Class 11 Commerce Maths - Limits And Derivatives

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) : $\sin (x + a)$

Let

$f (x) = \displaystyle \sin (x +a)$
Thus using first principle

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

$=\displaystyle \lim_{x\rightarrow 0}\frac{\sin \left ( x+h+a \right )-\sin \left ( x+a \right )}{h}$

$=\displaystyle \lim_{x\rightarrow 0}\frac{1}{h}\left [ 2 \cos \left ( \frac{x+h+a+x+a}{2} \right )\sin \left ( \frac{x+h+a-x-a}{2} \right ) \right ]$

$=\displaystyle \lim_{x\rightarrow 0}\frac{1}{h}\left [ 2 \cos \left ( \frac{2x +2a + h}{2} \right )\sin \left ( \frac{h}{2} \right ) \right ]$

$=\displaystyle \lim_{x\rightarrow 0}\left [ \cos \left ( \frac{2x+2a+h}{2} \right )\left \{ \frac{\sin \left ( \frac{h}{2} \right )}{\left ( \frac{h}{2} \right )} \right \} \right ]$

$=\displaystyle \lim_{x\rightarrow 0}\cos \left ( \frac{2x+2a+h}{2} \right )\lim_{x\rightarrow 0}\left \{ \frac{\sin \left ( \frac{h}{2} \right )}{\left ( \frac{h}{2} \right )} \right \}$

$=\displaystyle \cos \left ( \frac{2x+2a}{2} \right )\times 1$

$=\displaystyle \cos \left ( x + a \right )$

$\dfrac{d}{dx}\left\{\cos x^0\right\}=?$

$=-\sin x^0 \times (0) \times x^{0-1}$

$=0$

Find dy/dx=? If, x= $\cos(\log t )$ and y= $\log(\cos t)$

1339971_1281414_ans_7c0c6dc768a94951b4cf915421b332a6.jpg

Find the derivatives of $x \cos x$

$y=x\cos{x}$

By using product rule of differentiation,

$\dfrac{d(u \ . v)}{dx}=v\dfrac{du}{dx}+u\dfrac{dv}{dx}$

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

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

Find the differentiation of $\sec \left( {{{\tan }^{ - 1}}{\rm{x}}} \right)$ w.r.t. $x$ .

Let

$y=\sec (\tan^{-1}x)$

On differentiating w.r.t

$x$ , we get

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

$\dfrac{dy}{dx}=\sec (\tan^{-1}x). \tan (\tan^{-1}x)\times \dfrac{d}{dx}(\tan^{-1} x)$

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

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

Hence, this is the answer.

Find the derivative of the following functions:
$\displaystyle 5\sec x+4\cos x$

Let

$\displaystyle f\left ( x \right )=5\sec x+4\cos x$
Thus using first principle,

$\displaystyle f^{'}\left ( x \right )=\lim_{h\rightarrow 0}\frac{f\left ( x+h \right )-f\left ( x \right )}{h}$
=

$\displaystyle \lim_{h\rightarrow 0}\frac{5\sec \left ( x+h \right )+4\cos \left ( x+h \right )-\left [ 5\sec x+4\cos x \right ]}{h}$
=

$\displaystyle 5\lim_{h\rightarrow 0}\frac{\left [ \sec \left ( x+h \right ) -\sec x\right ]}{h}+4\lim_{h\rightarrow 0}\frac{\left [ \cos \left ( x+h \right ) -\cos x\right ]}{h}$
=

$\displaystyle 5\lim_{h\rightarrow 0}\frac{1}{h}\left [ \frac{1}{\cos \left ( x+h \right )}-\frac{1}{\cos x} \right ]+4\lim_{h\rightarrow 0}\frac{1}{h}\left [ \cos \left ( x+h \right )-\cos x \right ]$
=

$\displaystyle 5\lim_{h\rightarrow 0}\left [ \frac{\cos x-\cos \left ( x+h \right )}{\cos x\cos \left ( x+h \right )} \right ]+4\lim_{h\rightarrow 0}\frac{1}{h}\left [ \cos x\cos h-\sin x\sin h-\cos x \right ]$
=

$\displaystyle \frac{5}{\cos x}\cdot \lim_{h\rightarrow 0}\frac{1}{h}\left [ \frac{-2\sin \left ( \frac{2x+h}{2} \right )\sin \left ( -\frac{h}{2} \right )}{\cos \left ( x+h \right )} \right ]+4\left [ -\cos x\lim_{h\rightarrow 0}\frac{\left ( 1-\cos h \right )}{h}-\sin x\lim_{h\rightarrow 0}\frac{\sin h}{h} \right ]$
=

$\displaystyle \frac{5}{\cos x}\cdot \lim_{h\rightarrow 0}\left [ \frac{\sin \left ( \frac{2x+h}{2} \right )\cdot\frac{\sin \left ( \frac{h}{2} \right )}{\frac{h}{2}} }{\cos \left ( x+h \right )} \right ]+4\left [ \left ( -\cos x \right ) \cdot \left ( 0 \right )-\left ( \sin x \right )\cdot 1\right ]$
=

$\displaystyle \frac{5}{\cos x}\cdot \left [ \lim_{h\rightarrow 0} \frac{\sin \left ( \frac{2x+h}{2} \right )}{\cos \left ( x+h \right )}\cdot \lim_{h\rightarrow 0}\frac{\sin \left ( \frac{h}{2} \right )}{\frac{h}{2}}\right ]-4\sin x$
=

$\displaystyle \frac{5}{\cos x}\cdot \frac{\sin x}{\cos x}\cdot 1-4\sin x$
=

$\displaystyle 5\sec x\tan x\cdot -4\sin x$

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

Let

$\displaystyle f\left ( x \right )=\cos ec\,x$
Thus using first principle,

$\displaystyle f^{'}\left ( x \right )=\lim_{h\rightarrow 0}\frac{f\left ( x+h \right )-f\left ( x \right )}{h}$

$\displaystyle f^{'}\left ( x \right )=\lim_{h\rightarrow 0}\frac{1}{h}\left [ \cos ec\left ( x+h \right )-\cos ecx \right ]$
=

$\displaystyle \lim_{h\rightarrow 0}\frac{1}{h}\left [ \frac{1}{\sin \left ( x+h \right )}-\frac{1}{\sin x} \right ]$
=

$\displaystyle \lim_{h\rightarrow 0}\frac{1}{h}$

$\displaystyle \left [ \frac{\sin x-\sin \left ( x+h \right )}{\sin \left ( x+h \right )\sin x} \right ]$
=

$\displaystyle \lim_{h\rightarrow 0}\frac{1}{h}\left [ \frac{2\cos \left ( \frac{x+x+h}{2} \right )\cdot \sin \left ( \frac{x-x-h}{2} \right )}{\sin \left ( x+h \right )\sin x} \right ]$
=

$\displaystyle \lim_{h\rightarrow 0}\frac{1}{h}\left [ \frac{2\cos \left ( \frac{2x+h}{2} \right )\sin \left ( -\frac{h}{2} \right )}{\sin \left ( x+h \right )\sin x} \right ]$
=

$\displaystyle \lim_{h\rightarrow 0}\frac{-\cos \left ( \frac{2x+h}{2} \right ),\frac{\sin \left ( \frac{h}{2} \right )}{\left ( \frac{h}{2} \right )}}{\sin \left ( x+h \right )\sin x}$
=

$\displaystyle \lim_{h\rightarrow 0}\left ( \frac{-\cos \left ( \frac{2x+h}{2} \right )}{\sin \left ( x+h \right )\sin x} \right ).\lim_{\frac{h}{2}\rightarrow 0}\frac{\sin \left ( \frac{h}{2} \right )}{\left ( \frac{h}{2} \right )}$
=

$\displaystyle \left ( \frac{-\cos x}{\sin x\sin x} \right ).1= - \cos ec x\cot x$

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) : $\displaystyle \sin ^{n}x$

Let

$y = \displaystyle \sin ^{n}x$

$\displaystyle \frac{dy}{dx}=\frac{d}{dx}$

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

Find the differential coefficient of $\sin x$ by first principle.

Find the differential coefficient of

$\sin x$ by first principle. Suppose that

$f(x) = \sin x$

$\therefore f(x + h) = \sin (x + h)$

$\therefore f(x + h) - f(x) = \sin (x + h) - \sin x$

$\Rightarrow \dfrac {f(x + h) - f(x)}{h} = \dfrac {\sin (x + h) - \sin x}{h}$

$\Rightarrow \displaystyle \lim_{h\rightarrow 0} \dfrac {f(x + h) - f(x)}{h} = \displaystyle \lim_{h\rightarrow 0} \dfrac {\sin (x + h) - \sin x}{h}$

$\Rightarrow \dfrac {d}{dx}f(x) = {\displaystyle \lim_{h\rightarrow 0}} \left [\dfrac {2\cos \left (x + \dfrac {h}{2}\right ). \sin \dfrac {h}{2}}{h}\right ]$

$\Rightarrow \dfrac {d}{dx}\sin x = \displaystyle \lim_{h\rightarrow 0}\cos \left (x + \dfrac {h}{2}\right ) . \displaystyle \lim_{h\rightarrow 0} \left (\dfrac {\sin \dfrac {h}{2}}{\dfrac {h}{2}}\right )$

$= \cos (x + 0) . \displaystyle \lim_{\tfrac {h}{2}\rightarrow 0}\left (\dfrac {\sin \dfrac {h}{2}}{\dfrac {h}{2}}\right ) \because h\rightarrow 0$

$\therefore \dfrac {h}{2}\rightarrow 0$

$= \cos x . 1$

$\Rightarrow \dfrac {d}{dx}\sin x = \cos x$ .

Prove that the following functions are increasing.
$y \, = \, 2x \, + \, sin \, x \, for \, x \, \epsilon \, R$

$y=2x+\sin x$

$\implies \dfrac{dy}{dx}=y'=2+\cos x$

Since for all

$x\in R$ , we have

$-1\leq \cos x\leq 1$

$1\leq 2+\cos x\leq 3$

Hence, we see that

$y'>0 \forall x\in R$

And, hence

$y=2x+\sin x$ is increasing for

$x\in R$

Find the derivative of the following functions from the first principals w.r.t to $x$ .
$\tan 2x$

We know, first principal is,

$f'(x)=\lim\limits_{h\to0}\dfrac{f(x+h)-f(x)}{h}$

So with

$f(x)=\tan\,2x$ we have

$f'(x)=\lim\limits_{h\to0}\dfrac{\tan(2x+2h)-\tan\,2x}{h}$

$=\lim\limits_{h\to0}\dfrac{\dfrac{\sin(2x+2h)}{\cos(2x+2h)}-\dfrac{\sin\,2x}{\cos\,2x}}{h}$ [

$\tan\,A=\dfrac{\sin\,A}{\cos\,A}$ ]

$=\lim\limits_{h\to0}\dfrac{1}{h}\left[\dfrac{\sin(2x+2h)}{\cos(2x+2h)}-\dfrac{\sin\,2x}{\cos\,2x}\right]$

$=\lim\limits_{h\to0}\dfrac{1}{h}\left[\dfrac{\sin(2x+2h)\cos\,2x-\cos(2x+2h)\sin\,2x}{\cos(2x+2h)\cos\,2x}\right]$

$=\lim\limits_{h\to0}\dfrac{1}{h}\left[\dfrac{\sin(2x+2h-2x)}{\cos(2x+2h)\cos\,2x}\right]$ [

$\sin(A-B)=\sin\,A.\cos\,B-\cos\,B.\sin\,A$ ]

$=\lim\limits_{h\to0}\dfrac{1}{h}\left[\dfrac{\sin(2h)}{\cos(2x+2h)\cos\,2x}\right]$

$=\lim\limits_{h\to0}\dfrac{1}{\cos(2x+2h)\cos\,2x}.\dfrac{\sin\,2h}{h}$

$=\lim\limits_{h\to0}\dfrac{2}{\cos(2x+2h)\cos\,2x}.\dfrac{\sin\,2h}{2h}$

$=\lim\limits_{h\to0}\dfrac{2}{\cos(2x+2h)\cos\,2x}.\lim\limits_{h\to0}\dfrac{\sin\,2h}{2h}$

We know,

$\lim\limits{h\to0}\dfrac{\sin\,h}{h}=1\Rightarrow \lim\limits_{h\to 0}\dfrac{\sin\,2h}{2h}=1$

$f'(x)=\dfrac{2}{\cos\,2x.\cos\,2x}.1$

$=\dfrac{2}{\cos^22x}$

$=2\sec^22x$

Solve : $I_n=\displaystyle \int_{0}^{\dfrac{\pi}{2}}e^{-x}\sin ^nxdx$

$I_n=\int_{0}^{\frac{\pi}{2}}e^{-x}\sin ^nxdx$

show that $I_n=-e^{\frac{-\pi}{2}}+n\int_{0}^{\frac{\pi}{2}}e^{-x}\sin ^{n-1}xdx$

$I_n=\int_{0}^{\frac{\pi}{2}}e^{-x}\sin ^nxdx$

substitute $u=\sin ^nx,\dfrac{dv}{dx}=e^{-x}\Rightarrow v=-e^{-x}$

$I_n=[-e^{-x}\sin ^nx]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}-e^{-x}n\sin ^{n-1}x\cos xdx$

$I_n=-e^{\frac{-\pi}{2}}+n\int_{0}^{\frac{\pi}{2}}-e^{-x}\cos x\sin ^{n-1}xdx$

--------------------------------------------------

$\int_{0}^{\frac{\pi}{2}}e^{-x}\cos x\sin ^{n-1}xdx$

$=[-e^{-x}\cos x\sin ^{n-1}x]_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}e^{-x}((n-1)\sin ^{n-2}x-n\sin ^nx)dx$

$=(n-1)\int_{0}^{\frac{\pi}{2}}e^{-x}\sin ^{n-2}xdx-n\int_{0}^{\frac{\pi}{2}}e^{-x}\sin ^nxdx$

$=(n-1)I_{n-2}-nI_n$

$\therefore I_n=-e^{\frac{-\pi}{2}}+n(n-1)I_{n-2}-n^2I_n$

$\Rightarrow (n^2+1)I_n=-e^{\frac{-\pi}{2}}+n(n-1)I_{n-2}$

Solve: $\dfrac{d}{dx}(cosec x)=?$

Given,

$\dfrac { d }{ dx } (cosecx)\\ =\dfrac { d }{ dx } (\dfrac { 1 }{ \sin { x } } )$

We know the quotient rule of differentiation,

$\dfrac { d }{ dx } (\dfrac { u }{ v } )=\dfrac { v\dfrac { du }{ dx } -u\dfrac { dv }{ dx } }{ { v }^{ 2 } }$

Now, let

$u=1,\sin { x } =v$

$\therefore \dfrac { d }{ dx } (\dfrac { 1 }{ \sin { x } } )\\ =\dfrac { \sin { x } \dfrac { d }{ dx } (1)-1\dfrac { d }{ dx } (\sin { x } ) }{ { (\sin { x } ) }^{ 2 } } \\ =\dfrac { \sin { x } (0)-\cos { x } }{ { (\sin { x } ) }^{ 2 } } \\ =\dfrac { -\cos { x } }{ { (\sin { x } ) }^{ 2 } } \\ =\dfrac { -\cos { x } }{ \sin { x } } \cdot \dfrac { 1 }{ \sin { x } } \\ =-\cot { x } cosecx\\ \therefore \dfrac { d }{ dx } (cosecx)=-\cot { x } cosecx.$

Find the derivative of tan x using first principle of derivatives

From the first principle of derivatives,

$f'(x)=\overset{lim}{h\rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$=\overset{lim}{h\rightarrow 0} \dfrac{tan(x+h)-tanx}{h}$

$=\overset{lim}{h\rightarrow 0} \dfrac{\dfrac{sin(x+h)}{cos(x+h)}-\dfrac{sinx}{cosx}}{h}$

$=\overset{lim}{h\rightarrow 0} \dfrac{cos\:x\:sin(x+h)-sin\:x\:cos(x+h)}{hcos\:x\:cos(x+h)}$

$=\overset{lim}{h\rightarrow 0} \dfrac{\dfrac{sin(2x+h)+sin\:h}{2}-\dfrac{sin(2x+h)-sin\:h}{2}}{hcos\:x\:cos(x+h)}$

$=\overset{lim}{h\rightarrow 0} \dfrac{sin\:h}{hcos\:x\:cos(x+h)}$

$=\overset{lim}{h\rightarrow 0}\dfrac{sin\:h}{h} \times \overset{lim}{h\rightarrow 0} \dfrac{1}{cos\:x\:cos(x+h)}$

$=1\times \dfrac{1}{cos\:x \times cos\:x}$

$=\dfrac{1}{cos^2x}$

$f'(x)=sec^2x$

Find the derivative of $cos^2\:x$ , by using first principle of derivatives.

Increase from

$y$ to

$y+\delta y$ correspondingly

$x$ to

$x+\delta x$ in the above equation

$\left(1\right)$

$\Rightarrow y+\delta y={\cos}^{2}{\left(x+\delta x\right)}$ .....

$\left(2\right)$

Eqn

$\left(2\right)$ -Eqn

$\left(1\right)$

$\Rightarrow y+\delta y-y={\cos}^{2}{\left(x+\delta x\right)}-{\cos}^{2}{x}$

$\Rightarrow \delta y={\cos}^{2}{\left(x+\delta x\right)}-{\cos}^{2}{x}$

Divide both sides by

$\delta x$ we get

$\Rightarrow \dfrac{\delta y}{\delta x}=\dfrac{{\cos}^{2}{\left(x+\delta x\right)}-{\cos}^{2}{x}}{\delta x}$

$\Rightarrow \dfrac{\delta y}{\delta x}=-\dfrac{\sin{\left(2x+\delta x\right)}\sin{\delta x}}{\delta x}$ by using

${\cos}^{2}{B}-{\cos}^{2}{A}=\sin{\left(A+B\right)}\sin{\left(A-B\right)}$

$\Rightarrow {\lim}_{x\rightarrow 0}\dfrac{\delta y}{\delta x}={\lim}_{x\rightarrow 0}-\dfrac{\sin{\left(2x+\delta x\right)}\sin{\delta x}}{\delta x}$

$\Rightarrow \dfrac{dy}{dx}={\lim}_{x\rightarrow 0}-\dfrac{\sin{\left(2x+\delta x\right)}\sin{\delta x}}{\delta x}$

$\Rightarrow \dfrac{dy}{dx}=-\dfrac{\sin{\left(2x+0\right)}\times 1$$ since $${\lim}_{x\rightarrow 0}\sin{\delta x}}{\delta x}=1$

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

Dfferentiate w.r.t x:
$tan^2\, 7x$

Consider the given expression,

$\Rightarrow {{\tan }^{2}}7x$

Differentiate with respect to x,

$\Rightarrow \dfrac{d}{dx}{{\tan }^{2}}7x$

$\Rightarrow 2\tan 7x\dfrac{d}{dx}7x$

$\Rightarrow 2\left( \tan 7x \right).7$

$\Rightarrow \Rightarrow 14\tan 7x$

Hence, this is the answer.

Find the derivation of $\sqrt{tan x}$ with respect to x using first principle.

$a \rightarrow f'(a)=\underset{h\rightarrow 0}{lim}\dfrac{f(a+h)-f(a)}{h}=\underset{h\rightarrow 0}{lim}\dfrac{\sqrt{\tan(a+h)}-\sqrt{\tan a}}{h}\times \dfrac{\sqrt{\tan(a+h)}+\sqrt{\tan a}}{\sqrt{\tan(a+h)}+\sqrt{\tan a}}$

$=\underset{h \rightarrow 0}{lim}\dfrac{\tan(a+h)-\tan a}{h(2\sqrt{\tan a})}=\underset{h\rightarrow 0}{lim}\left(\dfrac{\tan a+\tan h-\tan a}{1-\tan a\tan h}\right)\dfrac{1}{2h\sqrt{\tan a}}$

$=\underset{h\rightarrow 0}{lim}\dfrac{\tan h+\tan^2a\tan h}{2h\sqrt{\tan a}}=\underset{h\rightarrow 0}{\lim}\dfrac{\tan h(1+\tan^2a)}{4\, 2\sqrt{\tan a}}=\dfrac{\sec^2a}{2\sqrt{\tan a}}$

If $y=x^{ \cos x }+{ \left( \tan x \right) }^{ \cot x },find\dfrac { dy }{ dx }$

We have,

$y={{x}^{\cos x}}+{{\left( \tan x \right)}^{\cot x}}$

Taking log both side and we get,

$\log y=\log {{\left( x \right)}^{\cos x}}+\log {{\left( \tan x \right)}^{\cot x}}$

$\log y=\cos x\log x+\cot x\log \tan x$

On differentiating with respect to x and we get,

$\dfrac{d}{dx}\log y=\cos x\dfrac{d}{dx}\log x+\log x\dfrac{d}{dx}\cos x+\cot x \dfrac{d}{dx}\log \tan x+\operatorname{logtan}x\dfrac{d}{dx}\cot x$

$\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=\cos x\dfrac{1}{x}+\log x\left( -\sin x \right)+\cot x\dfrac{1}{\tan x}\left( {{\sec }^{2}}x \right)+\log \tan x\left( -{{\csc }^{2}}x \right)$

$\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{\cos x}{x}-\sin x\log x-\dfrac{\cot x}{\tan x}{{\sec }^{2}}x-{{\csc }^{2}}x\cos x\log \tan x$

$\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{\cos x}{x}-{{\csc }^{2}}x-\sin x\log x-{{\csc }^{2}}x\cos x\log \tan x$

$\Rightarrow \dfrac{dy}{dx}=y\left[ \dfrac{\cos x}{x}-{{\csc }^{2}}x-\sin x\log x-{{\csc }^{2}}x\cos x\log \tan x \right]$

$\Rightarrow \dfrac{dy}{dx}={{x}^{\cos x}}+{{\left( \tan x \right)}^{\cot x}}\left[ \dfrac{\cos x}{x}-{{\csc }^{2}}x-\sin x\log x-{{\csc }^{2}}x\cos x\log \tan x \right]$

Hence, this is the answer.

Find $\frac{dy}{dx},$ if $y=\sqrt{cos(3x+1)}$

$y=\sqrt{\cos (3x+1)}$

then

$\dfrac{dy}{dx}=\dfrac{d}{dx}\sqrt{\cos (3x+1)}$

$\dfrac{dy}{dx}=\dfrac{1.(-(\sin 3x+1)\times 3)}{2\sqrt{\cos (3x+1)}}$

$\dfrac{dy}{dx}=-\dfrac{3}{2}\dfrac{\sin (3x+1)}{\sqrt{\cos (3x+1)}}$ .

1179641_1351821_ans_c9083a1212b54c4ebe26c3eddbd49623.JPG

If $y = \tan \left( {2x + 3} \right)$ . Find $\dfrac{{dy}}{{dx}}$ .

We have,

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

On differentiating w.r.t

$x$ , we get

$\dfrac{dy}{dx}=\dfrac{d}{dx}\tan(2x+3)$

$\dfrac{dy}{dx}=\sec^2(2x+3)\dfrac{d}{dx}(2x+3)$

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

$\dfrac{dy}{dx}=2\sec^2(2x+3)$

Hence, this is the answer.

$2\dfrac{dy}{dx}-y\sec x=y^3\tan x$ .

$2 \dfrac{dy}{dx} = y^{3} \tan x + y \sec x$

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

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

$2 \log y = y^{2} \sec^{2} x + \ln |\sec x + \tan x|+ C$

$\dfrac{2}{y^2} \cdot \log y = \sec^{2} x + \ln |\sec x + \tan x|$

$\dfrac{2}{y^2} \cdot \log y = \log e^{\sec^2 x} + \ln |\sec x + \tan x|$

$\dfrac{-2}{y^2} = -1 + (\pi + x) \cot \left (\dfrac{\pi}{4} + \dfrac{x}{2} \right )$

Differentiate: $y=\sin{\left(2x+3\right)}$ w.r.t $x$

$y=\sin{\left(2x+3\right)}$

$\Rightarrow\,\dfrac{dy}{dx}=\cos{\left(2x+3\right)}\dfrac{d}{dx}\left(2x+3\right)$

$\therefore\,\dfrac{dy}{dx}=2\cos{\left(2x+3\right)}$

$f\left( x \right) =\left( \sin x+\cos x \right)$ Find $f^{\prime} (x)$

$f\left( x \right) =\left( \sin x+\cos x \right) \\f^{\prime}(x)=\dfrac d{dx}(\sin x+\cos x)\\f^{\prime}(x)=\cos x-\sin x$

Find the derivative of $\dfrac { x ^ { 5 } - \cos x } { \sin x }$ with respect to $x.$

1894583_1487447_ans_a2e19b54d5fc4810b965d3735b3af1b0.jpg

Differentiate the function with respect to x.
$\cos x\cdot \cos 2x\cdot \cos 3x$ .

1315517_1584923_ans_26afa66a2b114e029f8328abface6e7a.jpg

Differentiate the following from first principle.
$\sin(x+1)$ .

1445498_1666119_ans_b330502d0f8d4752a007d0d1222d8eff.jpg

Differentiate the following from first principle.
$f(x)=\cos\left(x-\dfrac{\pi}{8}\right)$

We have,

$f(x)=\cos\left(x-\dfrac{\pi}{8}\right)$

$\therefore f(x+h)=\cos\left(x+h-\dfrac{\pi}{8}\right)=\cos\left\{\left(x-\dfrac{\pi}{8}\right)+h\right\}$

So,

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

$\Rightarrow f'(x)=\displaystyle\lim_{h\rightarrow 0}\dfrac{\cos\left\{\left(x-\dfrac{\pi}{8}\right)+h\right\}-\cos\left(x-\dfrac{\pi}{8}\right)}{h}$

$\Rightarrow f'(x)=\displaystyle\lim_{h\rightarrow 0}\dfrac{-2\sin\left\{\left(x-\dfrac{\pi}{8}\right)+\dfrac{h}{2}\right\}\sin\left(\dfrac{h}{2}\right)}{h}$ [

$\because \cos (a)-\cos(b)=-2\sin\dfrac{a+b}{2} \sin\dfrac{a-b}{2}$ ]

$\Rightarrow f'(x)=-\displaystyle\lim_{h\rightarrow 0}\sin\left\{\left(x-\dfrac{\pi}{8}\right)+\dfrac{h}{2}\right\}\dfrac{\sin\left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}$ As

$\displaystyle \lim_{x\rightarrow0}\dfrac{sinx}x=1$

$=-\sin\left(x-\dfrac{\pi}{8}\right)$ .

Differentiate the following
$\cos \sqrt{x}$ .

1422602_1668954_ans_25678de4af82451585f038756f74569a.jpg

Differentiate: $2\sqrt{\cot (x^{2})}$ w.e.t.X

Let

$y = 2\sqrt{\cot (x^{2})}$ Differentiating w.r.t.x both sides, we get

$\dfrac{dy}{dx} = 2 \times \dfrac{1}{2\sqrt{\cot (x^{2})}} \times -\ cosec^{2} (x^{2}) \times 2x$

$= \dfrac{-2x \ cosec^{2}x^{2}}{\sqrt{\cot (x^{2})}}$

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

Given,

$y + \sin y = \cos x.$

Differentiating w.r.t. x, we get

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

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

$\Rightarrow \dfrac{dy}{dx} = \dfrac{-\sin x}{1 + \cos y}\left [ \because y \neq (2n + 1) \pi \right ]$

If $y = f \left (\frac{2x - 1} {x^{2} + 1} \right )$ and $f^{'}(x) = \sin x^{2}$ , then $\frac{dy} {dx} =$ ___________. (IIT-JEE, 1982)

$y = f \left (\frac{2x - 1} {x^{2} + 1} \right )$ ;

$f^{'}(x) = \sin x^{2}$

$\dfrac{dy} {dx} = f^{'} \left (\frac{2x - 1} {x^{2} + 1} \right ) \frac{d} {dx} \left ( \frac{2x - 1} {x^{2} + 1} \right )$

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

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

Differentiable the function w.r.t .x.
$x ^{\sin x} + (\sin x) ^{\cos x}$

Let

$y = x^{\sin x} + (\sin x)^{\cos x}$

$= u + v ,$

where u and v

$u = x^{\sin x} and v = (\sin x)^{\cos x}$

$\log u = \sin x (\log x)$

$v = (\sin x)^{\cos x}$

$\log v = \cos x \log (sin x )$

$\dfrac{1}{v} \times \dfrac{dv}{dx} = \cos x \times \dfrac{1}{\sin x} \cos x + \log (\sin x)$

$\dfrac{dv}{dx} = (\sin x)^{\cos x} \left \{ \cos x \cot x - \sin x \log (\sin x) \right \}$

If $y = (\tan ^{-1} x)^{2}$ show that $(x ^{2} + 1)^{2} y_{2} + 2x (x^{2}+ 1)^{2} y_{2} + 2x ( x^{2} + 1) y_{1} = 2$ where $y_{1},y_{2}$ have their usual meaning.

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

$Differentiating \space wrt \space x$

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

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

$( x^{2} + 1) y_{1} = 2\tan^{-1} x$

$(x^{2}+ 1) y_{2} + y_{1} \times 2x = 2 \times \dfrac{1}{(x^{2} + 1)}$

$(x^{2}+ 1)^{2} y_{2} + 2x (x^{2}+ 1) y_{1} = 2$

prove that , $\dfrac{dy}{dx} = \dfrac{(1 + y) \cos x + y \sin x}{1 + 2y + \cos x - \sin x}$

$y$ can be rewritten as

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

$y = \dfrac{(1+ y)\sin x}{1 +y + \cos x}$

$y ^{2} + y + y \cos x = \sin x + y \sin x$

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

Find whether function is increasing or decreasing in given domain
$f (x) = sin^4 X + cos^4 x, x \epsilon (0, \frac{\pi }{4})$

$f (x) = sin^4 X + cos^4 x$
Diff. w.r.t. x,

$f'(x) = 4 sin^3 \times cos X + 4 cos^3 X (-sin X)$

$\Rightarrow f'(x) = 4 sin X cos X (sin^2 X - cos^2 X)$

$\Rightarrow f'(x) = -2.2 sin X cos X (cos^2 x - sin^2 X)$

$\Rightarrow f' (x) = -2 sin 2x cos 2x$

$f' (x) = - sin 4 x$
In Interval

$(0, \frac{\pi }{4}), - sin 4x < 0$
or

$f' (x) < 0$
Hence, function is decreasing in internal

$(0, \frac{\pi }{4})$ .

If $y=(\sin x - \cos x)^{(\sin x- \cos x)}$ then find $\dfrac{dy}{dx}$

Let

$y-(\sin x - \cos x){(\sin x- \cos )}$
Taking

$\log$ on both sides,

$\log y= (\sin x - \cos x) \log{(\sin x- \cos x)}$
Diff.w.r.t.

$x$ of both sides,

$\dfrac{1}{y}\dfrac{dy}{dx}=(\sin x- \cos x) \times \dfrac{d}{dx} [\log (\sin x- \cos x)]+ \log (\sin x - \cos x) \times \dfrac{d}{dx} (\sin x - \cos x)$

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

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

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

$\Rightarrow \dfrac{dy}{dx}=y (\cos x+ \sin x) \times [1+ \log (\sin x - \cos x)]$
Hence,

$\dfrac{dy}{dx} (\sin x - \cos x)^{(\sin x- \cos x)} \times (\cos x+ \sin x) \times [1+ \log (\sin x - \cos x)]$

Differentiate given functions w.r.t. $x$ :
$x^3e^x \sin x$

Let

$y=x^3e^x\sin x$
Diff.w.r.t.

$x$ of both sides,

$\dfrac{dy}{dx}=\dfrac{d}{dx} (x^3 e^x \sin x)$

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

$\Rightarrow \dfrac{dy}{dx}=x^2 [e^x \cos x+ \sin x e^x]+e^x \sin 3x^2$
Hence,

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

Find the derivatives of the following:
$cosec x$ .

Let

$f(x) = cosec (x)$
Then, derivative of

$f(x)$

$\dfrac {d}{dx} f(x) = \dfrac {d}{dx} (cosec x)$

$= \dfrac {d}{dx} \left (\dfrac {1}{\sin x}\right )$

$= \dfrac {\sin x \times \dfrac {d}{dx} (1) - 1\times \dfrac {d}{dx} (\sin x)}{(\sin x)^{2}}$
[From the formula of derivative of quotient of two functions]

$= \dfrac {\sin x\times 0 - 1\times \cos x}{(\sin x)^{2}} \left (\because \dfrac {d}{dx} \sin x = \cos x\right )$

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

$= -\dfrac {\cos x}{\sin x}\times \dfrac {1}{\sin x}$

$= -\cot x\times cosec x$

$= -cosec x \cot x$
Hence, derivative of the given function

$cosec x$

$= -cosec x \cot x$ .

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

2040757_1285874_ans_5ad828f367a34aec8fd5e79e68c6dc01.jpeg

If $cos^{-1}{(\dfrac{x^2-y^2}{x^2+y^2})}=2k$ , show that $y\dfrac{dy}{dx}=x tan^2k$

2042784_1533963_ans_3dd791c43549468983cfd3d2bb94bb02.png

$\sum _{ 0 }^{ \pi }{ \dfrac { (ax+b) \sec x \tan x }{ 4+{ tan }^{ 2 }x } } dx(a,b>0)$

2044527_1565752_ans_e6059096bdba4b89b6f43a3eddf93e97.jpg

Limits And Derivatives - Class 11 Commerce Maths - Extra Questions

Find the 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) : sin(x+a)\sin (x + a)

ddx{cosx0}=?\dfrac{d}{dx}\left\{\cos x^0\right\}=?

Find dy/dx=? If, x=cos(logt)\cos(\log t ) and y=log(cost)\log(\cos t)

Find the derivatives of xcosxx \cos x

Find the differentiation of sec(tan−1x)\sec \left( {{{\tan }^{ - 1}}{\rm{x}}} \right) w.r.t. xx.

Find the derivative of the following functions:5secx+4cosx\displaystyle 5\sec x+4\cos x

Find the derivative of the following functions: cosecx\displaystyle \cos ec\,x

Find the 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) : sinnx\displaystyle \sin ^{n}x

Find the differential coefficient of sinx\sin x by first principle.

Prove that the following functions are increasing.y=2x+sinxforxϵRy \, = \, 2x \, + \, sin \, x \, for \, x \, \epsilon \, R

Find the derivative of the following functions from the first principals w.r.t to xx.tan2x\tan 2x

Solve : In=∫π20e−xsinnxdxI_n=\displaystyle \int_{0}^{\dfrac{\pi}{2}}e^{-x}\sin ^nxdx

Solve:ddx(cosecx)=?\dfrac{d}{dx}(cosec x)=?

Find the derivative of tan x using first principle of derivatives

Find the derivative of cos2xcos^2\:x, by using first principle of derivatives.

Dfferentiate w.r.t x:tan27xtan^2\, 7x

Find the derivation of √tanx\sqrt{tan x} with respect to x using first principle.

If y=xcosx+(tanx)cotx,finddydxy=x^{ \cos x }+{ \left( \tan x \right) }^{ \cot x },find\dfrac { dy }{ dx }

Find dydx,\frac{dy}{dx}, if y=√cos(3x+1)y=\sqrt{cos(3x+1)}

If y=tan(2x+3)y = \tan \left( {2x + 3} \right) . Find dydx\dfrac{{dy}}{{dx}}.

2dydx−ysecx=y3tanx2\dfrac{dy}{dx}-y\sec x=y^3\tan x.

Differentiate:y=sin(2x+3)y=\sin{\left(2x+3\right)} w.r.t xx

f(x)=(sinx+cosx)f\left( x \right) =\left( \sin x+\cos x \right) Find f′(x)f^{\prime} (x)

Find the derivative of x5−cosxsinx\dfrac { x ^ { 5 } - \cos x } { \sin x } with respect to x.x.

Differentiate the function with respect to x.cosx⋅cos2x⋅cos3x\cos x\cdot \cos 2x\cdot \cos 3x.

Differentiate the following from first principle.sin(x+1)\sin(x+1).

Differentiate the following from first principle.f(x)=cos(x−π8)f(x)=\cos\left(x-\dfrac{\pi}{8}\right)

Differentiate the following cos√x\cos \sqrt{x}.

Differentiate: 2√cot(x2)2\sqrt{\cot (x^{2})} w.e.t.X

Find dydx\dfrac{dy}{dx} if y+siny=cosx.y + \sin y = \cos x.

If y=f(2x−1x2+1)y = f \left (\frac{2x - 1} {x^{2} + 1} \right ) and f′(x)=sinx2f^{'}(x) = \sin x^{2}, then dydx=\frac{dy} {dx} = ___________. (IIT-JEE, 1982)

Differentiable the function w.r.t .x.xsinx+(sinx)cosx x ^{\sin x} + (\sin x) ^{\cos x}

If y=(tan−1x)2 y = (\tan ^{-1} x)^{2} show that (x2+1)2y2+2x(x2+1)2y2+2x(x2+1)y1=2 (x ^{2} + 1)^{2} y_{2} + 2x (x^{2}+ 1)^{2} y_{2} + 2x ( x^{2} + 1) y_{1} = 2 wherey1,y2y_{1},y_{2} have their usual meaning.

prove that , dydx=(1+y)cosx+ysinx1+2y+cosx−sinx \dfrac{dy}{dx} = \dfrac{(1 + y) \cos x + y \sin x}{1 + 2y + \cos x - \sin x}

Find whether function is increasing or decreasing in given domainf(x)=sin4X+cos4x,xϵ(0,π4)f (x) = sin^4 X + cos^4 x, x \epsilon (0, \frac{\pi }{4})

If y=(sinx−cosx)(sinx−cosx)y=(\sin x - \cos x)^{(\sin x- \cos x)} then find dydx\dfrac{dy}{dx}

Differentiate given functions w.r.t. xx:x3exsinxx^3e^x \sin x

Find the derivatives of the following:cosecxcosec x.

ddx\dfrac{d}{dx}(cos2^2 x sin x)

If cos−1(x2−y2x2+y2)=2kcos^{-1}{(\dfrac{x^2-y^2}{x^2+y^2})}=2k, show that ydydx=xtan2ky\dfrac{dy}{dx}=x tan^2k

π∑0(ax+b)secxtanx4+tan2xdx(a,b>0)\sum _{ 0 }^{ \pi }{ \dfrac { (ax+b) \sec x \tan x }{ 4+{ tan }^{ 2 }x } } dx(a,b>0)

Class 11 Commerce Maths Extra Questions

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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) : $\sin (x + a)$

$\dfrac{d}{dx}\left\{\cos x^0\right\}=?$

Find dy/dx=? If, x= $\cos(\log t )$ and y= $\log(\cos t)$

Find the derivatives of $x \cos x$

Find the differentiation of $\sec \left( {{{\tan }^{ - 1}}{\rm{x}}} \right)$ w.r.t. $x$ .

Find the derivative of the following functions:
$\displaystyle 5\sec x+4\cos x$

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

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) : $\displaystyle \sin ^{n}x$

Find the differential coefficient of $\sin x$ by first principle.

Prove that the following functions are increasing.
$y \, = \, 2x \, + \, sin \, x \, for \, x \, \epsilon \, R$

Find the derivative of the following functions from the first principals w.r.t to $x$ .
$\tan 2x$

Solve : $I_n=\displaystyle \int_{0}^{\dfrac{\pi}{2}}e^{-x}\sin ^nxdx$

Solve: $\dfrac{d}{dx}(cosec x)=?$

Find the derivative of $cos^2\:x$ , by using first principle of derivatives.

Dfferentiate w.r.t x:
$tan^2\, 7x$

Find the derivation of $\sqrt{tan x}$ with respect to x using first principle.

If $y=x^{ \cos x }+{ \left( \tan x \right) }^{ \cot x },find\dfrac { dy }{ dx }$

Find $\frac{dy}{dx},$ if $y=\sqrt{cos(3x+1)}$

If $y = \tan \left( {2x + 3} \right)$ . Find $\dfrac{{dy}}{{dx}}$ .

$2\dfrac{dy}{dx}-y\sec x=y^3\tan x$ .

Differentiate: $y=\sin{\left(2x+3\right)}$ w.r.t $x$

$f\left( x \right) =\left( \sin x+\cos x \right)$ Find $f^{\prime} (x)$

Find the derivative of $\dfrac { x ^ { 5 } - \cos x } { \sin x }$ with respect to $x.$

Differentiate the function with respect to x.
$\cos x\cdot \cos 2x\cdot \cos 3x$ .

Differentiate the following from first principle.
$\sin(x+1)$ .

Differentiate the following from first principle.
$f(x)=\cos\left(x-\dfrac{\pi}{8}\right)$

Differentiate the following
$\cos \sqrt{x}$ .

Differentiate: $2\sqrt{\cot (x^{2})}$ w.e.t.X

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

If $y = f \left (\frac{2x - 1} {x^{2} + 1} \right )$ and $f^{'}(x) = \sin x^{2}$ , then $\frac{dy} {dx} =$ ___________. (IIT-JEE, 1982)

Differentiable the function w.r.t .x.
$x ^{\sin x} + (\sin x) ^{\cos x}$

If $y = (\tan ^{-1} x)^{2}$ show that $(x ^{2} + 1)^{2} y_{2} + 2x (x^{2}+ 1)^{2} y_{2} + 2x ( x^{2} + 1) y_{1} = 2$ where $y_{1},y_{2}$ have their usual meaning.

prove that , $\dfrac{dy}{dx} = \dfrac{(1 + y) \cos x + y \sin x}{1 + 2y + \cos x - \sin x}$

Find whether function is increasing or decreasing in given domain
$f (x) = sin^4 X + cos^4 x, x \epsilon (0, \frac{\pi }{4})$

If $y=(\sin x - \cos x)^{(\sin x- \cos x)}$ then find $\dfrac{dy}{dx}$

Differentiate given functions w.r.t. $x$ :
$x^3e^x \sin x$

Find the derivatives of the following:
$cosec x$ .

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

If $cos^{-1}{(\dfrac{x^2-y^2}{x^2+y^2})}=2k$ , show that $y\dfrac{dy}{dx}=x tan^2k$

$\sum _{ 0 }^{ \pi }{ \dfrac { (ax+b) \sec x \tan x }{ 4+{ tan }^{ 2 }x } } dx(a,b>0)$