(x)=\log x+\sin x\\f^{\prime} (x)=\dfrac d{dx}\log x+\sin x\\f^{\prime}(x)=\dfrac 1x+\cos x$

Differentiate with respect to $x$ :
$\ e^{-3x} \log (1+x)$

$y=e^{-3x}\log(1+x)$

By using the product rule

$\dfrac{dy}{dx}=-3e^{-3x} \log(1+x)+\dfrac{e^{-3x}}{1+x}$

If $\sin\theta + 2\cos\theta =1$ then prove that $2\sin\theta -\cos\theta=0$ .

Given,

$\sin\theta + 2\cos\theta =1$ .

Now differentiating both sides with respect to

$\theta$ we get,

$\cos \theta-2\sin \theta=0$

or,

$2\sin \theta-\cos \theta=0$ .

Differentiate with respect to $x$ :
$e^x \log \sin 2x$

$y=e^x \log \sin 2x$

$y'=e^x \dfrac{\cos 2x}{\sin 2x} \times 2 + e^x \log \sin 2x$

$y'=2e^x \cot 2x + e^x \log \sin 2x$

Find $\dfrac{dy}{dx}$ in the following:
$4x+3y=\log(4x-3y)$

$4x+3y=\log(4x-3y)$

Differentiating w.r.t. x, we get,

$4+3\dfrac{dy}{dx}=\dfrac{1}{4x-3y}(4-3\dfrac{dy}{dx})$

$4+3\dfrac{dy}{dx}=\dfrac{4}{4x-3y}-\dfrac{3}{4x-3y} \dfrac{dy}{dx}$

$\dfrac{dy}{dx}(3+\dfrac{3}{4x-3y})=\dfrac{4}{4x-3y}-4$

$\dfrac{dy}{dx}=\dfrac{(\dfrac{4}{4x-3y}-4)}{(3+\dfrac{3}{4x-3y})}$

$\dfrac{dy}{dx}=\dfrac {4-16x+12y}{3+12x-9y}$

Differentiate the following w.r.t. x:
$\cos x .e^{4x+5}$

$y=\cos x .e^{4x+5}$

Differentiating w.r.t. x, we get,

$\dfrac{dy}{dx}=-\sin x e^{4x+5}+4e^{4x+5}\cos x$

Let $y=\log(1+\sin x)$ for $0\le x\le \dfrac{\pi}{2}$ , find $\dfrac{dy}{dx}$ .

Given,

$y=\log(1+\sin x)$ for

$0\le x\le \dfrac{\pi}{2}$ .

Now differentiating both sides we get,

$\dfrac{dy}{dx}=\dfrac{\cos x}{1+\sin x}.$

Differentiate with respect to $x$ :
$e^x\tan x$

$y=e^x \tan x$

$\dfrac{dy}{dx}=e^x \tan x + e^x \sec^2x$

Differentiate $x^{2}$ with respected to $x^{3}$ .

Let

$u=x^2$ and

$v=x^3$ .

$\dfrac{du}{dx}=2x$ and

$\dfrac{dv}{dx}=3x^2$

Thus,

$\dfrac{du}{dv}=\dfrac{2x}{3x^2}=\dfrac{2}{3x}$

Differentiate $e^x \tan x$ with respect to $x$ .

$y=e^x\tan x$

Differentiating w.r.t. x, we get,

$\dfrac{dy}{dx}=e^x \tan x + e^x \sec^2 x$

Differentiate the following functions with respect to $x$ :
If $y=\tan^{-1}\left (\dfrac {2x}{1-x^2}\right), x > 0$ , prove that $\dfrac {dy}{dx} =\dfrac {4}{1+x^2}$

$y=\tan^{-1}\left(\dfrac{2x}{1-x^2}\right)$

Let

$x=\tan \theta$

$y=\tan^{-1}\left(\dfrac{2\tan \theta}{1-\tan^2 \theta}\right)$

$y=\tan^{-1} \tan 2 \theta$

$y=2\theta = 2\tan^{-1}x$

$\dfrac{dy}{dx}=\dfrac{2}{1+x^2}\times 2=\dfrac{4}{1+x^2}$

$[Hence \ proved]$

If $y=\log \sin x +e^x \tan x$ , then find $\dfrac{dy}{dx}$ .

$y=\log \sin x +e^x \tan x$

Differentiating w.r.t. x, we get,

$\dfrac{dy}{dx}=\dfrac{\cos x}{\sin x}+e^x \tan x +e^x \sec^2x$

$\dfrac{dy}{dx}=\cot x +e^x \tan x +e^x \sec^2x$

Differentiate w.r.t. $x$ :
$\sin 5x \cos 3x$

$y=\sin 5x \cos 3x$

Differentiating w.r.t x, we get,

$\dfrac{dy}{dx}=5\cos 5x \cos 3x-3\sin 3x\sin 5x$

Find $\dfrac{dy}{dx}$ , if $y=e^x 5^x$ .

$y=e^x 5^x$

Differentiating w.r.t. x, we get,

$\dfrac{dy}{dx}=e^x 5^x + e^x 5^x \log 5$

Differentiate w.r.t $x$ :
$y=x^2 \sin 2x$

$y=x^2 \sin 2x$

Differentiating w.r.t. x, we get,

$\dfrac{dy}{dx}=2x \sin 2x +2x^2 \cos 2x$

Differentiate w.r.t. $x$ :
$y=e^{3x-2} \sin 3x$

$y=e^{3x-2} \sin 3x$

Differentiating w.r.t. x, we get,

$\dfrac{dy}{dx}=3\cos 3x e^{3x-2} +3e^{3x-2} \sin 3x$

Find $\dfrac{dy}{dx}$ , if $y=(3x+2)^x$ .

$y=(3x+2)^x$

By Taking log on Both sides

$\log y = x \log (3x+2)$

$\dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{3x}{3x+2} +\log (3x+2)$

$\dfrac{dy}{dx}=(3x+2)^x[\dfrac{3x}{3x+2}+\log(3x+2)]$

Find $\dfrac{dy}{dx}$ , if $2x+3y=\sin x$ .

$2x+3y=\sin x$

Differentiating w.r.t x, we get,

$2+3\dfrac{dy}{dx}=\cos x$

$3\dfrac{dy}{dx}=\cos x-2$

$\dfrac{dy}{dx}=\dfrac{\cos x-2}{3}$

If $y=x\sin (a+y),$ then prove that $\dfrac {dy}{dx}=\dfrac {\sin^2 (a+y)}{\sin (a+y)-y \cos (a+y)}$ .

$y=x \sin(a+y)$

$\dfrac{dy}{dx}=\sin(a+y)+x\cos(a+y) \dfrac{dy}{dx}$

$\dfrac{dy}{dx}=\dfrac{\sin(a+y)}{1-x\cos (a+y)}$

Multiplying numerator and denominator by

$\sin (a+y)$ , we get,

$\dfrac{dy}{dx}=\dfrac{\sin^2 (a+y)}{\sin(a+y)-y \cos (a+y)}$

Differentiate w.r.t. $x$ :
$\dfrac{x}{y^3}=1$

$\dfrac{x}{y^3}=1$

$x=y^3$

Differentiating w.r.t x, we get,

$1=3y^2 \dfrac{dy}{dx}$

$\dfrac{dy}{dx}=\dfrac{1}{3y^2}$

Differentiate w.r.t. $x$ :
$y=e^{3x} \sin 4x$

$y=e^{3x}\sin 4x$

$\dfrac{dy}{dx}=3 e^{3x}\sin 4x+4e^{3x}\cos 4x$

Differentiate w.r.t. $x$ :
$y=\sin x \sin 2x$

$y=\sin x \sin 2x$

$\dfrac{dy}{dx}=\cos x \sin 2x+2\sin x \cos 2x$

If $y=x^2 \tan x$ , find $\dfrac{dy}{dx}$ .

$y=x^2 \tan x$

Differentiating w.r.t. x, we get,

$\dfrac{dy}{dx}=2x \tan x+x^2 \sec^2x$

Find $\dfrac{dy}{dx}$ , if $ax + by^2 =\cos y$ .

$ax +by^2 =\cos y$

Differentiating w.r.t. x, we get,

$a+2by \dfrac{dy}{dx}=-\sin y \dfrac{dy}{dx}$

$\dfrac{dy}{dx}(2by+\sin y)=-a$

$\dfrac{dy}{dx}=-\dfrac{a}{2by +\sin y}$

Differentiate w.r.t. $x$ :
$y=(\cos x)(1-\sin^2 x)$

$y=(\cos x)(1-\sin^2x)$

$y=\cos x \times \cos^2x$

$y=\cos^3x$

Differentiating w.r.t x, we get,

$\dfrac{dy}{dx}=3\cos^2x \times (-\sin x)$

$\dfrac{dy}{dx}=-3\sin x \cos^2x$

If $y=e^x+\sin x -4x^3$ , find $\dfrac{dy}{dx}$ .

$y=e^x+\sin x -4x^3$

Differentiating w.r.t. x, we get,

$\dfrac{dy}{dx}=e^x+\cos x-12x^2$

Solve the following differential equation.
$\dfrac{dy}{dx}=3x$

Consider,

$\dfrac{dy}{dx}=3x$

Integrating on both sides

$\displaystyle \int dy=\int 3xdx$

$y=\dfrac{3x^2}2+c$

If $x^3+y^3=3axy$ , find $\dfrac{dy}{dx}$ .

$x^3+y^3=3axy$

Differentiating w.r.t. x, we get,

$\Rightarrow 3x^2+3y^2 \dfrac{dy}{dx}=3a[x \dfrac{dy}{dx}+y]$

$\Rightarrow 3x^2+3y^2 \dfrac{dy}{dx}=3ax \dfrac{dy}{dx}+3ay$

$\Rightarrow (3y^2-3ax)\dfrac{dy}{dx}=3ay-3x^2$

$\Rightarrow 3(y^2-ax)\dfrac{dy}{dx}=3(ay-x^2)$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{ay-x^2}{y^2-ax}$

If $x^2+2xy+y^3=42$ , find $\dfrac{dy}{dx}$ .

$x^2+2xy+y^3=42$

Differentiating w.r.t. x, we get,

$\Rightarrow 2x+2(x \dfrac{dy}{dx}+y)+3y^2 \dfrac{dy}{dx}=0$

$\Rightarrow 2x+2y+\dfrac{dy}{dx}(2x+3y^2)=0$

$\Rightarrow \dfrac{dy}{dx}(2x+3y^2)=-2(x+y)$

$\therefore \dfrac{dy}{dx}=-\dfrac{2(x+y)}{(2x+3y^2)}$

If $x^2+y^2=t-\dfrac{1}{t}$ and $x^4+y^4=t^ 2+\dfrac{1}{t^2}$ , then prove that $\dfrac{dy}{dx}=\dfrac{1}{x^3y}$ .

We have,

$x^2+y^2=t-\dfrac{1}{t}$ and

$x^4+y^4=t^ 2+\dfrac{1}{t^2}$

$\Rightarrow (x^2+y^2)^2=(t-\dfrac{1}{t})^2$

$\Rightarrow x^4+y^4+2x^2y^2=t^2+\dfrac{1}{t^2}-2$

$\Rightarrow x^4+y^4+2x^2y^2=x^4+y^4-2$ [given]

$\Rightarrow 2x^2y^2=-2$

$\Rightarrow x^2y^2=-1$

$\Rightarrow y^2=-\dfrac{1}{x^2}$

$\Rightarrow y^2=-x^{-2}$

Differentiating w.r.t. x, we get,

$\Rightarrow 2y \dfrac{dy}{dx}=-(-2)x^{-3}$

$\Rightarrow y \dfrac{dy}{dx}=\dfrac{1}{x^3}$

$\therefore\ \dfrac{dy}{dx}=\dfrac{1}{x^3y}$

If $y=x\sin y$ , then prove that $\dfrac {dy}{dx}=\dfrac {y}{x(1-x\cos y)}$ .

$y=x \sin y$

$\dfrac{dy}{dx}=x \cos y \dfrac{dy}{dx}+\sin y$

$\dfrac{dy}{dx}=\dfrac{\sin y}{1-x\cos y}$

Multiplying numerator and denominator by x, we get,

$\dfrac{dy}{dx}=\dfrac{y}{x(1-x\cos y)}$

Is every differentiable function continuous? is every continuous function differentiable?

Yes, every differentiable function is continuous.

No, some continuous function may not be differentiable.

Find absolute value of $dy/dx$ at $x=1$ if $\displaystyle y^{x}.x^{y}=1$

$x^y. y^x=1$
Taking logarithm, we get,

$x\log y+y \log x=\log 1=0.$
Differentiating w.r.t.

$x$ ,

we get

$\displaystyle x\cdot \frac{1}{y}\frac{dy}{dx}+1.\log y+y.\frac{1}{x}+\frac{dy}{dx}\log x=0$

$\displaystyle \therefore \frac{dy}{dx}=-\frac{\log y+y/x}{x/y+\log x}\therefore |y'(1)| = |-1|=1$

If $f(x)$ is differentiable everywhere, then ${ \left| f(x) \right| }^{ 2 }$ is differentiable everywhere.( Enter 1 if true or 0 otherwise)

$f\left( x \right)$ is differentiable everywhere then

${ \left| f\left( x \right) \right| }^{ 2 }$ is differentiable everywhere because

${ \left| f\left( x \right) \right| }^{ 2 }=f\left( x \right)\times f\left( x \right)$ (The product of the two differential functions are differntiable)

$\therefore |f(x)|^2$ is differential everywhere

Given a function g which has derivative ${g}'\left ( x \right )$ for all x satisfying ${g}'\left ( 0 \right )=2$ and $g\left ( x+y \right )=e^{y}g\left ( x \right )+e^{x}g\left ( y \right )$ for all $x, y\epsilon R$ , $g(5)=32.$ The value of $\dfrac{{g}'\left ( 5 \right )-2e^{5}}{16}$ is

Putting

$x=y=0$ in

$g\left ( x+y \right )=e^{y}g\left ( x \right )+e^{x}g\left ( y \right )$
we get

$g(0)=g(0)+g(0)=2g(0) \Rightarrow g(0)=0$
So

$\displaystyle 2={g}'\left ( 0 \right )=\lim_{h\rightarrow 0}\dfrac{g\left ( h \right )-g\left ( 0 \right )}{h}=\lim_{h\rightarrow 0}\dfrac{g\left ( h \right )}{h}$
Also

$\displaystyle {g}'\left ( x \right )=\lim_{h\rightarrow 0}\dfrac{g\left ( x+h \right )-g\left ( x \right )}{h}$

$\displaystyle =\lim_{h\rightarrow 0}\dfrac{e^{x}g\left ( h \right )+e^{h}g\left ( x \right )-g\left ( x \right )}{h}$

$\displaystyle =\lim_{h\rightarrow 0}\left ( e^{x}\dfrac{g\left ( h \right )}{h}+\frac{e^{h}-1}{h}g\left ( x \right ) \right )$

$\displaystyle =e^{x}\lim_{h\rightarrow 0}\dfrac{g\left ( h \right )}{h}+1.g\left ( x \right )$

$=g\left ( x \right )+2e^{x}$
Thus

${g}'\left ( 5 \right )-2e^{5}=g\left ( 5 \right )=32$

Differentiate the function with respect to $x$
$\cos x^3 . \sin^2 (x^5)$

Let

$f(x)=\cos x^3 . \sin^2 (x^5)$

$\displaystyle\therefore f'(x)= \frac{d}{dx} [\cos x^3 . \sin^2 (x^5)] = \sin^2 (x^5) \times \frac{d}{dx} (\cos x^3) + \cos x^3 \times \frac{d}{dx} [\sin^2 (x^5)]$

$\displaystyle = \sin^2 (x^5) \times (-\sin x^3) \times \frac{d}{dx} ( x^3) + \cos x^3 \times 2 \sin (x^5) . \frac{d}{dx} [\sin x^5]$

$\displaystyle = -\sin x^3 \sin^2 (x^5) \times 3 x^2 + 2 \sin x^5 \cos x^3 . \cos x^5 \times \frac{d}{dx} (x^5)$

$\displaystyle = -3 x^2 \sin x^3 . \sin^2 (x^5) + 2 \sin x^5 \cos x^5 \cos x^3 \times 5 x^4$

$\displaystyle = 10 x^4 \sin x^5 \cos x^5 \cos x^3 - 3 x^2 \sin x^3 \sin^2 (x^5)$

Find the derivative of the following:
$(5x^{3} + 3x - 1)(x - 1)$ .

Let

$\displaystyle f\left ( x \right )=\left ( 5x^{3}+3x-1 \right )\left ( x-1 \right )$
Thus using product rule of differentiation,

$\displaystyle f^{'}\left ( x \right )=\left ( 5x^{3}+3x-1 \right )\frac{d}{dx}\left ( x-1 \right )+\left ( x-1 \right )\frac{d}{dx}\left ( 5x^{3}+3x-1 \right )$

$=\displaystyle \left ( 5x^{3} +3x-1\right )\left ( 1 \right )+\left ( x-1 \right )\left ( 5.3x^{2}+3-0 \right )$

$=\displaystyle \left ( 5x^{3}+3x-1 \right )+\left ( x-1 \right )\left ( 15x^{2} +3\right )$

$=\displaystyle 5x^{3}+3x-1+15x^{3}+3x-15x^{2}-3$

$=\displaystyle 20x^{3}-15x^{2}+6x-4$

Find the derivative of the following:
$x^{-4} (3 - 4x^{-5})$ .

Let

$\displaystyle f\left ( x \right )=x^{-4}\left ( 3-4x^{-5} \right )$
By Leibnitz product rule

$\displaystyle f^{'}\left ( x \right )=x^{-4}\frac{d}{dx}\left ( 3-4x^{-5} \right )+\left ( 3-4x^{-5} \right )\frac{d}{dx}\left ( x^{-4} \right )$

$=\displaystyle x^{-4}\left \{ 0-4\left ( -5 \right )x^{-5-3} \right \}+\left ( 3-4x^{-5} \right )\left ( -4 \right )x^{-4-1}$

$=\displaystyle x^{-4}\left ( 20x^{-6} \right )+\left ( 3-4x^{-5} \right )\left ( -4x^{-5} \right )$

$=\displaystyle 20x^{-10}-12x^{-5}+16x^{-10}$

$=\displaystyle 36x^{-10}-12x^{-5}$

Find $\displaystyle \frac{dy}{dx}$ of $\sin^2 x + \cos^2 y = 1$

We have

$\sin^2 x + \cos^2 y = 1$
Differentiating both sides w.r.t.

$x,$ we obtain

$\displaystyle \frac{d}{dx} (\sin^2 x + \cos^2 y) = \frac{d}{dx} (1)$

$\Rightarrow \displaystyle 2 \sin x . \frac{d}{dx} (\sin x) + 2 \cos y . \frac{d}{dx} (\cos y) = 0$

$\Rightarrow \displaystyle 2 \sin x \cos x + 2 \cos y (-\sin y) . \frac{dy}{dx} = 0$

$\Rightarrow \displaystyle \sin 2 x - \sin 2y \frac{dy}{dx} = 0$

$\therefore \displaystyle \frac{dy}{dx} = \frac{\sin 2x}{\sin 2y}$

Find $\displaystyle \frac{dy}{dx}$ of $x^3 + x^2 y + xy^2 + y^3 = 81$

$x^3 + x^2 y + xy^2 + y^3 = 81$
Differentiating both sides w.r.t

$x$ we get,

$\displaystyle \frac{d}{dx} (x^3 + x^2 y + xy^2 + y^3) = \frac{d}{dx} (81)$

$\Rightarrow \displaystyle 3x^2 + \left[ y . 2x + x^2 . \frac{dy}{dx} \right] + \left[ y^2 . 1 + x . 2y . \frac{dy}{dx} \right] + 3y^2 \frac{dy}{dx} = 0$

$\Rightarrow \displaystyle (x^2 + 2xy + 3y^2) \frac{dy}{dx} + (3x^2 + 2xy + y^2)= 0$

$\therefore \displaystyle \frac{dy}{dx} = \frac{-(3x^2 + 2xy + y^2)}{(x^2 + 2xy + 3y^2}$

Find the derivative of the following functions (it is to be understood that $a, b, c, d, p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers) : $(ax + b) (cx + d)^{2}$

Let

$f(x) = (ax + b)(cx + d)^{2}$
Thus by Leibnitz product rule

$f'(x) = (ax + b) \dfrac{d}{dx}\left ( cx + d\right )^{2}+\left ( cx +d \right )^{2}\dfrac{d}{dx}\left ( ax+b \right )$

$\displaystyle =\left ( ax + b \right )\frac{d}{dx}\left ( c^{2}x^{2}+2cdx+d^{2} \right )+\left ( cx+d \right )^{2}\frac{d}{dx}\left ( ax+b \right )$

$=\displaystyle \left ( ax+b \right )\left [ \frac{d}{dx}\left ( c^{2}x^{2} \right )+\frac{d}{dx}\left ( 2cdx \right )+\frac{d}{dx}d^{2} \right ]+\left ( cx+d \right )^{2}\left [ \frac{d}{dx}ax+\frac{d}{dx} b\right ]$

$= \displaystyle \left ( ax + b \right )\left ( 2c^{2}x+2cd \right )+\left ( cx+d^{2} \right )a$

$= 2a (ax + b)(cx + d)+ a (cx + d)^{2}$

Find the derivative of $\displaystyle x^{-3}\left ( 5+3x \right )$

$\dfrac{d}{dx}\left({x}^{−3}\left(5+3x\right)\right)$

$=\dfrac{d}{dx}\left(5{x}^{−3}+3{x}^{1−3}\right)$

$=\dfrac{d}{dx}\left(5{x}^{−3}+3{x}^{-2}\right)$

$=\dfrac{d}{dx}\left(5{x}^{−3}\right)+\dfrac{d}{dx}\left(3{x}^{-2}\right)$

$=5\dfrac{d}{dx}\left({x}^{−3}\right)+3\dfrac{d}{dx}\left({x}^{-2}\right)$

$=5\left(-3{x}^{−3-1}\right)+3\left(-2{x}^{-2-1}\right)$

$=-15{x}^{-4}-6{x}^{-3}$

$=\dfrac{15}{{x}^{4}}-\dfrac{6}{{x}^{3}}$

Find $\displaystyle \frac{dy}{dx}$ of $\sin^2 y + \cos xy = k$

We have,

$\sin^2 y + \cos xy = k$
Differentiating both sides with respect to

$x$ , we obtain

$\Rightarrow \displaystyle \frac{d}{dx} (\sin^2 y) + \frac{d}{dx} (\cos xy) = \frac{d(\pi)}{dx}=0$ .....(1)
Using chain rule, we obtain

$\displaystyle \frac{d}{dx} (\sin^2 y) = 2 \sin y \frac{d}{dx} (\sin y) = 2 \sin y \cos y \frac{dy}{dx}$ ...... (2)
and

$\displaystyle \frac{d}{dx} (\cos xy) = - \sin xy \frac{d}{dx} (xy) = - \sin xy \left[ y \frac{d}{dx} (x) + x \frac{dy}{dx} \right]$

$= - \sin xy \left[ y. 1 + x \dfrac{dy}{dx} \right] = -y \sin xy - x \sin xy \dfrac{dy}{dx}$ .....(3)
From (1), (2) and (3), we obtain

$\displaystyle 2 \sin y \cos y \frac{dy}{dx} - y \sin xy - x \sin xy \frac{dy}{dx} = 0$

$\Rightarrow \displaystyle (2 \sin y \cos y - x \sin xy) \frac{dy}{dx} = y \sin xy$

$\Rightarrow \displaystyle (\sin 2 y - x \sin xy) \frac{dy}{dx} = y \sin xy$

$\therefore \displaystyle \frac{dy}{dx} = \frac{y \sin xy}{\sin 2 y - x \sin xy}$

Find $\displaystyle \frac{dy}{dx}$ of $x^2 + xy + y^2 = 100$

$x^2 + xy + y^2 = 100$
Differentiating both sides w.r.t

$x$ we get,

$\displaystyle \frac{d}{dx} (x^2 + xy + y^2) = \frac{d}{dx} (100)=0$

$\Rightarrow \displaystyle 2x + y .1 + x . \frac{dy}{dx} + 2 y \frac{dy}{dx} = 0$

$\Rightarrow \displaystyle 2x + y + (x + 2y) \frac{dy}{dx} = 0$

$\therefore \displaystyle \frac{dy}{dx} = -\frac{2x + y}{x + 2y}$

Find the derivative of the following:
$x^{5} (3 - 6x^{-9})$ .

Let

$\displaystyle f\left ( x \right )=x^{5}\left ( 3-6x^{-9} \right )$
By Leibnitz product rule

$\displaystyle f^{'}\left ( x \right )=x^{5}\frac{d}{dx}\left ( 3-6x^{-9} \right )+\left ( 3-6x^{-9} \right )\frac{d}{dx}\left ( x^{5} \right )$

$=\displaystyle x^{3}\left \{ 0-6\left ( -9 \right )x^{-9-1} \right \}+\left ( 3-6x^{-9} \right )\left ( 5x^{4} \right )$

$=\displaystyle x^{3}\left ( 54x^{-10} \right )+15x^{4}-30x^{-5}$

$=\displaystyle 54x^{-7}+15x^{4}-30x^{-5}$

Find the derivative of the following functions (it is to be understood that $a, b, c, d, p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers) : $(px + q)\displaystyle \left ( \frac{r}{x}+s \right )$

Let

$f (x) = (px + q) \displaystyle \left ( \frac{r}{x}+s \right )$
Thus using product rule

$f'(x) = (px + q) \displaystyle \left ( \frac{r}{x}+s \right )' + \displaystyle \left ( \frac{r}{x}+s \right ) (px + q)'$

$=(px + q) \displaystyle \left ( rx^{-1}+s \right )'+\left ( \frac{r}{x}+s \right )\left ( P \right )$

$\displaystyle =\left ( px+q \right )\left ( -rx^{-2} \right )+\left ( \frac{r}{x}+s \right )p$

$=\displaystyle \left ( px + q \right )\left ( \frac{-r}{x^{2}} \right )+\left ( \frac{r}{x}+s \right )p$

$\displaystyle =\frac{-pr}{x}-\frac{qr}{x^{2}}+\frac{pr}{x}+ps= ps \displaystyle -\frac{qr}{x^{2}}$

Find the derivative of the following functions (it is to be understood that $a, b, c, d, p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers) : $\displaystyle \cos ec\, x\,\cot \,x$

let

$f(x) = \displaystyle \cos ec\, x\,\cot \,x$
By Leibnitz product rule

$f' (x) =\displaystyle \cos ec \times \left ( \cot x \right )' + \cot x\left ( \cos ec \,x \right )'$

$\displaystyle = \cos ex\,x\left ( -\cos ec^{2} \,x\right )+\cot \,x\left ( -\cos ec \,x \cot \,x \right )$

$=-\displaystyle \cos ec^{3}x-\cot ^{2}x \cos ec x$

Find the derivative of the following functions (it is to be understood that $a, b, c, d, p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers) : $\displaystyle \left ( ax^{2}+\sin \,x \right )\left ( p + q \,\cos \,x \right )$

Let

$f (x) = \displaystyle \left ( ax^{2}+\sin \,x \right )\left ( p + q \,\cos \,x \right )$
Thus by product rule,

$f'(x) = \displaystyle \left ( ax^{2}+\sin x \right )\frac{d}{dx}\left ( p+q\cos x \right )+\left ( p+q \cos x \right )\frac{d}{dx}\left ( ax^{2}+\sin x \right )$

$=\displaystyle \left ( ax^{2}+\sin x \right )\left ( -q\sin x \right )+\left ( p+q\cos x \right )\frac{d}{dx}\left ( ax^{2} +\sin x\right )$

$= - q \displaystyle \sin x\left ( ax^{2}+\sin x \right )+\left ( p+q\cos x \right )\left ( 2ax+\cos x \right )$

Differentiate the given function w.r.t. $x$ .
$y=\displaystyle e^x + e^{x^2} + ... + e^{x^5}$

$\displaystyle \frac{d}{dx} (e^x + e^{x^2} + ... + e^{x^5})$

$=\displaystyle \frac{d}{dx} (e^x) + \frac{d}{dx} (e^{x^2}) + \frac{d}{dx} (e^{x^3}) + \frac{d}{dx} (e^{x^4}) + \frac{d}{dx} (e^{x^5})$

$=\displaystyle e^x + \left[ e^{x^2} . \frac{d}{dx} (x^2) \right] + \left[ e^{x^3} . \frac{d}{dx} (x^3) \right] + \left[ e^{x^4} . \frac{d}{dx} (x^4) \right] + \left[ e^{x^5} . \frac{d}{dx} (x^5) \right]$

$=\displaystyle e^x + ( e^{x^2} \times 2x) + (e^{x^3} \times 3x^2) + ( e^{x^4} \times 4x^3) + ( e^{x^5} \times 5x^4)$

$=\displaystyle e^x + 2x e^{x^2} + 3x^2 e^{x^3} + 4x^3 e^{x^4} + 5x^4 e^{x^5}$

Find $\displaystyle \frac{dy}{dx}$ of $\displaystyle y = \sin^{-1} ( 2 x \sqrt{1 - x^2} ) , -\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$

Put

$x=\sin\theta\Rightarrow \theta=\sin^{-1}x$

Then we have,

$y = \sin^{-1} (2\sin\theta\sqrt{1-\sin^2\theta})=\sin^{-1}(2\sin\theta\cos\theta)$

$\quad =\sin^{-1}(\sin2\theta)=2\theta=2\sin^{-1}x$

$\therefore \displaystyle \dfrac{dy}{dx}=2\cdot \dfrac{1}{\sqrt{1-x^2}}=\dfrac{2}{\sqrt{1-x^2}}$

Differentiate the function w.r.t. $x$ .
$\displaystyle \sqrt{ \frac{(x - 1)(x - 2)}{(x - 3)(x - 4) (x - 5)} }$

Let

$\displaystyle y = \sqrt{ \frac{(x - 1)(x - 2)}{(x - 3)(x - 4) (x - 5)} }$
Taking logarithm on both the sides, we obtain

$\displaystyle \log y = \log \sqrt{ \frac{(x - 1)(x - 2)}{(x - 3)(x - 4) (x - 5)} }$

$\Rightarrow \displaystyle \log y = \frac{1}{2} log \left[ \frac{(x - 1)(x - 2)}{(x - 3)(x - 4) (x - 5)} \right]$

$\Rightarrow \displaystyle \log y = \frac{1}{2}[ \log [(x - 1)(x - 2)] - \log[(x - 3)(x - 4) (x - 5)] ]$

$\Rightarrow \displaystyle \log y = \frac{1}{2}[ \log (x - 1) + \log (x - 2) - \log(x - 3) - \log (x - 4) - \log (x - 5) ]$
Differentiating both sides with respect to

$x$ , we obtain

$\Rightarrow \displaystyle \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{x - 1} . + \frac{1}{x - 2} - \frac{1}{x - 3} - \frac{1}{x - 4}- \frac{1}{x - 5} \right]$

$\Rightarrow \displaystyle \frac{dy}{dx} = \frac{y}{2} \left( \frac{1}{x - 1} + \frac{1}{x - 2} - \frac{1}{x - 3} - \frac{1}{x - 4} - \frac{1}{x - 5} \right)$

$\therefore \displaystyle \frac{dy}{dx} = \frac{1}{2} \sqrt{ \frac{(x - 1)(x - 2)}{(x - 3)(x - 4) (x - 5)} } \left[ \frac{1}{x - 1} + \frac{1}{x - 2} - \frac{1}{x - 3} - \frac{1}{x - 4} - \frac{1}{x - 5} \right]$

Find the derivative of the following functions (it is to be understood that $a, b, c, d, p, q, r, s$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers) : $\displaystyle x^{4}\left ( 5\,\sin \,x-3\,\cos \,x \right )$

Let

$f (x) = \displaystyle x^{2}\left ( 5 \sin x-3\cos x \right )$
Thus by product rule

$f'(x) = \displaystyle x^{4}\frac{d}{dx}\left ( 5\sin x-3\cos x \right )+\left ( 5\sin x-3\cos x \right )\frac{d}{dx}\left ( x^{4} \right )$
=

$\displaystyle x^{4}\left [ 5\frac{d}{dx}\left ( \sin x \right ) -3\frac{d}{dx}\left ( \cos x \right )\right ]+\left ( 5\sin x-3\cos x \right )\frac{d}{dx}\left ( x^{4} \right )$
=

$\displaystyle x^{4}\left [ 5\cos x-3\left ( -\sin x \right ) \right ]+(5\sin x-3\cos x)\left ( 4x^{3} \right )$
=

$\displaystyle x^{3}\left [ 5x\cos x+3x\sin x+20\sin x-12\cos x \right ]$

Find the derivative of the following functions (it is to be understood that $a, b, c, d, p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers) : $\displaystyle \left ( x+\cos \,x \right )\left ( x-\tan \,x \right )$

Let

$f (x) = \displaystyle \left ( x+\cos x \right )\left ( x-\tan x \right )$
Thus by product rule,

$f'(x) = \displaystyle \left ( x+\cos x \right )\frac{d}{dx}\left ( x-\tan x \right )+\left ( x-\tan x \right )\frac{d}{dx}\left ( x+\cos x \right )$

$=\displaystyle \left ( x+\cos x \right )\left ( 1-\sec ^{2}x \right )+\left ( x-\tan x \right )\left ( 1-\sin x \right )$

$=\displaystyle \left ( x+\cos x \right )\left ( -\tan ^{2}x \right )+\left ( x-\tan x \right )\left ( 1-\sin x \right )$

$=\displaystyle -\tan ^{2}x\left ( x+\cos x \right )+\left ( x-\tan x \right )\left ( 1-\sin x \right )$

Differentiate the function w.r.t. $x$ .
$(x + 3)^2 . (x + 4)^3 . (x + 5)^4$

Let

$y = (x + 3)^2 . (x + 4)^3 . (x + 5)^4$
Taking logarithm on both the sides, we obtain

$\log y = \log (x + 3)^2 + \log (x + 4)^3 + \log (x + 5)^4$

$\Rightarrow \log y = 2 \log (x + 3) + 3 \log (x + 4) + 4 \log (x + 5)$
Differentiating both sides with respect to

$x$ , we obtain

$\displaystyle \frac{1}{y} \frac{dy}{dx} = 2 . \frac{1}{ x + 3} + 3 . \frac{1}{ x + 4} + 4 . \frac{1}{ x + 5}$

$\Rightarrow \displaystyle \frac{dy}{dx} = y \left[ \frac{2}{ x + 3} + \frac{3}{ x + 4} + \frac{4}{ x + 5} \right]$

$\Rightarrow \displaystyle \frac{dy}{dx} = (x + 3)^2 . (x + 4)^3 . (x + 5)^4 . \left[ \frac{2}{ x + 3} + \frac{3}{ x + 4} + \frac{4}{ x + 5} \right]$

$\Rightarrow \displaystyle \frac{dy}{dx} = (x + 3)^2 . (x + 4)^3 . (x + 5)^4 . \left[ \frac{2(x + 4)(x + 5) + 3(x+ 3)(x + 5) + 4(x + 3)(x + 4)}{(x + 3)(x + 4)(x + 5)} \right]$

$\Rightarrow \displaystyle \frac{dy}{dx} = (x + 3)^2 . (x + 4)^3 . (x + 5)^4 . \left[ 2 (x^2 + 9x + 20) + 3 (x^2 + 8x + 15) + 4 (x^2 + 7x + 12) \right]$

$\therefore \displaystyle \frac{dy}{dx} = (x + 3)^2 . (x + 4)^3 . (x + 5)^4 . (9x^2 + 70x + 133)$

Find the derivative of the following functions (it is to be understood that $a, b, c, d, p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are fixed integers) : $\displaystyle \left ( x+\sec \,x \right )\left ( x-\tan \,x \right )$

Let

$f (x) =\displaystyle \left ( x+\sec \,x \right )\left ( x-\tan \,x \right )$
Thus by product rule

$f'(x) = \displaystyle \left ( x+\sec x \right )\frac{d}{dx}\left ( x-\tan x \right )+\left ( x-\tan x \right )\frac{d}{dx}\left ( x+\sec x \right )$

$=\displaystyle \left ( x+\sec x \right )\left [ \frac{d}{dx}\left ( x \right ) -\frac{d}{dx}\tan x\right ]+\left ( x-\tan x \right )\left [ \frac{d}{dx} \left ( x \right )+\frac{d}{dx}\sec x\right ]$

$=\displaystyle \left ( x+\sec x \right )\left [ 1-\frac{d}{dx}\tan x \right ]+\left ( x-\tan x \right )\left [ 1+\frac{d}{dx}\sec x \right ]$

$\displaystyle =\left ( x+\sec x \right )\left ( 1-\sec ^{2}x \right )+\left ( x-\tan x \right )\left ( 1+\sec x\tan x \right )$

Differentiate $(x^2 - 5x + 8)(x^3 + 7x + 9)$ in three ways mentioned below,
(i) By using product rule
(ii) By expanding the product to obtain a single polynomial
(iii) By logarithmic differentiation

Do they all give the same answer?

Let

$y = (x^2 - 5x + 8)(x^3 + 7x + 9)$
(i) Let

$x^2 - 5x + 8 = u$ and

$x^3 + 7x + 9 = v$

$\therefore y = uv$

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{du}{dx} . v + u . \frac{dv}{dx}$

By using product rule,

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{d}{dx} (x^2 - 5x + 8) . (x^3 + 7x + 9) + (x^2 - 5x + 8) . \frac{d}{dx} (x^3 + 7x + 9)$

$\displaystyle \Rightarrow \frac{dy}{dx} = (2x - 5)(x^3 + 7x + 9) + (x^2 - 5x + 8) (3x^2 + 7)$

$\displaystyle \Rightarrow \frac{dy}{dx} = 2x (x^3 + 7x + 9) - 5 (x^3 + 7x + 9) + x^2 (3x^2 + 7) - 5x (3x^2 + 7) + 8 (3x^2 + 7)$

$\displaystyle \Rightarrow \frac{dy}{dx} = (2x^4 + 14x^2 + 18x) - 5x^3 - 35x - 45 + (3x^4 + 7x^2) - 15x^3 - 35x + 24x^2 + 56$

$\displaystyle \therefore \frac{dy}{dx} = 5x^4 - 20 x^3 + 45 x^2 - 52 x + 11$

(ii)

$y = (x^2 - 5x + 8)(x^3 + 7x + 9)$

$\displaystyle = x^2 (x^3 + 7x + 9) - 5x (x^3 + 7x + 9) + 8(x^3 + 7x + 9)$

$\displaystyle = x^5 + 7x^3 + 9x^2 - 5x^4 - 35x^2 - 45x + 8x^3 + 56x + 72$

$\displaystyle = x^5 - 5x^4 + 15x^3 - 26x^2 + 11x + 72$

$\displaystyle \therefore \frac{dy}{dx} = \frac{d}{dx} (x^5 - 5x^4 + 15x^3 - 26x^2 + 11x + 72)$

$\displaystyle = \frac{d}{dx} (x^5) - 5 \frac{d}{dx} (x^4) + 15 \frac{d}{dx} (x^3) - 26 \frac{d}{dx} (x^2) + 11 \frac{d}{dx} (x) + \frac{d}{dx} (72)$

$\displaystyle = 5 x^4 - 5 \times 4x^3 + 15 \times 3x^2 - 26 \times 2x + 11 \times 1 + 0$

$\displaystyle = 5 x^4 - 20 x^3 + 45 x^2 - 52 x + 11$

(iii)

$y = (x^2 - 5x + 8)(x^3 + 7x + 9)$
Taking logarithm on both the sides, we obtain

$\log y = \log (x^2 - 5x + 8) + \log (x^3 + 7x + 9)$
Differentiating both sides with respect to x, we obtain

$\displaystyle \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \log (x^2 - 5x + 8) + \frac{d}{dx} \log (x^3 + 7x + 9)$

$\Rightarrow \displaystyle \frac{1}{y} \frac{dy}{dx} = \frac{1}{x^2 - 5x + 8} . \frac{d}{dx} (x^2 - 5x + 8) + \frac{1}{x^3 + 7x + 9} . \frac{d}{dx} (x^3 + 7x + 9)$

$\Rightarrow \displaystyle \frac{dy}{dx} = y \left[ \frac{1}{x^2 - 5x + 8} \times (2x - 5) + \frac{1}{x^3 + 7x + 9} \times (3x^2 + 7) \right]$

$\Rightarrow \displaystyle \frac{dy}{dx} = (x^2 - 5x + 8)(x^3 + 7x + 9) \left[ \frac{2x - 5}{x^2 - 5x + 8} + \frac{3x^2 + 7}{x^3 + 7x + 9} \right]$

$\Rightarrow \displaystyle \frac{dy}{dx} = (x^2 - 5x + 8)(x^3 + 7x + 9) \left[ \frac{(2x - 5)(x^3 + 7x + 9) + (3x^2 + 7)(x^2 - 5x + 8)}{(x^2 - 5x + 8)(x^3 + 7x + 9)} \right]$

$\Rightarrow \displaystyle \frac{dy}{dx} = 2x (x^3 + 7x + 9) - 5(x^3 + 7x + 9) + 3x^2 (x^2 - 5x + 8) + 7(x^2 - 5x + 8)$

$\Rightarrow \displaystyle \frac{dy}{dx} = (2x^4 +14 x^2 + 18x) - 5x^3 - 35x - 45 + (3x^4 - 15x^3 + 24x^2) + (7x^2 - 35x + 56)$

$\Rightarrow \displaystyle \frac{dy}{dx} = 5 x^4 - 20 x^3 + 45 x^2 - 52 x + 11$
From the above three observations, it can be concluded that all the

$\displaystyle \frac{dy}{dx}$ results of are same.

Find $\dfrac{dy}{dx}$ of the function

$x^y + y^x = 1$

The given function is

$x^y + y^x = 1$

Let

$x^y = u$ and

$y^x = v$

Then, the function becomes

$u + v = 1$

$\displaystyle \frac{du}{dx} + \frac{dv}{dx} = 0$ ...(1)

$u = x^y$

$\Rightarrow \log u = \log (x^y)$

$\Rightarrow \log u = y \log x$

Differentiating both sides with respect to

$x$ , we obtain

$\Rightarrow \displaystyle \frac{1}{u} \frac{du}{dx} = \log x \frac{dy}{dx} + y . \frac{d}{dx} (\log x)$

$\Rightarrow \displaystyle \frac{du}{dx} = x^y \left( \log x \frac{dy}{dx} + \frac{y}{x} \right)$ ...(2)

$v = y^x$

$\Rightarrow \log v = \log (y^x)$

$\Rightarrow \log v = x \log y$

Differentiating both sides with respect to

$x$ , we obtain

$\displaystyle \frac{1}{v} \frac{dv}{dx} = \log y . \frac{d}{dx} (x) + x . \frac{d}{dx} (\log y)$

$\Rightarrow \displaystyle \frac{dv}{dx} = v \left( \log y . 1 + x . \frac{1}{y} . \frac{dy}{dx} \right)$

$\Rightarrow \displaystyle \frac{dv}{dx} = y^x \left( \log y + \frac{x}{y} . \frac{dy}{dx} \right)$ ...(3)

From (1), (2) and (3) we obtain

$\displaystyle x^y \left( \log x \frac{dy}{dx} + \frac{y}{x} \right) + y^x \left( \log y + \frac{x}{y} . \frac{dy}{dx} \right) = 0$

$\Rightarrow \displaystyle (x^y \log x + xy^{x - 1} ) \frac{dy}{dx} = - ( yx^{y - 1} + y^x \log y)$

$\therefore \displaystyle \frac{dy}{dx} = - \frac{yx^{y - 1} + y^x \log y}{x^y \log x + xy^{x - 1}}$

Find $\displaystyle \frac{dy}{dx}$ of $y^x = x^y$

The given function is

$y^x = x^y$
Taking logarithm on both the sides, we obtain

$x \log y = y \log x$
Differentiating both sides with respect to

$x$ , we obtain

$\displaystyle \log y . \frac{d}{dx} (x) + x . \frac{d}{dx} (\log y) = \log x . \frac{d}{dx} (y) + y . \frac{d}{dx} (\log x)$

$\Rightarrow \displaystyle \log y . 1 + x . \frac{1}{y} . \frac{dy}{dx} = \log x . \frac{dy}{dx} + y . \frac{1}{x}$

$\Rightarrow \displaystyle \log y + \frac{x}{y} \frac{dy}{dx} = \log x . \frac{dy}{dx} + \frac{y}{x}$

$\Rightarrow \displaystyle \left( \frac{x}{y} - \log x \right) \frac{dy}{dx} = \frac{y}{x} - \log y$

$\Rightarrow \displaystyle \left( \frac{x - y \log x}{y} \right) \frac{dy}{dx} = \frac{y - x \log y}{x}$

$\therefore \displaystyle \frac{dy}{dx} = \frac{y}{x} \left( \frac{y - x \log y}{x - y \log x} \right)$

If $x\sqrt {1+y}+y\sqrt{1+x}=0$ , for $-1< x< 1$ , prove that $\cfrac{dy}{dx}=\cfrac { 1 }{ { \left( 1+x \right) }^{ 2 } }$

Given,

$x\sqrt {1+y}+y \sqrt {1+x}=0$

If we differentiate this function with respect to

$x$ , then we get

$\sqrt { 1+y } +\dfrac { x }{ 2\sqrt { 1+y } } \dfrac { dy }{ dx } +\dfrac { y }{ 2\sqrt { 1+x } } +\dfrac { dy }{ dx } \sqrt { 1+x } =0$

Taking out

$\dfrac { dy }{ dx }$ one side, we get

$\dfrac { dy }{ dx } =\dfrac { 1 }{ { (1+x) }^{ 2 } }$

If $\sin y = x \sin (a + y)$ , then prove that $\dfrac{dy}{dx} = \dfrac{\sin^2 (a+y)}{\sin a}$ .

Given,

$\sin { y } =x\sin { \left( a+y \right) }$ .....(1)

$\sin { y=x\left[ \sin { a } \cos { y } +\cos { a } \sin { y } \right] }$

$I=x\left[ \cfrac { \sin { a } \cos { y } +\cos { a } \sin { y } }{ \sin { y } } \right]$

$I=x\left[ \sin { a } \cot { y } +\cos { a } \right]$

$x\sin { a } \cot { y } =-x\cos { a }$
Differentiate both sides with respect to

$x$

$\sin { a } \cot { y } +x\sin { a } \left[ -\csc ^{ 2 }{ y } \right] \cfrac { dy }{ dx } =-\cos { a }$

$\cfrac { dy }{ dx } =\cfrac { \cos { a } +\sin { a } \cot { y } }{ x\sin { a } \csc ^{ 2 }{ y } }$

$\cfrac { dy }{ dx } =\cfrac { \cos { a } +\cfrac { \sin { a } \cos { y } }{ \sin { y } } }{ x\cfrac { \sin { a } }{ \sin ^{ 2 }{ y } } }$

$\cfrac { dy }{ dx } =\left[ \cfrac { \cos { a\sin { y } + } \sin { a } \cos { y } }{ x\sin { a } } \right] \sin { y }$

$\cfrac { dy }{ dx } =\cfrac { \sin { \left( a+y \right) } }{ \sin { a } } \left[ \cfrac { \sin { y } }{ x } \right]$

$\cfrac { dy }{ dx } =\cfrac { \sin ^{ 2 }{ \left( a+y \right) } }{ \sin { a } }$
[ From equation (1) :

$\cfrac { \sin { y } }{ x } =\sin { \left( a+y \right) }$ ]

If $x^y + y^x = a^b$ , then find $\dfrac{dy}{dx}$ .

Given

${ x }^{ y }+{ y }^{ x }={ a }^{ b }$

$\Rightarrow y\times { x }^{ y-1 }+x\times { y }^{ x-1 }\times \cfrac { dy }{ dx } =0$ ...Differentiating on both sides

$\Rightarrow \cfrac { dy }{ dx } =\cfrac { -y\times { x }^{ y-1 } }{ x\times { y }^{ x-1 } } \\ \Rightarrow \cfrac { dy }{ dx } =\cfrac {-y\times { x }^{ y-2 } }{ x\times{ y }^{ x-2 } }$

Find $\cfrac { dy }{ dx }$ where $y=\log { \left( \sec { x } \right) \ \ } 0\le x\le \cfrac { \pi }{ 2 }$

$0\le x\le \frac { \pi }{ 2 }$

$\sec x$ is positive

Hence, the function

$y$ is defined on that interval

$y=\log { (\sec x) }$

Differentiate with respect to

$x$ on both sides

$\Longrightarrow \dfrac { dy }{ dx } =\dfrac { 1 }{\sec x } \dfrac { d }{ dx } (\sec x)\\ \Longrightarrow \dfrac { dy }{ dx } =\dfrac { 1 }{\sec x } \tan x \sec x$

$\Longrightarrow \dfrac { dy }{ dx } =\tan x$

If $x^y = a^x$ , prove that $\dfrac{dy}{dx}=\dfrac{x\log_ea-y}{x\log_ex}$ .

Given :

$x^y = a^x$

Taking log on both sides we get

$\log (x^y)=\log (a^x)$

$y \log x=x \log a$

Differentiating both sides we get

$\log x\dfrac{dy}{dx}+y\dfrac{1}{x}=\log a$

$\implies \log x\dfrac{dy}{dx}=\log a-\dfrac{y}{x}$

$\implies \log_e x\dfrac{dy}{dx}=\dfrac{x\log_e a-y}{x}$

$\implies \dfrac{dy}{dx}=\dfrac{x\log_e a-y}{x \log_e a}$

Solve: $\dfrac {dy}{dx} = \cos (x + y)$

Let

$x+y=2v$

$\Rightarrow 1+\dfrac{dy}{dx}=2\dfrac{dv}{dx}$

So given differential equation becomes,

$2\dfrac{dv}{dx}-1=\cos 2v$

$\Rightarrow 2\dfrac{dv}{dx}=1+\cos 2v=2\cos^2v$

$\Rightarrow \dfrac{dv}{dx}=\cos^2v$

$\Rightarrow \sec^2vdv=dx$

Now integrate both sides,

$\tan v=x+C$

$\Rightarrow \tan\dfrac{x+y}{2}=x+C$

If $y=f(x)$ is a differentiable function of $x$ , then show that
$\cfrac { { d }^{ 2 }x }{ d{ y }^{ 2 } } =-{ \left( \cfrac { dy }{ dx } \right) }^{ -3 }.\cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } }$

$y = f(x)$ is a differentiable function of x such that inverse function

$x = f^{-1}(y)$ exists, then

$\frac{dx}{dy} = \frac{1}{(\frac{dy}{dx})}$ . where

$\frac{dy}{dx} \neq 0$

$\therefore \frac{d^2 x}{dy^2} = \frac{d}{dx} \left ( \frac{dx}{dy} \right )$

$= \frac{d}{dy} \left [ \frac{1}{\left ( \frac{dy}{dx} \right )} \right ]$

$= \frac{d}{Dx} \left ( \frac{dy}{dx} \right )^{-1} \times \frac{dx}{dy}$

$= -1 \left ( \frac{dy}{dx} \right )^{-2} \cdot \frac{d}{dx} \left ( \frac{dy}{dx} \right ) \times \frac{1}{\left ( \frac{dy}{dx} \right )}$

$= - \left ( \frac{dy}{dx} \right )^{-2} \cdot \frac{d^2y}{dx^2} \cdot \left ( \frac{dy}{dx} \right )^{-1}$

$\therefore \frac{d^2x}{dy^2} = - \left ( \frac{dy}{dx} \right )^{-3} \cdot \frac{d^2 y}{dx^2}$

Solution of differential equation $x^2=1+(\dfrac{x}{y})^{-1}\, \dfrac{dy}{dx}+\dfrac{(\dfrac{x}{y})^{-2}\, (\dfrac{dy}{dx})^2}{2!}+\dfrac{(\dfrac{x}{y})^{-3}\,(\dfrac{dy}{dx})^3}{3!}+....$ is

Given,

${ x }^{ 2 }=1+{ \left( \dfrac { x }{ y } \right) }^{ -1 }\dfrac { dy }{ dx } +\dfrac { { \left( \dfrac { x }{ y } \right) }^{ -1 }\dfrac { dy }{ dx } }{ 2! } +\dfrac { { \left( \dfrac { x }{ y } \right) }^{ -1 }\dfrac { dy }{ dx } }{ 3! } +....$

$\Rightarrow { x }^{ 2 }=1+{ \left( \dfrac { x }{ y } \right) }^{ -1 }\dfrac { dy }{ dx } +\dfrac { { { \left( { \left( \dfrac { x }{ y } \right) }^{ -1 }\dfrac { dy }{ dx } \right) }^{ 2 } } }{ 2! } +\dfrac { { { \left( { \left( \dfrac { x }{ y } \right) }^{ -1 }\dfrac { dy }{ dx } \right) }^{ 3 } } }{ 3! } +....$

$\Rightarrow { x }^{ 2 }={ e }^{ { \left( \dfrac { x }{ y } \right) }^{ -1 }\dfrac { dy }{ dx } }$ $($ Using ${ e }^{ x }=1+x+\dfrac { { { x }^{ 2 } } }{ 2! } +\dfrac { { { x }^{ 3 } } }{ 3! } +....$ $)$

Taking log of both sides,

$\Rightarrow \log _{ e }{ { x }^{ 2 } } ={ \left( \dfrac { x }{ y } \right) }^{ -1 }\dfrac { dy }{ dx }$

$\Rightarrow \log _{ e }{ { x }^{ 2 } } .xdx=ydy$

Integrating both sides, we get,

$\Rightarrow \int { \log _{ e }{ { x }^{ 2 } } .xdx } =\int { ydy }$

Putting ${ x }^{ 2 }=t$ on Left-hand side, we get,

$\Rightarrow \dfrac { 1 }{ 2 } \int { \log _{ e }{ t } dt } =\int { y } dy$ $(Since \quad dt=2xdx)$

$\Rightarrow \dfrac { 1 }{ 2 } t\log _{ e }{ t } -\dfrac { 1 }{ 2 } t=\dfrac { { y }^{ 2 } }{ 2 } +C$ $(Using \int { \log _{ e }{ x } } dx =x\log _{ e }{ x } -x+C)$

$\Rightarrow \dfrac { 1 }{ 2 } { x }^{ 2 }\log _{ e }{ { x }^{ 2 } } -\dfrac { 1 }{ 2 } { x }^{ 2 }=\dfrac { { y }^{ 2 } }{ 2 } +C$

$\Rightarrow { y }^{ 2 }=2{ x }^{ 2 }\log _{ e }{ { x } } -{ x }^{ 2 }+C$

Hence the solution of given differential equation is ${ y }^{ 2 }=2{ x }^{ 2 }\log _{ e }{ { x } } -{ x }^{ 2 }+C$

If $(x^2+y^2)^2=xy$ , find $\displaystyle\frac{dy}{dx}$ .

Now the given equation is

$(x^2 +y^2)^2 =xy$

Or,

$(x^2)^2 + 2xy + (y^2)^2 = xy$

$\rightarrow x^4+xy+y^4 = 0$

Differentiating the above expression wrt

$x$ , we get

$4x^3+x\dfrac{dy}{dx} + y + 4y^3 \dfrac{dy}{dx}=0$

$\dfrac{dy}{dx} (x+4y^3) = -(y+4x^3)$

$\therefore$

$\dfrac{dy}{dx} = -\dfrac{(y+4x^3)}{(x+4y^3)}$

Differentiate the following functions w.r.t. $x$ .
$\sqrt{x+\displaystyle\frac{1}{x}}$ .

Given

$y=\sqrt{x+\frac{1}{x}}$

$\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x+\dfrac{1}{x}}}(1-\dfrac{1}{x^{2}})$

$=\dfrac{1}{2}\dfrac{\sqrt{x}}{\sqrt{x^{2}+1}}\dfrac{(x^{2}-1)}{x^{2}}$

$=\dfrac{1}{2}\dfrac{(x^{2}-1)}{\sqrt{x^{2}+1}.x^{2-\dfrac{1}{2}}}=\dfrac{1}{2}\dfrac{(x^{2}-1)}{(\sqrt{x^{2}+1})(x^{\dfrac{3}{2}})}$

If $y=1+x.{e}^{y}$ , show that $\dfrac { dy }{ dx } =\dfrac { { e }^{ y } }{ 2-y }$

Given

$y=1+x.{ e }^{ y }$ ,

$\Longrightarrow \dfrac { y-1 }{ { e }^{ y } } =x$ ,

Now differentiating the equation with respect to

$y$ on both sides gives,

$\dfrac { { e }^{ y }-\left( y-1 \right) { e }^{ y } }{ { e }^{ 2y } } =\dfrac { dx }{ dy } \\ \Longrightarrow \dfrac { 2-y }{ { e }^{ y } } =\dfrac { dx }{ dy } \\ \Longrightarrow \dfrac { dy }{ dx } =\dfrac { { e }^{ y } }{ 2-y } \\$

Find the equation of the tangent to the curve $y=\dfrac { x-y }{ \left( x-2 \right) \left( x-3 \right) }$ at the point where it cuts the x-axis.

The point at which it cuts

$x$ -axis is at

$y=0$ i.e

$(x,y)=(x,0)$

On substituting

$y=0$ in the equation of the curve we get

$x=0$

$\therefore$ the point is

$(0,0)$ .

Slope of tangent is

$\dfrac{dy}{dx}$ at

$(0,0)$ is

$\dfrac { dy }{ dx } =\dfrac { d }{ dx } \left\{\dfrac { x-y }{ (x-2)(x-3) } \right\}=\dfrac { (x-2)(x-3)(1-{ y }_{ 1 })-(x-y)(2x-5) }{ { \{ (x-2)(x-3)\} }^{ 2 } }$

On substituting

$x=0$ and

$y=0$ we get

${ y }_{ 1 }=\dfrac { 1 }{ 7 }$

$\therefore$ slope is

$\dfrac{1}{7}$

$\therefore$ the equation of the tangent is

$y=\dfrac{1}{7}x$ .

If $u$ and $v$ are differentiable functions of $x$ and if $y=u+v$ then $\cfrac { dy }{ dx } =\cfrac { du }{ dx } +\cfrac { dv }{ dx }$

Let

$y=u+v$ where

$u,v$ are differentiable fuctions of

$x$ .

Consider

$y=u+v$

Differentiate both sides with respect to

$x$ we get

$\dfrac{dy}{dx}=\dfrac{du}{dx}+\dfrac{dv}{dx}$

Calculate $\displaystyle\, f\, (x) + f'\, (x) + 2$ , if $f\, (x) = x\, \sin\, 2x$

Here, Given

$f(x) = x\sin 2x$

$f'(x) = \dfrac {d}{dx}(x\sin 2x)$

$= x\dfrac {d}{dx} (\sin 2x) + \sin 2x \dfrac {d}{dx} (x)$ [Product Rule of differentiation]

$= 2x \cos 2x + \sin 2x (1)$

$= 2x \cos 2x + \sin 2x$

So,

$f(x) + f'(x) + 2 = x\sin 2x + 2x \cos 2x + \sin 2x + 2$

$= \sin 2x (x + 1) + 2x \cos 2x + 2$

$= 2x \cos 2x + \sin 2x (x + 1) + 2$ .

Prove the following identities
$\displaystyle\, 2f'\, \left ( x +\tfrac{\pi }{3} \right )\, f'\, \left ( x -\tfrac{\pi }{6} \right )\, = f'\, (0)\, -\, f\, \left ( 2x +\tfrac{\pi }{6} \right )$ , where $f\, (x) = \cos\, x$

$2f'\left (x + \dfrac {\pi}{3}\right ) f' \left (x - \dfrac {\pi}{6}\right ) = f'(0) - f\left (2x + \dfrac {\pi}{6}\right )$ where

$f(x) = \cos x$

$f(x)= \cos x \Rightarrow f'(x) = -\sin x$

$f'\left (x + \dfrac {\pi}{3}\right ) = -\sin \left (x + \dfrac {\pi}{3}\right )= - \left (\sin x \cos \dfrac {\pi}{3} + \cos x \sin \dfrac {\pi}{3}\right )$

$= -\left (\dfrac {1}{2}\sin x + \dfrac {\sqrt {3}}{2} \cos x\right )$

$= \dfrac {-1}{2} [\sin + \sqrt {3}\cos x]$

$f'\left (x - \dfrac {\pi}{6}\right ) = -\sin \left (x - \dfrac {\pi}{6}\right ) = - (\sin x \cos \dfrac {\pi}{6} - \cos x \sin \dfrac {\pi}{6})$

$= -\left (\dfrac {\sqrt {3}}{2} \sin x - \dfrac {1}{2} \cos x\right ) = \dfrac {-1}{2} (\sqrt {3}\sin x - \cos x)$

$2f' \left (x + \dfrac {\pi}{3}\right ) f' \left (x - \dfrac {\pi}{6}\right ) = \dfrac {1}{2} [(\sin x + \sqrt {3}\cos x)(\sqrt {3}\sin x -\cos x)]$

$= \dfrac {1}{2} [\sqrt {3} \sin^{2}x + 2\sin x \cos x - \sqrt {3} \cos^{2} x]$

$= \dfrac {1}{2} [\sin 2x - \sqrt {3} \cos 2x]$

Now,

$f'(0) = 0$

$f\left (2x + \dfrac {\pi}{6}\right ) = \cos \left (2x + \dfrac {\pi}{6}\right ) = \left (\cos 2x \cos \dfrac {\pi}{6} - \sin 2x \sin \dfrac {\pi}{6}\right ) = \left (\dfrac {\sqrt {3}}{2} \cos 2x - \dfrac {1}{2} \sin 2x \right )$

$f'(0) - f\left (2x + \dfrac {\pi}{6}\right ) = \dfrac {-1}{2} (\sqrt {3}\cos 2x - \sin 2x) = \dfrac {1}{2} (\sin 2x - \sqrt {3} \cos 2x)$

$LHS = RHS$

If $y=(x^2+1)^5\sin 4x$ , find $\dfrac{dy}{dx}$ .

$\dfrac{dy}{dx} = 5(x^{2}_{}+1).2x.sin4x + 4cos4x(x^{2}_{}+1)$

$\dfrac{dy}{dx} = 10x(x^{2}_{}+1)sin4x + 4cos4x(x^{2}_{}+1)$

Find the derivative of $\displaystyle\, y \, =\, x\sqrt{1 \, +\, x^2}$

$y = x \sqrt {1 + x^{2}}$

$\dfrac {dy}{dx} = x \dfrac {d}{dx} (\sqrt {1 + x^{2}}) + \sqrt {1 + x^{2}} \dfrac {d}{dx} (x)$

$= x\times \dfrac {1}{2\sqrt {1 + x^{2}}}(2x) + \sqrt {1 + x^{2}}$

$= \dfrac {x^{2}}{\sqrt {1 + x^{2}}} + \sqrt {1 + x^{2}} \left [\because \dfrac {d}{dx} \sqrt {x} = \dfrac {1}{2\sqrt {x}}\right ]$

$= \dfrac {x^{2}}{\sqrt {1 + x^{2}}} + \sqrt {1 +x^{2}}$

$= \dfrac {x^{2} + 1 + x^{2}}{\sqrt {1 + x^{2}}} = \dfrac {2x^{2} + 1}{\sqrt {1 + x^{2}}}$ .

If $y=x^4+4^x+\log x-1$ , find $\cfrac{dy}{dx}$ .

Given,

$y=x^{4}+4^{x}+\log x-1$

Differentiate on both sides w.r.t x

$\dfrac{{d} y}{{d} x}=4x^{3}+4^{x}\log 4+\dfrac{1}{x}$

$\left [ \because \dfrac{{d} }{{d} x}(x^{n})=nx^{n-1} , \dfrac{{d} }{{d} x}(a^{x})=a^{x}\log a \right ]$

$\therefore \dfrac{{d} y}{{d} x}=4x^{3}+\dfrac{1}{x}+(\log 4)4^{x}$

If $x^3-y^3+\tan (xy)= \log (x+y)$ , find $\cfrac{dy}{dx}$

$x^3-y^3+\tan (xy)= \log (x+y)$

Differentiating w.r.t.

$x$

$3x^2-3y^2\large{\cfrac{dy}{dx}}$

$+\sec^2(xy)\left(y+x\large{\cfrac{dy}{dx}}\right)$

$=\large{\cfrac{1}{x+y}}$

$\left(1+\large{\cfrac{dy}{dx}}\right)$

$\Rightarrow 3x^2+y\sec^2(xy)-\large{\cfrac{1}{x+y}}$

$=\left(3y^2-x\sec^2(xy)+\large{\cfrac{1}{x+y}}\right)\cfrac{dy}{dx}$

$\Rightarrow \large{\cfrac{dy}{dx}}=\cfrac{3x^2+y\sec^2(xy)-\cfrac{1}{x+y}}{3y^2-x\sec^2(xy)+\cfrac{1}{x+y}}$

Find the derivative of the following
(i) $x^{x^n}+x^{n^x}$
(ii) $\ log{\sqrt{x^2+1}}$

$(i)\ y=x^{x^n}+x^{n^x}$

$y'=x^n(x^{x^n-1})nx^{n-1}+n^xx^{n^x-1}n^x\ln n$

$(ii)y=\ log{\sqrt{x^2+1}}$

$y'=\dfrac{1}{\sqrt{x^2+1}}\times \dfrac{2x}{2\sqrt{x^2+1}}$

$y'=\dfrac{x}{x^2+1}$

If $x = at^{4}, y = bt^{3}$ , then find $\dfrac {dy}{dx}$ .

This is called implicit differentiation or indirect differentiation. Here, both the given terms are differentiated independently w.r.t. a third variable and then combined together. Differentiating

$x$ and

$y$ w.r.t.

$t$ , we get

$\dfrac {dx}{dt} = 4at^{3}, \dfrac {dy}{dt} = 3bt^{2}$
Dividing

$\dfrac {dy}{dt} by \dfrac {dx}{dt}$ , we get

$\dfrac {dy}{dx} = \dfrac {dy}{dt} \times \dfrac {dt}{dx} = \dfrac {3bt^{2}}{4at^{3}} = \dfrac {3b}{4at}$ .

If $y = 4x^{4} + 2x^{3} + \dfrac {5}{x} + 9$ , then find $dy/dx$ .

$\dfrac {dy}{dx} = \dfrac {d}{dx} \left [4x^{4} + 2x^{3} + \dfrac {5}{x} + 9\right ]$
(Differentiating both sides w.r.t.

$x$ )

$= \dfrac {d}{dx} [4x^{4}] + \dfrac {d}{dx} [2x^{3}] + \dfrac {d}{dx}\left [\dfrac {5}{x}\right ] + \dfrac {d(9)}{dx}$ (Using the property of linearity)

$= \dfrac {d}{dx} [4x^{4}] + \dfrac {d}{dx} [2x^{3}] + \dfrac {d}{dx} \left [\dfrac {5}{x}\right ]$
(As differentiation of constant

$= 0, \dfrac {d9}{dx} = 0)$

$= 4\dfrac {d}{dx}, [x]^{4} + 2\dfrac {d}{dx}[x^{3}] + 5\dfrac {d}{dx} [x^{-1}]$ (Separating constant coefficients)

$= 4\times 4x^{3} + 2\times 3x^{2} + 5(-1)x^{-2}$

$= 16x^{3} + 6x^{2} - \dfrac {5}{x^{2}}$ .

If ${1 \over 2}\left( {{e^y} - {e^{ - y}}} \right) = x$ , prove that
$\cfrac { dy }{ dx } =\cfrac { 1 }{ \sqrt { { x }^{ 2 }+1 } }$

$\cfrac { { e }^{ y }-{ e }^{ -y } }{ 2 } =x\rightarrow 1$

${ e }^{ y }-{ e }^{ -y }=2x$

${ \left( { e }^{ y }+{ e }^{ -y } \right) }^{ 2 }={ \left( { e }^{ y }-{ e }^{ -y } \right) }^{ 2 }+4$

$=4{ x }^{ 2 }+4$

$\therefore { e }^{ y }+{ e }^{ -y }=2\sqrt { { x }^{ 2 }+1 }$

Differentiate

$1$ w.r.t. x

$\cfrac { { e }^{ y }+{ e }^{ -y } }{ 2 } \left( \cfrac { dy }{ dx } \right) =1$

$\cfrac { 2\sqrt { { x }^{ 2 }+1 } }{ 2 } \left( \cfrac { dy }{ dx } \right) =1$

$\therefore \cfrac { dy }{ dx } =\cfrac { 1 }{ \sqrt { { x }^{ 2 }+1 } }$

$y={ e }^{ \left\{ \sin ^{ 2 }{ x } +\sin ^{ 4 }{ x } +\sin ^{ 6 }{ x } +... \right\} }$ , then $\cfrac { dy }{ dx } =$

$\cfrac { a }{ 1-r }$ , where

$a=\sin ^{ 2 }{ x }$

$r=\sin ^{ 2 }{ x }$

$=\cfrac { \sin ^{ 2 }{ x } }{ 1-\sin ^{ 2 }{ x } } =\tan ^{ 2 }{ x }$

$y={ e }^{ \tan ^{ 2 }{ x } }$

Differentiating with respect to x,

$\cfrac { dy }{ dx } =2{ e }^{ \tan ^{ 2 }{ x } }\tan { x } \sec ^{ 2 }{ x }$

Differentiate the following
${ x }^{ 2 }{ (3x-2) }^{ 4 } \cos x$

Let

$y=x^{2}(3x-2)^{4}\cos x$

Differentiate on both sides w.r.t

$x$

$\dfrac{{d} y}{{d} x}=2x(3x-2)^{4}\cos x+x^{2}(4(3x-2)^{3})3\cos x-x^{2}(3x-2)^{4}\sin x$

$\therefore \dfrac{{d} y}{{d} x}=x(3x-2)^{3}[(6x-4)\cos x+12x\cos x-x(3x-2)\sin x]$

$\therefore \dfrac{{d} y}{{d} x}=x(3x-2)^{3}[(18x-4)\cos x-(3x^{2}-2x)\sin x]$

Differentiate
$f\left( x \right) = \ln \left( {1 + {x^2}} \right)$

Given

$f(x)=ln(1+x^{2})$

Differentiate on both sides w.r.t x

$\frac{\mathrm{d} f(x)}{\mathrm{d} x}=\frac{1}{1+x^{2}}\frac{\mathrm{d} }{\mathrm{d} x}(1+x^{2})$

${f}'(x)=\frac{2x}{1+x^{2}}$

$\therefore {f}'(x)=\frac{2x}{1+x^{2}}$

Let $x^2y^3=(x+y)^5$ then find $\dfrac{dy}{dx}$

Given,

$x^2y^3=(x+y)^5$

Now taking logarithm with respect to the base

$e$ both sides we get,

$2\log x+3\log y=5\log(x+y)$

Now differentiating both sides with respect to

$x$ we get,

$\dfrac{2}{x}+\dfrac{3}{y}\dfrac{dy}{dx}=\dfrac{5}{x+y}\left(1+\dfrac{dy}{dx}\right)$

or,

$\left(\dfrac{2}{x}-\dfrac{5}{x+y}\right)=\left(\dfrac{5}{x+y}-\dfrac{3}{y}\right)\dfrac{dy}{dx}$

or,

$\left(\dfrac{2y-3x}{x(x+y)}\right)=\left(\dfrac{2y-3x}{y(x+y)}\right)\dfrac{dy}{dx}$

or,

$\dfrac{dy}{dx}=\dfrac{y}{x}$ .

If $x^2+y^2+ 2gx + 2fy + 2hxy +c =0$ find $\frac{dy}{dx}$

Solution:-

${x}^{2} + {y}^{2} + 2gx + 2fy + 2hxy + c = 0$

Differentiating the given equation w.r.t. x, we have

$2x + 2y \cfrac{dy}{dx} + 2g + 2f \cfrac{dy}{dx} + 2h \left( \left( 1 \times y \right) + x\cfrac{dy}{dx} \right) + 0 = 0$

$\Rightarrow \; \left( 2x + 2g + 2hy \right) + \cfrac{dy}{dx} \left( 2y + 2f + 2hx \right) = 0$

$\Rightarrow \; \cfrac{dy}{dx} = \cfrac{- \left( 2x + 2g + 2hy \right)}{\left( 2y + 2f + 2hc \right)}$

If $y = \log \left[ {\dfrac{{x + \sqrt {{x^2} + 25} }}{{\sqrt {{x^2} + 25 - x} }}} \right]$ , find $\dfrac{{dy}}{{dx}}$

$y=\log\left[\dfrac{x+\sqrt{x^2+25}}{\sqrt{x^2+25}-x}\right]$

$\Rightarrow e^y=\dfrac{x+\sqrt{x^2+25}}{\sqrt{x^2+25}-x}$

$\Rightarrow \dfrac{e^y+1}{e^y-1}=\dfrac{2\sqrt{x^2+25}}{2x}=\sqrt{1+\dfrac{25}{x^2}}$

$\Rightarrow \left[\dfrac{e^y(e^y-1)-e^y(e^y+1)}{(e^y-1)^2}\right]\dfrac{dy}{dx}=\dfrac{\dfrac{-50}{x^3}}{2\sqrt{1+\dfrac{25}{x^2}}}$

$\Rightarrow \dfrac{-2e^y}{(e^y-1)^2}\dfrac{dy}{dx}=\dfrac{\dfrac{-25}{x^3}}{\dfrac{\sqrt{x^2+25}}{x}}$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{25}{x^2\sqrt{x^2+25}}.\dfrac{(e^y-1)^2}{2e^y}$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{25}{x^2\sqrt{x^2+25}}=\dfrac{\left(\dfrac{x+\sqrt{x^2+25}}{\sqrt{x^2+25-x}}-1\right)^2}{2.\dfrac{x+\sqrt{x^2+25}}{\sqrt{x^2+25}-x}}$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{25(x^2+25)-x}{2x^2\sqrt{x^2+25}(x+\sqrt{x^2+25})}.\left(\dfrac{2x}{x^2+25}\right)^2$

$=\dfrac{25\times 2}{(\sqrt{x^2+25})(x+\sqrt{x^2+25})(\sqrt{x^2+25}-x)}$

$=\dfrac{25\times 2}{\sqrt{x^2+25(x^2+25-x^2)}}$

$=\boxed{\dfrac{2}{\sqrt{x^2+25}}}$

If $y = sin^2(\log (2x + 3))$ , find $\dfrac{dy}{dx}$ .

Given,

$y = \sin^2 (log(2x + 3))$

$y = [\sin (log (2x + 3))]^2$

$\therefore \dfrac{dy}{dx} = \dfrac{d}{dx}[\sin (log (2x + 3))]^2$

$= 2\sin(log (2x + 3)) \dfrac{d}{dx}(\sin(log(2x + 3)))$

$= 2\sin(log (2x + 3)) (\cos(log(2x + 3)))\left(\dfrac{2}{2x+3}\right)$

If ${\left( {{x^2} + {y^2}} \right)^2} = xy,$ then $\left (\cfrac{dy}{dx} \right )$ is,

We have,

$(x^2+y^2)^2=xy$

On differentiating w.r.t

$x$ , we get

$\dfrac{d}{dx}(x^2+y^2)^2=\dfrac{d}{dx}(xy)$

$2(x^2+y^2)\dfrac{d}{dx}(x^2+y^2)=x\dfrac{d}{dx}(y)+y\dfrac{d}{dx}(x)$

$2(x^2+y^2)(2x+2y\dfrac{dy}{dx})=x\dfrac{dy}{dx}+y\times 1$

$4(x^2+y^2)(x+y\dfrac{dy}{dx})=x\dfrac{dy}{dx}+y$

$4x(x^2+y^2)+4xy(x^2+y^2)\dfrac{dy}{dx}=x\dfrac{dy}{dx}+y$

$4xy(x^2+y^2)\dfrac{dy}{dx}-x\dfrac{dy}{dx}=y-4x(x^2+y^2)$

$\left[4xy(x^2+y^2)-x\right]\dfrac{dy}{dx}=y-4x(x^2+y^2)$

$\dfrac{dy}{dx}=\dfrac{y-4x(x^2+y^2)}{[4xy(x^2+y^2)-x]}$

Hence, this is the answer.

Let $y=xe^x$ then prove that $x\dfrac{dy}{dx}=(1+x)e^x$ .

Given,

$y=xe^x$

Now differentiating both sides with respect to

$x$ we get,

$\dfrac{dy}{dx}=xe^x+e^x$

or,

$x\dfrac{dy}{dx}=x^2e^x+xe^x$

or,

$x\dfrac{dy}{dx}=(x+1)e^x$

Find $\dfrac{{dy}}{{dx}}\,,\,if\,\,{x^{\dfrac{2}{3}}}\, + \,{y^{\dfrac{2}{3}\,}}\, = \,{a^{\dfrac{2}{3}}}.$

Let

$x^{\dfrac{2}{3}}+y^{\dfrac{2}{3}}=a^{\dfrac{2}{3}}$

Now differentiating both sides with respect to

$x$ we get,

$\dfrac{2}{3}x^{-\large{\frac{1}{3}}}$

$+\dfrac{2}{3}y^{-\large{\frac{1}{3}}}\dfrac{dy}{dx}=0$

or,

$\dfrac{dy}{dx}=-\dfrac{y^{\large{\frac{1}{3}}}}{x^{\large{\frac{1}{3}}}}$

$\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}=4$ , Then $\dfrac{dy}{dx}=$

$\sqrt{\dfrac{x}{y}} + \sqrt{\dfrac{y}{x}} = 4$

$\implies \dfrac{x + y}{\sqrt{xy}} = 4$

$\implies x + y = 4\sqrt{xy}$

$\implies (x + y)^2 = (4\sqrt{xy} )^2$

$\implies x^2 + y^2 + 2xy = 16xy$

$\implies x^2 + y^2 = 14xy$

$\implies 2x + 2y \dfrac{dy}{dx} = 14[x\dfrac{dy}{dx} + y ]$

$\implies 2x - 14y = (14x - 2y)\dfrac{dy}{dx}$

$\implies \dfrac{dy}{dx} = \dfrac{2x - 14y}{14x - 2y}$

$\therefore \dfrac{dy}{dx} = \dfrac{x - 7y}{7x - y}$

If $\sin(x+y)=\log(x+y)$ , then: $\dfrac{dy}{dx}$ =

$\sin(x+y)=\log(x+y)$

$\dfrac{d}{dx}[\sin(x+y)]=\dfrac{d}{dx}[\log(x+y)]$

$\cos(x+y)\dfrac{d}{dx}(x+y)=\dfrac{1}{x+y}\dfrac{d}{dx}(x+y)$

$\cos(x+y)\left(1+\dfrac{dy}{dx}\right)=\dfrac{1}{x+y}\left(1+\dfrac{dy}{dx}\right)$

$\cos(x+y)\left(1+\dfrac{dy}{dx}\right)-\dfrac{1}{x+y}\left(1+\dfrac{dy}{dx}\right)=0$

$\left(1+\dfrac{dy}{dx}\right)\left(\cos(x+y)-\dfrac{1}{x+y}\right)=0$

$\Rightarrow$

$1+\dfrac{dy}{dx}=0$

$\therefore$

$\dfrac{dy}{dx}=-1$

Find $f'(x)$
$f\left( x \right) = \sqrt {{4^x} - {{\left( {44} \right)}^x}} + \frac{3}{{{{\log }_4}\left( {1 - x} \right)}}$

Given ,

$f(x)=\sqrt{{{4}^{x}}+{{44}^{x}}}+\dfrac{3}{{{\log }_{4}}\left( 1-x \right)}$

Differentiate both side w.r. to x , we get

$f'(x)=\dfrac{1}{2\sqrt{{{4}^{x}}+{{44}^{x}}}}\dfrac{d}{dx}\left( {{4}^{x}}+{{44}^{x}} \right)+3\dfrac{d}{dx}\dfrac{1}{{{\log }_{4}}\left( 1-x \right)}$

$=\dfrac{1}{2\sqrt{{{4}^{x}}+{{44}^{x}}}}\left( {{4}^{x}}{{\log }_{e}}4+{{44}^{x}}{{\log }_{e}}44 \right)+3\left( -1 \right)\dfrac{1}{{{\left( {{\log }_{4}}\left( 1-x \right) \right)}^{2}}}\dfrac{d}{dx}{{\log }_{4}}\left( 1-x \right)$

$=\dfrac{{{4}^{x}}{{\log }_{e}}4+{{44}^{x}}{{\log }_{e}}44}{2\sqrt{{{4}^{x}}+{{44}^{x}}}}-\dfrac{3}{{{\left( {{\log }_{4}}\left( 1-x \right) \right)}^{2}}}\times \dfrac{1}{\left( 1-x \right){{\log }_{e}}4}\dfrac{d}{dx}\left( 1-x \right)$

$=\dfrac{{{4}^{x}}{{\log }_{e}}4+{{44}^{x}}{{\log }_{e}}44}{2\sqrt{{{4}^{x}}+{{44}^{x}}}}+\dfrac{3}{{{\left( {{\log }_{4}}\left( 1-x \right) \right)}^{2}}}\times \dfrac{1}{\left( 1-x \right){{\log }_{e}}4}$

Differentiate the following w.r.t. x:
$e^{x^3}$ .

let

$y=e^{x^3}$

how take

$\log_e$ on both side

$\log_e y=x^3$

now different above wrt

$x.$ we get,

$\dfrac {1}{y} \dfrac {dy}{dx}=3x^2$

$\dfrac {dy}{dx}=y.(3x^2)$

$\boxed {\dfrac {dy}{dx}=e^{x^3} (3x^2)}$

Let $xy=x+y$ then prove that $\dfrac{dy}{dx}+\dfrac{1}{(x-1)^2}=0$ .

Given,

$xy=x+y$

or,

$(x-1)y=x$

or,

$y=\dfrac{x}{x-1}$

Now differentiating both sides with respect to

$x$ we get,

$\dfrac{dy}{dx}=\dfrac{(x-1).1-x.1}{(x-1)^2}=-\dfrac{1}{(x-1)^2}$

or,

$\dfrac{dy}{dx}+\dfrac{1}{(x-1)^2}=0$ .

Find $\dfrac{dy}{dx}$ for $y=\log\left(\dfrac{1-x^2}{1+x^2}\right)$ .

Given,

$y=\log\left(\dfrac{1-x^2}{1+x^2}\right)$

or,

$y=\log(1-x^2)-\log(1+x^2)$

Now, differentiating both sides with respect to

$x$ we get,

$\dfrac{dy}{dx}=\dfrac{-2x}{1-x^2}-\dfrac{2x}{1+x^2}$

or,

$\dfrac{dy}{dx}=\dfrac{-4x}{1-x^4}$ .

If $y = \dfrac{{{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}$ , prove that $\left( {1 - {x^2}} \right)\dfrac{{dy}}{{dx}} - xy = 1$

Consider the given function.

$y=\dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}$

On differentiating both sides w.r.t $x$ , we get

$\dfrac{dy}{dx}=\dfrac{\left( \sqrt{1-{{x}^{2}}} \right)\times \dfrac{1}{\sqrt{1-{{x}^{2}}}}-{{\sin }^{-1}}x\times \dfrac{1}{2\sqrt{1-{{x}^{2}}}}\times -2x}{{{\left( \sqrt{1-{{x}^{2}}} \right)}^{2}}}$

$\dfrac{dy}{dx}=\dfrac{1+\dfrac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}}{1-{{x}^{2}}}$

$\dfrac{dy}{dx}=\dfrac{1+xy}{1-{{x}^{2}}}$

$\left( 1-{{x}^{2}} \right)\dfrac{dy}{dx}=1+xy$

$\left( 1-{{x}^{2}} \right)\dfrac{dy}{dx}-xy=1$

Hence, proved

If $y={(2x+3)}^{(3x-5)},$ find $\dfrac{dy}{dx}$ .

Solution:-

Given:-

$y = {\left( 2x + 3 \right)}^{\left( 3x - 5 \right)}$

To find:-

$\cfrac{dy}{dx} = ?$

$y = {\left( 2x + 3 \right)}^{\left( 3x - 5 \right)}$

Taking log both sides, we have

$\log{y} = \log{ \left( {\left( 2x + 3 \right)}^{\left( 3x - 5 \right)} \right)}$

$\log{y} = \left( 3x - 5 \right) \log{\left( 2x + 3 \right)}$

Now, differentiating both sides w.r.t. x, we have

$\cfrac{d}{dx} \left( \log{y} \right)= \left( 3x - 5 \right). \cfrac{d}{dx} \left( \log{\left( 2x + 3 \right)} \right) + \left( \log{\left( 2x + 3 \right)} \right). \cfrac{d}{dx} \left( 3x - 5 \right)$

$\cfrac{1}{y} \cfrac{dy}{dx} = \left( 3x - 5 \right) \left( \cfrac{1}{2x + 3} . 2 \right) + \left( \log{\left( 2x + 3 \right)} \right) \left( 3 \right)$

$\cfrac{dy}{dx} = y \left( \cfrac{2\left( 3x - 5 \right)}{2x + 3} + 3 \log{\left( 2x + 3 \right)} \right)$

$\Rightarrow \; \cfrac{dy}{dx} = \left( {\left( 2x + 3 \right)}^{\left( 3x - 5 \right)} \right) \left( \cfrac{\left( 6x - 10 \right)}{2x + 3} + 3 \log{\left( 2x + 3 \right)} \right)$

Let $f:\left[ {1,\infty } \right] \to \left[ {2,\infty } \right]$ if $f\left( 1 \right) = 2.$ be differentiable function such that $6\int_1^x {f\left( t \right)dt = 3xf\left( x \right) - {x^3}}$ then the value of $f\left( 2 \right)$ is....

$6\displaystyle\int_{1}^{2}f(t) dt=3x f(x)-x^{3}$

Diff. both sides.

$6f(x)=3f(x)+3x f(x)-3x^{2}$

$y=\dfrac{x\ dy}{dx}-x^{2}$

$\dfrac{dy}{dx}-\dfrac{y}{x}=x$

$I.E=e^{\displaystyle\int\dfrac{1}{x}\ dx}=\dfrac{1}{x}$

$\dfrac{y}{x}=\displaystyle\int\dfrac{1}{x}\times x\ dx$

$y=x^{2}+C$

$f(1)=2$

$C=1$

$y=x^{2}+1$

$f(2)=2^{2}+1=5$

If $y={y}^{\sqrt{x}}+{x}^{\sqrt{e}}$ , find $\cfrac{dy}{dx}$

$y=y^{\sqrt x}+x^{\sqrt e}$ if

$y^{\sqrt x}=\mu$

$\Rightarrow \ y=\mu +x^{\sqrt x}$

$\dfrac {dy}{dx}=\dfrac {d\mu}{dx}=\sqrt e (x^{\sqrt e-1})---(1)$

$y^{\sqrt x}=\mu \Rightarrow \ \sqrt x \log y=\log u$

$\Rightarrow \sqrt x+\dfrac {1}{y} \dfrac {dy}{dx}+\dfrac {1}{2\sqrt x} \log y =\dfrac {1}{\mu} \dfrac {d\mu }{dx}$

$\Rightarrow \ \dfrac {d\mu}{dx} =\mu \left (\dfrac {\sqrt x}{y} \dfrac {dy}{dx}+\dfrac {1}{2\sqrt x}\log y \right)----(2)$

$\therefore \ \dfrac {dy}{dx}=\dfrac {\mu \sqrt x}{y} \dfrac {dy}{dx}+\dfrac {\mu}{2\sqrt x} \log y +\sqrt e (x^{\sqrt e -1})$

$\dfrac {dy}{dx} \left (\dfrac {y-\mu \sqrt x}{y}\right)=\dfrac {\mu}{2\sqrt x} \log y+\sqrt e(x^{\sqrt e -1})$

$\Rightarrow \ \dfrac {dy}{dx}=\left (\dfrac {y}{y-\mu \sqrt x}\right) \left (\dfrac {\mu}{2\sqrt x} \log y+\sqrt e (x^{\sqrt e-1})\right)$

where

$\mu =y^{\sqrt x}$ Answer

If $y+\sin { y } =\cos { x }$ , find $\cfrac { dy }{ dx }$

Given,

$y+\sin y=\cos x$

differentiating on both sides with respect to

$x$

$(1+\cos y)\dfrac{d{y}}{d{x}}=-\sin x$

$\dfrac{d{y}}{d{x}}=-\dfrac{\sin x}{1+\cos y}$

Find $\dfrac {dy}{dx}$ if $y=\cos^{-1}(e^{x})+\sqrt {\sqrt {e^{\sqrt {x}}}}$

Consider the following equation.

$y={{\cos }^{-1}}\left( {{e}^{x}} \right)+\sqrt{\sqrt{{{e}^{\sqrt{x}}}}}$

$y={{\cos }^{-1}}\left( {{e}^{x}} \right)+{{\left( {{e}^{\sqrt{x}}} \right)}^{\frac{1}{4}}}$

$\dfrac{dy}{dx}=\dfrac{-1}{\sqrt{1-{{x}^{2}}}}{{e}^{x}}+\dfrac{1}{4({{e}^{\sqrt{x}}})^{3/4}}{{e}^{\sqrt{x}}}\dfrac{1}{2\sqrt{x}}$

$=\dfrac{-1}{\sqrt{1-{{x}^{2}}}}{{e}^{x}}+\dfrac{(e^{\sqrt{x}})^{1/4}}{8\sqrt{x}}$

Hence, this is the required answer..

If ${ x }^{ y }={ a }^{ x }.Prove \ that\ \cfrac { dy }{ dx } =\cfrac { x\log _{ e }{ a-y } }{ x\log _{ e }{ x } }$

$x^{y}=a^{x}$

taking logarith on both sides , we get

$y$ log

$x=x$ log

$a$

differentiating both sides with respect to

$x$

$\dfrac{dy}{dx}$ log

$x + y.\dfrac{1}{x}=$ log

$a$

$\Rightarrow \dfrac{dy}{dx}\times x$ log

$x+y=x$ log

$a$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{x log a -y}{xlogx}$

Solve: $\dfrac{d}{dx}x\sin^2 x$

$\quad \dfrac { d }{ dx } \left[ x\sin ^{ 2 }{ x } \right] \\ =\dfrac { d }{ dx } \left[ x \right] .\sin ^{ 2 }{ x } +x.\dfrac { d }{ dx } \left[ \sin ^{ 2 }{ x } \right] \\ =\sin ^{ 2 }{ x+2\sin { x.\dfrac { d }{ dx } \left[ sinx \right] .x } } \\ =\sin ^{ 2 }{ x+2x\cos { x } \sin { x } }$

Differentiate w.r.t $x$ :
$\dfrac { 3 }{ 4 } \log { { x }^{ 2 } }$

$y=\dfrac 34\log x^2$

$\dfrac{dy}{dx}=\dfrac d{dx}\dfrac 34\log x^2$

$=\dfrac 34 \dfrac 1{x^2}\dfrac d{dx} x^2$

$=\dfrac 3{4x^2} 2x$

$=\dfrac 3{2x}$

If $y = \dfrac{1}{{\sqrt {{a^2} - {x^2}} }}$ , find $\dfrac{{dy}}{{dx}}$

$y=\dfrac{1}{\sqrt{a^2-x^2}}$

$\dfrac{dy}{dx}=\dfrac{d}{dx}\left(\dfrac{1}{\sqrt{a^2-x^2}}\right)$

$=\dfrac{\sqrt{a^2-x^2}\cdot \dfrac{d}{dx}(1)-1\cdot \dfrac{d}{dx}(\sqrt{a^2-x^2})}{(\sqrt{(a^2-x^2})^2}$

$=\dfrac{0-\dfrac{-2x}{2\sqrt{a^2-x^2}}}{(a^2-x^2)}$

$=\dfrac{x}{(a^2-x^2)^{3/2}}$ .

If y=tan x.log{sin x}
then $y'$ at $x=\dfrac{\pi}{4}$ is?

$\begin{matrix} y=\log \sin x\tan x,{ y^{ 1 } }\, \, at\, \, x=\frac { \pi }{ 4 } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ { \because \log { a^{ b } } =b\log a } \right] \\ \Rightarrow y=\tan x\log \sin x \\ \frac { { dy } }{ { dx } } =\tan x\left[ { \frac { { d\left( { \log \sin x } \right) } }{ { d\left( { \sin x } \right) } } } \right] \times \frac { { d\left( { \sin x } \right) } }{ { dx } } +\log \sin x\times \frac { { d\tan x } }{ { dx } } \\ \frac { { dy } }{ { dx } } =\tan x\left[ { \frac { 1 }{ { \sin x } } \times \cos x } \right] +\log \sin x\times \left( { { { \sec }^{ 2 } }x } \right) \\ Put\, \, x=\frac { \pi }{ 4 } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ { \because a\log b=\log { b^{ a } } } \right] \\ \frac { { dy } }{ { dx } } \int { x=\frac { \pi }{ 4 } =\left[ { \cot \frac { \pi }{ 4 } } \right] } ,\, \, \tan \frac { \pi }{ 4 } +{ \sec ^{ 2 } }\frac { \pi }{ 4 } \log \sin \frac { \pi }{ 4 } \\ [\frac { { dy } }{ { dx } } \int { x= } \frac { \pi }{ 4 } =1+2\log \frac { 1 }{ { \sqrt { 2 } } } =1+\log { \left( { \frac { 1 }{ { \sqrt { 2 } } } } \right) ^{ 2 } } =\, \, \boxed { 1+\log \frac { 1 }{ 2 } } \, \, \, \underline { Ans. } \\ \end{matrix}$

If $\sqrt{x}+\sqrt{y}=\sqrt{a}$ prove that $\dfrac{dy}{dx}=-\sqrt{\dfrac{y}{x}}$ .

$\sqrt { x } +\sqrt { y } =\sqrt { a }$

On differentiating , we get

$\cfrac { 1 }{ 2\sqrt { x } } +\cfrac { 1 }{ 2\sqrt { y } } \cfrac { dy }{ dx } =0\\ \cfrac { dy }{ dx } =-\sqrt { \cfrac { y }{ x } }$

If $y=\log _{ 5 }{ \left( \log _{ 7 }{ x } \right) }$ , find $\dfrac {dy}{dx}$

$y=\log_5 (\log_7 x)$

$\Rightarrow \ \dfrac {dy}{dx} =\dfrac {1}{\log_7 x \log_e 5}\times \dfrac {1}{x \log_e 7}$

$=\dfrac {1}{x\log_7 x \log_e 5.\log_e 7}$

Find $\dfrac {dy}{dx}$ at $x=1,y=\dfrac {\pi}{4}$ if $\sin^{2}y+\cos xy=k$ .

$\sin^{2}y+\cos xy=k$

$2 \sin y \cos y\dfrac{dy}{dx}+(-\sin xy)\left(y+x\dfrac{dy}{dx}\right)=0$

Put

$y=\dfrac{\pi }{4},x=1$

$2\times \dfrac{1}{\sqrt{2}}\times \dfrac{1}{\sqrt{2}}\dfrac{dy}{dx}-\dfrac{1}{\sqrt{2}}\left(\dfrac{\pi }{4}+\dfrac{dy}{dx}\right)=0$

$\displaystyle \frac{dy}{dx}-\frac{1}{\sqrt{2}}\frac{dy}{dx}=\frac{\pi }{4\sqrt{2}}$

$\displaystyle \frac{dy}{dx}=\frac{\pi }{4(\sqrt{2}-1)}$

Differentiate the following w.r.t.x:
$\sqrt{x} \cdot cosec^{-1}\left(\dfrac{1}{4x}\right)$

1337007_1124585_ans_f5598cc143a841a9bb6aff88137b9916.jpg

find $\dfrac{dy}{dx}$ .

We have,

$y^{2}=(x-c)^{3}$

On differentiating w.r.t

$x$ , we get

$2y\dfrac{dy}{dx}=3(x-c)^2$

$\dfrac{dy}{dx}=\dfrac{3(x-c)^2}{2y}$

Hence, this is the answer.

If $y=\dfrac{\sin^2x}{1+\cot x}+\dfrac{\cos^2x}{1+\tan x}$ then $y^{'}$ is equal to?

$y = \cfrac{\sin^2x}{1+\cot x} + \cfrac{\cos^2x}{1+\tan x}$

$y = \cfrac{\sin^3x}{\sin x+\cos x}+\cfrac{\cos^3x}{\sin x+\cos x}$

$\cfrac{(\sin x+\cos x)^3 - 3\sin x\cos x(\sin x+\cos x)}{\sin x + \cos x}$

$(\cos x + \sin x)^2 - 3 \sin x \cos x$

$y = 1 - \sin x \cos x$

$y^{'} = -(\cos^2 x - \sin^2x)$

$y^{'} = -\cos 2x$

Find derivative of $(ax + b)^n (cx + d)^n$

$f(x)=(ax+b)^n(cx+d)^n$

$u=(ax+b)^n\Rightarrow u'=an(ax+b)^{n-1}$

$v=(cx+d)^n\Rightarrow v'=cn(cx+d)^{n-1}$

$f'(x)=(uv)'$

$=uv'+vu'$

$=(ax+b)^n cn(cx+b)^{n-1}+(cx+b)^{n} an(ax+b)^{n-1}$

If $x^y - y^x= 1$ , then the value of $\dfrac {dy}{dx}$ is :

$\begin{array}{l} { x^{ y } }-{ y^{ x } }=1 \\ { y_{ 1 } }={ x^{ y } } \\ \log { y_{ 1 } } =y\log x \\ \frac { { dy' } }{ { dx } } ={ x^{ y } }\left( { \frac { y }{ x } +\log x.\frac { { dy } }{ { dx } } } \right) \\ { y_{ 2 } }={ y^{ x } } \\ \log { y_{ 2 } } =x\log y \\ \frac { { d{ y_{ 2 } } } }{ { dx } } ={ y^{ x } }\left( { \frac { x }{ y } .\frac { { dy } }{ { dx } } +\log y } \right) \\ \therefore { x^{ y } }\left( { \frac { y }{ x } +\log x.\frac { { dy } }{ { dx } } } \right) ={ y^{ x } }\left( { \frac { x }{ y } .\frac { { dy } }{ { dx } } +\log y } \right) \\ y.{ x^{ 4-1 } }+{ x^{ y } }\log x.\frac { { dy } }{ { dx } } =x{ y^{ x-1 } }\frac { { dy } }{ { dx } } +{ y^{ x } }\log y \\ \left( { { x^{ y } }\log x-x{ y^{ x-1 } } } \right) \frac { { dy } }{ { dx } } ={ y^{ x } }\log y-4{ x^{ y-1 } } \\ \frac { { dy } }{ { dx } } =\left( { \frac { { { y^{ x } }\log y-y{ x^{ y-1 } } } }{ { { x^{ y } }.\log x-x{ y^{ x-1 } } } } } \right) \end{array}$

Differentiate $f(x) = \dfrac{{\log \left( {1 + 3x} \right) - \log x}}{{\left( {1 - 2x} \right)}}$

Consider the function

$f(x) = \dfrac{{\log \left( {1 + 3x} \right) - \log x}}{{\left( {1 - 2x} \right)}}$

Apply the quotient rule

$\begin{array}{c} \dfrac { { \dfrac { d }{ { dx } } \left( { \log \left( { 1+3x } \right) -\log x } \right) \times \left( { 1-2x } \right) -\left( { \log \left( { 1+3x } \right) -\log x } \right) \dfrac { d }{ { dx } } \left( { 1-2x } \right) } }{ { { { \left( { 1-2x } \right) }^{ 2 } } } } \\ =\dfrac { { \left( { \dfrac { 1 }{ { 1+3x } } \cdot 3-\dfrac { 1 }{ x } } \right) \times \left( { 1-2x } \right) -\left( { \log \left( { 1+3x } \right) -\log x } \right) \left( { -2 } \right) } }{ { 1+4{ x^{ 2 } }-4x } } \end{array}$

Hence, $f'\left( x \right) = \dfrac{{\left( {\dfrac{1}{{1 + 3x}} \cdot 3 - \dfrac{1}{x}} \right) \times \left( {1 - 2x} \right) - \left( {\log \left( {1 + 3x} \right) - \log x} \right)\left( { - 2} \right)}}{{1 + 4{x^2} - 4x}}$

If $x=at^2$ and $y=2at$ , then find $\dfrac{d^2y}{dx^2}$ .

Given,

$x=at^2$ and

$y=2at$

On differentiating both sides w.r.t. t, we get,

$\dfrac{dx}{dt}=2at$ and

$\dfrac{dy}{dt}=2a$

Therefore,

$\dfrac{dy}{dx}=\dfrac{2a}{2at}=\dfrac{1}{t}$

Now,

$\dfrac{d^2y}{dx^2}=\dfrac{d}{dt}(\dfrac{dy}{dx})\times \dfrac{dt}{dx}$

$=\dfrac{d}{dt}(\dfrac{1}{t}) \times \dfrac{1}{2at} =-\dfrac{1}{t^2} \times \dfrac{1}{2at}=-\dfrac{1}{2at^3}$

${\text{if}}\;{\text{y = }}{{\text{e}}^{\text{x}}}{\text{ cos x,prove}}\;{\text{that}}\;$ $\dfrac{{dy}}{{dx}} = \sqrt 2 {e^x}\cos \left( {x + \dfrac{\pi }{4}} \right)$

Given

$y=e^x\cos x$

Differentiating w.r.t. x

$\dfrac{dy}{dx}=e^x(-\sin x)+\cos xe^x$ [Product rule]

$=e^x[\cos x-\sin x]$

$=\sqrt{2}e^x\left[\dfrac{1}{\sqrt{2}}\cos x-\dfrac{1}{\sqrt{2}}\sin x\right]$

$=\sqrt{2}e^x\left[\cos\dfrac{\pi}{4}\cos x-\sin\dfrac{\pi}{4}\sin x\right]$

$=\sqrt{2}e^x\cos \left(x+\dfrac{\pi}{4}\right)$ .

Differentiate $\sin^{2}\ 3x. \tan^{3}\ 2x$

$\displaystyle \frac{d}{dx}(sin^{2}3x.tan^{3}2x) = sin^{2}3x\times \frac{d}{dx}(tan^{3}2x)+tan^{3}2x\times \frac{d}{dx}(sin^{2}3x)$

$\displaystyle= sin^{2}3x(3\ tan^{2}2x\sec^{2}2x.2)+tan^{3}2x(2\sin{3}x.cos{3}x.3)$

$\displaystyle = 6 \sin^{2}3x\tan^{2}2x \sec ^{2} 2x+6\sin\,3x \cos {3}x\tan^{3}2x$

If $y=sin^{-1}\left[2tan^{-1}\sqrt{\dfrac{1-x}{1+x}}\right]$ then find $\dfrac{dy}{dx}$ .

$y=\sin^{-1}\left[2\tan^{-1}\sqrt{\dfrac{1-x}{1+x}}\right]$

$\sin y=2\tan^{-1}\sqrt{\dfrac{1-x}{1+x}}$

$\cos y\left(\dfrac{dy}{dx}\right)=\dfrac{2}{1+\left(\sqrt{\dfrac{1-x}{1+x}}\right)^2}\times\dfrac{1}{2} \sqrt{\dfrac{1+x}{1-x}}\times \dfrac{[(-1)(1+x)-1(1-x)]}{(1-x)^2}$

$\cos y\left(\dfrac{dy}{dx}\right)=\dfrac{-2(1+x)}{(1+x+1-x}\times \dfrac{1}{2}\sqrt{\dfrac{1+x}{(1-x)}}\times \dfrac{[1+x+1-x]}{(1+x)^2}$

$=\dfrac{(1+x)}{(1+x)^2}\times \sqrt{\dfrac{1+x}{(1-x)}}=\dfrac{-1}{\sqrt{(1+x)(1-x)}}$

$\cos y\left(\dfrac{dy}{dx}\right)=\dfrac{-1}{\sqrt{(1+x(1-x)}}$

$\left(\dfrac{dy}{dx}\right)=\dfrac{-secy}{\sqrt{(1+x)(1-x)}}$

$\dfrac{dy}{dx}$ =

$\dfrac{-\sec (sin^{-1}\left[2\tan^{-1}\sqrt{\dfrac{1-x}{1+x}}\right])}{\sqrt{(1+x)(1-x)}}$

$y = 6{x^4} - 5{x^3} - 4{x^2}$
Diff. with respect to $x$

given,

$y = 6x^2 - 5x^3 - 4x^2$

$\dfrac{dy}{dx} = \dfrac{d}{dx} (6x^4 - 5x^3 - 4x^2)$

$= 24x^3 - 15x^2 - 8x$

Find $\dfrac{{dy}}{{dx}}$ if $y = {\left( {x + \dfrac{1}{x}} \right)^x}$

$y={\left(x+\dfrac{1}{x}\right)}^{x}$

$\log{y}=x\log{\left(x+\dfrac{1}{x}\right)}$

Differentiating both sides, w.r.t

$x$ we get

$\dfrac{1}{y}\dfrac{dy}{dx}=\log{\left(x+\dfrac{1}{x}\right)}+x\times\dfrac{1}{x+\dfrac{1}{x}}\dfrac{d}{dx}\left(x+\dfrac{1}{x}\right)$

$\dfrac{1}{y}\dfrac{dy}{dx}=\log{\left(x+\dfrac{1}{x}\right)}+\dfrac{{x}^{2}}{{x}^{2}+1}\left(1-\dfrac{1}{{x}^{2}}\right)$

$\dfrac{1}{y}\dfrac{dy}{dx}=\log{\left(x+\dfrac{1}{x}\right)}+\dfrac{{x}^{2}}{{x}^{2}+1}\left(\dfrac{{x}^{2}-1}{{x}^{2}}\right)$

$\dfrac{1}{y}\dfrac{dy}{dx}=\log{\left(x+\dfrac{1}{x}\right)}+\dfrac{{x}^{2}-1}{{x}^{2}+1}$

$\dfrac{dy}{dx}=y\log{\left(x+\dfrac{1}{x}\right)}+y\dfrac{{x}^{2}-1}{{x}^{2}+1}$

If $y = \dfrac{{{x^{\sin x}}}}{{1 + x + {x^2}}}$ , Find the $\dfrac{{dy}}{{dx}}$

$y=\dfrac{{x}^{\sin{x}}}{1+x+{x}^{2}}$

$\Rightarrow\,\left(1+x+{x}^{2}\right)y={x}^{\sin{x}}$

Let

$u={x}^{\sin{x}}$

$\Rightarrow\,\log{u}=\log{{x}^{\sin{x}}}$ by applying

$\log$ to both sides.

$\Rightarrow\,\log{u}=\sin{x}\log{x}$ using

$\log{{a}^{m}}=m\log{a}$

$\Rightarrow\,\dfrac{1}{u}\dfrac{du}{dx}=\dfrac{\sin{x}}{x}+\log{x}\cos{x}$

$\Rightarrow\,\dfrac{du}{dx}=u\left(\dfrac{\sin{x}}{x}+\cos{x}\log{x}\right)$

$\therefore\,\dfrac{d}{dx}\left({x}^{\sin{x}}\right)={x}^{\sin{x}}\left(\dfrac{\sin{x}}{x}+\cos{x}\log{x}\right)$

$\Rightarrow\,\left(1+x+{x}^{2}\right)\dfrac{dy}{dx}+\left(1+2x\right)y=\dfrac{d}{dx}\left({x}^{\sin{x}}\right)$

$\Rightarrow\,\left(1+x+{x}^{2}\right)\dfrac{dy}{dx}={x}^{\sin{x}}\left(\dfrac{\sin{x}}{x}+\cos{x}\log{x}\right)-\left(1+2x\right)y$

$\therefore\,\dfrac{dy}{dx}=\dfrac{\sin{x}{x}^{\sin{x}-1}+{x}^{\sin{x}}\cos{x}\log{x}-y-2xy}{1+x+{x}^{2}}$

${y^x} = {x^y}$

Solution : -

$x^{y} = y^{x}$

then

Taking log on both sides

$y\, log\, x = x\,log\, y$

Now differentiation w.r.t x

$y.\dfrac{1}{x}+log\,x.\dfrac{dy}{dx} = x.\dfrac{1}{y}\dfrac{dy}{dx}+log\,y$

$(log{x}-\dfrac{x}{y}) \dfrac{dy}{dx} = logy+\dfrac{-y}{x}$

$\dfrac{dy}{dx} = \dfrac{log\,y-y/x}{log{x}-x/y}\times \dfrac{xy}{xy} = \dfrac{xy\log\,y-y^{2}}{xy\,log\,y-x^{2}}$

If $y=\dfrac{{\sin 4x}}{{{x^2} + 16}}$ , then find $\dfrac {dy}{dx}$

Given that:

$y=\dfrac{\sin4x}{x^2+16}$

Differentiate with respect to x,

$\dfrac{dy}{dx}=\dfrac{d\left(\dfrac{\sin4x}{x^2+16}\right)}{dx}$

$\dfrac{dy}{dx}=\dfrac{\cos4x\times4\times(x^2+16)-\sin4x(2x)}{(x^2+16)^2}$

$\dfrac{dy}{dx}=\dfrac{4\cos4x(x^2+16-2x\sin4x)}{(x^2+16)^2}$

If $y=\log (\dfrac{x}{a+bx})^x$ , prove that $x^3\dfrac{d^2y}{dx^2}=(x\dfrac{dy}{dx}-y)^2$

$y=\log\left(\dfrac{x}{a+bx}\right)^{x}\Rightarrow y=x\log\left(\dfrac{x}{a+bx}\right)$

$\dfrac{dy}{dx}=\log\left(\dfrac{x}{a+bx}\right)+x\times \dfrac{\left(a+bx\right)}{x}\times \left[\dfrac{\left(a+bx\right)-bx}{\left(a+bx\right)^{2}}\right]$

$\dfrac{dy}{dx}=\dfrac{y}{x}+\dfrac{a+bx-bx}{\left(a+bx\right)}$

$\dfrac{dy}{dx}=\dfrac{y}{x}+\dfrac{a}{\left(a+bx\right)}\Rightarrow \dfrac{d^{2}y}{dx^{2}}=\left[\dfrac{\dfrac{dy}{dx}x-y}{x^{2}}\right]+\dfrac{d}{dx}\left[\dfrac{a}{a+bx}\right]$

$\Rightarrow \dfrac{d^{2}y}{dx^{2}}=\left[\dfrac{x\dfrac{dy}{dx}-y}{x^{2}}\right]-\dfrac{ab}{\left(a+bx\right)^{2}}$

from

$(1)$

$\dfrac{\left(x\dfrac{dy}{dx}-y\right)}{x}=\dfrac{a}{(a+bx)}$

$\dfrac{d^{2}y}{dx^{2}}=\left[\dfrac{x\dfrac{dy}{dx}-y}{x^{2}}\right]-\left[\dfrac{x\dfrac{dy}{dx}-y}{x}\right]\dfrac{b}{\left(a+bx\right)}$

$\dfrac{d^{2}y}{dx^{2}}=\left[\dfrac{x\dfrac{dy}{dx}-y}{x^{2}}\right]-\left[\dfrac{x\dfrac{dy}{dx}-y}{x^{2}}\right]\dfrac{\left(bx+a-a\right)}{\left(a+bx\right)}$

$\dfrac{d^{2}y}{dx^{2}}=\left[\dfrac{x\dfrac{dy}{dx}-y}{x^{2}}\right]-\left[\dfrac{x\dfrac{dy}{dx}-y}{x^{2}}\right]\times \left[1-\dfrac{a}{\left(a+bx\right)}\right]$

$\dfrac{d^{2}}{dx^{2}}=\left[\dfrac{x\dfrac{dy}{dx}-y}{x^{2}}\right]\left[1-1+\dfrac{a}{a+bx}\right]$

$\dfrac{d^{2}y}{dx^{2}}=\left[\dfrac{x\dfrac{dy}{dx}-y}{x^{2}}\right]\left[\dfrac{x\dfrac{dy}{dx}-y}{x}\right]\left[\dfrac{a}{\left(a+bx\right)}=\dfrac{x\dfrac{dy}{dx}-y}{x}\right]$

$\Rightarrow x^{3}\dfrac{d^{2}y}{dx^{2}}=\left(x\dfrac{dy}{dx}-y\right)^{2}$

Find $\cfrac{dy}{dx}$ , if $y={ e }^{ \sin ^{ 2 }{ x } }\left\{ 2\tan ^{ -1 }{ \sqrt { \cfrac { 1+x }{ 1-x } } } \right\}$ .

$y = e^{sin^{2}x} \left\{ 2 tan^{-1} \, \sqrt{\dfrac{1 + x}{1 - x}}\right \}$

$\dfrac{dy}{dx} = e^{sin^2 x} . sin 2x . 2tan^{-1} \sqrt{\dfrac{1 + x}{1 - x}} + e^{sin^2 x} \dfrac{d}{dx} \left(2 tan^{-1} \sqrt{\dfrac{1 + x}{1 - x}}\right)$

$\dfrac{d}{dx} \left(2 tan^{-1} \sqrt{\dfrac{1 + x}{1 - x}} \right)$

Put

$x = cos 2 \theta$

$\therefore 2 tan^{-1} \sqrt{\dfrac{1 + x}{1 - x}} = 2 tan^{-1} (cot \theta) = 2 [\dfrac{\pi}{2} - \theta]$

$= \pi - 2 \times \dfrac{1}{2} cos^{-1} x$

$= \pi - cos^{-1} x$

$\dfrac{dy}{dx} = e^{sin^2 x} . sin 2x . 2 tan^{-1} \sqrt{\dfrac{1 + x}{1 - x}} + e^{sin ^2x} . \dfrac{1}{\sqrt{1 - x^2}}$

The degree and order of differential equation $\left(\dfrac{d^2y}{dx^2} \right)^2 =\left(y + \dfrac{dy}{dx} \right)^{\frac{1}{2}}$ is which of the following?

We have,

$\left(\dfrac{d^2y}{dx^2} \right)^2 =\left(y + \dfrac{dy}{dx} \right)^{\frac{1}{2}}$

On squaring both sides, we get

$\left(\dfrac{d^2y}{dx^2} \right)^4 =\left(y + \dfrac{dy}{dx} \right)$

So,

Order

$=2$

Degree

$=4$

Hence, this is the answer.

Find $\dfrac{dy}{dx}$ for $y = {e^x}\left( {a\cos x + b\sin x} \right)$ .

Given

$y = e^{x} ( a\,cosx+b\,sinx)$

Now differentiating both sides

with respect to 'x' we get,

$\dfrac{dy}{dx} = e^{x}(a\,cosx+b\, sinx)+$

$e^{x}(-a\,sinx+b\,cosx)$

Compute the derivative of $6x^{100}-x^{55}+x$ .

$=6x^{100}-x^{55}+x$

$=600x^{99}-55x^{54}+1$

Find the derivative of $\sin 2 x \cos 3 x$

2173769_1195642_ans_aac333530e384a8aa33eca011b642fce.jpg

Find $\dfrac{dy}{dx}$ for $y=\dfrac{\tan x}{x}\log x^x$ .

Now,

$y=\dfrac{\tan x}{x}\log x^x$

or,

$y=\dfrac{\tan x}{x}(x\log x)$

or,

$y=\tan x.\log x$ .

Now differentiating both sides with respect to

$x$ we get,

$\dfrac{dy}{dx}=\sec^2 x.\log x+\dfrac{\tan x}{x}$ .

If $f (x) = 2017 \times tan x + 3x^{3}sinx+x^{2018}+2x^{3}$ then find $f'(x)$ .

Given,

$y=2017\tan x+3x^3\sin x+x^{2018}+2x^3$

Let,

$y=f(x)$ ;

Now, differentiation both sides w.r.t. 'x', we get,

$f'(x)=2017\sec^2x+9x^2\sin x+3x^3\cos x+2018x^{2077}+6x^2$ .

1180394_1195595_ans_eb6c08e50ab6476d93bef11db0076957.jpg

Differentiate $\dfrac{{e}^{x}-tan{x}}{cot{x}-{x}^{n}}$ w.r.t $x$ .

$y=\dfrac {e^x-\tan x}{\cot x-x^x}$

$\dfrac { dy }{ dx } =\dfrac { \left( \cot { x } -{ x }^{ n } \right) \dfrac { d }{ dx } \left( { e }^{ x }-\tan { x } \right) -\left( { e }^{ x }-\tan { x } \right) \dfrac { d }{ dx } \left( \cot { x } -{ x }^{ n } \right) }{ { \left( \cot { x } -{ x }^{ x } \right) }^{ 2 } }$

$=\dfrac { \left( \cot { x } -{ x }^{ x } \right) \left( { e }^{ x }-\sec ^{ 2 }{ x } \right) -\left( { e }^{ x }-\tan { x } \right) \left( -\csc ^{ 2 }{ x } -n{ x }^{ n-1 } \right) }{ { \left( \cot { x } -{ x }^{ x } \right) }^{ 2 } }$

$=\dfrac { { e }^{ x }\cot { x } -{ x }^{ n }{ e }^{ x }-\cot { x } \sec ^{ 2 }{ x } +{ x }^{ n }\sec ^{ 2 }{ x } +{ e }^{ x }\csc ^{ 2 }{ x } +n{ e }^{ x }{ x }^{ n-1 }-\tan { x\csc ^{ 2 }{ x } -n{ x }^{ n-1 }\tan { x } } }{ { \left( \cot { x } -{ x }^{ x } \right) }^{ 2 } }$

Find $\cfrac{dy}{dx}$ , if ${x}^{2}+xy+{y}^{2}=100$ .

Given

$x^2+xy+y^2=100$

Differentiating it w.r.t x

$2x+\dfrac{d(xy)}{dx}+2y.\dfrac{dy}{dx}=0$

Using product rule in

$xy=x'y+y'x$

$2x+\left[1.y+x.\dfrac{dy}{dx}\right]+2y.\dfrac{dy}{dx}=0$

$(2x+y)+\dfrac{dy}{dx}.(x+2y)=0$

$\dfrac{dy}{dx}.(x+2y)=-(2x+y)$

$\dfrac{dy}{dx}=\dfrac{-(2x+y)}{(x+2y)}$

Find $\dfrac{dy}{dx}$ of the following implicit functions : $y^{3}-3y^{2}x=x^{3}+3x^{2}y$

${ y }^{ 3 }-3{ y }^{ 2 }x={ x }^{ 3 }+3{ x }^{ 2 }y$

Differentiating both sides w.r.t

$x$

$3{ y }^{ 2 }\cfrac { dy }{ dx } -3\times 2y\cfrac { dy }{ dx } .x-3{ y }^{ 2 }=3{ x }^{ 2 }\cfrac { dy }{ dx } +3{ x }^{ 2 }\\ 3{ y }^{ 2 }\cfrac { dy }{ dx } -3{ x }^{ 2 }\cfrac { dy }{ dx } -6xy\cfrac { dy }{ dx } =3{ x }^{ 2 }+3{ y }^{ 2 }\\ \cfrac { dy }{ dx } ({ 3y }^{ 2 }-3{ x }^{ 2 }-6xy)=3{ x }^{ 2 }+{ 3y }^{ 2 }\\ \cfrac { dy }{ { dx }^{ 2 } } =\cfrac { 3{ x }^{ 2 }+3{ y }^{ 2 } }{ { 3y }^{ 2 }-{ 3x }^{ 2 }-6xy } =\cfrac { { x }^{ 2 }+{ y }^{ 2 } }{ { y }^{ 2 }-{ x }^{ 2 }-2xy }$

If $y=\dfrac{\sin x+\cos x}{\sin x-\cos x}$ find $\dfrac{dy}{dx}$ .

$y=\cfrac { \sin { x } +\cos { x } }{ \sin { x } -\cos { x } } =\cfrac { \tan { x } +1 }{ \tan { x } -1 }$

$\cfrac { dy }{ dx } =\cfrac { \left( \tan { x } -1 \right) \left( \sec ^{ 2 }{ x } \right) -\left( \tan { x } +1 \right) \left( \sec ^{ 2 }{ x } \right) }{ { \left( \tan { x } -1 \right) }^{ 2 } }$

$=\cfrac { 2\sec ^{ 2 }{ x } }{ { \left( \tan { x } -1 \right) }^{ 2 } } =\cfrac { 2\sec ^{ 2 }{ x } }{ \sec ^{ 2 }{ x } -2\tan { x } }$

$=\cfrac { \cfrac { 2 }{ \cos ^{ 2 }{ x } } }{ \cfrac { 1 }{ \cos ^{ 2 }{ x } } -\cfrac { 2\sin { x } }{ \cos { x } } }$

$=\cfrac { 2 }{ 1-2\sin { x } \cos { x } } =\cfrac { 2 }{ 1-\sin { 2x } }$

If $\sqrt { x } + \sqrt { y } = \sqrt { a } ,$ prove that $\frac { d y } { d x } = - \sqrt { \frac { y } { x } }$

$\begin{array}{l} \sqrt { x } +\sqrt { y } =\sqrt { a } \\ Differentiate\, with\, respect\, to\, x \\ \frac { 1 }{ { 2\sqrt { x } } } +\frac { 1 }{ { 2\sqrt { y } } } \frac { { dy } }{ { dx } } =0 \\ \frac { 1 }{ { 2\sqrt { y } } } \frac { { dy } }{ { dx } } =-\frac { 1 }{ { 2\sqrt { x } } } \\ \frac { { dy } }{ { dx } } =-\sqrt { \frac { y }{ x } } \end{array}$

If sin $^2$ x + cos $^2$ y = 1 find $\dfrac{dy}{dx}$

$\textbf{Hint: Use Chain rule}$

Solution:

$\textbf{Step 1: Use chain rule}$

${\rm{si}}{{\rm{n}}^2}x + {\rm{co}}{{\rm{s}}^2}y = 1$

Differentiating both sides with respect to $\ x.$

$\dfrac{d}{{dx}}\left( {{\rm{si}}{{\rm{n}}^2}x + {\rm{co}}{{\rm{s}}^2}y} \right) = \dfrac{d}{{dx}}\left( 1 \right)$ $\left(\because\dfrac{d\ c}{dx}=0\right)$

$\dfrac{d}{{dx}}\left( {{\rm{si}}{{\rm{n}}^2}x} \right) + \dfrac{d}{{dx}}({\rm{co}}{{\rm{s}}^2}y) = 0$

$\Rightarrow2\ {\rm{sin}}\ x\ {\rm{cos}}\ x - 2\ {\rm{sin}}\ y\ {\rm{cos}}\ y\dfrac{{dy}}{{dx}} = 0$ $\ \left[\because \dfrac{d\sin\theta}{d\theta}=cos\theta\ \quad and \quad\dfrac{d\cos\theta}{d\theta}=-sin\ \theta \right]$

$\textbf{Step 2: Simplify the above equation}$

$\therefore\ {\rm{sin}}\ 2x = {\rm{sin}}\ 2y\dfrac{{dy}}{{dx}}$ $\left[\because sin\left( 2\theta\right) =2\ sin\ \theta\ cos \ \theta\right]$

$\Rightarrow\dfrac{{dy}}{{dx}} = \dfrac{{{\rm{sin}}\ 2x}}{{{\rm{sin}}\ 2y}}$

$\textbf{ Hence, }$ $\dfrac{{dy}}{{dx}} = \dfrac{{{\rm{sin}}\ 2x}}{{{\rm{sin}}\ 2y}}$

Differentiate the following:
(i) $x ^ { 2 } \sin x$
(ii) $x ^ { 2 } e ^ { x }$
(iii) $e ^ { \sqrt { a x + b } }$

(i)

${ x }^{ 2 }sinx$

${ f }^{ 1 }\left( x \right) =2xsinx+{ x }^{ 2 }cosx$

(ii)

${ x }^{ 2 }{ e }^{ x }$

${ f }^{ 1 }\left( x \right) ={ x }^{ 2 }{ e }^{ x }+2x{ e }^{ x }$

(iii)

${ e }^{ \sqrt { ax+b } }$

${ f }^{ 1 }\left( x \right) ={ e }^{ \sqrt { ax+b } }.\dfrac { 1.a }{ 2\sqrt { ax+b } }$

$=\dfrac { a }{ 2\sqrt{ax+b} } { e }^{ \sqrt { ax+b } }$

Differentiate w.r.t $x$ in $nx (1 + x)^{n - 1}$ .

1212466_1249323_ans_ebfbefcb39a247daac47968dd7df0948.jpeg

If $(x^2+y^2)^2=xy$ , then $\dfrac {dy}{dx}$ .

$(x^{2}+y^{2})^{2} = xy$

$x^{4}+2x^{2}y^{2}+y^{4}-xy = 0$

Differentiating w.r.t x,

$4x^{3}+2[x^{2}.2y\left(\dfrac{dy}{dx}\right)+y^{2}(2x)]+4y^{3}\left(\dfrac{dy}{dx}\right)-[\dfrac{xdy}{dx}+y(1)] = 0$

$\left(\dfrac{dy}{dx}\right)[4x^{2}y+4y^{3}-x]+4x^{3}+4xy^{2}-y = 0$

$\left(\dfrac{dy}{dx}\right)[4x^{2}y+4y^{3}-x] = y-4x^{3}-4xy^{2}$

$\left(\dfrac{dy}{dx}\right) = \dfrac{y-4x^{3}-4xy^{2}}{4x^{2}y+4y^{3}-x}$

Find $\dfrac{dy}{dx}$ for $y=\tan\sqrt{\sin x}$ .

Given,

$y=\tan \sqrt{\sin x}$

Now differentiating both sides with respect to

$x$ we get,

$\dfrac{dy}{dx}=\sec^2{\sqrt{\sin x}}.\dfrac{\cos x}{2\sqrt{\sin x}}$ .

If $\cos y=x\cos\left(a+y\right)$ , with $\cos a\neq\pm1$ , prove that $\dfrac{dy}{dx}=\dfrac{\cos^{2}\left(a+y\right)}{\sin a}$ .

Given

$\cos y = x \cos (a + y)$

$\dfrac{\cos y}{\cos (a + y)} = x$

Diff w.r.t 'x'

$\dfrac{d(x)}{dx} = \dfrac{d}{dx} \left(\dfrac{\cos y}{\cos (a + y)} \right)$

$1 = \dfrac{d}{dx} \left(\dfrac{\cos y}{\cos (a + y)} \right) . \dfrac{dy}{dy}$

$1 = \dfrac{d}{dy} \left(\dfrac{\cos y}{\cos (a + y) } \right) . \dfrac{dy}{dx}$

$1 = \left(\dfrac{\dfrac{d(\cos y)}{dy} . cos (a + y) - \dfrac{d(\cos (a +y))}{dy} . \cos y}{(cos (a + y))^2} \right). \dfrac{dy}{dx}$

$1 = \left(\dfrac{-\sin y. \cos (a + y) - (-sin (a + y)) \dfrac{d (a + y)}{dy} . \cos y}{\cos^2 (a + y)} \right) . \dfrac{dy}{dx}$

$1 = \left(\dfrac{\sin (a + y) . \cos y - \cos (a + y) . \sin y}{\cos^2 (a + y)} \right) . \dfrac{dy}{dx}$

$1 = \dfrac{\sin .((a + y) - y)}{\cos^2 (a + y)} . \dfrac{dy}{dx}$

$1 = \dfrac{\sin a}{\cos^2 (a + y)} . \dfrac{dy}{dx}$

$\dfrac{\cos^2 (a + y)}{\sin (a)} = \dfrac{dy}{dx}$

Solve
$\int \dfrac{x^{3}}{x^{3}+1}dx$

We have,

$I= \int { \dfrac { { { x^{ 3 } } } }{ { { x^{ 3 } }+1 } } dx }$

$I= \int { \dfrac { { { x^{ 3 } +1-1} } }{ { { x^{ 3 } }+1 } } dx }$

$I =-\int { \dfrac { 1 }{ { { x^{ 3 } }+1 } } dx } +\int { 1dx }$

$\\ By\, converting\, in\, partial\, fraction\, and\, putting\, the\, { { int } }egral\, sign,\, we\, get$

$\\ =\int { \dfrac { { 2-x } }{ { 3\left( { { x^{ 2 } }-x+1 } \right) } } dx } +\int { \dfrac { 1 }{ { 3\left( { x+1 } \right) } } dx } +\int { 1dx } \\ =\dfrac { 1 }{ 6 } \left( { -\ln { \left| { \dfrac { 4 }{ 3 } +\dfrac { { 4{ x^{ 2 } }-4x } }{ 3 } } \right| } +2\sqrt { 3 } \arctan \left( { \dfrac { { 2x-1 } }{ { \sqrt { 3 } } } } \right) } \right) +\dfrac { 1 }{ 3 } \ln { \left| { x+1 } \right| } +x+C \\$

$Hence,\, solved.$

If ${ 3x }^{ 2 }+4xy-{ 7y }^{ 2 }=0$ Find $(a)\dfrac { dy }{ dx }$ and $(b)\dfrac { { d }^{ 2 }y }{ { dx }^{ 2 } }$

$(a)3\dfrac{d}{dx}\left({x}^{2}\right)+4\dfrac{d}{dx}\left(xy\right)-7\dfrac{d}{dx}\left({y}^{2}\right)=0$

$3\left(2x\right)+4\left(x\dfrac{dy}{dx}+y\right)-7\left(2y\dfrac{dy}{dx}\right)=0$

$\left(4x-14y\right)\dfrac{dy}{dx}=-6x-4y$

$2\left(2x-7y\right)\dfrac{dy}{dx}=-2\left(3x+2y\right)$

$\dfrac{dy}{dx}=\dfrac{-\left(3x+2y\right)}{2x-7y}$

$(b)\dfrac{{d}^{2}y}{d{x}^{2}}=-\dfrac{d}{dx}\left(\dfrac{3x+2y}{2x-7y}\right)$

$=-\left[\dfrac{\left(2x-7y\right)\dfrac{d}{dx}\left(3x+2y\right)-\left(3x+2y\right)\dfrac{d}{dx}\left(2x-7y\right)}{{\left(2x-7y\right)}^{2}}\right]$

$=-\left[\dfrac{\left(2x-7y\right)\left(3+2\dfrac{dy}{dx}\right)-\left(3x+2y\right)\left(2-7\dfrac{dy}{dx}\right)}{{\left(2x-7y\right)}^{2}}\right]$

Since

$\dfrac{dy}{dx}=\dfrac{-\left(3x+2y\right)}{2x-7y}$

$=-\left[\dfrac{\left(2x-7y\right)\left(3-2\dfrac{\left(3x+2y\right)}{2x-7y}\right)-\left(3x+2y\right)\left(2+7\dfrac{\left(3x+2y\right)}{2x-7y}\right)}{{\left(2x-7y\right)}^{2}}\right]$

$\dfrac{-1}{{\left(2x-7y\right)}^{3}}\left[\left(2x-7y\right)\left(6x-21y-6x-4y\right)-\left(3x+2y\right)\left(4x-14y+21x+14y\right)\right]$

$\dfrac{-1}{{\left(2x-7y\right)}^{3}}\left[\left(2x-7y\right)\left(-25y\right)-\left(3x+2y\right)\left(25x\right)\right]$

$\dfrac{25}{{\left(2x-7y\right)}^{3}}\left[\left(2x-7y\right)\left(y\right)+\left(3x+2y\right)\left(x\right)\right]$

$=\dfrac{25}{{\left(2x-7y\right)}^{3}}\left[2xy-7{y}^{2}+3{x}^{2}+2xy\right]$

$=\dfrac{25}{{\left(2x-7y\right)}^{3}}\left[3{x}^{2}+4xy-7{y}^{2}\right]$

$=\dfrac{25}{{\left(2x-7y\right)}^{3}}\left[3{x}^{2}+7xy-3xy-7{y}^{2}\right]$

$=\dfrac{25}{{\left(2x-7y\right)}^{3}}\left[3x\left(x-y\right)+7y\left(x-y\right)\right]$

$=\dfrac{25}{{\left(2x-7y\right)}^{3}}\left[\left(x-y\right)\left(3x+7y\right)\right]$

$x\sqrt{1+y}+y\sqrt{1+x} = 0$ Show that: $\dfrac{dy}{dx} = \dfrac{-1}{(1+x)^{2}}$

1257612_1323716_ans_db78f8ad5cd745c7b64e16d76decaaa0.jpg

Find $\dfrac{dy}{dx}$ if $y = \left(x + \dfrac{1}{x}\right)^x + {x}^{\left(x + \frac{1}{x}\right)}$

$y=\left(x+\dfrac{1}{x}\right)^x+x^{x+\dfrac{1}{x}}$

Let

$u=\left(x+\dfrac{1}{x}\right)^x$

$\Rightarrow log x=x log\left(x+\dfrac{1}{x}\right)$

$\Rightarrow \dfrac{1}{u}\dfrac{dy}{dx}=\dfrac{x}{\left(x+\dfrac{1}{x}\right)}\left(1-\dfrac{1}{x^2}\right)+log\left(x+\dfrac{1}{x}\right)$

$\Rightarrow \dfrac{1}{u}\dfrac{dy}{dx}=\dfrac{\left(x-\dfrac{1}{x}\right)}{\left(x+\dfrac{1}{x}\right)}+log\left(x+\dfrac{1}{x}\right)$

$\Rightarrow \dfrac{dy}{dx}=u\left[\dfrac{\left(x-\dfrac{1}{x}\right)}{\left(x+\dfrac{1}{x}\right)}+log \left(x+\dfrac{1}{x}\right)\right]$

Let

$v=x^{x+\dfrac{1}{x}}$

$log v=\left(x+\dfrac{1}{x}\right)log x$

$\Rightarrow \dfrac{1}{v}\dfrac{dv}{dx}=\left(1-\dfrac{1}{x^2}\right)log x+\left(x+\dfrac{1}{x}\right)\dfrac{1}{x}$

$\Rightarrow\dfrac{dv}{dx}=v\left[\left(1-\dfrac{1}{x^2}\right)log x+\left(1+\dfrac{1}{x^2}\right)\right]$

$\dfrac{dy}{dx}=\dfrac{du}{dx}+\dfrac{dv}{dx}$

$=u\left[\dfrac{x-\dfrac{1}{x}}{x+\dfrac{1}{x}}+log \left(x+\dfrac{1}{x}\right)\right]+v\left[\left(1-\dfrac{1}{x^2}\right)log x+\left(1+\dfrac{1}{x^2}\right)\right]$

$\left(x+\dfrac{1}{x^2}\right)^x\left[\dfrac{x-\dfrac{1}{x}}{x+\dfrac{1}{x}}+log\left(x+\dfrac{1}{x}\right)\right]+x^{x+\dfrac{1}{x}}\left[\left(1-\dfrac{1}{x^2}\right) log x+x^{x+\dfrac{1}{x}}\left(1+\dfrac{1}{x^2}\right)\right]$ .

1257503_1326909_ans_8d5e75acb3684b9aa1409f468e2eb927.PNG

If $y=\tan(2x+3)$ find $\dfrac { dy }{ dx }$

$y=\tan(2x+3)$

Differentiating with respect to

$x,$ we get

$\dfrac{dy}{dx}=\sec^2(2x+3).2$

$\dfrac{dy}{dx}=2\sec^2(2x+3)$ .

If $x\sqrt { 1+y } +y\sqrt { 1+x } =0,$ prove that $\dfrac { dy }{ dx } =\dfrac { -1 }{ { (1+x) }^{ 2 } } .$

$x\sqrt{1+y}+y\sqrt{1+x}=0$

$\Rightarrow x\sqrt{1+y}=y\sqrt{1+x}$

Squaring both sides we get

$x^2(1+y)=y^2(1+x)$

$\Rightarrow x^2+x^2y=y^2+xy^2$

$\Rightarrow x^2-y^2=xy^2-x^2y$

$\Rightarrow (x-y)(x+y)=xy(y-x)$

$\Rightarrow (x-y)(x+y)=-xy(x-y)$

$\Rightarrow (x+y)=-xy$

$\Rightarrow x=-xy-y$

$\Rightarrow x=(-x-1)y$

$\Rightarrow y=\dfrac{x}{-x-1}$

$\Rightarrow y=\dfrac{-x}{x+1}$

Differentiating with respect to x, we get

$\dfrac{dy}{dx}=\dfrac{(x+1)(-1)-(-x)}{(x+1)^2}$

$=\dfrac{-x-1+x}{(x+1)^2}$

$=\dfrac{-1}{(x+1)^2}$ .

1190059_1330032_ans_5bf2f101c47d4329a5b48bcae82467cd.JPG

Differentiate
$x^{x\cos x}+\dfrac{x^{2}+1}{x^{2}-1}$ w.r.t.x.

Let

$y={x}^{x\cos{x}}+\dfrac{{x}^{2}+1}{{x}^{2}-1}$

$\Rightarrow y=u+v$

$\therefore u={x}^{x\cos{x}}$

$\Rightarrow \log{u}=x\cos{x}\log{x}$

$\Rightarrow \dfrac{1}{u}\dfrac{du}{dx}=x\cos{x}\dfrac{d}{dx}\left(\log{x}\right)+x\log{x}\dfrac{d}{dx}\left(\cos{x}\right)+\cos{x}\log{x}\dfrac{d}{dx}\left(x\right)$

$=x\cos{x}\times \dfrac{1}{x}-x\log{x}\sin{x}+\cos{x}\log{x}$

$\dfrac{1}{u}\dfrac{du}{dx}=\cos{x}-x\log{x}\sin{x}+\cos{x}\log{x}$

$\dfrac{du}{dx}=u\left(\cos{x}-x\log{x}\sin{x}+\cos{x}\log{x}\right)$

$\therefore \dfrac{du}{dx}={x}^{x\cos{x}}\left(\cos{x}-x\log{x}\sin{x}+\cos{x}\log{x}\right)$

$v=\dfrac{{x}^{2}+1}{{x}^{2}-1}$

$\dfrac{dv}{dx}=\dfrac{\left({x}^{2}-1\right)\dfrac{d}{dx}\left({x}^{2}+1\right)-\left({x}^{2}+1\right)\dfrac{d}{dx}\left({x}^{2}-1\right)}{{\left({x}^{2}-1\right)}^{2}}$

$\dfrac{dv}{dx}=\dfrac{\left({x}^{2}-1\right)\left(2x\right)-\left({x}^{2}+1\right)\left(2x\right)}{{\left({x}^{2}-1\right)}^{2}}$

$\dfrac{dv}{dx}=\dfrac{2x\left({x}^{2}-1-{x}^{2}-1\right)}{{\left({x}^{2}-1\right)}^{2}}$

$\dfrac{dv}{dx}=\dfrac{-4x}{{\left({x}^{2}-1\right)}^{2}}$

$\therefore \dfrac{dy}{dx}=\dfrac{du}{dx}+\dfrac{dv}{dx}$

$={x}^{x\cos{x}}\left(\cos{x}-x\log{x}\sin{x}+\cos{x}\log{x}\right)-\dfrac{4x}{{\left({x}^{2}-1\right)}^{2}}$

Differentiate $\log (x+\sqrt{a^{2}+x^{2}})$

Let

$y= log (x+ \sqrt{a^2+x^2})$

On diffrently wrt 'x' we get

$\Rightarrow \frac {d}{dx} = \frac {d}{dx} [log (x+ \sqrt {a^2+x^2})]$

$\frac {d}{dx} =\frac {1}{x+\sqrt {a^2+x^2}} \,\,\,\,\, [1+\frac{2x}{2\sqrt {a^2+x^2}}]$

$\frac {d}{dx} = \frac {1} {\sqrt {a^2+x^2}} \times \frac{x+\sqrt {a^2+x^2}}{\sqrt {a^2+x^2}}=\frac{1}{\sqrt {a^2+x^2}}$

$\frac {d}{dx}=\frac{1}{\sqrt {a^2+x^2}}$

Let $x^{3}+y^{2}=24$ . What is the value of $\dfrac{d^{2}y}{dx^{2}}$ at the point $(2,4)$ ?

$x^{2}+y^{2}=2y$

$3x^{2}+2y \frac{dy}{dx} = 0$

$\frac{dy}{dx} = \frac{-3x^{2}}{2y}; \frac{dy}{dx}]_{2,4} =\frac{-3}{2}$

$\Rightarrow 2y \frac{dy}{dx}= -3x^{2}$

$\Rightarrow y \frac{d^{2}y}{dx^{2}}+(\frac{dy}{dx})^{2} = \frac{-3}{2}(2x)$

$\Rightarrow y \frac{d^{2}y}{ax^{2}}]$

$=-3x-(\frac{dy}{dx})^{2}$

$\Rightarrow \frac{d^{2}y}{dx^{2}}]_{2,y} = \frac{-3(2)-(\frac{-3}{2})^{2}}{4}$

$(\frac{d^{2}y}{dx^{2}})_{2,4} =\frac{-33}{16}$

1215211_1354275_ans_00daa5c32fb540d68e4d3aeff8ba020b.jpg

Find $\dfrac{d}{dx} (x^{3} \log_{e}x)$

Let

$y=\frac {d}{dx} \, (x^3log_{e}x)$

then using multilication law

$y = x^3 \frac{d}{dx}(log_{e^x}) + log_{e}x \frac{d}{dx} (x^3)$

$y =\frac{x^3}{x}+ log_{e}x \cdot 2x^2$

$y= x^2 +2x^2 log_{e}x$

$y= x^2 [1+log_{e}x^2]$

Find $\dfrac{dy}{dx}$
$8x^{2}+9y=6x+2$

$8x^{2}+9y=6x+2$

Differentiating w.r.t x on both sides,

$16x+9\dfrac{dy}{dx}=6$

$9\dfrac{dy}{dx}=6-16x$

$\dfrac{dy}{dx}=\dfrac{6-16x}{9}$

Find $\dfrac{dy}{dx}$ : $4x^{2}-\log x =y^{2}+\sin x -\cos x$

$4x^{2}-\log x = y^{2}+\sin x -\cos x$

$8x-\dfrac{1}{x}=2y\dfrac{dy}{dx}+\cos x + \sin x$

$2y\dfrac{dy}{dx}=8x-\dfrac{1}{x}-\sin x-\cos x$

$\dfrac{dy}{dx} = \dfrac{1}{2y}(8x-\dfrac{1}{x}-\cos x-\sin x)$

Find $\frac{dy}{dx}, if x^{y}=e^{x-y}$

$x^{y}=e^{x-y}$

$ylogx=x-y$

$\dfrac{dy}{dx}logx+\dfrac{y}{x}=1-\dfrac{dy}{dx}$

$\dfrac{dy}{dx}logx+\dfrac{dy}{dx}=1-\dfrac{y}{x}$

$\dfrac{dy}{dx}(1+logx)=\dfrac{x-y}{x}$

$\dfrac{dy}{dx}=\dfrac{x-y}{x(1+logx)}$

If $x=y\sqrt {1-y^{2}}$ , then $\dfrac {dy}{dx}$ equals to?

$x=y\sqrt{1-{y}^{2}}$

$\Rightarrow \dfrac{dx}{dx}=y\dfrac{d}{dx}\left(\sqrt{1-{y}^{2}}\right)+\sqrt{1-{y}^{2}}\dfrac{d}{dx}\left(y\right)$

$\Rightarrow 1=y\left(\dfrac{-2y}{\sqrt{1-{y}^{2}}}\dfrac{dy}{dx}\right)+\sqrt{1-{y}^{2}}\dfrac{dy}{dx}$

$\Rightarrow 1=\left(\dfrac{-2{y}^{2}}{\sqrt{1-{y}^{2}}}+\sqrt{1-{y}^{2}}\right)\dfrac{dy}{dx}$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\dfrac{-2{y}^{2}}{\sqrt{1-{y}^{2}}}+\sqrt{1-{y}^{2}}}$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{\sqrt{1-{y}^{2}}}{-2{y}^{2}+1-{y}^{2}}$

$\therefore \dfrac{dy}{dx}=\dfrac{\sqrt{1-{y}^{2}}}{1-3{y}^{2}}$

Differentiate: $y=k{x}^{n}$

$y=k{x}^{n}$

$\Rightarrow\,\dfrac{dy}{dx}=k\dfrac{d}{dx}\left({x}^{n}\right)$

$\therefore\,\dfrac{dy}{dx}=kn{x}^{n-1}$

Find $\dfrac{dy}{dx}$ : $\sin x +7x^{2}+y^{2}=5$

$\sin x +7x^{2}+y^{2}=5$

$-\cos x +14x+2y \dfrac{dy}{dx}=0$

$2y \dfrac{dy}{dx}=\cos x -14x$

$\dfrac{dy}{dx}=\dfrac{\cos x-14x}{2y}$

If $\sqrt{1-x^{2}}+\sqrt{1-y^{2} }=a$ find $\dfrac{d y}{d x}$

$\sqrt {1-x^2}+\sqrt {1-y^2}=a\\\dfrac{d}{dx}(\sqrt {1-x^2}+\sqrt {1-y^2})=0\\\dfrac {-2x}{2\sqrt {1-x^2}}+\dfrac{-2y}{\sqrt 2{1-y^2}}\dfrac {dy}{dx}=0\\\dfrac{-y}{\sqrt {1-y^2}}\dfrac {dy}{dx}=\dfrac {x}{\sqrt {1-x^2}}\\\dfrac{dy}{dx}=-\dfrac{x\sqrt {1-y^2}}{y\sqrt {1-x^2}}$

If $y=3x^2-4x$ find $\dfrac{dy}{dx}$

$y=3x^2-4x\\\dfrac{dy}{dx}=\dfrac{d}{dx}3x^2-4x\\\dfrac{dy}{dx}=6x-4$

Solve $\frac{dy}{dx}=\dfrac{1}{cos (x+y)}$

$\dfrac{dy}{dx}=\dfrac{1}{cos(x+y)}$

Let

$x+y=v$

$1+\dfrac{dy}{dx}=\dfrac{dv}{dx}$

$\dfrac{dy}{dx}=\dfrac{dv}{dx}-1$

The original differential equation becomes :

$\dfrac{dv}{dx}-1=\dfrac{1}{cos v}$

$\dfrac{dv}{dx}=\dfrac{1+\cos v}{\cos v}$

$\int \dfrac{\cos v dv}{1+\cos v}=\int dx$

$\int \dfrac{cos v +1-1}{1+cosv} dv=x+C$

$\int dv - \int \dfrac{1}{1+cos v }dv=x+C$

$v-\int \dfrac{1}{2 cos^{2}(\dfrac{v}{2})}dv=x+C$

$v-\dfrac{1}{2}\int sec^{2} (\dfrac{v}{2})dv=x+C$

$v-\dfrac{1}{2} tan(\dfrac{v}{2})=x+C$

$x+y-\dfrac{1}{2} tan \dfrac{x+y}{2}=x+C$

$y-\dfrac{1}{2} tan \dfrac{x+y}{2}=C$

differentiate
$y=log\left( { 3x }^{ 2 }+2x+1 \right)$

$y=\log{\left(3{x}^{2}+2x+1\right)}$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\left(3{x}^{2}+2x+1\right)}\dfrac{d}{dx}\left(3{x}^{2}+2x+1\right)$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\left(3{x}^{2}+2x+1\right)}\left(6x+2\right)$

$\therefore \dfrac{dy}{dx}=\dfrac{2\left(3x+1\right)}{\left(3{x}^{2}+2x+1\right)}$

Find derivative of f(x)
$f(x)=x\sin x$

$f(x)=x\sin x\\f^{\prime}(x)=\dfrac{d}{dx}x \sin x\\\sin x+x\cos x$

Find $\dfrac{dy}{dx}$ , if ${x}^{3}+8xy+{y}^{3}=64$ .

${x}^{3}+8xy+{y}^{3}=64$

Differentiating both sides w.r.t

$x$ we get

$3{x}^{2}+8\left[x\dfrac{dy}{dx}+y\right]+3{y}^{2}\dfrac{dy}{dx}=0$

$\Rightarrow 8\left[x\dfrac{dy}{dx}+y\right]+3{y}^{2}\dfrac{dy}{dx}=-3{x}^{2}$

$\Rightarrow 8x\dfrac{dy}{dx}+8y+3{y}^{2}\dfrac{dy}{dx}=-3{x}^{2}$

$\Rightarrow 8x\dfrac{dy}{dx}+3{y}^{2}\dfrac{dy}{dx}=-3{x}^{2}-8y$

$\Rightarrow \left(8x+3{y}^{2}\right)\dfrac{dy}{dx}=-3{x}^{2}-8y$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{-\left(3{x}^{2}+8y\right)}{\left(8x+3{y}^{2}\right)}$

If $x^3y^5 = (x + y)^8$ , then show that $\dfrac{dy}{dx} = \dfrac{y}{x}$

${x}^{3}{y}^{5}={\left(x+y\right)}^{8}$

$\Rightarrow {x}^{3}\left(5{y}^{4}\dfrac{dy}{dx}\right)+{y}^{5}\left(3{x}^{2}\right)=8{\left(x+y\right)}^{7}\left(1+\dfrac{dy}{dx}\right)$

$\Rightarrow 5{x}^{3}{y}^{4}\dfrac{dy}{dx}+3{x}^{2}{y}^{5}=8{\left(x+y\right)}^{7}+8{\left(x+y\right)}^{7}\dfrac{dy}{dx}$

$\Rightarrow \left(5{x}^{3}{y}^{4}-8{\left(x+y\right)}^{7}\right)\dfrac{dy}{dx}=8{\left(x+y\right)}^{7}-3{x}^{2}{y}^{5}$

We have

${x}^{3}{y}^{5}={\left(x+y\right)}^{8}\Rightarrow {x}^{3}{y}^{4}=\dfrac{{\left(x+y\right)}^{8}}{y}$

and

${x}^{2}{y}^{5}={\left(x+y\right)}^{8}\Rightarrow {x}^{2}{y}^{5}=\dfrac{{\left(x+y\right)}^{8}}{x}$

$\Rightarrow \left(\dfrac{{5\left(x+y\right)}^{8}}{y}-8{\left(x+y\right)}^{7}\right)\dfrac{dy}{dx}=8{\left(x+y\right)}^{7}-\dfrac{{3\left(x+y\right)}^{8}}{x}$

$\Rightarrow {\left(x+y\right)}^{7}\left(\dfrac{5x+5y}{y}-8\right)\dfrac{dy}{dx}={\left(x+y\right)}^{7}\left(8-\dfrac{3x+3y}{x}\right)$

$\Rightarrow \dfrac{5x+5y-8y}{y}\dfrac{dy}{dx}=\dfrac{8x-3x-3y}{x}$

$\Rightarrow \dfrac{5x-3y}{y}\dfrac{dy}{dx}=\dfrac{5x-3y}{x}$

$\therefore \dfrac{dy}{dx}=\dfrac{y}{x}$

Find $\dfrac{dy}{dx}$ , if $x^3 + y^2 + xy = 10$

${x}^{3}+{y}^{3}+xy=10$

$\Rightarrow 3{x}^{2}+3{y}^{2}\dfrac{dy}{dx}+x\dfrac{dy}{dx}+y=0$ by differentiating w.r.t

$x$

$\Rightarrow \left(3{y}^{2}+x\right)\dfrac{dy}{dx}=-\left(3{x}^{2}+y\right)$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{-\left(3{x}^{2}+y\right)}{\left(3{y}^{2}+x\right)}$

If ${ xy }^{ 2 }=1$ , then prove that $2\dfrac { dy }{ dx } +{ y }^{ 3 }=0$ .

We have,

$xy^{2}=1 \Rightarrow x=\dfrac{1}{y^{2}} \Rightarrow \dfrac{dx}{dy}=-\dfrac{2}{y^{3}} \Rightarrow \dfrac{dy}{dx}=-\dfrac{y^{3}}{2}\Rightarrow 2\dfrac{dy}{dx}+y^{3}=0$

If $y={ e }^{ { x+e }^{ { x+e }^{ { x+e }^{ x+...\infty } } } }$ , then prove that $\dfrac { dy }{ dx } =\dfrac { y }{ 1-y }$ .

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If $y = {e^x}\cos x,$ then prove that $\dfrac{{dy}}{{dx}} = \sqrt 2 {e^x}\cos \left( {x + \dfrac{\pi }{4}} \right)$ .

1315186_1538533_ans_9848e20e77244cf7a66b5f7f4c2d5cb5.jpeg

Find $\dfrac { dy }{ dx }$ , if $y=\sqrt { 1+\sin 2x }$ .

$y = \sqrt{1 + \sin 2 x }$

We know that,

$1=\sin^2 x+\cos^2 x$ and

$\sin2x=2\sin x \cos x$

$y = \sqrt{\sin^{2}x + \cos^{2} + 2 \sin x \cos x }$

$y= \sqrt{(\sin x + \cos x)^{2}}$

$y = \sin x + \cos x$

Differentiating w.r.t

$x$

$\dfrac{dy}{dx} = \cos x - \sin x$

If $f(x) = 1 + x + x^{2} + ..... + x^{100}$ then find $f'(1)$ .

We need to find the derivative of

$f(x)$ at

$x=1$

i.e.

$f'(1)$

$f(x)=1+x+x^2+......+x^{100}$

$\Rightarrow f'(x)=(1+x+x^2+.....+x^{100})'$

$\Rightarrow f'(x)=0+1+2x+.....+100x^{99}$

$\therefore f'(x)=1+2x+.....+100x^{99}$

Putting

$x=1$ in

$f'(x)$

$\Rightarrow f'(1) = 1+2 \times 1+.....+100 \times 1^{99}$

$\Rightarrow f'(1) = 1+2+....+100$

$\Rightarrow f'(1) =\dfrac {100(100+1)}{2}$ =

$5050$

$\Rightarrow f'(1)$ =

$5050$

If $y=\frac{a-x}{a+x}(x\neq -a)$ , find $\frac{dy}{dx}$

1321960_1586421_ans_dfd44bcaec7447bcb1e3f03c490a8f0c.jpg

Find $\dfrac{{dy}}{{dx}}$ ,if ${e^x} + {e^y} = {e^{x - y}}$ .

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Differentiate the following functions with respect to $x$ :
$\dfrac {\sqrt {x^2+1}+\sqrt {x^2-1}}{\sqrt {x^2+1}-\sqrt {x^2-1}}$

Let's first Rationalize the given expression to get a simpler expression to deal with,

$\dfrac {\sqrt {x^2+1}+\sqrt {x^2-1}}{\sqrt {x^2+1}-\sqrt {x^2-1}} = \dfrac{2x^2+2\sqrt{x^4-1}}{2} = x^2 + \sqrt{x^4-1}$

So,

$\dfrac{d}{dx}\left(\dfrac {\sqrt {x^2+1}+\sqrt {x^2-1}}{\sqrt {x^2+1}-\sqrt {x^2-1}}\right) = \dfrac{d}{dx} (x^2+\sqrt{x^4-1})$

$=2x + \dfrac{1}{2}\times \dfrac{1}{\sqrt{x^4+1}}\times\dfrac{d(x^4)}{dx}$

$=2x+\dfrac{2x^3}{\sqrt{x^4-1}}$

If $xy=1,$ prove that $\dfrac{dy}{dx}+y^{2}=0$ .

We have,

$xy=1 \Rightarrow \dfrac{dy}{dx}=-\dfrac{1}{x^{2}}$

Therefore,

$\dfrac{dy}{dx}+y^{2}=-\dfrac{1}{x^{2}}+\dfrac{1}{x^{2}}=0$

If $xy \log(x+y)=1,$ prove that $\dfrac{dy}{dx}=-\dfrac{y(x^{2}y+x+y)}{x(xy^{2}+x+y)}$ .

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Find $\dfrac{dy}{dx}$ in the following:
$x^{5}+y^{5}=5\ xy$

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If $y= x \sin (a+y),$ prove that $\dfrac{dy}{dx}= \dfrac{\sin^{2}(a+y)}{\sin (a+y) -y \cos(a+y)}$ .

$y=x \sin (a+y)$

$\dfrac{dy}{dx}=\sin(a+y)+x \cos (a+y)\dfrac{dy}{dx}$

$\dfrac{dy}{dx}=\dfrac{\sin(a+y)}{1-x\cos(a+y)}$

Multiplying numerator and denominator by

$\sin(a+y)$ , we get,

$\dfrac{dy}{dx}=\dfrac{\sin^2(a+y)}{\sin(a+y)-y\cos(a+y)}$

$[Hence \ proved]$

If $\sin (xy)+\dfrac{y}{x}=x^{2}-y^{2},$ find $\dfrac{dy}{dx}$ .

$\sin(xy)+\dfrac{y}{x}=x^2-y^2$

Differentiating w.r.t. x, we get,

$\cos(xy)[y+x\dfrac{dy}{dx}]-\dfrac{y}{x^2}+\dfrac{1}{x}\dfrac{dy}{dx}=2x-2y\dfrac{dy}{dx}$

$\dfrac{dy}{dx}[x\cos (xy)+\dfrac{1}{x}+2y]=2x-y\cos (xy)+\dfrac{y}{x^2}$

$\dfrac{dy}{dx}=\dfrac{2x-y\cos(xy)+\dfrac{y}{x^2}}{x\cos (xy)+\dfrac{1}{x}+2y}$

A circular disc of radius 3 cm is being heated. Due to expansion, its radius increases at the rate of 0.05 cm/sec. Find the rate at which its area is increasing when radius is 3.2 cm.

Let r be the radius and A be the area of the disc at any time t. Then,

$A = \pi r^{2}.$
It is given that

$\dfrac{dr}{dt} = 0.05\,cm/sec$ .
Now,

$A = \pi r^{2}$

$\Rightarrow \dfrac{dA}{dt} = 2\pi r\frac{dr}{dt}\Rightarrow \left ( \dfrac{dA}{dt} \right )_{r = 3.2} = 2\pi \times 3.2\times 0.05 = 0.320\,\pi \,cm^{2}/sec.$

Find $\dfrac{dy}{dx}$ , if $2x+3y=\sin y$ .

$2x+3y=\sin y$

Differentiating w.r.t x, we get,

$2+3\dfrac{dy}{dx}=\cos y \dfrac{dy}{dx}$

$\dfrac{dy}{dx}(\cos y-3)=2$

$\dfrac{dy}{dx}=\dfrac{2}{\cos y-3}$

Find $\dfrac {dy}{dx},$ when
$x=a\left(1-\cos {\theta}\right)$ and $y=a \left(\theta + \sin {\theta} \right)$ at $\theta =\dfrac {\pi}{2}$

We have,

$x=a(1- \cos {\theta}),\ y=a({\theta} + \sin {\theta})$

Differentiating w.r.t.

$\theta$ , we get,

$\Rightarrow \dfrac { dx }{ d\theta } =a\sin { \theta } ,\dfrac { dy }{ d\theta } =a\left( 1+\cos { \theta } \right)$

$\therefore \dfrac { dy }{ dx } =\dfrac { dy/d\theta }{ dx/d\theta } =\dfrac { a\left( 1+\cos { \theta } \right) }{ a\sin { \theta } } =\dfrac { 2\cos ^{ 2 }{ \dfrac { \theta }{ 2 } } }{ 2\sin { \dfrac { \theta }{ 2 } } \cos { \dfrac { \theta }{ 2 } } } =\cot { \dfrac { \theta }{ 2 } }$

$\Rightarrow { \left( \dfrac { dy }{ dx } \right) }_{ \theta =\dfrac { \pi }{ 2 } }=\cot { \dfrac { \pi }{ 4 } } =1.$

Differentiate the following function with respect to x.
$x^3\sin x$ .

Given expression

$x^3 \sin x$

differentiating w.r.t.

$x$ , we get

$\dfrac d{dx}(x^3 \sin x)$

$\implies x^3\dfrac d{dx}\sin x+\dfrac d{dx}(x^3) \sin x$

$\implies x^3\cos x+3x^2\sin x$

$\implies x^2(x\cos x +3\sin x)$

Differentiate the following function with respect to x.
$(1+x^2)\cos x$ .

Given expression

$(1+x^2)\cos x$

differentiating w.r.t.

$x$ , we get

$=\dfrac d{dx}((1+x^2)\cos x)$

$=(1+x^2)\dfrac d{dx}\cos x+\cos x\dfrac {d}{dx}(1+x^2)$

$=(1+x^2)(-\sin x)+\cos x(2x)$

$=2x\cos x-(1+x^2)\sin x$

Differentiate the following function with respect to x.
For the function $f(x)=\dfrac{x^{100}}{100}+\dfrac{x^{99}}{99}+.....+\dfrac{x^2}{2}+x+1$ . Prove that $f'(1)=100f'(0)$ .

Given function

$f(x)=\dfrac {x^{100}}{100}+\dfrac {x^{99}}{99}+.....\dfrac {x^2}2+x+1$

$\implies f^{\prime}(x)=\dfrac d{dx}\left(\dfrac {x^ {100}}{100}+\dfrac {x^{99}}{99}+.........+\dfrac {x^2}2+x+1\right)$

$f^{\prime}(x)=x^{99}+x^{98}+.....+x^2+x+1$

$f^{\prime}(1)=1+1+1+.....+1=100$

$f^{\prime}(0)=0+0+0+....+0+0+1=1$

$\implies f^{\prime}(1)=100f^{\prime}(0)$

Differentiate the following function with respect to x.
$\dfrac{a+b\sin x}{c+d\cos x}$ .

Given expression,

$\dfrac{a+b\sin x}{c+d\cos x}$ .

Using,

$\dfrac{d}{dx}\left(\dfrac{u}{v}\right)=\dfrac{(u)\dfrac{d}{dx}(v)-v\dfrac{d}{dx}(u)}{(v)^2}$

we get,

$\dfrac{d}{dx}\left(\dfrac{a+b\sin x}{c+d\cos x}\right)$

$=\dfrac{(c+d\cos x)\dfrac{d}{dx}(a+b\sin x)-(a+b\sin x)\dfrac{d}{dx}(c+d\cos x)}{(c+d\cos x)^2}$

$=\dfrac{(c+d\cos x)(0+b\cos x)-(a+b\sin x)(0-d\sin x)}{(c+d\cos x)^2}$

$=\dfrac{bc\cos x+ad\sin x+bd}{(c+d\cos x)^2}$ .

Find $\dfrac {dy}{dx}$ , where $xy^2-x^2y-5=0$ .

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Differentiate the following function :
$(2x+3)(3x-5)$

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Find $\dfrac {dy}{dx}$ , where $(x^2 +y^2)^2 =xy$

Given,

$(x^2 +y^2)^2 =xy$

$x^4 +y^4 +2x^2 y^2 =xy$

Differentiate with respect to

$x$

$\Rightarrow \ 4x^3+4y^3.\dfrac {dy}{dx}+4x^2 y. \dfrac {dy}{dx}+4xy^2=x \dfrac {dy}{dx}+y$

$\Rightarrow \ (4y^3 +4x^2 y-x)\dfrac {dy}{dx}=(y-4xy^2 -4x^3)$

$\Rightarrow \ \dfrac {dy}{dx}=\dfrac{(y-4xy^2 -4x^3)}{(4y^3 +4x^2 y-x)}$

Differentiate the following function :
$ax^3+bx^2+cx+d$ , where $a, b, c, d$ are constant.

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Differentiate the following function :
$x(1+x)^3$

1879436_1710547_ans_c887eedf8c52471cb7bd0f83ee8b8750.jpg

Differentiate the following function :
$\left (x-\dfrac {1}{x}\right)^2$

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Differentiate the following function with respect to x.
$x^4(5\sin x-3\cos x)$ .

$\dfrac{d}{dx}\left\{x^4(5\sin x-3\cos x)\right\}$

$=(5\sin x-3\cos x)\dfrac{d}{dx}(x^4)+x^4\dfrac{d}{dx}(5\sin x-3\cos x)$

$=4x^3(5\sin x-3\cos x)+x^4(5\cos x+3\sin x)$ .

$=20 x^3\sin x+5x^4\cos x-12x^3\cos x+3x^4\sin x$ .

Let f(x) = ( x - 1) (x - 2) ( x - 3)..........( x - n), $n \space \epsilon \space N$ and $f'(n) = 5040$ , then the value of n is

$\ln(f(x)) = \ln(x-1) + \ln(x-2) +...........+\ln(x-n)$

$\Rightarrow f'(x) = f(x)\begin{bmatrix}\dfrac {1}{x-2} + \dfrac{1}{x-2} +........\dfrac{1}{x-n}\end{bmatrix}$

$\Rightarrow f(x) = (x-2) (x-3) (x-n) + (x-1)(x-3).....(x-n) +......+ (x-1) (x - 2)....(x-(n - 1))$

$\Rightarrow f'(x) = (n-1)(n-2)(n-3). 3. 2. 1$ (all other except the last vanishes when x = n)

$\Rightarrow 5040 = (n-1)!$

$\Rightarrow n = 8$

Differentiate the following with respect to $x$ :
$e^{3x}\cos 2x$

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If $x^{2}+y^{2}=R^{2}(\text { where } R>0)$ and $k=\dfrac{y^{\prime \prime}}{\sqrt{\left(1+y^{\prime 2}\right)^{3}}}$
then find $k$ in terms of $R$ alone.

$x^{2}+y^{2}=R^{2}$

Differentiating w.r.t.x,

$\Rightarrow 2 x+2 y y^{\prime}=0 \quad\quad\quad(i)$

$\Rightarrow y^{\prime}=-\dfrac{x}{y} \quad\quad\quad(ii)$

Differentiating (1) w.r.t.

$x, \Rightarrow 1+y y^{\prime \prime}+(y)^{2}=0$

$\Rightarrow y^{\prime \prime}=-\dfrac{1+\left(y^{\prime}\right)^{2}}{y}$

Given

$k=\dfrac{y^{\prime \prime}}{\left(1+\left(y^{\prime}\right)^{2}\right)^{3 / 2}}=-\dfrac{1+\left(y^{\prime}\right)^{2}}{y\left(1+\left(y^{\prime}\right)^{2}\right)^{3 / 2}}$

$=-\dfrac{1}{y \sqrt{1+\left(y^{\prime}\right)^{2}}}=-\dfrac{1}{y \sqrt{1+\dfrac{x^{2}}{y^{2}}}}$

$=-\dfrac{1}{\sqrt{y^{2}+x^{2}}}=-\dfrac{1}{R}$

Let $f(x)=x+\dfrac{1}{2x+\dfrac{1}{2x+ \dfrac{1}{2x + \cdots \infty}}}$
Compute the value of $f(50) . f^{\prime}(50)$

$f(x)=x+\dfrac{1}{x+x+\dfrac{1}{2 x+\dfrac{1}{2 x+\cdots \infty}}}=x+\dfrac{1}{x+f(x)}$

$\Rightarrow f(x)-x=\dfrac{1}{x+f(x)}$

$\Rightarrow f^{2}(x)-x^{2}=1$

differentiating w.r.t.x

$\Rightarrow 2 f(x) \cdot f^{\prime}(x)-2 x=0$

$\Rightarrow f(x) \cdot f^{\prime}(x)=x$

$\Rightarrow f(50) \cdot f^{\prime}(50)=50$

If $f$ is a real function such that $f(x) > 0, f'(x)$ is continuous for all real $x$ and $ax \ge 2 \sqrt {f(x)}-2a\ f(x), (ax\neq 2)$ show that $\sqrt {f(x)}\ge \dfrac {\sqrt {f(1)}}{x}, x\ge 1$ .

We have to prove that

$\sqrt {f(x)} \ge \dfrac {\sqrt {f(1)}}{x}, x \ge 1$ or

$x\sqrt {f(x)}\ge 1\sqrt {f(1)}$ .
This suggests that we have to consider function

$x\sqrt {f(x)}$ .
Now, given that

$ax f'(x) \ge 2 \sqrt {f(x)}-2af (x)$ ,
dividing both sides by

$2\sqrt {f(x)}$ we have

$ax \dfrac {f'(x)}{2\sqrt {f(x)}}+a\sqrt {f(x)}-1 \ge 0$

$\Rightarrow \ \dfrac {d}{dx}(ax \sqrt {f(x)}-x)\ge 0$

$\Rightarrow \$ Hence,

$ax \sqrt {f(x)}-x$ is an increasing function,

$\Rightarrow \ x\ge 1$ then

$f(x) \ge f(1)$

$\Rightarrow \ ax \sqrt {f(x)}-x \ge a\sqrt {f(1)}-1$

$\Rightarrow \ ax \sqrt {f(x)} \ge a\sqrt {f(1)} +x-1\ge a\sqrt {f(1)}$ (as

$x \ge 1$ )

$\Rightarrow \ \sqrt {f(x)}\ge \dfrac {\sqrt {f(1)}}{x}$ .

Use the function $f(x)=x^{1/x}, x > 0$ , to determine the bigger of the two number $e^{\pi}$ and $\pi^{e}$

Let

$f(x)=x^{1/x}$

$\Rightarrow \log f(x)=(1/x)\log x$

$\dfrac{f'(x)}{f(x)}=\dfrac{1}{x}\dfrac{1}{x}-\dfrac{1}{x^{2}}\log x=\dfrac{1}{x^{2}}(1-\log x)$

$\Rightarrow f'(x)=\dfrac{x^{1/x}}{x^{2}}(1-\log x)$
Obviously for

$x > e, \log x > 1$ so

$f'(x) < 0$

$\therefore f(x)$ is a monotonically decreasing function of

$x$ for

$x\ge e$
Also,

$\pi > e\Rightarrow f(\pi) < f(e)$

$\Rightarrow \pi^{1/\pi}< e^{1/e}\Rightarrow (\pi^{1/\pi})^{\pi}(e^{1/e})^{\pi}\Rightarrow \pi < (e^{\pi})^{1/e}\Rightarrow \pi^{e} < e^{\pi}$

Differentiate the following w.r.t.x: $\log \left[\tan ^{3} x \cdot \sin ^{4} x \cdot\left(x^{2}+7\right)^{7}\right]$

Let

$y=\log \left[\tan ^{3} x \cdot \sin ^{4} x \cdot\left(x^{2}+7\right)^{7}\right]$

$=\log \tan ^{3} x+\log \sin ^{4} x+\log \left(x^{2}+7\right)^{7}$

$=3 \log \tan x+4 \log \sin x+7 \log \left(x^{2}+7\right)$
Differentiating w.r.t.

$x,$ we get

$\dfrac{\mathrm{dy}}{\mathrm{dx}}=\dfrac{\mathrm{d}}{\mathrm{dx}}\left[3 \log \tan x+4 \log \sin x+7 \log \left(x^{2}+7\right)\right]$

$=3 \dfrac{d}{d x}(\log \tan x)+4 \dfrac{d}{d x}(\log \sin x)+7 \dfrac{d}{d x}\left[\log \left(x^{2}+7\right)\right]$

$=3 \times \dfrac{1}{\tan x} \cdot \dfrac{\mathrm{d}}{\mathrm{dx}}(\tan x)+4 \times \dfrac{1}{\sin x} \cdot \dfrac{\mathrm{d}}{\mathrm{dx}}(\sin x)+7 \times \dfrac{1}{x^{2}+7} \cdot \dfrac{\mathrm{d}}{\mathrm{dx}}\left(x^{2}+7\right)$

$=3 \times \dfrac{1}{\tan x} \cdot \sec ^{2} x+4 \times \dfrac{1}{\sin x} \cdot \cos x+7 \times \dfrac{1}{x^{2}+7} \cdot(2 x+0)$

$=3 \times \dfrac{\cos x}{\sin x} \times \dfrac{1}{\cos ^{2} x}+4 \cot x+\dfrac{14 x}{x^{2}+7}$

$=\dfrac{6}{2 \sin x \cos x}+4 \cot +\dfrac{14 x}{x^{2}+7}$

$=\dfrac{6}{\sin 2 x}+4 \cot x+\dfrac{14 x}{x^{2}+7}$

$=6 \operatorname{cosec} 2 x+4 \cot x+\dfrac{14 x}{x^{2}+7}$

If u, v and w are function of x , then show that
Undefined control /cfrac
in two ways - first by repeated aplication of product rule , second by logarithmic differentiation .

1910401_1858630_ans_66417dbc59634e93841c6ddf3dabf6f9.jpg

If $ax^{2}+\dfrac{b}{x}\ge c$ for all positive $x$ where $a > 0$ and $b > 0$ , show that $27ab^{2}\ge 4c^{3}$

Given

$ax^{2}+\dfrac{b}{x}\ge c$

$\forall x > 0, a > 0, b > 0$
To show that

$27ab^{2}\ge 4c^{3}$
Let us consider the function

$f(x)=ax^{2}+\dfrac{b}{x}-c$ ,
then

$f'(x)=2ax-\dfrac{b}{x^{2}}=0\Rightarrow x^{3}=b/2a\Rightarrow x=(b/2a)^{1/3}$
Also,

$f"(x)=2a+\dfrac{2b}{x^{3}}\Rightarrow f"\left(\left(\dfrac{b}{2a}\right)^{1/3}\right)=6a > 0$
Therefore,

$f$ is minimum at

$x=\left(\dfrac{b}{2a}\right)^{1/3}$
As

$(1)$ is true

$\forall x\therefore$ is so for

$x=\left(\dfrac{b}{2a}\right)^{1/3}$

$\Rightarrow a\left(\dfrac{b}{2a}\right)^{2/3}+\dfrac{b}{(b/2a)^{1/3}}\ge c$

$\Rightarrow \dfrac{a\left(\dfrac{b}{2a}\right)+b}{(b/2a)^{1/3}}\ge c\Rightarrow \dfrac{3b}{2}\left(\dfrac{2a}{b}\right)^{1/3}\ge c$
As

$a, b$ are

$+ve$ , cubing both sides, we get

$\dfrac{27b^{3}}{8}\dfrac{2a}{b}\ge c^{3}$

$\Rightarrow 27ab^{2}\ge 4c^{3}$

If $y=y(x)$ and it follows the relation $2e^{xy^{2}}+y \cos (x^{2})=4,$ then $|y '(0)|$ is equal to

$x=0 \Rightarrow 2+ycos0=4$

$\Rightarrow y=2$
Differentiate it w.r.t. x

$\Rightarrow 2e^{xy^{2}}(y^{2}+2xy.y')+y'.cos x^{2}-2xy sin(x^{2})=0$
put

$x=0,y=2$

$y'(0)=-8$

$\Rightarrow |y'(0)|=8$

Let $\displaystyle \left ( a-b\cos y \right )\left ( a+b\cos x \right )=a^{2}-b^{2}$ and $\displaystyle \frac{dy}{dx}=\frac{\sin xf\left ( y ) \right )}{\left ( a+b\cos x \right )^{2}}$ . If $a^{2}-b^{2}=192, then f\left ( \pi /2 \right ).$

Expanding the above expression, we get

$a^{2}+ab(\cos(x)-\cos(y))-b^{2}\cos x\cos y=a^{2}-b^{2}$

$a[\cos(x)-\cos(y)]=b[\cos(x)\cos(y)-1]$

Differentiating with respect to x.

$a[\sin(y).y'-\sin(x)]=-b[\sin(x)\cos(y)+\cos(x)\sin(y).y']$

$y'[a\sin(y)+b\cos(x)\sin(y)]=a\sin(x)-b\sin(x)\cos(y)$

$y'=\dfrac{a\sin(x)-b\sin(x)\cos(y)}{a\sin(y)+b\cos(x)\sin(y)}$

$=\dfrac{\sin(x)}{a+b\cos(x)}.\dfrac{a-b\cos(y)}{\sin(y)}$

Now

$(a-b\cos y)=\dfrac{a^{2}-b^{2}}{a+b\cos x}$

Substituting above,we get

$\dfrac{\sin(x)}{a+b\cos(x)}.\dfrac{a-b\cos(y)}{\sin(y)}$

$=\dfrac{\sin(x)}{(a+b\cos(x))^{2}}.\dfrac{a^{2}-b^{2}}{\sin(y)}$

Hence

$f(y)=\dfrac{a^{2}-b^{2}}{\sin(y)}$

Now

$f(\dfrac{\pi}{2})=a^{2}-b^{2}$

$=192$ .

If $\displaystyle y=\left ( x^{2}+1 \right )\sin x$ then $\left ( \dfrac{\pi}2 \right )^{2}-y_{20}\left ( \dfrac{\pi}2 \right )$ is equal to

$y=\left( x^{ 2 }+1 \right) \sin x$

${ y }_{ 1 }=\left( x^{ 2 }+1 \right) \cos { x } +2x\sin { x }$

${y }_{ 2 }=-\left( x^{ 2 }+1 \right) \sin { x } +4x\cos { x } +2\sin { x }$

$\Rightarrow { y }_{ 2 }=-y+4x\cos { x } +2\sin { x }$

${ y }_{ 3 }=-{ y }_{ 1 }-4x\sin { x } +6\cos { x }$

$\Rightarrow { y }_{ 3 }=-\left( x^{ 2 }+1 \right) \cos { x } -6x\sin { x } +6\cos { x }$

${ y }_{ 4 }=-{ y }_{ 2 }-4x\cos { x } -10\sin { x }$

$\Rightarrow { y }_{ 4 }=\left( x^{ 2 }+1 \right) \sin { x } -8x\cos { x } -12\sin { x }$

$\Rightarrow { y }_{ 5 }=\left( x^{ 2 }+1 \right) \cos { x } +10x\sin { x } -20\cos { x }$

$\Rightarrow { y }_{ 6 }=-\left( x^{ 2 }+1 \right) \sin { x } +12x\cos { x } +30\sin { x }$

$\Rightarrow { y }_{ 7 }=-{ y }_{ 1 }-12x\sin { x } +42\cos { x }$

$\Rightarrow { y }_{ 8 }=\left( x^{ 2 }+1 \right) \sin { x } -16x\cos { x } -56\sin { x }$

So, we can conclude,

${y}_{4n}=\left( x^{ 2 }+1 \right) \sin { x }-2(4n)\cos { x } -(4n)(4n-1)\sin { x }$
Put

$n=5$ in above eqn

${y}_{20}=\left( x^{ 2 }+1 \right) \sin { x }-40\cos { x } -380\sin { x }$

${y}_{20}(\dfrac{\pi}{2})=\left( (\dfrac{\pi}{2})^{ 2 }+1 \right) -380$

$\Rightarrow (\dfrac{\pi}{2})^{ 2 }-{y}_{20}(\dfrac{\pi}{2})=379$

Differentiate the following function with respect to $x$ :
$(\log x)^x + x^{\log x}$ .

Let

$y=(\log x)^x+x^{\log x}$
Then

$\dfrac{dy}{dx}=\dfrac{d((\log x)^x+x^{\log x})}{dx}$

$=\dfrac{d(\log x)^x}{dx}+\dfrac{d(x^{\log x})}{dx}$

$=(\log x)^x\dfrac{d{(x\log (\log x))}}{dx}+x^(\log x)\dfrac{d(\log x \log x)}{dx}$

From

$\left(\dfrac{d(u^v)}{dx}=u^v\dfrac{d(v\log u)}{dx} \right )$

$=(\log x)^x\left \{ x\left ( \dfrac{1}{\log x}\dfrac{1}{x}+\log (\log x) \right ) \right \}+x^{\log x}\left ( 2(\log x)\dfrac{1}{x} \right )$

Therefore,

$\left ( \dfrac{d(\log x \log x)}{dx}=\dfrac{d(\log x)^2}{dx} \right )$

$=(\log x)^x\left \{ \dfrac {1}{\log x}+\log (\log x)\right \}+2\left ( \dfrac{\log x}{x}x^{\log x} \right )$

If $y = f(u)$ is a differentiable function of $u$ and $u = g(x)$ is a differentiable function of $x$ , then prove that $y = f(g(x))$ is a differentiable function of $x$ and $\dfrac {dy}{dx} = \dfrac {dy}{du}\times \dfrac {dy}{dx}$

Let

$\delta x, \delta y$ and

$\delta u$ be a small change in

$x, y$ and

$u$ respectively.
As,

$\delta x \rightarrow 0, \delta y \rightarrow 0, \delta u\rightarrow 0$ .
As

$u$ is differentiable function, it is continuous.
Consider the incrementary ratio

$\dfrac {\delta y}{\delta x}$
We have,

$\dfrac {\delta y}{\delta x} = \dfrac {\delta y}{\delta u} \times \dfrac {\delta u}{\delta x}$
Taking limit as

$\delta x \rightarrow 0$ on both sides

$\displaystyle \lim_{\delta x\rightarrow 0} \dfrac {\delta y}{\delta x} = \displaystyle \lim_{\delta x\rightarrow 0} \left (\dfrac {\delta y}{\delta u}\times \dfrac {\delta u}{\delta x}\right )$

$\displaystyle \lim_{\delta x\rightarrow 0} \dfrac {\delta y}{\delta x} = \displaystyle \lim_{\delta x\rightarrow 0} \dfrac {\delta y}{\delta u} \times \displaystyle \lim_{\delta x\rightarrow 0} \dfrac {\delta u}{\delta x}$

$\displaystyle \lim_{\delta x\rightarrow 0} \dfrac {\delta y}{\delta x} = \displaystyle \lim_{\delta x\rightarrow 0} \dfrac {\delta y}{\delta y}\times \displaystyle \lim_{\delta x\rightarrow 0} \dfrac {\delta u}{\delta x} ..... (i)$
Since

$y$ is a differentiable function of

$u, \displaystyle \lim_{\delta x\rightarrow 0} \left (\dfrac {\delta y}{\delta u}\right )$ exists and

$\displaystyle \lim_{\delta x\rightarrow 0} \left (\dfrac {\delta u}{\delta x}\right )$ exists as

$u$ is a differentiable function of

$x$ .
R.H.S. of equation (i) exists.
Now,

$\displaystyle \lim_{\delta x\rightarrow 0} \dfrac {\delta y}{\delta u} = \dfrac {dy}{du}$ and

$\displaystyle \lim_{\delta x\rightarrow 0} \dfrac {\delta u}{\delta x} = \dfrac {du}{dx}$

$\therefore \displaystyle \lim_{\delta x\rightarrow 0} \dfrac {\delta y}{\delta x} = \dfrac {dy}{du}\times \dfrac {du}{dx}$
Since R.H.S. exists.

$\therefore$ L.H.S. also exists and

$\displaystyle \lim_{\delta x\rightarrow 0} \dfrac {\delta y}{\delta x} = \dfrac {dy}{dx}$

$\therefore \dfrac {dy}{dx} = \dfrac {dy}{du} \cdot \dfrac {du}{dx}$

If $x^x + x^y + y^x = a^b$ , then find $\dfrac {dy}{dx}$ .

Given,

${ x }^{ x }+{ x }^{ y }+{ y }^{ x }={ a }^{ b }$
Let

$p={ x }^{ x },q={ x }^{ y },r={ y }^{ x }$
Let us take

$p={ x }^{ x }$
Take

$\log$ on both sides

$\log { p=x\log { x } }$

$\Rightarrow \cfrac { 1 }{ p } \cfrac { dp }{ dx } =\log { x } +\cfrac { x }{ x } =\log { x } +1=\log { x+\log { e } }$

$\Rightarrow \cfrac { dp }{ dx } ={ x }^{ x }\log { ex }$ ....(1)
Now lets take

$q={ x }^{ y }$
Take

$\log$ on both sides

$\log { q } =y\log { x }$

$\Rightarrow \cfrac { 1 }{ q } \cfrac { dq }{ dx } =\cfrac { y }{ x } +\log { x } \cfrac { dy }{ dx }$

$\Rightarrow \cfrac { dq }{ dx } =y{ x }^{ y-1 }+\left( { x }^{ y }\log { x } \right) \cfrac { dy }{ dx }$ ....(2)
and

$r={ y }^{ x }$
Take

$\log$ on both sides

$\log { r } =x\log { y }$

$\Rightarrow \cfrac { 1 }{ r } \cfrac { dr }{ dx } =\cfrac { x }{ y } \cfrac { dy }{ dx } +\log { y }$

$\Rightarrow \cfrac { dr }{ dx } ={ y }^{ x }\log { y } +x{ y }^{ x-1 }\cfrac { dy }{ dx }$ ....(3)

${ x }^{ x }+{ x }^{ y }+{ y }^{ x }={ a }^{ b }\Longrightarrow p+q+r={ a }^{ b }$
Differentiate both sides with respect to

$x$

$\cfrac { dp }{ dx } +\cfrac { dq }{ dx } +\cfrac { dr }{ dx } =0$

${ x }^{ x }\log { ex } +y{ x }^{ y-1 }+{ y }^{ x }\log { y } +\cfrac { dy }{ dx } \left[ { x }^{ y }\log { x } +x{ y }^{ x-1 } \right] =0$

$\cfrac { dy }{ dx } =-\left[ \cfrac { { x }^{ x }\log { ex } +y{ x }^{ y-1 }+{ y }^{ x }\log { y } }{ { x }^{ y }\log { x } +x{ y }^{ x-1 } } \right]$

If $y = \tan^{2} (\log x^{3})$ , find $\dfrac {dy}{dx}$ .

Given

$y=tan^2(log\:x^3)$

We need to find

$\dfrac{dy}{dx}$

Consider

$y=tan^2(log\:x^3)$

$\Rightarrow y=tan^2(3\:log\:x)$

$\Rightarrow y=[tan\:(3\:log\:x)]^2$

Differentiate with respect to

$x$ on both sides we get

$\Rightarrow \dfrac{dy}{dx}=2[tan\:(3\:log\:x)] \cdot sec^2(3\:log\:x)\cdot \dfrac{3}{x}$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{6}{x}\cdot [tan\:(3\:log\:x)] \cdot sec^2(3\:log\:x)$

$\therefore \dfrac{dy}{dx}=\dfrac{6}{x}\cdot [tan\:(log\:x^3)] \cdot sec^2(log\:x^3)$

Find $\dfrac {dy}{dx}$ if $x\sin y + y\sin x = 0$

Given $x siny+ysinx=0$

Now differentiate the equation with respective to $x$

We get $siny+xcosy \cfrac{dy}{dx}+\cfrac{dy}{dx}sinx+ycosx=0$

$\Rightarrow \cfrac{dy}{dx}(xcosy+sinx)+(siny+ycosx)=0$

$\Rightarrow \cfrac{dy}{dx}=-(\cfrac{siny+ycosx}{xcosy+sinx})$

Find the derivative of $\displaystyle\, y = (x \, +\, 1)(x \, +\, 2)^2$

$y = (x + 1)(x + 2)^{2}$

$\dfrac {dy}{dx} = (x + 1) \dfrac {d}{dx} (x + 2)^{2} + (x + 2)^{2} \dfrac {d}{dx} (x + 1)$ [Apply product Rule of differentiable]

$\left [\dfrac {d}{dx} f(x) \cdot g(x) = \dfrac {d}{dx} f(x) . q (x) + \dfrac {d}{dx} g(x)\right ]$

$= (x + 1) \cdot 2(x + 2) + (x + 2)^{2} \cdot 1$

$= (x + 2) [2(x + 1) + x + 2]$

$= (x + 2) [2x + 2 + x + 2]$

$= (x + 2) [3x + 4]$

$= (x + 2) (3x + 4)$ .

Find the derivative of $\displaystyle\, y = e^x(x^2 \, -\, 2x \, +\, 2)$

$y = e^{x} (x^{2} - 2x + 2)$

Applying product Rule of differentiation

$\dfrac {dy}{dx} = e^{x} \cdot \dfrac {d}{dx} (x^{2} - 2x + 2) + (x^{2} - 2x + 2) \dfrac {d}{dx} (e^{x})$

$= e^{x} (2x - 2) + (x^{2} - 2x + 2) e^{x}$

$= 2xe^{x} - 2e^{x} + x^{2}e^{x} - 2xe^{x} + 2e^{x}$

$= x^{2}e^{x}$ .

$\dfrac {dy}{dx} = x^{2}e^{x}$ .

Find the sum of $\dfrac {1}{1 + x} + \dfrac {2x}{1 + x^{2}} + \dfrac {4x^{3}}{1 + x^{4}} + .... + \dfrac {2^{n} x^{2^{n} - 1}}{1 + x^{2^{n}}}$ .

Let

$f(x) = \dfrac{1}{1+x} + \dfrac{2x}{1+x^2} + \dfrac{4x^3}{1+x^4}+ ..... + \dfrac{2^nx^{2^n-1}}{1+x^{2^n}}$

We see that in every term of

$f(x)$ , the numerator is the derivative of the denominator.

Taking advantage of this, we integrate

$f(x)$

$\Rightarrow \int f(x) dx = ln(1+x) + ln (1+x^2) + ln(1+x^4) + ... + ln(1+x^{2^n})$

$\therefore \int f(x) dx = ln[(1+x)(1+x^2)...(1+x^{2^n})]$

Now, we multiply and divide by (1-x) inside the logarithm

$\therefore \int f(x) dx = ln\bigg[\dfrac{(1-x)(1+x)(1+x^2)...(1+x^{2^n})}{(1-x)}\bigg]$

Using

$(a+b)(a-b) = a^2-b^2$ repetitively in the numerator, we get

$\therefore \int f(x) dx = ln\dfrac{(1-x^{2^{n+1}})}{(1-x)} = ln(1-x^{2^{n+1}}) - ln(1-x)$

Now, differentiating on both sides w.r.t. x, we get

$\therefore f(x) = \dfrac{1}{(1-x^{2^{n+1}})}(-x^{2^{n+1}-1}) + \dfrac{1}{(1-x)}$

This is the value of the required sum. If limits are asked, remember to put them at the end, after the differentiation.

If y = $x^x$ + $(sinx)^{cotx}$ . find $\frac{dy}{dx}$

$\dfrac{dy}{dx}= \dfrac{d(x)^{x}}{dx} + \dfrac{(sinx)^{cotx}}{dx}$

$\Rightarrow$ Let

$z= x^{x}$

Take log on both sides

$\Rightarrow$

$logz =xlogx$

Now differentiate above equation with respect to

$x$

$\Rightarrow$

$\dfrac{1}{z}\times\dfrac{dz}{dx}$ =

$1+logx$

$\Rightarrow$

$\dfrac{dz}{dx}= x^{x}(1+logx)$

$\Rightarrow$ Let

$t=(sinx)^{cotx}$

Take log on both sides

$\Rightarrow$

$logt=cotx\times log(sinx)$

Now differentiate with respect to

$x$

$\Rightarrow$

$\dfrac{1}{t}\times \dfrac{dt}{dx}=-cosec^{2}x\times log(sinx) + cotx\times\dfrac{1}{sinx}\times cosx$

$\Rightarrow$

$\dfrac{dt}{dx}$ =

$(sinx)^{cotx}\times(-cosec^{2}x\times log(sinx) +cot^{2}x)$

$\Rightarrow$

$\dfrac{dy}{dx} = \dfrac{dt}{dx} +\dfrac{dz}{dx}$

Put the values of

$\dfrac{dz}{dx} and \dfrac{dt}{dx}$

$\dfrac{dy}{dx}= x^{x}(1+logx)$

$+$

$(sinx)^{cotx}\times(-cosec^{2}x\times log(sinx) +cot^{2}x)$

If $y = {\tan ^{ - 1}}\left( {\dfrac{a}{x}} \right) + \log \sqrt {\dfrac{{x - a}}{{x + a}}} \,$ , prove that $\dfrac{{dy}}{{dx}} = \dfrac{{2{a^3}}}{{{x^4} - {a^4}}}$

Given that ,

$y={{\tan }^{-1}}\left( \dfrac{a}{x} \right)+\log \sqrt{\dfrac{x-a}{x+a}}$

$={{\tan }^{-1}}\left( \dfrac{a}{x} \right)+\log \sqrt{x-a}-\log \sqrt{x+a}$

$={{\tan }^{-1}}\left( \dfrac{a}{x} \right)+\dfrac{1}{2}\log \left( x-a \right)-\dfrac{1}{2}\log \left( x+a \right)$

$={{\tan }^{-1}}\left( \dfrac{a}{x} \right)+\dfrac{1}{2}\left( \log \left( x-a \right)-\log \left( x+a \right) \right)$

Differentiate with respect. to x ,we get

$\dfrac{dy}{dx}=\dfrac{1}{1+{{\left( \dfrac{a}{x} \right)}^{2}}}\left( -a.\dfrac{1}{{{x}^{2}}} \right)+\dfrac{1}{2}\left( \dfrac{1}{x-a}-\dfrac{1}{x+a} \right)$

$=\dfrac{{{x}^{2}}}{{{x}^{2}}+{{\left( a \right)}^{2}}}\left( -a.\dfrac{1}{{{x}^{2}}} \right)+\dfrac{1}{2}\left( \dfrac{x+a-\left( x-a \right)}{\left( x-a \right)\left( x+a \right)} \right)$

$=\dfrac{1}{{{x}^{2}}+{{a}^{2}}}\left( -a \right)+\dfrac{1}{2}\left( \dfrac{x+a-x+a}{\left( x-a \right)\left( x+a \right)} \right)$

$=\dfrac{-a}{{{x}^{2}}+{{a}^{2}}}+\dfrac{a}{\left( x-a \right)\left( x+a \right)}$

$=\dfrac{-a}{{{x}^{2}}+{{a}^{2}}}+\dfrac{a}{{{x}^{2}}-{{a}^{2}}}$

$=\dfrac{-a\left( {{x}^{2}}-{{a}^{2}} \right)+a\left( {{x}^{2}}+{{a}^{2}} \right)}{\left( {{x}^{2}}-{{a}^{2}} \right)\left( {{x}^{2}}+{{a}^{2}} \right)}$

$=\dfrac{2{{a}^{3}}}{\left( {{x}^{4}}-{{a}^{4}} \right)}$

Differentiate $x^{\sin x}, x > 0$ with respect to $x$ .

Let

$y = x^{\sin x}$
Apply logarithms on both sides

$\log y = \sin x \cdot \log x$
Diff w.r.t

$x$

$\dfrac {1}{y} \dfrac {dy}{dx} = \sin x\cdot \dfrac {d}{dx} (\log x)+ \log x \cdot \dfrac {d}{dx}(\sin x)$

$\dfrac {1}{y} \dfrac {dy}{dx} = \sin x \cdot \dfrac {1}{x} + \log x \cdot \cos x$

$\dfrac {dy}{dx} = y\left [\dfrac {\sin x}{x} + \log x\cdot \cos x\right ]$

$\therefore \dfrac {dy}{dx} = x^{\sin x} \left [\dfrac {\sin x}{x} + \log x\cdot \cos x\right ]$ .

Differentiate the function with respect to x:
$x^{\sin x} + (\sin x)^{\cos x}$

Let

$y=x^{\sin x} + (\sin x)^{\cos x}$

Taking Log both sides

$\log y={\sin x}\log x + {\cos x}\log(\sin x)$

Now Differentiate wrt x

$\dfrac1y\dfrac{dy}{dx}=\dfrac{d}{dx}({\sin x}\log x) + \dfrac{d}{dx}({\cos x}\log(\sin x))\\\Rightarrow \dfrac1y\dfrac{dy}{dx}=(\sin x.\dfrac1x+\cos x.\log x)+(\cos x.\dfrac1{\sin x}.(\cos x)-\sin x\log \sin x)\\\Rightarrow \dfrac{dy}{dx}= [\sin x.\dfrac1x+\cos x.\log x+\dfrac{\cos^2x}{\sin x}-\sin x\log \sin x][x^{\sin x} + (\sin x)^{\cos x}]$

If $y=x^x+x^7+7^x+7^7$ , then $\dfrac{dy}{dx}=?$

$y={ x }^{ x }+{ x }^{ 7 }+{ 7 }^{ x }+{ 7 }^{ 7 }\\ Let\quad { x }^{ x }=u$

Taking log both the sides,

$\\ \log { { x }^{ x } } =\log { u } \\ =>x\log { x } =\log { u }$

Differentiating both sides with respect to 'x',

$\cfrac { 1 }{ u } \cfrac { du }{ dx } =\cfrac { x }{ x } +\log { x } \\ =>\cfrac { du }{ dx } =u\times \left( 1+\log { x } \right) \\ =>\cfrac { du }{ dx } ={ x }^{ x }\times \left( 1+\log { x } \right)$

Now,

$y={ x }^{ x }+{ x }^{ 7 }+{ 7 }^{ x }+{ 7 }^{ 7 }\\ =>y=u+{ x }^{ 7 }+{ 7 }^{ x }+{ 7 }^{ 7 }\\ =>\cfrac { dy }{ dx } =\cfrac { du }{ dx } +7{ x }^{ 6 }+{ 7 }^{ x }\log { 7 } +0\\ =>\cfrac { dy }{ dx } ={ x }^{ x }\times \left( 1+\log { x } \right) +7{ x }^{ 6 }+{ 7 }^{ x }\log { 7 }$

Establish the following
If $\sqrt{1-{x}^{2}}+\sqrt{1-{y}^{2}}=a(x-y)$ then $\cfrac{dy}{dx}=\sqrt { \cfrac { 1-{ y }^{ 2 } }{ 1-{ x }^{ 2 } } }$

1328673_1081548_ans_14add20385964a958e9efd75199c823d.jpg

$x \cos (a + y) = \cos y$
then prove that
$\dfrac { dy }{ dx } =\dfrac { { \cos }^{ 2 }(a+y) }{ { \sin }_{ a } }$

It is given that

$cosy=xcos\left( a+y \right)$

$\frac { d }{ dx } \left[ cosy \right] =\frac { d }{ dx } \left[ xcos\left( a+y \right) \right]$

$-siny\frac { dy }{ dx } =cos\left( a+y \right) \ast \frac { d }{ dx } \left( x \right) \ast x\frac { d }{ dx } \left[ cos\left( a+y \right) \right]$

$-siny\frac { dy }{ dx } =cos\left( a+y \right) +x\left[ -sin\left( a+y \right) \right] \frac { d }{ dx }$

$\left[ xsin\left( a+y \right) -siny \right] \frac { dy }{ dx } =cos\left( a+y \right) \Longrightarrow (1)$

Since

$cosy=xcos\left( a+y \right) ,x=\frac { cosy }{ cos(a+y) }$

The eqa (1) reduces to

$\left[ \frac { cosy }{ cos(a+y) } \ast sin\left( a+y \right) -siny \right] \frac { dy }{ dx } =cos\left( a+y \right)$

$\left[ cosy\ast sin\left( a+y \right) -siny\ast cos\left( a+y \right) \right] \frac { dy }{ dx } ={ cos }^{ 2 }\left( a+y \right)$

$sin\left[ a+y-y \right] \frac { dy }{ dx } ={ cos }^{ 2 }\left( a+b \right)$ \\

$\frac { dy }{ dx } =\frac { { cos }^{ 2 }\left( a+b \right) }{ sina }$

If $y=xlog\left(\dfrac{x}{a+bx}\right)$ prove that $x^3y_2=(xy_1-y)^2$ .

1185674_1062880_ans_0bc2f4f0d7814321b5ff310fc32c6e4b.jpg

Evaluate:
$x^x-2^{\sin x}$ .

given

$y=x^{x}-2^{\sin x}$

Let

$y=x^{x}$ and

$y_{2}=2^{\sin x}$

then

$\dfrac{dy}{dx}=\dfrac{dy}{dx^{1}}-\dfrac{dy}{dx^{2}}$

$\log y_{1}=x\log x; \log y_{2}=\sin x\log 2$

$\dfrac{1}{y_{1}}\dfrac{dy}{dx^{1}}=\log x+x\times \dfrac{1}{x}; \dfrac{1}{y_{2}}\dfrac{dy}{dx^{2}}=\cos x.\log 2$

$\dfrac{1}{y_{1}}\dfrac{dy}{dx^{1}}=\log x+1; \dfrac{1}{y_{2}}\dfrac{dy}{dx^{2}}=\cos \log 2$

$\dfrac{dy}{dx^{1}}=y_{1}(1+\log x); \dfrac{dy}{dx^{2}}=2^{\sin x}.\cos x.\log2$

$\dfrac{dy}{dx}=\dfrac{dy}{dx^{1}}-\dfrac{dy}{dx^{2}}$

$\dfrac{dy}{dx}=x^{x}(1+\log x)-2^{\sin x}\cos x\log 2$

If $x\sqrt {1+y}+y\sqrt {1+x}=0$ and $x \neq y$ , show that $\dfrac {dy}{dx}=\dfrac {-1}{(1+x)^{2}}$

$x\sqrt{1+y}+y\sqrt{1+x}=0$

$\Rightarrow x\sqrt{1+y}=-y\sqrt{1+x}$

Squaring both sides, we get

${x}^{2}\left(1+y\right)={y}^{2}\left(1+x\right)$

$\Rightarrow {x}^{2}+{x}^{2}y={y}^{2}+{y}^{2}x$

$\Rightarrow {x}^{2}-{y}^{2}={y}^{2}x-{x}^{2}y$

$\Rightarrow \left(x+y\right)\left(x-y\right)=xy\left(y-x\right)$

$\Rightarrow x+y=-xy$

$\Rightarrow x+y+xy=0$

$\Rightarrow y\left(1+x\right)=-x$

$\Rightarrow y=\dfrac{-x}{x+1}$

Differentiating both sides w.r.t

$x$ we get

$\Rightarrow \dfrac{dy}{dx}=-\dfrac{\left(1+x\right)\dfrac{d}{dx}\left(x\right)-x\dfrac{d}{dx}\left(1+x\right)}{{\left(1+x\right)}^{2}}$

$=-\dfrac{1+x-x}{{\left(1+x\right)}^{2}}$

$=\dfrac{-1}{{\left(1+x\right)}^{2}}$

Hence proved.

Differentiate:
${ xy + y}^{ 2 } = \tan x + y$ ?

$xy+y²=tanx+y$

$\dfrac { δx }{ δy } =y-sec^2x$

$\dfrac { δx }{ δy } =x+2y-1$

$now,\dfrac { δx }{ δy } =\dfrac { -\dfrac { δx }{ δy } }{ \dfrac { δy }{ δx } }$

$=\dfrac { sec^2x-y }{ x+2y-1 }$

Differentiate $y=\dfrac { 4\sin { \theta } }{ 2+\cos { \theta } }$ than $\dfrac { dy }{ dx }=$ ?

1356723_1128692_ans_ef02cc6d527a4af5ad47015962aeff3d.jpg

$f(x) = \sqrt {\log_{1/2}\left (\dfrac {5x - x^{2}}{4}\right )}$ .

$f(x)=\sqrt{\log_{1/2}(\dfrac{5x-x^2}{4})}$

$\log_{\dfrac{1}{2}}(\dfrac{5x-x^2}{4})\ge 0$

$\dfrac{5x-x^2}{4} \le (\dfrac{1}{2})^0$

$\because{\begin{bmatrix}b<a\\a>1\\b>a\end{bmatrix}}{}$

$\dfrac{5x-x^2}{4} \le 1$

$5x-x^2 \le 4$

$x^2-5x \le -4$

$x^2-5x+4 \ge 0$

$x^2-4x-x+4 \ge 0$

$x(x-4)-1(x-4) \ge 0$

$(x-1)(x-4) \ge 0$

$x=(-\infty,1)\cup (4,\infty)$ Refer image(1)

another condition

$\dfrac{5x-x^2}{4} >0 \Rightarrow 5x-x^2 >0\Rightarrow x^2-5c < 0$

$x(x-5) < 0$ Refer image.2

They value of

$x$

$x\in (0,1)\cup (4,5)$

1351867_1130562_ans_1de0968735f44ff3848a0d9c1a1f2272.png

Implict function ${ x }^{ 3 }+{ y }^{ 3 }=3axy$

${ x }^{ 3 }+{ y }^{ 3 }=axy$

By differentiating,

$3{ x }^{ 2 }+3{ y }^{ 2 }\dfrac { dy }{ dx } =a\left[ x\dfrac { dy }{ dx } +y\left( 1 \right) \right]$

$3{ x }^{ 2 }+3{ y }^{ 2 }\dfrac { dy }{ dx } =ax\dfrac { dy }{ dx } +ya$

$\dfrac { dy }{ dx } \left( 3{ y }^{ 2 }-ax \right) =ay-3{ x }^{ 2 }$

$\dfrac { dy }{ dx } =\dfrac { ay-3{ x }^{ 2 } }{ 3{ y }^{ 2 }-ax }$

If, $\sin ( x y ) + \frac { y } { x } = x ^ { 2 } - y ^ { 2 }$ .
Find $\dfrac { d y } { d x }$ .

$\sin{xy}+\dfrac{y}{x}={x}^{2}-{y}^{2}$

$\Rightarrow\,\cos{xy}\left(x\dfrac{dy}{dx}+y\right)+\dfrac{x\dfrac{dy}{dx}-y}{{x}^{2}}=2x-2y\dfrac{dy}{dx}$

$\Rightarrow\,x\cos{xy}\dfrac{dy}{dx}+y\cos{xy}+\dfrac{1}{x}\dfrac{dy}{dx}-\dfrac{y}{{x}^{2}}=2x-2y\dfrac{dy}{dx}$

$\Rightarrow\,\left(x\cos{xy}+\dfrac{1}{x}+2y\right)\dfrac{dy}{dx}=2x-y\cos{xy}+\dfrac{y}{{x}^{2}}$

$\Rightarrow\,\dfrac{dy}{dx}=\dfrac{2x-y\cos{xy}+\dfrac{y}{{x}^{2}}}{\left(x\cos{xy}+\dfrac{1}{x}+2y\right)}$

$\Rightarrow\,\dfrac{dy}{dx}=\dfrac{2{x}^{3}-y{x}^{2}\cos{xy}+y}{{x}^{2}\left({x}^{2}\cos{xy}+1+2y{x}^{2}\right)}$

Differentiate w.r.t $x$ : $x^y +y^x =1$

Let

${x}^{y}=t$ ......

$(1)$

$\Rightarrow y\log{x}=\log{t}$ by taking

$\log$ both sides.

Differentiating w.r.t

$x$ we get

$\Rightarrow \dfrac{y}{x}+\log{x}\dfrac{dy}{dx}=\dfrac{1}{t}\dfrac{dt}{dx}$

$\Rightarrow \dfrac{dt}{dx}=t\left( \dfrac{y}{x}+\log{x}\dfrac{dy}{dx}\right)$

$\Rightarrow \dfrac{dt}{dx}={x}^{y}\left( \dfrac{y}{x}+\log{x}\dfrac{dy}{dx}\right)$ where

$t={x}^{y}$

$\Rightarrow \dfrac{dt}{dx}=y{x}^{y-1}+{x}^{y}\log{x}\dfrac{dy}{dx}$ ......

$(2)$

Let

${y}^{x}=w$ .....

$(3)$

$\Rightarrow x\log{y}=\log{w}$ by taking

$\log$ both sides.

Differentiating w.r.t

$x$ we get

$\Rightarrow \dfrac{x}{y}\dfrac{dy}{dx}+\log{y}=\dfrac{1}{w}\dfrac{dw}{dx}$

$\Rightarrow \dfrac{dw}{dx}=w\left(\dfrac{x}{y}\dfrac{dy}{dx}+\log{y}\right)$

$\Rightarrow \dfrac{dw}{dx}={y}^{x}\left(\dfrac{x}{y}\dfrac{dy}{dx}+\log{y}\right)$ where

$w={y}^{x}$

$\Rightarrow \dfrac{dw}{dx}=x{y}^{x-1}\dfrac{dy}{dx}+{y}^{x}\log{y}$ ......

$(4)$

From the question we have

${x}^{y}+{y}^{x}=1$

$\Rightarrow t+w=1$ from

$(1)$ and

$(3)$

$\Rightarrow \dfrac{dt}{dx}+\dfrac{dw}{dx}=0$

$\Rightarrow y{x}^{y-1}+{x}^{y}\log{x}\dfrac{dy}{dx}+x{y}^{x-1}\dfrac{dy}{dx}+{y}^{x}\log{y}=0$ from

$(2)$ and

$(4)$

$\Rightarrow \left({x}^{y}\log{x}+x{y}^{x-1}\right)\dfrac{dy}{dx}=-\left(y{x}^{y-1}+{y}^{x}\log{y}\right)$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{-\left(y{x}^{y-1}+{y}^{x}\log{y}\right)}{\left({x}^{y}\log{x}+x{y}^{x-1}\right)}$

Differentiate wrt to x.
${ tan }^{ -1 }( \sqrt { \frac { 1-sin 4x }{ 1+sin 4x }})$ = $\dfrac {psin8x}{1-sin4x}$ then p=

2029843_1234086_ans_73871454e1f047dd9ad841dc899e4f9c.jpg

If $x$ = $\dfrac{ t^2}{1 + t^2}$ , $y$ = $\dfrac{2at}{1 + t^2}$ then $\dfrac{dy}{dx}$ =

$x=\dfrac{{t}^{2}}{1+{t}^{2}}$

$\dfrac{dx}{dt}=\dfrac{\left(1+{t}^{2}\right)\dfrac{d}{dt}\left({t}^{2}\right)-{t}^{2}\dfrac{d}{dt}\left(1+{t}^{2}\right)}{{\left(1+{t}^{2}\right)}^{2}}$

$=\dfrac{\left(1+{t}^{2}\right)\left(2t\right)-{t}^{2}\left(2t\right)}{{\left(1+{t}^{2}\right)}^{2}}$

$=\dfrac{\left(1+{t}^{2}-{t}^{2}\right)\left(2t\right)}{{\left(1+{t}^{2}\right)}^{2}}$

$\therefore \dfrac{dx}{dt}=\dfrac{2t}{{\left(1+{t}^{2}\right)}^{2}}$

$y=\dfrac{2at}{1+{t}^{2}}$

$\dfrac{dy}{dt}=\dfrac{\left(1+{t}^{2}\right)\dfrac{d}{dt}\left(2at\right)-\left(2at\right)\dfrac{d}{dt}\left(1+{t}^{2}\right)}{{\left(1+{t}^{2}\right)}^{2}}$

$\dfrac{dy}{dt}=\dfrac{\left(1+{t}^{2}\right)\left(2a\right)-\left(2at\right)\left(2t\right)}{{\left(1+{t}^{2}\right)}^{2}}$

$\therefore \dfrac{dy}{dt}=\dfrac{\left(1+{t}^{2}\right)\left(2a\right)-\left(2at\right)\left(2t\right)}{{\left(1+{t}^{2}\right)}^{2}}$

$\therefore \dfrac{dy}{dt}=\dfrac{\left(1+{t}^{2}-2{t}^{2}\right)\left(2a\right)}{{\left(1+{t}^{2}\right)}^{2}}$

$\therefore \dfrac{dy}{dt}=\dfrac{\left(1-{t}^{2}\right)\left(2a\right)}{{\left(1+{t}^{2}\right)}^{2}}$

$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{\dfrac{2t}{{\left(1+{t}^{2}\right)}^{2}}}{\dfrac{\left(1-{t}^{2}\right)\left(2a\right)}{{\left(1+{t}^{2}\right)}^{2}}}=\dfrac{2t}{2a\left(1-{t}^{2}\right)}$

Find $\dfrac {dy}{dx}$ while:
$x^{y}+y^{x}=a^{b}$

Let

$u={x}^{y}$ and

$v={y}^{x}$

$\Rightarrow u+v={a}^{b}$

$\Rightarrow \dfrac{du}{dx}+\dfrac{dv}{dx}=\dfrac{d}{dx}\left({a}^{b}\right)=0$ since differential of a constant is

$0$

$\Rightarrow \dfrac{du}{dx}+\dfrac{dv}{dx}=0$

$u={x}^{y}\Rightarrow \log{u}=y\log{x}$

$\Rightarrow \dfrac{1}{u}\dfrac{du}{dx}=y\dfrac{d}{dx}\left(\log{x}\right)+\log{x}\dfrac{d}{dx}\left(y\right)$

$\Rightarrow \dfrac{1}{u}\dfrac{du}{dx}=\dfrac{y}{x}+\log{x}\dfrac{dy}{dx}$

$\Rightarrow \dfrac{du}{dx}=u\left(\dfrac{y}{x}+\log{x}\dfrac{dy}{dx}\right)$

$v={y}^{x}\Rightarrow \log{v}=x\log{y}$

$\Rightarrow \dfrac{1}{v}\dfrac{dv}{dx}=x\dfrac{d}{dx}\left(\log{y}\right)+\log{y}\dfrac{d}{dx}\left(x\right)$

$\Rightarrow \dfrac{1}{v}\dfrac{dv}{dx}=\dfrac{x}{y}\dfrac{dy}{dx}+\log{y}$

$\Rightarrow \dfrac{dv}{dx}=v\left(\dfrac{x}{y}\dfrac{dy}{dx}+\log{y}\right)$

$\therefore \dfrac{dy}{dx}=\dfrac{du}{dx}+\dfrac{dv}{dx}$

$\Rightarrow \dfrac{dy}{dx}=u\left(\dfrac{y}{x}+\log{x}\dfrac{dy}{dx}\right)+v\left(\dfrac{x}{y}\dfrac{dy}{dx}+\log{y}\right)$

$\Rightarrow \dfrac{dy}{dx}=u\dfrac{y}{x}+u\log{x}\dfrac{dy}{dx}+v\dfrac{x}{y}\dfrac{dy}{dx}+v\log{y}$

$\Rightarrow\left(1- u\log{x}-v\dfrac{x}{y}\right)\dfrac{dy}{dx}=u\dfrac{y}{x}+v\log{y}$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{u\dfrac{y}{x}+v\log{y}}{1- u\log{x}-v\dfrac{x}{y}}$

$\therefore \dfrac{dy}{dx}=\dfrac{{x}^{y}\dfrac{y}{x}+{y}^{x}\log{y}}{1- {x}^{y}\log{x}-{y}^{x}\dfrac{x}{y}}$ where

$u={x}^{y}$ and

$v={y}^{x}$

$\dfrac{dy}{dx}=\dfrac{y{x}^{y-1}+{y}^{x}\log{y}}{1-{x}^{y}\log{x}-x{y}^{x-1}}$

If $l^{2}+m^{2}=1$ , then find the maximum value of $l+m$ .

${l}^{2}+{m}^{2}=1$

$\Rightarrow {m}^{2}=1-{l}^{2}$

$\Rightarrow m=\pm\sqrt{1-{l}^{2}}$

Let

$N=l+m\Rightarrow N=l\pm\sqrt{1-{l}^{2}}$

$\Rightarrow \dfrac{dN}{dl}=1\pm\dfrac{1}{2\sqrt{1-{l}^{2}}}\times -2l$

$=1\pm\dfrac{l}{\sqrt{1-{l}^{2}}}$

For maximum value we have

$\dfrac{dN}{dl}=0$

$\Rightarrow 1\pm\dfrac{l}{\sqrt{1-{l}^{2}}}=0$

$\Rightarrow 1-{l}^{2}={l}^{2}$

$\Rightarrow 2{l}^{2}=1$

$\therefore l=\pm\dfrac{1}{\sqrt{2}}$

$\Rightarrow \dfrac{{d}^{2}N}{d{l}^{2}}=\pm\dfrac{\sqrt{1-{l}^{2}}-l\left(\dfrac{-2l}{2\sqrt{1-{l}^{2}}}\right)}{\sqrt{1-{l}^{2}}}$

$=\pm\dfrac{1-{l}^{2}+{l}^{2}}{\left(1-{l}^{2}\right)\sqrt{1-{l}^{2}}}$

$=\pm\dfrac{1}{{\left(1-{l}^{2}\right)}^{\frac{3}{2}}}$ from above

$l=\pm\dfrac{1}{\sqrt{2}}$

$=\pm\dfrac{1}{{\left(1-{\left(\dfrac{1}{\sqrt{2}}\right)}^{2}\right)}^{\frac{3}{2}}}=2\sqrt{2}>0$

$N=l+\sqrt{1-{l}^{2}}$

$=\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}\times\dfrac{\sqrt{2}}{\sqrt{2}}=\sqrt{2}$

Hence the maximum value of

$l+m=\sqrt{2}$

$y=\sin^{-1}\left(\dfrac{1-x^2}{1+x^2}\right), 0 < x < 1$ .
$Find\ \dfrac{dy}{dx}$

1293633_1579616_ans_3a57926b6de347f080eaaaaf524ecc12.jpg

$\frac{d}{dx}(log_{3}x-3 log_{e}x+2 tan x)$

1293785_1585236_ans_5614bf7e2b774c708ef3542fcf208774.jpg

If $\log \sqrt{x^{2}+y^{2}}= \tan^{-1} \left(\dfrac{y}{x}\right),$ then prove that $\dfrac{dy}{dx} = \dfrac{x+y}{x-y}$ .

1403903_1660134_ans_8fde15cb73f34affbf4a7c8dfa4c94a2.jpg

Differentiate the following with respect to $x$ :
$e^{-5x}\cot 4x$

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1982543_1710607_ans_d6e0233631974674a036d01671fe45de.jpg

Find $\dfrac{dy}{dx}$ in the following:

$\tan^{-1}(x^{2}+y^{2})=a$

1402929_1659955_ans_432e4d5735084afba5b7f6566713ccf2.jpg

If $x \sqrt{1+y}+y \sqrt{1+x}=0,$ then prove that $(1+x^{2}) \dfrac{dy}{dx}+1=0.$

1403900_1660087_ans_ad6227ce804d40d380df6ac0f7f0d646.jpg

Find the derivative of the function given by
Differentiate $( x^{2} - 5x + 8) (x^{3} + 7x + 9)$ in three ways mentioned below :
by loagarithmic differentiation
Do they all give the same answer.

1910394_1858625_ans_5f6554fb57b044038402b63af0253333.jpg

If $\displaystyle y = \frac{ax^{2}} {\left (x-a\right )\left (x-b\right )\left (x-c\right )} + \frac{bx} {\left (x-b\right )\left (x-c\right )} + \frac{c} {x-c} + 1$
prove that $\displaystyle \frac{y^{'}} {y} = \frac{1} {x} \left (\frac{a} {a-x} + \frac{b} {b-x} + \frac{c} {c-x} \right )$ (IIT-JEE, 1998)

$y = \dfrac{ax^{2}} {\left (x-a\right )\left (x-b\right )\left (x-c\right )} + \dfrac{bx} {\left (x-b\right )\left (x-c\right )} + \dfrac{c} {x-c} + 1$

$\Rightarrow y = \dfrac{ax^{2}} {\left (x-a\right )\left (x-b\right )\left (x-c\right )} + \dfrac{bx} {\left (x-b\right )\left (x-c\right )} + \left (\dfrac{c+x-c} {x-c} \right )$

$\displaystyle \Rightarrow y = \dfrac{ax^{2}} {\left (x-a\right )\left (x-b\right )\left (x-c\right )} + \frac{bx} {\left (x-b\right )\left (x-c\right )} + \frac{x} {x-c}$

$\displaystyle \Rightarrow y = \frac{ax^{2}} {\left (x-a\right )\left (x-b\right )\left (x-c\right )} + \frac{bx + x\left (x-b\right ) } {\left (x-b\right )\left (x-c\right )}$

$\displaystyle \Rightarrow y = \frac{x^{3}} {\left (x-a\right )\left (x-b\right )\left (x-c\right )}$

$\displaystyle \Rightarrow \log y = \log \left \{ \frac{x^{3}} {\left (x-a\right )\left (x-b\right )\left (x-c\right )} \right \}$

$\displaystyle \Rightarrow \log y = 3 \log x - \left \{\log\left (x-a \right ) \log\left (x-b\right ) \log \left (x-c \right ) \right \}$

On differentiating w.r.t.

$x$ , we get

$\displaystyle \Rightarrow \frac{dy} {dx} = y\left \{\left (\frac{1} {x} - \frac{1} {x-a}\right ) + \left (\frac{1} {x} - \frac{1} {x-b}\right ) + \left (\frac{1} {x} - \frac{1} {x-c} \right ) \right \}$

$\displaystyle \Rightarrow \frac{dy} {dx} = y \left \{-\frac{a} {x \left ( x-a \right )} - \frac{b} {x \left ( x-b \right ) } - \frac{c} {x \left ( x-c \right )} \right \}$

$\displaystyle \Rightarrow \frac{dy} {dx} = \frac{y} {x} \left \{\frac{a} {a-x} + \frac{b} {b-x } + \frac{c} {x-c} \right \}$

$\displaystyle \frac{y^{'}} {y} = \frac{1} {x} \left (\frac{a} {a-x} + \frac{b} {b-x} + \frac{c} {c-x} \right )$

Differentiation - Class 11 Commerce Applied Mathematics - Extra Questions

Find dydx \displaystyle \frac{dy}{dx} of ax+by2=cosyax + by^2 = \cos y

Find the derivative of the following functions: \displaystyle \sin x \cos x \displaystyle \sin x \cos x

Find \displaystyle \frac{dy}{dx} \displaystyle \frac{dy}{dx} of 2x + 3y = \sin y2x + 3y = \sin y

Find \displaystyle \frac{dy}{dx} \displaystyle \frac{dy}{dx} of xy + y^2 = \tan x + yxy + y^2 = \tan x + y

Find the derivative of the function given by f(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)f(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8) and hence find f'(1).f'(1).

Find derivative of the following functions (it is to be understood that a, b, c, d, p, q, ra, b, c, d, p, q, r and ss are fixed non-zero constants and mm and nn are integers) : \displaystyle \left ( x^{2}+ 1 \right )\cos \, x\displaystyle \left ( x^{2}+ 1 \right )\cos \, x

Differentiate the given function w.r.t. xx:\sin^3x + \cos^6 x\sin^3x + \cos^6 x

If y = x^{x}y = x^{x}, then find \dfrac {dy}{dx}\dfrac {dy}{dx}

Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r a, b, c, d, p, q, r and ss are fixed non-zero constants and mm and nn are integers) : (ax + b)\displaystyle ^{n} (cx + d)\displaystyle ^{m}(ax + b)\displaystyle ^{n} (cx + d)\displaystyle ^{m}

If f(x)=\tan { 2x } \tan { 3x } \tan { 5x } f(x)=\tan { 2x } \tan { 3x } \tan { 5x } , then prove that f'(x)=5\sec ^{ 2 }{ 5x } -3\sec ^{ 2 }{ 3x } -2\sec ^{ 2 }{ 2x } f'(x)=5\sec ^{ 2 }{ 5x } -3\sec ^{ 2 }{ 3x } -2\sec ^{ 2 }{ 2x }

Find the \dfrac{dy}{dx}\dfrac{dy}{dx} by implicit differentiation. x^2 - 8xy + y^2 = 8x^2 - 8xy + y^2 = 8

Find \cfrac { dy }{ dx } \cfrac { dy }{ dx } where { x }^{ 3 }+{ y }^{ 3 }+3xy=7 { x }^{ 3 }+{ y }^{ 3 }+3xy=7.

Differentiable \log_7(log x)\log_7(log x) with respect to xx.

Find the value of f'(e)f'(e) where f(x)=\log_e(\log_ex)f(x)=\log_e(\log_ex).

y = \sqrt{log \,x+\sqrt{log \,x + \sqrt{log \,x}}} +y = \sqrt{log \,x+\sqrt{log \,x + \sqrt{log \,x}}} +_____ \infty\infty show that \dfrac{dy}{dx} = \dfrac{1}{x(2y - 1)}\dfrac{dy}{dx} = \dfrac{1}{x(2y - 1)}.

If \sqrt{x}+\sqrt{y}=4\sqrt{x}+\sqrt{y}=4 then find \dfrac{dy}{dx}\dfrac{dy}{dx} at x=1x=1.

Find \dfrac{dy}{dx}\dfrac{dy}{dx} :x+y^2=\log y + x^2x+y^2=\log y + x^2

If f(x)= \left| x-2 \right| +\left| x-5 \right|f(x)= \left| x-2 \right| +\left| x-5 \right| then find f'(4)f'(4).

If y=3x^{2}y=3x^{2},Find \dfrac {dy}{dx}=?\dfrac {dy}{dx}=?

If e^{y}+xy=ee^{y}+xy=e, find y''(0)y''(0)

Find \dfrac{dy}{dx}\dfrac{dy}{dx} when x^2 + y^2 = c^2x^2 + y^2 = c^2.

\dfrac{{dy}}{{dx}}\, + \,\dfrac{y}{x}\,\,{\rm I}n\,y\, = \,\dfrac{y}{{{x^2}}}{\left( {{\rm I}n\,y} \right)^2}\dfrac{{dy}}{{dx}}\, + \,\dfrac{y}{x}\,\,{\rm I}n\,y\, = \,\dfrac{y}{{{x^2}}}{\left( {{\rm I}n\,y} \right)^2}

Differentiate: \sqrt { \dfrac { \left( x-3 \right) \left( { x }^{ 2 }+4 \right) }{ { 3 }x^{ 2 }+4x+5 } } \sqrt { \dfrac { \left( x-3 \right) \left( { x }^{ 2 }+4 \right) }{ { 3 }x^{ 2 }+4x+5 } }

If \sin y = x\sin (a + y)\sin y = x\sin (a + y) , then show \dfrac{dy}{dx}=\dfrac{\sin \left( a+y \right)}{\cos y-x.\cos \left( a+y \right)} \dfrac{dy}{dx}=\dfrac{\sin \left( a+y \right)}{\cos y-x.\cos \left( a+y \right)}

Find \dfrac{{dy}}{{dx}},x = a\left( {\cos t + \log \tan \frac{t}{2}} \right),y = a\sin t \dfrac{{dy}}{{dx}},x = a\left( {\cos t + \log \tan \frac{t}{2}} \right),y = a\sin t

If e^x + e^y = e^{x+y}e^x + e^y = e^{x+y}, find \dfrac{dy}{dx} \dfrac{dy}{dx} .

Find \dfrac{dy}{dx}\dfrac{dy}{dx} for y=e^{\sqrt{x}}y=e^{\sqrt{x}}.

Differentiate {\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + {x^2}} - 1}}{x}} \right){\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + {x^2}} - 1}}{x}} \right) with respect to {\sin ^{ - 1}}x{\sin ^{ - 1}}x

Differentiate with respect to xx.y = x\tan xy = x\tan x

Differentiate the following w.r.t.x:5^x\cdot \sec^{-1}2x5^x\cdot \sec^{-1}2x

Differentiate x = \tan (e^{-y})x = \tan (e^{-y})

Find \dfrac {dy}{dx}\dfrac {dy}{dx}\dfrac{{dx}}{{dt}} = ap\cos pt,\dfrac{{dy}}{{dt}} = - bp\sin pt\dfrac{{dx}}{{dt}} = ap\cos pt,\dfrac{{dy}}{{dt}} = - bp\sin pt

Differentiate the following w.r.t.x:\sin^{-1}x+\cos^{-1}x\sin^{-1}x+\cos^{-1}x

If y = {3^{2\log x + 7}}y = {3^{2\log x + 7}} then \dfrac{{dy}}{{dx}}\dfrac{{dy}}{{dx}}

Differentiate the following function:{x}^{\log{x}}+{(\log{x})}^{x}{x}^{\log{x}}+{(\log{x})}^{x}

If y=\sin {\sqrt {1-x^{2}}}y=\sin {\sqrt {1-x^{2}}}, then find \dfrac{dy}{dx}\dfrac{dy}{dx}.

Find:\dfrac{d}{{dx}}\left\{ {\dfrac{{5 + 4x}}{{x + 3}}} \right\}\dfrac{d}{{dx}}\left\{ {\dfrac{{5 + 4x}}{{x + 3}}} \right\}

If y = {a^{\dfrac{1}{2}{{\log }_a}\cos x}}.y = {a^{\dfrac{1}{2}{{\log }_a}\cos x}}. Find \dfrac{{dy}}{{dx}}\dfrac{{dy}}{{dx}}

Find the derivative of f\left ( x \right )=1+x+x^{2}+--+x^{50}atx=1f\left ( x \right )=1+x+x^{2}+--+x^{50}atx=1.

Find the derivate of \ {e^{\sqrt {2x + 1} }}\ {e^{\sqrt {2x + 1} }} with respect to xx at x=12x=12

If y = \sqrt x + \dfrac{1}{{\sqrt x }},y = \sqrt x + \dfrac{1}{{\sqrt x }}, then show that 2x\dfrac{{dy}}{{dx}} + y = 2\sqrt x 2x\dfrac{{dy}}{{dx}} + y = 2\sqrt x

Find the derivative of sin^{2}xsin^{2}x with respect to x using product rule.

Let y=\log(\sin x)y=\log(\sin x) find \dfrac{d^2y}{dx^2}\dfrac{d^2y}{dx^2}.

Find \dfrac{dy}{dx} \dfrac{dy}{dx} , if x\ \sin\ y+y\ \sin\ x=0x\ \sin\ y+y\ \sin\ x=0.

y=x^3-x^2-x+7y=x^3-x^2-x+7 find x =? when \dfrac{dy}{dx}=0\dfrac{dy}{dx}=0

Differentiate \log(sin\ x)(log\ x)\log(sin\ x)(log\ x)

Differentiate: {\sin}^{2}{3x}.{\tan}^{3}{2x}{\sin}^{2}{3x}.{\tan}^{3}{2x}

Find \dfrac {dy}{dx}\dfrac {dy}{dx} for y= \left( \sin 20x+{{a}^{2x}}+10 \right) \\y= \left( \sin 20x+{{a}^{2x}}+10 \right) \\

If x\sqrt { 1-{ y }^{ 2 } } +y\sqrt { 1-{ x }^{ 2 } } =1x\sqrt { 1-{ y }^{ 2 } } +y\sqrt { 1-{ x }^{ 2 } } =1, prove that \dfrac { dy }{ dx } =-\sqrt { \frac { { 1-y }^{ 2 } }{ { 1-x }^{ 2 } } . } \dfrac { dy }{ dx } =-\sqrt { \frac { { 1-y }^{ 2 } }{ { 1-x }^{ 2 } } . }

y=x^{3}-3x+2y=x^{3}-3x+2 Find \frac{dy}{dx}\frac{dy}{dx} if the given function is continuous.

If {x^y} + {y^x} = {a^b}{x^y} + {y^x} = {a^b} then show that \dfrac{{dy}}{{dx}} = - \left[ {\dfrac{{y{x^{y - 1}} + {y^x}\log y}}{{{x^y}\log x + x{y^{x - 1}}}}} \right]\dfrac{{dy}}{{dx}} = - \left[ {\dfrac{{y{x^{y - 1}} + {y^x}\log y}}{{{x^y}\log x + x{y^{x - 1}}}}} \right]

Find \dfrac{dy}{dx}\dfrac{dy}{dx} if xy=e^{x-y}xy=e^{x-y}

{\left( {{x^2} + {y^2}} \right)^2} = xy,{\left( {{x^2} + {y^2}} \right)^2} = xy, then \dfrac{dy}{dx}\dfrac{dy}{dx}

Find derivative of f(x)=x \cos xf(x)=x \cos x

If y\sqrt { { x }^{ 2 }+1 } =log\{ \sqrt { { x }^{ 2 }+1 } -x\} ,y\sqrt { { x }^{ 2 }+1 } =log\{ \sqrt { { x }^{ 2 }+1 } -x\} , show that { (x }^{ 2 }+1)\dfrac { dy }{ dx } +xy+1=0.{ (x }^{ 2 }+1)\dfrac { dy }{ dx } +xy+1=0.

Find f'(-3) using definition if f\left( x \right) =\frac { 2x }{ x-3 } f\left( x \right) =\frac { 2x }{ x-3 }

Find \dfrac { d }{ dx } { \sin }^{ 3 }4x{ \cos }^{ 8 }5x\dfrac { d }{ dx } { \sin }^{ 3 }4x{ \cos }^{ 8 }5x

Differentiate xe^xxe^x from first principles.

If \cos (x+y)=y \sin x\cos (x+y)=y \sin x, then find \dfrac { dy }{ dx } .\dfrac { dy }{ dx } .

Differentiate the following w.r.t.xw.r.t.x.\sqrt{x}.cotx\sqrt{x}.cotx

Find \dfrac{dy}{dx}\dfrac{dy}{dx} :\sin x +\cos y = 5x+4\sin x +\cos y = 5x+4

Differentiate of the function x^{3}+y^{3}+3x^{2}y=a^{3}x^{3}+y^{3}+3x^{2}y=a^{3} with respect to xx.

Find \dfrac{dy}{dx}\dfrac{dy}{dx} :\sin x - 3x = 5y\sin x - 3x = 5y

If x^py^q=(x+y)^{p+q}x^py^q=(x+y)^{p+q} then prove that \dfrac{dy}{dx}=\dfrac{y}{x}\dfrac{dy}{dx}=\dfrac{y}{x}.

If f(x)=x+\log xf(x)=x+\log x find f^{\prime}(x)f^{\prime}(x)

Find drivative of y=(2-\sin x)(e^{x}+x^{3}+2)y=(2-\sin x)(e^{x}+x^{3}+2) with respect to xx.

Find \dfrac{dy}{dx}\dfrac{dy}{dx} if 3x+4y=93x+4y=9

Find the derivative of \dfrac{x+\cos x}{\tan x}\dfrac{x+\cos x}{\tan x}

Differentiate \sin^{2} x\sin^{2} x wrt xx

Find \dfrac{dy}{dx}\dfrac{dy}{dx}y=\sin x.\ \cos xy=\sin x.\ \cos x

f((x)=\log x+\sin xf((x)=\log x+\sin x find f^{\prime }(x)f^{\prime }(x)

Differentiate with respect to xx:\ e^{-3x} \log (1+x)\ e^{-3x} \log (1+x)

If \sin\theta + 2\cos\theta =1\sin\theta + 2\cos\theta =1 then prove that 2\sin\theta -\cos\theta=02\sin\theta -\cos\theta=0.

Differentiate with respect to xx:e^x \log \sin 2xe^x \log \sin 2x

Find \dfrac{dy}{dx}\dfrac{dy}{dx} in the following:4x+3y=\log(4x-3y)4x+3y=\log(4x-3y)

Differentiate the following w.r.t. x:\cos x .e^{4x+5}\cos x .e^{4x+5}

Let y=\log(1+\sin x)y=\log(1+\sin x) for 0\le x\le \dfrac{\pi}{2}0\le x\le \dfrac{\pi}{2}, find \dfrac{dy}{dx}\dfrac{dy}{dx}.

Differentiate with respect to xx:e^x\tan xe^x\tan x

Differentiate x^{2}x^{2} with respected to x^{3}x^{3}.

Find $\displaystyle \frac{dy}{dx}$ of $ax + by^2 = \cos y$

Find the derivative of the following functions: $\displaystyle \sin x \cos x$

Find $\displaystyle \frac{dy}{dx}$ of $2x + 3y = \sin y$

Find $\displaystyle \frac{dy}{dx}$ of $xy + y^2 = \tan x + y$

Find the derivative of the function given by $f(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)$ and hence find $f'(1).$

Find derivative of the following functions (it is to be understood that $a, b, c, d, p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers) : $\displaystyle \left ( x^{2}+ 1 \right )\cos \, x$

Differentiate the given function w.r.t. $x$ :
$\sin^3x + \cos^6 x$

If $y = x^{x}$ , then find $\dfrac {dy}{dx}$

Find the derivative of the following functions (it is to be understood that $a, b, c, d, p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers) : $(ax + b)\displaystyle ^{n} (cx + d)\displaystyle ^{m}$

If $f(x)=\tan { 2x } \tan { 3x } \tan { 5x }$ , then prove that $f'(x)=5\sec ^{ 2 }{ 5x } -3\sec ^{ 2 }{ 3x } -2\sec ^{ 2 }{ 2x }$

Find the $\dfrac{dy}{dx}$ by implicit differentiation. $x^2 - 8xy + y^2 = 8$

Find $\cfrac { dy }{ dx }$ where ${ x }^{ 3 }+{ y }^{ 3 }+3xy=7$ .

Differentiable $\log_7(log x)$ with respect to $x$ .

Find the value of $f'(e)$ where $f(x)=\log_e(\log_ex)$ .

$y = \sqrt{log \,x+\sqrt{log \,x + \sqrt{log \,x}}} +$ _____ $\infty$ show that $\dfrac{dy}{dx} = \dfrac{1}{x(2y - 1)}$ .

If $\sqrt{x}+\sqrt{y}=4$ then find $\dfrac{dy}{dx}$ at $x=1$ .

Find $\dfrac{dy}{dx}$ :

$x+y^2=\log y + x^2$

If $f(x)= \left| x-2 \right| +\left| x-5 \right|$ then find $f'(4)$ .

If $y=3x^{2}$ ,Find $\dfrac {dy}{dx}=?$

If $e^{y}+xy=e$ , find $y''(0)$

Find $\dfrac{dy}{dx}$ when $x^2 + y^2 = c^2$ .

$\dfrac{{dy}}{{dx}}\, + \,\dfrac{y}{x}\,\,{\rm I}n\,y\, = \,\dfrac{y}{{{x^2}}}{\left( {{\rm I}n\,y} \right)^2}$

Differentiate: $\sqrt { \dfrac { \left( x-3 \right) \left( { x }^{ 2 }+4 \right) }{ { 3 }x^{ 2 }+4x+5 } }$

If $\sin y = x\sin (a + y)$ , then show
$\dfrac{dy}{dx}=\dfrac{\sin \left( a+y \right)}{\cos y-x.\cos \left( a+y \right)}$

Find $\dfrac{{dy}}{{dx}},x = a\left( {\cos t + \log \tan \frac{t}{2}} \right),y = a\sin t$

If $e^x + e^y = e^{x+y}$ , find $\dfrac{dy}{dx}$ .

Find $\dfrac{dy}{dx}$ for $y=e^{\sqrt{x}}$ .

Differentiate ${\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + {x^2}} - 1}}{x}} \right)$ with respect to ${\sin ^{ - 1}}x$

Differentiate with respect to $x$ .
$y = x\tan x$

Differentiate the following w.r.t.x:
$5^x\cdot \sec^{-1}2x$

Differentiate $x = \tan (e^{-y})$

Find $\dfrac {dy}{dx}$
$\dfrac{{dx}}{{dt}} = ap\cos pt,\dfrac{{dy}}{{dt}} = - bp\sin pt$

Differentiate the following w.r.t.x:
$\sin^{-1}x+\cos^{-1}x$

If $y = {3^{2\log x + 7}}$ then $\dfrac{{dy}}{{dx}}$

Differentiate the following function:
${x}^{\log{x}}+{(\log{x})}^{x}$

If $y=\sin {\sqrt {1-x^{2}}}$ , then find $\dfrac{dy}{dx}$ .

Find:
$\dfrac{d}{{dx}}\left\{ {\dfrac{{5 + 4x}}{{x + 3}}} \right\}$

If $y = {a^{\dfrac{1}{2}{{\log }_a}\cos x}}.$ Find $\dfrac{{dy}}{{dx}}$

Find the derivative of $f\left ( x \right )=1+x+x^{2}+--+x^{50}atx=1$ .

Find the derivate of $\ {e^{\sqrt {2x + 1} }}$ with respect to $x$ at $x=12$

If $y = \sqrt x + \dfrac{1}{{\sqrt x }},$ then show that $2x\dfrac{{dy}}{{dx}} + y = 2\sqrt x$

Find the derivative of $sin^{2}x$ with respect to x using product rule.

Let $y=\log(\sin x)$ find $\dfrac{d^2y}{dx^2}$ .

Find $\dfrac{dy}{dx}$ , if $x\ \sin\ y+y\ \sin\ x=0$ .

$y=x^3-x^2-x+7$ find x =? when $\dfrac{dy}{dx}=0$

Differentiate $\log(sin\ x)(log\ x)$

Differentiate: ${\sin}^{2}{3x}.{\tan}^{3}{2x}$

Find $\dfrac {dy}{dx}$ for
$y= \left( \sin 20x+{{a}^{2x}}+10 \right) \\$

If $x\sqrt { 1-{ y }^{ 2 } } +y\sqrt { 1-{ x }^{ 2 } } =1$ , prove that $\dfrac { dy }{ dx } =-\sqrt { \frac { { 1-y }^{ 2 } }{ { 1-x }^{ 2 } } . }$

$y=x^{3}-3x+2$
Find $\frac{dy}{dx}$ if the given function is continuous.

If ${x^y} + {y^x} = {a^b}$ then show that $\dfrac{{dy}}{{dx}} = - \left[ {\dfrac{{y{x^{y - 1}} + {y^x}\log y}}{{{x^y}\log x + x{y^{x - 1}}}}} \right]$

Find $\dfrac{dy}{dx}$ if $xy=e^{x-y}$

${\left( {{x^2} + {y^2}} \right)^2} = xy,$ then $\dfrac{dy}{dx}$

Find derivative of $f(x)=x \cos x$

If $y\sqrt { { x }^{ 2 }+1 } =log\{ \sqrt { { x }^{ 2 }+1 } -x\} ,$ show that ${ (x }^{ 2 }+1)\dfrac { dy }{ dx } +xy+1=0.$

Find f'(-3) using definition if $f\left( x \right) =\frac { 2x }{ x-3 }$

Find $\dfrac { d }{ dx } { \sin }^{ 3 }4x{ \cos }^{ 8 }5x$

Differentiate $xe^x$ from first principles.

If $\cos (x+y)=y \sin x$ , then find $\dfrac { dy }{ dx } .$

Differentiate the following $w.r.t.x$ .
$\sqrt{x}.cotx$

Find $\dfrac{dy}{dx}$ :

$\sin x +\cos y = 5x+4$

Differentiate of the function $x^{3}+y^{3}+3x^{2}y=a^{3}$ with respect to $x$ .

Find $\dfrac{dy}{dx}$ :

$\sin x - 3x = 5y$

If $x^py^q=(x+y)^{p+q}$ then prove that $\dfrac{dy}{dx}=\dfrac{y}{x}$ .

If $f(x)=x+\log x$ find $f^{\prime}(x)$

Find drivative of $y=(2-\sin x)(e^{x}+x^{3}+2)$ with respect to $x$ .

Find $\dfrac{dy}{dx}$ if $3x+4y=9$

Find the derivative of $\dfrac{x+\cos x}{\tan x}$

Differentiate $\sin^{2} x$ wrt $x$

Find $\dfrac{dy}{dx}$
$y=\sin x.\ \cos x$

$f((x)=\log x+\sin x$ find $f^{\prime }(x)$

Differentiate with respect to $x$ :
$\ e^{-3x} \log (1+x)$

If $\sin\theta + 2\cos\theta =1$ then prove that $2\sin\theta -\cos\theta=0$ .

Differentiate with respect to $x$ :
$e^x \log \sin 2x$

Find $\dfrac{dy}{dx}$ in the following:
$4x+3y=\log(4x-3y)$

Differentiate the following w.r.t. x:
$\cos x .e^{4x+5}$

Let $y=\log(1+\sin x)$ for $0\le x\le \dfrac{\pi}{2}$ , find $\dfrac{dy}{dx}$ .

Differentiate with respect to $x$ :
$e^x\tan x$

Differentiate $x^{2}$ with respected to $x^{3}$ .

Differentiate $e^x \tan x$ with respect to $x$ .