Class 11 Engineering Maths - Limits And Derivatives

The value of the limit $\underset{x \rightarrow \dfrac{\pi}{2}}{\lim} \dfrac{4 \sqrt{2} (\sin 3x + \sin x)}{\left(2 \sin 2x \sin \dfrac{3x}{2} + \cos \dfrac{5x}{2} \right) - \left(\sqrt{2} + \sqrt{2} \cos 2x + \cos \dfrac{3x}{2}\right)}$ is ______.

$\underset{x \rightarrow \frac{\pi}{2}}{\lim} \dfrac{4 \sqrt{2} (\sin 3x + \sin x)}{\left(2 \sin 2x \sin \dfrac{3x}{2} + \cos \dfrac{5x}{2} \right) - \left(\sqrt{2} + \sqrt{2} \cos x + \cos \dfrac{3x}{2} \right)}$

$=\underset{x \rightarrow \frac{\pi}{2}}{\lim} \dfrac{8 \sqrt{2} \sin 2x . \cos x}{\cos \dfrac{x}{2} - \cos \dfrac{7x}{2} + \cos \dfrac{5x}{2} - \left(2 \sqrt{2} \cos^2 x + \cos \dfrac{3x}{2}\right)}$

$=\underset{x \rightarrow \frac{\pi}{2}}{\lim} \dfrac{4 \sqrt{2}. 2. \sin x \cos x. \cos x}{2 \sin x \sin \dfrac{x}{2} + 2 \sin 3x. \sin \dfrac{x}{2} - 2 \sqrt{2} \cos^2 x}$

$=\underset{x \rightarrow \frac{\pi}{2}}{\lim} \dfrac{16 \sqrt{2} \sin x. \cos^2 x}{2 \sin \dfrac{x}{2} (2 \sin 2x. \cos x ) - 2 \sqrt{2} \cos^2 x}$

$=\underset{x \rightarrow \frac{\pi}{2}}{\lim} \dfrac{16 \sqrt{2} \sin x}{2 \sin \dfrac{x}{2} . 4 \sin x - 2 \sqrt{2}} = \dfrac{16 \sqrt{2}}{4 \sqrt{2} - 2 \sqrt{2}} = 8$ .

State whether the following statement is true or false.
Enter $1$ for true and $0$ for false
$f(x)$ is differentiable at a point $P$ , if there exists a unique tangent at point $P$ .

The function

$f(x)$ is differentiable at a point

$P$ , if and only if there exists a unique tangent at point

$P$ . In other words,

$f(x)$ is differentiable at point

$P$ , if and only if the curve does not have

$P$ as the corner point i.e. the function is not differentiable at those points on which the function has sharp edges.

Find the derivative of the following functions from first principle:
$\displaystyle \cos \left ( x-\frac{\pi }{8} \right )$

Let

$f(x)= \displaystyle \cos \left ( x-\frac{\pi }{8} \right )$
Thus using first principle

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

$\displaystyle \lim_{x\rightarrow 0}\frac{1}{h}\left [ \cos \left ( x + h-\frac{\pi }{8} \right )-\cos \left ( x-\frac{\pi }{8} \right ) \right ]$
=

$\displaystyle \lim_{x\rightarrow 0}\frac{1}{h}\left [ -2\sin \frac{\left ( x+h-\frac{\pi }{8} +x-\frac{\pi }{8}\right )}{2}\sin \left ( \frac{x+h-\frac{\pi }{8}-x+\frac{\pi }{8}}{2} \right ) \right ]$

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

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

$=\displaystyle -\sin \left ( \frac{2x+0-\frac{\pi }{4}}{2} \right )\cdot 1= -\sin \left ( x-\frac{\pi }{8} \right )$

Evaluate the Given limit: $\displaystyle \lim_{x\rightarrow \pi}\frac{\sin \left ( \pi -x \right )}{\pi \left ( \pi -x \right )}$

$\displaystyle \lim_{x\rightarrow \pi }\frac{\sin \left ( \pi -x \right )}{\pi \left ( \pi -x \right )}=\frac{1}{\pi }\lim_{\left ( z-x \right )\rightarrow 0}\frac{\sin \left ( \pi -x \right )}{\left ( \pi -x \right )}$
=

$\displaystyle \frac{1}{\pi }\times 1,$

$\because \displaystyle \left [ \lim_{y\rightarrow 0}\frac{\sin \,y}{y}=1 \right ]$
=

$\displaystyle \frac{1}{\pi }$

Evaluate : $\displaystyle \underset{n\, \rightarrow \, \infty }{Lt} \frac{1}{2^n}$

As $n\, \rightarrow \, \infty,$

$a^n \, \rightarrow \, 0$ where 0 < a <1

$\displaystyle \therefore \underset{n\, \rightarrow \, \infty }{Lt}\frac{1}{2^n}\, =\, \left ( \frac{1}{2} \right )^n=0$

Evaluate the given limit: $\displaystyle \lim_{x\rightarrow 0}\frac{\cos \,x}{\pi -x}$

The value of

$\displaystyle \lim_{x\rightarrow 0}\frac{\cos \,x}{\pi -x}$

$=\dfrac{\cos \,0}{\pi -0}$

$=\dfrac{1}{\pi }$

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

$2x + 3y = sin x$
Differentiating both sides w.r.t.

$x$ , we obtain

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

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

$\Rightarrow \displaystyle 2 + 3 \frac{dy}{dx} = \cos x$

$\Rightarrow \displaystyle 3 \frac{dy}{dx} = \cos x - 2$

$\therefore \displaystyle \frac{dy}{dx} = \frac{\cos x - 2}{3}$

Find the derivative of the following functions from first principle:
$\sin (x + 1)$

Let

$f(x) = \sin (x + 1)$
Thus using first principle,

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

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

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

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

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

$\displaystyle \cos \left ( \frac{2x+0+2}{2} \right )\cdot 1 =\cos \left ( x+1 \right )$

If the derivative of the function $\displaystyle 4\sqrt{x}-2$ is $\dfrac{a}{\sqrt{x}}.$ Find the value of $a$ .

Let

$f(x) = \displaystyle 4\sqrt{x}-2$

$f'(x) = \displaystyle \cfrac { d }{ dx } (4\sqrt { x } -2)=\cfrac { d }{ dx } (4\sqrt { x } )-\cfrac { d }{ dx } (2)$

$\displaystyle = 4\frac{d}{dx}\left ( x^{\frac{1}{2}} \right )-0=4\left ( \frac{1}{2}x^{\frac{1}{2}-1} \right )$

$\displaystyle =\left ( 2x^{-\frac{1}{2}} \right )=\frac{2}{\sqrt{x}}$

Find the derivative of $\displaystyle 2x-\frac{3}{4}$

Let

$\displaystyle f\left ( x \right )=2x-\frac{3}{4}$

$\displaystyle\therefore f^{'}\left ( x \right )=\frac{d}{dx}\left ( 2x-\frac{3}{4} \right )$

$\displaystyle = 2\frac{d}{dx}\left ( x \right )-\frac{d}{dx}\left ( \frac{3}{4} \right )= 2 - 0 = 2$

$\lim_\limits{x\to\frac{\pi}{4}} \dfrac{\sqrt{1-\sqrt{\sin 2x}}}{\pi-4x}=-\dfrac{1}{m}$ .Find $m$

L= $\mathop {\lim }\limits_{x \to \pi /4} {{\sqrt {1 - \sqrt {\sin 2x} } } \over {x - 4x}}\left( {{0 \over 0}} \right)form$

$\mathop {\lim }\limits_{x \to \pi /4} {{{1 \over {2\sqrt {1 - \sqrt {\sin 2x} } }}} \over { - 4}}\left( { - {1 \over {2\sqrt {\sin 2x} }},\cos 2x} \right)$

$- {1 \over 4}$

$m = 4$

Evaluate $\displaystyle \lim_{x \rightarrow 1}\left( \dfrac{1}{x^2 + x - 2} - \dfrac{x}{x^3 - 1}\right)$

We need to find value of $\lim _{ x\rightarrow 1 }{ \left( \dfrac { 1 }{ { x }^{ 2 }+x-2 } -\dfrac { x }{ { x }^{ 3 }-1 } \right) }$

$= \lim _{ x\rightarrow 1 }{ \left( \dfrac { 1 }{ (x-1)(x+2) } -\dfrac { x }{ { (x-1)(x }^{ 2 }+1+x) } \right) } \\ =\lim _{ x\rightarrow 1 }{ \left\{ \dfrac { 1 }{ (x-1) } \left( \dfrac { 1 }{ x+2 } -\dfrac { x }{ { x }^{ 2 }+x+1 } \right) \right\} } \\= \lim _{ x\rightarrow 1 }{ \left\{ \dfrac { 1 }{ (x-1) } \left( \dfrac { { x }^{ 2 }+x+1-{ x }^{ 2 }-2x }{ (x+2)({ x }^{ 2 }+x+1) } \right) \right\} } \\= \lim _{ x\rightarrow 1 }{ \left\{ \dfrac { 1 }{ (x-1) } \left( \dfrac { 1-x }{ (x+2)({ x }^{ 2 }+x+1) } \right) \right\} } \\= \lim _{ x\rightarrow 1 }{ \dfrac { -1 }{ (x+2)({ x }^{ 2 }+x+1) } } \\ =\dfrac { -1 }{ (1+2)(1+1+1) } \\ =-\dfrac { 1 }{ 9 }$

Differentiate $\sec x$ by first principle.

Let $f(x)=\sec x$

$\therefore f(x+h)\sec (x+h)$

$\displaystyle\cfrac{d}{dx}f(x)=\displaystyle\lim_{h\rightarrow 0}\left[\displaystyle\cfrac{f(x+h)-f(x)}{h}\right]$

$\displaystyle\cfrac{d}{dx}(\sec x)=\displaystyle\lim_{h\rightarrow 0}\left[\displaystyle\cfrac{\sec (x+h)-\sec x}{h}\right]$

$\displaystyle\cfrac{d}{dx}(\sec x)=\displaystyle\lim_{h\rightarrow 0}\left[\displaystyle\cfrac{1}{h}\left(\displaystyle\cfrac{1}{\cos (x+h)}-\cfrac{1}{\cos x}\right)\right]$

$\displaystyle\cfrac{d}{dx}(\sec x)=\displaystyle\lim_{h\rightarrow 0}\left[\displaystyle\cfrac{1}{h}\left(\displaystyle\cfrac{\cos (x)-\cos (x+h)}{\cos (x+h)\cos x}\right)\right]$

$\displaystyle\cfrac{d}{dx}(\sec x)=\displaystyle\lim_{h\rightarrow 0}\left[\displaystyle\cfrac{1}{h}\cfrac{2\sin \left(\displaystyle\cfrac{2x+h}{2}\right)\sin \left(\displaystyle\cfrac{x+h-x}{2}\right)}{h.\cos (x+h)\cos x}\right]$

and $\displaystyle\cfrac{d}{dx}(\sec x)=\displaystyle\lim_{h\rightarrow 0}\left[\displaystyle\cfrac{\sin \left(\displaystyle x+\cfrac{h}{2}\right)}{\cos (x+h)\cos x}\right]$ $\cdot \displaystyle\lim_{h\rightarrow 0}\left[\displaystyle \cfrac{\displaystyle \sin \cfrac{h}{2}}{\displaystyle\cfrac{h}{2}}\right]$

$\displaystyle\cfrac{d}{dx}(\sec x)=\displaystyle\cfrac{\sin x}{\cos x\cdot \cos x}\times 1$

$\displaystyle\cfrac{d}{dx}(\sec x)=\sec x\cdot \tan x$ .

$\displaystyle\lim _{ x\rightarrow 0 }{ \left[ \dfrac { 8\sin { x } +x\sin { x } }{ 3\tan { x } +{ x }^{ 2 } } \right] }$ is equal to

$\displaystyle\lim _{ x\rightarrow 0 }{ \left[ \dfrac { 8\sin { x } +x\cos { x } }{ 3\tan { x } +{ x }^{ 2 } } \right] }$

$=\displaystyle\lim _{ x\rightarrow 0 }{ \left[ \dfrac { \dfrac { 8\sin { x } }{ x } +\cos { x } }{ \dfrac { 3\tan { x } }{ x } +x } \right] }$

$=\dfrac { \displaystyle\lim _{ x\rightarrow 0 }{ \left[ \dfrac { 8\sin { x } }{ x } +\cos { x } \right] } }{ \displaystyle\lim _{ x\rightarrow 0 }{ \left[ \dfrac { 3\tan { x } }{ x } +x \right] } }$

$=\dfrac { 8+1 }{ 3+0 } =\dfrac { 9 }{ 3 } =3$

If y = log(logx) + $2sinx$ , find $\frac{dy}{dx}$

$y= log(logx) +2sinx$

Differentiate with respect to x

$\dfrac{dy}{dx} = \dfrac{1}{logx} \times \dfrac {1}{x} +2cosx$

$\dfrac{dy}{dx} = \dfrac{1}{xlogx} +2cosx$

If $f(x)=\tan x$ , find $f'(x)$ and hence find $f'\left(\dfrac{\pi}{4}\right)$ .

$f\left(x\right)=\tan{x}$

${f}^{\prime}{\left(x\right)}={\sec}^{2}{x}$

${f}^{\prime}{\left(\dfrac{\pi}{4}\right)}={\sec}^{2}{\left(\dfrac{\pi}{4}\right)}={\left(\sqrt{2}\right)}^{2}=2$

If $x = a(\theta + \sin \theta)$ and $y = a(1 - \cos \theta)$ , find $dy/ dx$ .

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

$\Rightarrow \dfrac{dx}{d\theta}=a(1+cos\theta)$ (1)

$y=a(1-cos \theta)$

$\Rightarrow \dfrac{dy}{d\theta}=a(0-(-\sin \theta))=a\sin\theta$ (2)

Dividing (2) and (1)

$\dfrac{dy/d\theta}{dx/d\theta}=\dfrac{dy}{dx}=\dfrac{\sin \theta}{1+\cos \theta}= tan(\theta/2)$

Write the negation of "some continuous functions are differentiable".

Negation of "some continuous functions are differentiable" is "All non continuous functions are not differentiable ".

$Find \ \dfrac{{dy}}{{dx}}\,in\,the\,following:$
1) $2x + 3y = \sin x$
2) $ax + b{y^2} = \cos x$
3) ${x^3} + xy + y = 100$

$2x + 3y = \sin x$

Differentiating w.r.t

$x$ :

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

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

$ax + b{y^2} = \cos x$

Differentiating w.r.t

$x$ :

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

$\Rightarrow \dfrac{dy}{dx}=\dfrac{-\sin x-a }{2by}$

${x^3} + xy + y = 100$

Differentiating w.r.t

$x$ :

$3x^2+xy'+y=0$

$\Rightarrow y'=\dfrac{-y-3x^2}{x}$

Differentiate the function with respect to $x$ .
$f(x)=$ $x^{2/3} + 7e^x - \dfrac{5}{x} + 7\tan x$

Let,

$f(x)=$

$x^{2/3} + 7e^x - \dfrac{5}{x} + 7\tan x$

Then

$\dfrac{df}{dx}=\dfrac{2}{3}x^{-\dfrac{1}{3}}+7e^{x}+\dfrac{5}{x^2}+7\sec^2 x$ . [Differentiating with respect to

$x$ ]

Find $\dfrac{dy}{dx}$ ; $y=4-x^2$

$y=4-x^2$

$\dfrac{dy}{dx}=\dfrac{d}{dx}(4-x^2)$

$=\dfrac{d}{dx}(4)-\dfrac{d}{dx}(x^2)$

$=0-2x$

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

$\sin^{2}(2x+5)$

Let

$y=\sin^2(2x+5)$

On differentiating both sides with respect to

$x$ , we have

$\dfrac{dy}{dx}=2\sin(2x+5)\times\cos(2x+5)\times2$

$\dfrac{dy}{dx}=2\times2\sin(2x+5)\cos(2x+5)$

We know that

$\sin2A=2\sin A\cos A$

Therefore,

$\dfrac{dy}{dx}=2\sin 2(2x+5)$

Hence, this is the answer.

$\underset {x\rightarrow 0}{\lim}\dfrac{x-\sin\,x}{x+\cos^2\,x}=$

$\displaystyle \lim _{ x\rightarrow 0 }{ \dfrac { x-\sin { x } }{ x+\cos ^{ 2 }{ x } } } \\ =\dfrac { 0-0 }{ 0+1 } \\ =0$

If $y=\sqrt{x^2+1}$ , find $\dfrac{dy}{dx}$ at $x=-1$

1237658_996971_ans_b2dabdf69315471e89d5f489def7f198.jpeg

If $f\left( x \right) = \sin 2x - \cos 2x,$ find $f'\left( {\dfrac{\pi}{6}} \right)$

We have,

$f\left( x \right)=\sin 2x-\cos 2x$

On differentiating both sides w.r.t x, we get

$f'\left( x \right)=\cos 2x\times \left( 2 \right)-\left( -\sin 2x \right)\times \left( 2 \right)$

$f'\left( x \right)=2\left( \cos 2x+\sin 2x \right)$

Put $x=\dfrac{\pi }{6}$

Therefore,

$f'\left( \dfrac{\pi }{6} \right)=2\left( \cos \dfrac{\pi }{3}+\sin \dfrac{\pi }{3} \right)$

$f'\left( \dfrac{\pi }{6} \right)=2\left( \dfrac{1}{2}+\dfrac{\sqrt{3}}{2} \right)$

$f'\left( \dfrac{\pi }{6} \right)=2\left( \dfrac{1+\sqrt{3}}{2} \right)$

$f'\left( \dfrac{\pi }{6} \right)= 1+\sqrt{3}$

Hence, the value is $1+\sqrt{3}$ .

Evaluate the following: $\underset{n \rightarrow \infty}{\lim}\dfrac{\cos n + \sin n}{n^2}$ ,

We have,

$0\le \left|\dfrac{\cos n+\sin n}{n^2}\right|\le\dfrac{\sqrt 2}{n^2} \forall n\in \mathbb{N}$ ......(1) [ Since

$|\sin n|\le 1$ and

$|\cos n|\le 1$ for

$n$ being natural number]

Now from (1) it is clear as

$n\to \infty$ ,

$\dfrac{sin n+cos n}{n^2}\to 0$ .

So,

$\underset{n \rightarrow \infty}{\lim}\dfrac{\cos n + \sin n}{n^2}=0.$

Find $\dfrac{dy}{dx}$ , if $y = \sin^3 x \, \cos^5 x$ .

$y= \sin^{3}x \cos^{5}x$

$\displaystyle \frac{dy}{dx}= \sin^{3}x\frac{d}{dx}\cos^{5}x+\cos^{5}x\frac{d}{dx}\sin^{3}x$

$\displaystyle \frac{dy}{dx}=\sin^{3}x \ 5 \cos^{4}x(-\sin x)+ \cos^{5}x \ 3\sin^{2}x \cos x$

$\displaystyle \frac{dy}{dx}=-5 \sin^{4}x \cos^{4}x+3 \cos^{6}x \sin^{2}x$

$\displaystyle \frac{dy}{dx}= \sin^{2}x\cos^{4}x(-5 \sin^{2}x+3 \cos^{2}x)$

Evaluate :
$lim_{x\rightarrow \infty} \dfrac{3x^2+4x+5}{4x^2+7}$

$lim_{x\rightarrow \infty} \dfrac{3x^2+4x+5}{4x^2+7}$

Put

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

When

$x\rightarrow \infty, y\rightarrow 0$

$lim_{x\rightarrow \infty} \dfrac{3x^2+4x+5}{4x^2+7} = lim_{y\rightarrow 0} \dfrac{3.\dfrac{1}{y^2}+4.\dfrac{1}{y}+5}{4.\dfrac{1}{y^2}+7}$

$=lim_{y\rightarrow 0}\dfrac{3+4y+5y^2}{4+7y^2}=\dfrac{3}{4}$

$\lim\limits_{x \to 0}\dfrac{\sin x^o}{x}=$

Now,

$\lim\limits_{x \to 0}\dfrac{\sin x^o}{x}$

$=\lim\limits_{x \to 0}\dfrac{\sin \dfrac{x\pi}{180}}{\dfrac{x\pi}{180}}\times \dfrac{180}{\pi}$ [ Since

$180^o=\pi^c$ ]

$=\lim\limits_{\dfrac{x\pi}{180} \to 0}\dfrac{\sin \dfrac{x\pi}{180}}{\dfrac{x\pi}{180}}\times \dfrac{180}{\pi}$

$=1.\dfrac{180}{\pi}$

$=\dfrac{180}{\pi}$

Find $\displaystyle \lim_{x \rightarrow \frac {\pi}{2}}[\sin x]$ where $[x]$ indicates greatest integer function

$\displaystyle\lim_{x\rightarrow \frac{\pi}{2}}[\sin x ]$

Now,

$L.H.L=\displaystyle\lim_{x\rightarrow \frac{\pi^-}{2}}[\sin x ]=\left[\sin\left(\dfrac{\pi^-}{2}\right)\right]=0$

$R.H.L=\displaystyle\lim_{x\rightarrow \frac{\pi^+}{2}}[\sin x ]=\left[\sin\left(\dfrac{\pi^+}{2}\right)\right]=0$

$\therefore$

$L.H.L=R.H.L=0$

Find the derivative of $f(x)$ from the first principles, where $f(x)$ is
$\sin x+\cos x$

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

We know,

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

Now,

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

Therefore,

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

$=lim_{h\rightarrow 0} \dfrac{\sin x \cos h +\cos x \sin h +\cos x \cos h -\sin x \sin h -\sin x-\cos x}{h}$

$=lim_{h\rightarrow 0} \dfrac{\sin h(\cos x-\sin x)+\sin x(\cos h-1)+\cos x(\cos h-1)}{h}$

$=lim_{h\rightarrow 0} \dfrac{\sin h(\cos x-\sin x)}{h}+lim_{h\rightarrow 0} \dfrac{\sin x(\cos h-1)}{h}+lim_{h\rightarrow 0} \dfrac{\cos x(\cos h-1)}{h}$

$=(\cos x-\sin x)lim_{h\rightarrow 0} \dfrac{\sin h}{h}-\sin x lim_{h\rightarrow 0} \dfrac{1-\cos h}{h} -\cos x lim_{h\rightarrow 0} \dfrac{1-\cos h}{h}$

$=\cos x -\sin x -0-0$

$=\cos x -\sin x$

Hence,

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

$\mathop {\lim }\limits_{x \to 1} \left( {1 - x} \right)\tan \left( {\frac{{\pi x}}{2}} \right)$

$\lim_{x\rightarrow 1} f(x) = (1-x) tan(\frac{x\pi }{2}) = f(1)$

$\lim_{x\rightarrow 1} sin(\frac{\pi x}{2}) \frac{1-x}{\frac{cosx\pi }{2}} = \lim_{x\rightarrow 1} sin\frac{\pi x}{2} \lim_{x\rightarrow 1} \frac{1-x}{\frac{cos\pi x}{2}}$

$\Rightarrow (1) \lim_{x\rightarrow 1} \frac{-1}{\frac{-\pi }{2}sin\frac{\pi x}{2}} = \frac{2}{\pi } = f(1)$

1208442_1180041_ans_c80a1545f6704db3a8cdb13884348ccb.jpg

Derivative of :
$(x+1)^2$

$y=(x+1)^2$

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

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

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

Evaluate $\underset{x \rightarrow a}{\lim} \dfrac{\sin x - \sin a}{\sqrt{x} - \sqrt{a}}$

$\lim_{x\rightarrow a} \frac{sinx-sina}{\sqrt{x}-\sqrt{a}}$

$= \lim_{x\rightarrow a} \frac{(sinx-sina)(\sqrt{x}+\sqrt{a})}{(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})}$

$= \lim_{x\rightarrow a} \frac{(sinx-sina)(\sqrt{x}+\sqrt{a})}{x-a}$

$= \lim_{x\rightarrow a} \frac{2cos(\frac{x+a}{2})sin(\frac{x-a}{2})\sqrt{x}+\sqrt{a}}{2(\frac{x-a}{2})}$

$(sinx-siny = 2cos(\frac{x+y}{2})sin(\frac{x-y}{2}))$

$= \frac{2cosa2\sqrt{a}}{2} = 2\sqrt{a}cosa(\lim_{x\rightarrow a} \frac{sinx}{x} = 1)$

$\therefore \lim_{x\rightarrow a} \frac{sinx-sina}{\sqrt{x}-\sqrt{a}} = 2\sqrt{a}cosa$

1098152_1190113_ans_8f83fc33c13e4a76b587a36ba578f940.jpg

If $y = \dfrac {1}{4} (x\pm A)^{2}$
Hence prove: $y_{1}^{2} = y$ .

$y=\cfrac { 1 }{ 4 } { (x\pm A) }^{ 2 }\\ { y }_{ 1 }=\cfrac { dy }{ dx } =\cfrac { 1 }{ 4 } \times 2(x\pm A)\\ { y }_{ 1 }^{ 2 }=\cfrac { 1 }{ 4 } { (x\pm A) }^{ 2 }=y$

Hence proved.

Differentiate with respect to $x$ , where $y = x - x^2.$

We have,

$y=x-x^2$

On differentiating w.r.t

$x$ , we get

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

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

Hence, this is the answer.

Solve $\mathop {\lim }\limits_{x \to 0} \dfrac{{\cos x}}{{\pi - x}}$

We have,

$\underset{x\to 0}{\mathop{\lim }}\,\left( \dfrac{\cos x}{\pi -x} \right)$

Apply L-Hospital rule,

$\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\left( \dfrac{-\sin x}{0-1} \right)$

$\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\left( \sin x \right)$

$\Rightarrow \sin 0$

$\Rightarrow 0$

Hence, this is the answer.

Evaluate $\underset { x\rightarrow 0 }{ lim\quad } log\frac { sinx }{ x } \\$

$\underset { x\rightarrow 0 }{lim }\,\,\, log\left(\dfrac{\sin x}{x}\right)$

We know that

$\underset { x\rightarrow 0 }{lim }\,\,\, \dfrac{\sin x}{x}=1$

$=log(1)$

$=0$

Find $\dfrac{dy}{dx}$
$y = - x ^ { 3 } + x$ ...

Consider given the equation,

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

Differentiate with respect to x both sides,

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

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

Hence, this is the answer.

Find the derivatives of the following:
$\sin x\cos x$ .

Let

$y=\sin x\cos x$

On differentiating w.r.t

$x$ , we get

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

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

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

Hence, this is the answer.

Evaluate : $\mathop {\lim }\limits_{x \to 0} \cfrac{{{2^x} - 1}}{{{{(1 + x)}^{1/2}} - 1}}$

We have,

$\mathop {\lim }\limits_{x \to 0} \cfrac{{{2^x} - 1}}{{{{(1 + x)}^{1/2}} - 1}}$

This is

$\dfrac{0}{0}$ form.

So, apply L-Hospital rule,

$\mathop {\lim }\limits_{x \to 0} \cfrac{2^x \ln 2 - 0}{\dfrac{1}{2\sqrt {1 + x}} - 0}$

$\mathop {\lim }\limits_{x \to 0} 2^x \ln 2 (2\sqrt {1 + x})$

$=2^0 \ln 2 (2\sqrt {1 + 0})$

$=1\times \ln 2 (2\sqrt {1 })$

$= 2\ln 2$

Hence, this is the answer.

$\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - {e^5}}}{{x - 5}}$

$\underset{x\rightarrow 0}{lim}\dfrac{e^x-e^5}{x-5}$

$\Rightarrow \dfrac{e^0-e^5}{0-5}=\dfrac{1-e^5}{-5}$

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

Evaluate $\lim_{x\rightarrow \pi/2} \dfrac {\cos x}{\pi-2x}$

$\displaystyle\lim _{x\rightarrow \dfrac{\pi}{2}} \dfrac{\cos x}{\pi-2 x}=\lim_{x\rightarrow \pi/2} \dfrac{\sin \bigg(\dfrac{\pi}{2}-x\bigg)}{2\bigg(\dfrac{\pi}{2}-x\bigg)}=\dfrac{1}{2}$

Find the derivative of $x^{2}$ - 2 at x = 10

Let

$y=x^2-2$

$\dfrac{dy}{dx}=2x$

$\dfrac{dy}{dx}$ at

$x=10$ is

$2(10)=20$ .

1192236_1195794_ans_eeccf5ffa7084ba3bd14ce6e622f143a.jpg

Evaluate $\mathop {\lim }\limits_{x \to \infty } \dfrac{{5{x^2} + 4}}{{\sqrt {2{x^4} + 1} }}$

We have,

$\mathop {\lim }\limits_{x \to \infty } \dfrac{{5{x^2} + 4}}{{\sqrt {2{x^4} + 1} }}$

$\mathop {\lim }\limits_{x \to \infty } \dfrac{{x^2\left(5 + \dfrac{4}{x^2}\right)}}{x^2\sqrt {2 + \dfrac{1}{x^4}} }$

$\mathop {\lim }\limits_{x \to \infty } \dfrac{{\left(5 + \dfrac{4}{x^2}\right)}}{\sqrt {2 + \dfrac{1}{x^4}} }$

$\Rightarrow \dfrac{{\left(5 + \dfrac{4}{(\infty)^2}\right)}}{\sqrt {2 + \dfrac{1}{(\infty)^4}} }$

$\Rightarrow \dfrac{{\left(5 +0\right)}}{\sqrt {2 +0} }$

$\Rightarrow \dfrac{5}{\sqrt 2 }$

Hence, this is the answer.

Find the derivative of the following functions:
(i) tan $x$ cos $x$
(ii) sec $x$

Part

$(i)$

Let

$y=\tan x\cos x$

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

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

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

$\dfrac{dy}{dx}=\sec x-\tan x\sin x$

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

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

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

Part

$(ii)$

Let

$y= \sec x$

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

$\dfrac{dy}{dx}=\sec x.\tan x$

Hence, this is the answer.

Evaluate: $\displaystyle{\lim}_{x\rightarrow2}\dfrac{\sin(x^{2}-5x+6)}{x^{2}-7x+10}$

$= \lim_{x\rightarrow 2} \dfrac{sin(x^{2}-5x+6)}{x^{2}-7x+10}$

$= \lim_{x\rightarrow 2} \dfrac{sin(x-2)(x-3)}{(x-5)(x-2)}$

$= \lim_{x\rightarrow 2} \dfrac{1}{(x-5)} \lim_{x\rightarrow 2} \dfrac{sin(x-3)(x-2)}{(x-2)}$

$= -(\dfrac{1}{3})(-1) = \dfrac{1}{3}$

1113247_1204125_ans_cf2ffa7fd070417fbaaa97bc47647170.jpg

$\mathop {\lim }\limits_{x \to 0} \cfrac{{\sin \left( {\pi {{\cos }^2}x} \right)}}{{{x^2}}}$

We have,

$\mathop {\lim }\limits_{x \to 0} \cfrac{{\sin (\pi {{\cos }^2}x)}}{{{x^2}}}$

$\mathop {\lim }\limits_{x \to 0} \cfrac{{\sin (\pi (1-\sin^2 x))}}{{x^2}}$

$\mathop {\lim }\limits_{x \to 0} \cfrac{{\sin (\pi-\pi \sin^2 x)}}{{x^2}}$

$\mathop {\lim }\limits_{x \to 0} \cfrac{{\sin (\pi \sin^2 x)}}{{x^2}}$

$\mathop {\lim }\limits_{x \to 0} \cfrac{{\sin (\pi \sin^2 x)}}{{\pi \sin ^2 x}}\times \dfrac{\pi \sin^2 x}{x^2}$

We know that

$\lim_{x\to 0}\dfrac{\sin x}{x}=1$

Therefore,

$=1\times \pi \times 1$

$=\pi$

Hence, this is the answer.

Evaluate : $\displaystyle \lim _ { x \rightarrow 3 } \dfrac { \sqrt [ 4 ] { x } - \sqrt [ 4 ] { 3 } } { \sqrt [ 3 ] { x } - \sqrt [ 3 ] { 3 } }$

${ \lim }_{ x\to 3 }\dfrac { { \sqrt [ 4 ]{ x } -\sqrt [ 4 ]{ 3 } } }{ { \sqrt [ 3 ]{ x } -\sqrt [ 3 ]{ 3 } } } \\ ={ \lim }_{ x\to 3 }\dfrac { { { { \left( x \right) }^{ 1/4 } }-{ 3^{ 1/4 } } } }{ { { { \left( x \right) }^{ 1/3 } }-{ { \left( 3 \right) }^{ 1/3 } } } }$

Apply L-hospital rule,

$\\ ={ \lim }_{ x\to 3 }\dfrac { { \dfrac { 1 }{ 4 } { { \left( x \right) }^{ -3/4 } } } }{ { \dfrac { 1 }{ 3 } { { \left( x \right) }^{ -2/3 } } } } \\ =\dfrac { 3 }{ 4 } { x^{ 2/3-3/4 } } \\ ={ \lim }_{ x\to 3 }{ x^{ -1/12 } } \\ =\dfrac { 3 }{ 4 } { \left( 3 \right) ^{ -1/12 } } \\ =\dfrac { { { { \left( 3 \right) }^{ 11/12 } } } }{ 4 }$

Hence, this is the answer.

$\lim _{ x\rightarrow 0 }{ \dfrac { \tan { x }-\cos x }{ x} }$ is p then ${p}$ =

$lim_{x\rightarrow 0}{\dfrac{tan x- cos x}{x}}=p$

$lim_{x\rightarrow 0}\dfrac{\dfrac{sinx-cos^2x}{cosx}}{x}=p$

$lim_{x\rightarrow 0}{\dfrac{sinx-cos^2x}{xcosx}}=p$

$lim_{x\rightarrow 0}{\dfrac{sinx}{x}\times\dfrac{1}{cosx}}-lim_{x\rightarrow 0}\dfrac{cos^2x}{cosx}=p$

$1\times lim_{x\rightarrow 0}{\dfrac{1}{cosx}}-lim_{x\rightarrow 0}cosx=p$

Applying limit

$1\times1-1=p$

$p=0$

Solve:
$\lim_{x\to\ 3}\dfrac{x^2-9}{x-3}$

We have,

$\lim_{x\to\ 3}\dfrac{x^2-9}{x-3}$

$\lim_{x\to\ 3}\dfrac{x^2-3^2}{x-3}$

$\lim_{x\to\ 3}\dfrac{(x-3)(x+3)}{x-3}$

$\lim_{x\to\ 3}(x+3)$

$=(3+3)=6$

Hence, this is the answer.

Find the value of $\displaystyle x\xrightarrow { lim } 1\frac{\sqrt{x-1}+\sqrt{x}-1}{\sqrt{{x}^{2}-1}}$

We have,

$\lim _{ x\rightarrow 1 }{ \cfrac { \sqrt { x-1 } +\sqrt { x } -1 }{ \sqrt { { x }^{ 2 }-1 } } }$

Putting the limits we find it to be in

$\cfrac { 0 }{ 0 }$ form using L' Hospitals rule.

$=\lim _{ x\rightarrow 1 }{ \cfrac { \cfrac { 1 }{ 2\sqrt { x-1 } } +\cfrac { 1 }{ 2\sqrt { x } } }{ \cfrac { 2x }{ 2\sqrt { { x }^{ 2 }-1 } } } } \\ =\lim _{ x\rightarrow 1 }{ \cfrac { \cfrac { 1 }{ 2\sqrt { x-1 } } +\cfrac { 1 }{ 2\sqrt { x } } }{ \cfrac { x }{ \sqrt { { x }^{ 2 }-1 } } } } \\ =\lim _{ x\rightarrow 1 }{ \cfrac { \cfrac { \sqrt { { x }^{ 2 }-1 } }{ 2\sqrt { x-1 } } +\cfrac { \sqrt { { x }^{ 2 }-1 } }{ 2\sqrt { x } } }{ x } } \\ =\lim _{ x\rightarrow 1 }{ \cfrac { \cfrac { \sqrt { x-1 } \sqrt { x+1 } }{ 2\sqrt { x-1 } } +\cfrac { \sqrt { { x }^{ 2 }-1 } }{ 2\sqrt { x } } }{ x } } \\ =\lim _{ x\rightarrow 1 }{ \cfrac { \cfrac { \sqrt { x+1 } }{ 2 } +\cfrac { \sqrt { { x }^{ 2 }-1 } }{ 2\sqrt { x } } }{ x } }$

Putting the limits we get

$=\cfrac { \cfrac { \sqrt { 2 } }{ { 2 } } }{ 1 } \\ =\cfrac { 1 }{ \sqrt { 2 } }$

Hence, this is the answer.

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

1345594_1344942_ans_c273eda278ea417baa2a94d33b8c4960.jpg

Find the derivative of $f(x)$ from the first principle.
$\sin x\div \cos x$

$f{\left( x \right)} = \sin{x} \div \cos{x}$

$\Rightarrow f{\left( x \right)} = \cfrac{\sin{x}}{\cos{x}} = \tan{x}$

$\therefore f'{\left( x \right)} = \sec^{2}{x}$

Find the derivative by first principle
$\cos 5 x$

Let

$f(x)=\cos 5 x$

$f'(x)=\displaystyle\lim_{h\to 0} \dfrac{f(x+h)-f(x)}{h}=\lim _{h\to 0}\dfrac{\cos 5(x+h)-\cos 5 x}{h}=\lim_{h\to 0} \dfrac{-2\sin \bigg(\dfrac{10 x+5 h}{2}\bigg)\sin \bigg(\dfrac{5 h}{2}\bigg)}{h}$

$=(-2\sin 5{x})\lim_{h\to 0} \dfrac{\sin \dfrac {5 h}{2}}{h}=-5\sin 5 x$

Evaluate $\underset { x\rightarrow 0 }{ lim } f(x)$ , where $f(x)=\{ \overset { { |x| } ,\quad \quad x\neq 0 }{ \underset { 0,\quad \quad x=0 }{ x } }$

$\lim_{x\to 0^+}f(x)=+1\\\lim_{x\to 0^-}f(x)=-1$

The right and left hand limits are not equal so the limit doesnot exist

$\lim _{ x\rightarrow \pi /2 }{ \frac { \sin ^{ }{ x } }{ x } }$

$\lim_{\pi \to 2}\dfrac{\sin x}{x}\\\dfrac{\sin \pi/2}{\dfrac{\pi}{2}}\\\dfrac2{\pi}$

Find the derivative by first principle in examples to $( 8 )$
$\cos x$

$\lim_{h\to 0}\dfrac {\cos(x+h)-\cos x}{h}\\\lim_{h\to 0}\dfrac {\cos x\cos h-\sin x\sin h-\cos x}{h}\\\lim_{h\to 0}\dfrac{\cos x(\cos h-1)}{h}-\lim_{h\to 0}\dfrac{\sin x \sin h}{h}\\0-\sin x=-\sin x$

If $y=x^2+5x$ find $\dfrac {dy}{dx}$

$y=x^2+5x\\\dfrac {dy}{dx}=\dfrac {d}{dx}x^2+5x=2x+5$

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

$\dfrac {dy}{dx}=\cos x+\sin x\\y=\displaystyle \int \cos x+\sin x dx\\y=\sin x-\cos x+c$

Find $\dfrac {dy}{dx}$ if $y=\dfrac {x^2+x}2$

$y=\dfrac {x^2+x}{2}\\\dfrac{dy}{dx}=\dfrac{d}{dx}\dfrac {x^2+x}{2}\\\dfrac{dy}{dx}=\dfrac 12(2x+1)\\\dfrac{dy}{dx}=x+\dfrac 12$

$\displaystyle \lim_{x\rightarrow 0} \tan \left(\dfrac {\pi}{4}+x\right)$

$\displaystyle \lim_{x \to 0} \tan \left(\dfrac {\pi}{4}+x\right)\\\tan\left(\dfrac {\pi}{4}+0\right)=1$

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

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

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

$y=e^x+\sec x$

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

$\displaystyle \lim _{ x\rightarrow 0 } \left(\dfrac{a}{b}+\dfrac{\cos{x}}{b}\right)$

$\displaystyle \lim _{ x\rightarrow 0 } \left(\dfrac{a}{b}+\dfrac{\cos{x}}{b}\right)=\bigg(\dfrac{a}{b}+\dfrac{\cos 0}{b}\bigg)=\bigg(\dfrac{a}{b}+\dfrac{1}{b}\bigg)=\dfrac{a+1}{b}$

Differentiate with respect to $x$ :
$3x^2-e^{-3x}+\sec x$

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

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

$\dfrac{dy}{dx}=6x+3e^{-3x}+\sec x \tan x$

Find $\frac { d y } { d x } ,$ if $y = \log \left( \sqrt { x } - \frac { 1 } { \sqrt { x } } \right)$

Given

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

Differentiating on both sides

$\dfrac{d y}{d x}=\dfrac{1}{x-1}-\dfrac{1}{2 x}=\dfrac{2 x-x+1}{2 x (x-1)}=\dfrac{x+1}{2 x(x-1)}$

Differentiate with respect to $x$ :
$\cos x+\sin 2x$

$y=\cos x +\sin 2x$

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

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

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

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

$y=3x^2+2$

$\dfrac{dy}{dx}=\dfrac {d}{dx}(3x^2+2)\\\dfrac {dy}{dx}=6x$

A curve has equation $y=(2x-1)^{-1}+2x$ . $x\neq\dfrac{1}{2}$
Find $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ .

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

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

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

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

$\dfrac{d^2y}{dx^2}=8(2x-1)^{-3}$ .

$\lim_{x\rightarrow 3} x^2-3x+1$

The limit is calculated by direct substitution.

$\lim_{x\rightarrow 3} x^2-3x+1\\3^2-3(2)+1=1$

Evaluate $lim_{ x_\rightarrow 2} \,\dfrac { x ^ { 5 } - 32 } { x ^ { 3 } - 8 }$

1321560_1547190_ans_55fae64517044ace840d15e5f645312f.jpeg

Evaluate $\lim_{x\rightarrow 2}{(x)^{3}}$ .

$\lim_{x\rightarrow 2}{(x)^{3}}\\$

$2^3=8$ .

Differentiate $e^x+e^{-x}$ with respect to x.

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

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

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

$y=\dfrac{6}{x^2}+2x$ , then find $\dfrac{dy}{dx}$ .

$y=6x^{-2}+2x$

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

$\dfrac{dy}{dx}=-12x^{-3}+2$

$\dfrac{dy}{dx}=\dfrac{-12}{x^3}+2$

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

$y=2x^2-\log x +(x+1)^2$

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

$\dfrac{dy}{dx}=4x-\dfrac{1}{x}+2(x+1)$

$\dfrac{dy}{dx}=6x-\dfrac{1}{x}+2$

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

$y=a^x +\log x +3x^2$

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

$\dfrac{dy}{dx}=a^x \log a +\dfrac{1}{x}+6x$

Differentiate with respect to $x$ :
$y=e^{-3x}+\sin 2x$

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

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

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

If $y=e^x+e^{-x}+\log x^2$ , find $\dfrac{dy}{dx}$ .

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

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

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

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

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

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

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

$\dfrac{dy}{dx}=6x^2-6x^{-3}$

$\dfrac{dy}{dx}=6x^2-\dfrac{6}{x^3}$

Differentiate w.r.t. $x$ :
$\log e^x+\sin 3x-8$

$y=\log e^x +\sin 3x-8$

Differentiating w.r.t x, we get,

$\dfrac{dy}{dx}=\dfrac{1}{e^x}\times e^x+3\cos 3x$

$\dfrac{dy}{dx}=1+3\cos 3x$

If $y=e^x+x^2+\log x$ , then find $\dfrac{dy}{dx}$ .

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

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

Differentiate with respect to $x$ :
$y=\cos x + \sin 2x$

$y=\cos x +\sin 2x$

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

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

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

Differentiate $e^x+e^{x^2}+....+e^{x^5}$ w.r.t. $x$ .

Let

$y=e^x+e^{x^2}+...+e^{x^5}$

On differentiating both sides w.r.t. x, we get,

$\dfrac{dy}{dx}=e^x +2x e^{x^2}+3x^2 e^{x^3}+4x^3 e^{x^4}+5x^4e^{x^5}$

Evaluate the following limits.
$\displaystyle\lim_{x\rightarrow a}a^3x$

$\displaystyle\lim_{x\rightarrow a}a^3x$

Substituting the limit

$=a^3(a)=a^4$

Answer the following question in one word or one sentence or as per exact requirement of the question.
If the value of $\displaystyle\lim_{x\rightarrow -\infty}(3x+\sqrt{9x^2-x})$ $=\dfrac{1}{6b}$ . Find the value of $b$ ?

1498892_1668728_ans_39d9055cab46420db750fad229b333d5.jpg

Answer the following question in one word or one sentence or as per exact requirement of the question.
Write the value of $\displaystyle\lim_{n\rightarrow \infty}\dfrac{n!+(n-1)!}{(n+1)!+(n+2)!}$ .

We have,

$\Rightarrow \displaystyle\lim_{n\rightarrow \infty}\dfrac{n!+(n-1)!}{(n+1)!+(n+2)!}$

$\Rightarrow \displaystyle\lim_{n\rightarrow \infty}\dfrac{n(n-1)!+(n-1)!}{(n+1)!+(n+2)(n+1)!}$

$\Rightarrow \displaystyle\lim_{n\rightarrow \infty}\dfrac{(n-1)!(n+1)}{(n+1)!(1+n+2)}$

$\Rightarrow \displaystyle\lim_{n\rightarrow \infty}\dfrac{(n-1)!(n+1)}{(n-1)!(n+1)(n)(n+3)}$

$\Rightarrow \displaystyle\lim_{n\rightarrow \infty}\dfrac{(n+1)}{(n)(n+1)(n+3)}$

$\Rightarrow \displaystyle\lim_{n\rightarrow \infty}\dfrac{1}{(n)(n+3)}=0$

Evaluate the following:
$\displaystyle\lim_{x\rightarrow 0}x^2-3$

We have,

$\displaystyle\lim_{x\rightarrow 0} x^2-3$

By substituting the limit we get,

$=0^2-3$

$=-3$

Differentiate each of the following w.r.t. $x$ :
$\sin 4x$

1543039_1705856_ans_7d68dc7fdb9945e29e7d0587c0f5e8df.jpg

Evaluate the following question:
$\displaystyle\lim_{x\rightarrow 2}2x+5$

$\displaystyle\lim_{x\rightarrow 2} 2x+5$

By substituting the limit we get,

$=2(2)+5$

$=9$

Answer the following question in one word or one sentence or as per exact requirement of the question.
Write the value of $\displaystyle\lim_{x\rightarrow 0^-}[x]$ .

1498883_1668696_ans_28dc599769a141679e44912e9c592249.jpg

Evaluate the following limit:

$\displaystyle\lim_{h\rightarrow 0}\dfrac{\sqrt{x+h}-\sqrt{x}}{h}, x\neq 0$ .

$\displaystyle\lim_{h\rightarrow 0}\dfrac{\sqrt{x+h}-\sqrt{x}}{h}, x\neq 0$ .

Rationalizing ..

Multiplying and Divding by

$\dfrac{\sqrt {x+h}+\sqrt x}{\sqrt {x+h}+\sqrt x}$

$=\displaystyle \lim _{h \to 0} \dfrac{\sqrt {x+h}-\sqrt x}{h}\times \dfrac{\sqrt {x+h}+\sqrt x}{\sqrt {x+h}+\sqrt x}$

$=\displaystyle \lim _{h \to 0} \dfrac{x+h-x}{h(\sqrt{x+h}+\sqrt x)}$

$=\displaystyle \lim _{h \to 0} \dfrac{1}{(\sqrt{x+h}+\sqrt x)}$

Substituting the value

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

Differentiate
$\left ( \dfrac{sec\,x - 1}{sec\,x + 1} \right )$

$\dfrac{d}{dx} \left ( \dfrac{sec\,x - 1}{sec\,x + 1} \right )$

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

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

$= \dfrac{2 \, sec \,x tan\,x}{(sec \, x + 1)^2}$

Find derivative of $sec\, x$ by first principle

Let

$f(x) = sec\,x$

$f(x + h) = sec (x + h)$

$f^1(x) = \underset {h \rightarrow 0} {\lim} \dfrac{f(x + h) - f(x)}{h}$

$= \underset {h \rightarrow 0} {\lim}\dfrac{sec(x + h) - sec\,x}{h}$

$= \underset {h \rightarrow 0} {\lim} \dfrac{\dfrac{1}{cos(s + h)} - \dfrac{1}{cos\,x}}{h}$

$= \underset {h \rightarrow 0} {\lim} \dfrac{cos\,x - cos(x + h)}{cos(x + h)cos xh}$

$= \underset {h \rightarrow 0} {\lim} \dfrac{-2 \, sin \left [ \dfrac{2x + h}{2} \right ] sin \left [ \dfrac{-h}{2} \right ]}{cos(x + h) cos\,xh}$

$= \underset {h \rightarrow 0} {\lim} \dfrac{2\,sin \left [ \dfrac{2x + h}{2} \right ] sin \dfrac{h}{2}}{cos(x + h) cos \,xh}$

$[sin (- \theta = - sin \, \theta)]$

$= \underset {h \rightarrow 0} {\lim} \dfrac{2 \, sin \dfrac{h}{2}}{2 \dfrac{h}{2}} \times \dfrac{ \underset {h \rightarrow 0} {\lim} sin \dfrac{(2x + h)}{2}}{ \underset {h \rightarrow 0} {\lim} cos (x + h) cos x}$

$= 1 \times \dfrac{sin \left ( \dfrac{2x + 0}{2} \right )}{cos (x + 0) cos \,x} = \dfrac{sin\,x}{cos\, x \, cos \, x}$

$= tan \,x \, sec \, x$

Differentiate the following w.r.t.x: $\sin ^{2} x^{2}-\cos ^{2} x^{2}$

Let

$y=\sin ^{2} x^{2}-\cos ^{2} x^{2}$
Differentiating w.r.t. x, we get

$\dfrac{d y}{d x}=\dfrac{d}{d x}\left[\sin ^{2} x^{2}-\cos ^{2} x^{2}\right]$

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

$=2 \sin x^{2} \cdot \dfrac{d}{d x}\left(\sin x^{2}\right)-2 \cos x^{2} \cdot \dfrac{d}{d x}\left(\cos x^{2}\right)$

$=2 \sin x^{2} \cdot \cos x^{2} \cdot \dfrac{d}{d x}\left(x^{2}\right)-2 \cos x 2 \cdot(-\sin x 2) \cdot \dfrac{d}{d x}\left(x^{2}\right)$

$=2 \sin x^{2} \cdot \cos x^{2} \times 2 x+2 \sin x^{2} \cdot \cos x^{2} \times 2 x$

$=4 x\left(2 \sin x^{2} \cdot \cos x^{2}\right)$

$=4 x \sin \left(2 x^{2}\right)$

Find the derivative of following functions w.r.t. $x$ :
$\sec x^o$

Let

$y=\sec x^o$
We know

$180^o= \pi$ (radian)

$1^o=\dfrac{\pi}{180}$ radian
Then $$x^o=\dfrac{\pi x}{180} radian ...(i)

$Hence$ y=\sec x^o=\sec \left( \dfrac{\pi x}{180} \right)

$[From$ (i)$$]
or

$y=\sec u$ where

$u=\left( \dfrac{\pi x}{180} \right)$
Diff. w.r.t.

$x$ ,

$\dfrac{du}{dx}=\dfrac{\pi x}{180} ...(ii)$

$\dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx}$

$=\dfrac{d}{du} (\sec u).\dfrac{\pi x}{180}$ [From

$(ii)$ ]

$=\dfrac{\pi x}{180} \sec t \tan u$

$=\dfrac{\pi x}{180} \sec \left( \dfrac{\pi x}{180} \right) \tan \left( \dfrac{\pi x}{180} \right)$

$=\dfrac{\pi x}{180} \sec x^o \tan x^o$ [from

$(i)$ ]

If $\displaystyle \underset{x\rightarrow a}{lim} f\left ( x \right )g\left ( x \right )$ exists, then does it imply that $\displaystyle \underset{x\rightarrow a}{lim} f\left ( x \right )$ and $\displaystyle \underset{x\rightarrow a}{lim}\ g\left ( x \right )$ also exist? Yes =5 No=7

No. Let us choose

$\displaystyle f\left ( x \right )= x, g\left ( x \right )= \frac{1}{x}$

$\displaystyle \underset{x\rightarrow 0}{lim}f\left ( x \right )g\left ( x \right )= \underset{x\rightarrow 0}{lim} 1= 1.$

$\displaystyle \underset{x\rightarrow 0}{lim}f\left ( x \right )= \underset{x\rightarrow 0}{lim} x= 0.$ Limit exists.But

$\displaystyle \underset{x\rightarrow 0}{lim} g\left ( x \right )= \underset{x\rightarrow 0}{lim} \frac{1}{x}.$ This limit does not exist

as

$\displaystyle R.H.L.= \infty , L.H.L.= -\infty .$

Find $m+n$
$\displaystyle \underset{x\rightarrow \infty }{lim} \frac{2\sqrt{x}+3\sqrt[3]{x}+5\sqrt[5]{x}}{\sqrt{\left ( 5x-2 \right )}+\sqrt[3]{\left ( 3x-2 \right )}}$ is $\dfrac{m}{\sqrt{n}}$

${\lim}_{x\rightarrow\,\infty}\dfrac{2{x}^{\dfrac{1}{2}}+3{x}^{\dfrac{1}{3}}+5{x}^{\dfrac{1}{5}}}{\sqrt{x\left(5-\dfrac{2}{x}\right)}+\sqrt[3]{x\left(3-\dfrac{2}{x}\right)}}$

$={\lim}_{x\rightarrow\,\infty}\dfrac{{x}^{\dfrac{1}{2}}\left(2+3{x}^{\dfrac{1}{3}-\dfrac{1}{2}}+5{x}^{\dfrac{1}{5}-\dfrac{1}{2}}\right)}{{x}^{\dfrac{1}{2}}{\left(5-\dfrac{2}{x}\right)}^{\dfrac{1}{2}}+{x}^{\dfrac{1}{3}}{\left(3-\dfrac{2}{x}\right)}^{\dfrac{1}{3}}}$

$={\lim}_{x\rightarrow\,\infty}\dfrac{{x}^{\dfrac{1}{2}}\left(2+3{x}^{\dfrac{1}{3}-\dfrac{1}{2}}+5{x}^{\dfrac{1}{5}-\dfrac{1}{2}}\right)}{{x}^{\dfrac{1}{2}}\left[{\left(5-\dfrac{2}{x}\right)}^{\dfrac{1}{2}}+{x}^{\dfrac{1}{3}-\dfrac{1}{2}}{\left(3-\dfrac{2}{x}\right)}^{\dfrac{1}{3}}\right]}$

$={\lim}_{x\rightarrow\,\infty}\dfrac{2+3{x}^{\frac{2-3}{6}}+5{x}^{\frac{2-5}{10}}}{{\left(5-\dfrac{2}{x}\right)}^{\frac{1}{2}}+{x}^{\frac{2-3}{6}}{\left(3-\dfrac{2}{x}\right)}^{\frac{1}{3}}}$

$={\lim}_{x\rightarrow\,\infty}\dfrac{2+3{x}^{\dfrac{-1}{6}}+5{x}^{\dfrac{-3}{10}}}{{\left(5-\dfrac{2}{x}\right)}^{\dfrac{1}{2}}+{x}^{\dfrac{-1}{6}}{\left(3-\dfrac{2}{x}\right)}^{\dfrac{1}{3}}}$

$={\lim}_{x\rightarrow\,\infty}\dfrac{2+\dfrac{3}{{x}^{\dfrac{1}{6}}}+\dfrac{5}{{x}^{\dfrac{3}{10}}}}{{\left(5-\dfrac{2}{x}\right)}^{\dfrac{1}{2}}+\dfrac{1}{{x}^{\dfrac{1}{6}}}{\left(3-\dfrac{2}{x}\right)}^{\dfrac{1}{3}}}$

$x\rightarrow\,\infty,\,\,\dfrac{1}{x}\rightarrow\,0$

$={\lim}_{x\rightarrow\,\infty}\dfrac{2+0+0}{{\left(5-0\right)}^{\dfrac{1}{2}}+0{\left(3-0\right)}^{\dfrac{1}{3}}}$

$={\lim}_{x\rightarrow\,\infty}\dfrac{2}{\sqrt{5}}=\dfrac{m}{\sqrt{n}}$ (given)

where

$m=2,\,n=5$

$\therefore\,m+n=2+5=7$

Find the value of $\displaystyle \lim_{x\rightarrow 1}e\left ( 1+\sin \pi x \right )^{\cot \pi x}$

$\displaystyle \lim_{x\to 1}e (1+sin\pi x)^{cot\pi x}=e\cdot \displaystyle \lim_{x\to 1} (1+sin\pi x)^{(1/sin\pi x)cos\pi x}=e\cdot e^{cos\pi} = e \cdot e^{-1} =1$

Match the columns

$f\left( x \right) =x\cfrac { e^{ 1/x }-e^{ -1/x } }{ e^{ 1/x }+e^{ -1/x } }$

$f'\left( 0+ \right) =\lim _{ x\rightarrow 0+ }{ \cfrac { f\left( x \right) -f\left( 0 \right) }{ x-0 } } =\lim _{ x\rightarrow 0+ }{ \cfrac { e^{ 1/x }-e^{ -1/x } }{ e^{ 1/x }+e^{ -1/x } } } =1$

B)

$f\left( x \right) =x^{ 2 }\sin \cfrac { 1 }{ x }$

$f'\left( 0+ \right) =\lim _{ x\rightarrow 0+ }{ x } \sin \cfrac { 1 }{ x } =0$

C)

$f\left( x \right) =\left| x+1 \right| +\sin x$

$f'\left( 0+ \right) =\lim _{ x\rightarrow 0+ }{ \cfrac { x+1+\sin x }{ x } } =\lim _{ x\rightarrow 0+ }{ \cfrac { 1+0+\cos { x } }{ 1 } } =2$

D)

$f'\left( 0- \right) =\lim _{ x\rightarrow 0- }{ \cfrac { 1-x-x+x+1 }{ x } } =\lim _{ x\rightarrow 0- }{ \cfrac { -1 }{ 1 } } =-1$

Extending the functions in col. 1 as $\lim _{ x\rightarrow \pi }{ f(x)}$

$\lim _{ x\rightarrow \pi } f\left( x \right) =\lim _{ x\rightarrow \pi } \dfrac { 1-\cos (7(x-\pi )) }{ x-\pi } =\lim _{ x\rightarrow \pi } \dfrac { 2\sin ^{ 2 } ((7/2)(\pi -x)) }{ ((7/2)(\pi -x))^{ 2 } } .\dfrac { 49 }{ 4 } (\pi -x)=\left( 2 \right) \left( 1 \right) \left( \dfrac { 49 }{ 4 } (\pi -\pi ) \right) =0$

$\Rightarrow f(\pi )=0$

B)

$\lim _{ x\rightarrow \pi } f\left( x \right) =\lim _{ x\rightarrow \pi } \dfrac { 1-\cos (7(x-\pi )) }{ (x-\pi )^{ 2 } } =\lim _{ x\rightarrow \pi } \dfrac { 2\sin ^{ 2 } ((7/2)(\pi -x)) }{ ((7/2)(\pi -x))^{ 2 } } .\dfrac { 49 }{ 4 } =\left( 2 \right) \left( 1 \right) \left( \dfrac { 49 }{ 4 } \right) =\dfrac { 49 }{ 2 }$

$\Rightarrow f(\pi )=\cfrac { 49 }{ 2 }$

C)

$\lim _{ x\rightarrow \pi } f\left( x \right) =\lim _{ x\rightarrow \pi } \dfrac { \sin x }{ 1-x^{ 2 }/\pi ^{ 2 } } =\lim _{ x\rightarrow \pi } \dfrac { \sin (\pi -x) }{ (\pi -x)(\pi +x) } .\pi ^{ 2 }=\cfrac { \pi ^{ 2 } }{ 2\pi } =\cfrac { \pi }{ 2 }$

$\Rightarrow f(\pi )=\cfrac { \pi }{ 2 }$

D)

$\lim _{ x\rightarrow \pi } f\left( x \right) =\lim _{ x\rightarrow \pi } \dfrac { \sin 7x }{ \sin 2x } =\lim _{ u\rightarrow 0 } \dfrac { \sin 7(\pi -u) }{ \sin 2(\pi -u) } =\lim _{ u\rightarrow 0 } \dfrac { -\sin 7u }{ \sin 2u } =\dfrac { -7 }{ 2 }$

$\Rightarrow f(\pi )=\dfrac { -7 }{ 2 }$

Let $\displaystyle f(x)=x+\dfrac {1}{2x+\dfrac {1}{2x+\dfrac {1}{2x+.....\infty}}}$
Compute the value of $f(100).f'(100).$

Let

$\displaystyle y = f(x)=x+\dfrac {1}{2x+\dfrac {1}{2x+\dfrac {1}{2x+.....\infty}}}=x+\dfrac{1}{x+y}$

$\Rightarrow \displaystyle y(x+y)=x(x+y)+1\Rightarrow y^2 =x^2+1$
Differentiating both side w.r.t

$x$ , we get

$2y\cfrac{dy}{dx}=2x\Rightarrow f(x).f'(x) = x\therefore f(100).f'(100)=100$

If the function $f(x)$ satisfies $\displaystyle \lim_{x\rightarrow 1}\frac{f\left ( x \right )-2}{x^{2}-1}=\pi$ evaluate $\displaystyle \lim_{x\rightarrow 1}f\left ( x \right )$

$\displaystyle \lim_{x\rightarrow 1}\frac{f\left ( x \right )-2}{x^{2}-1}=\pi$

$\displaystyle \Rightarrow \frac{\lim_{x\rightarrow 1}\left ( f\left ( x \right )-2 \right )}{\lim_{x\rightarrow 1}\left ( x^{2} -1\right )}=\pi$

$\displaystyle \Rightarrow \lim_{x\rightarrow 1}\left ( f\left ( x \right )-2 \right )=\pi \lim_{x\rightarrow 1}\left ( x^{2}-1 \right )$

$\displaystyle \Rightarrow \lim_{x\rightarrow 1}\left ( f\left ( x \right )-2 \right )=\pi \left ( 1^{2}-1 \right )$

$\displaystyle \Rightarrow \lim_{x\rightarrow 1}\left ( f\left ( x \right )-2 \right )=0$

$\displaystyle \Rightarrow \lim_{x\rightarrow 1}f\left ( x \right )-\lim_{x\rightarrow 1}2=0$

$\displaystyle \Rightarrow \lim_{x\rightarrow 1}f\left ( x \right )-2=0$

$\displaystyle \therefore \lim_{x\rightarrow 1}f\left ( x \right )=2$

Evaluate the Given limit: $\displaystyle \lim_{x\rightarrow 0}\frac{\sin \,ax}{bx}$

$\displaystyle \lim_{x\rightarrow 0}\frac{\sin \,ax}{bx}=\lim_{x\rightarrow 0}\frac{\sin \,ax}{ax}\times \frac{ax}{bx}$

$=\displaystyle \lim_{x\rightarrow 0}\left ( \frac{\sin \,ax}{ax} \right )\times \left ( \frac{a}{b} \right )$

$=\displaystyle \frac{a}{b}\lim_{x\rightarrow 0}\left ( \frac{\sin \,ax}{ax} \right )$

$=\displaystyle \frac{a}{b}\times 1 = \frac{a}{b}$

Find the derivative of the following functions: $\displaystyle 2\tan x-7\sec x$

Let

$\displaystyle f\left ( x \right )=2\tan x-7\sec x$
Thus using first principle,

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

$\displaystyle \lim_{h\rightarrow 0}\frac{1}{h}\left [ 2\tan \left ( x+h \right )-7\sec \left ( x+h \right )-2\tan x+7\sec x \right ]$
=

$\displaystyle \lim_{h\rightarrow 0}\frac{1}{h}\left [ 2\left \{ \tan \left ( x+h \right )-\tan x \right \} -7\left \{ \sec \left ( x+h \right )-\sec x \right \}\right ]$
=

$\displaystyle 2\lim_{h\rightarrow 0}\frac{1}{h}\left [ \tan \left ( x+h \right )-\tan x \right ]-7\lim_{h\rightarrow 0}\frac{1}{h}\left [ \sec \left ( x+h \right )-\sec x \right ]$
=

$\displaystyle 2\lim_{h\rightarrow 0}\frac{1}{h}\left [ \frac{\sin \left ( x+h \right )}{\cos \left ( x+h \right )}-\frac{\sin x}{\cos x} \right ]-7\lim_{h\rightarrow 0}\frac{1}{h}\left [ \frac{1}{\cos \left ( x+h \right )} -\frac{1}{\cos x}\right ]$
=

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

$\displaystyle -7\lim_{h\rightarrow 0}\frac{1}{h}\left [ \frac{\cos x-\cos \left ( x+h \right )}{\cos x\cos \left ( x+h \right )} \right ]$
=

$\displaystyle 2\lim_{h\rightarrow 0}\frac{1}{h}\left [ \frac{\sin \left ( x+h-x \right )}{\cos x\cos \left ( x+h \right )} \right ]-7\lim_{h\rightarrow 0}\frac{1}{h}$

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

$\displaystyle 2\lim_{h\rightarrow 0}\left [ \left ( \frac{\sin h}{h} \right ) \frac{1}{\cos x\cos \left ( x+h \right )}\right ]-7\lim_{h\rightarrow 0}\frac{1}{h}$

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

$\displaystyle 2\left ( \lim_{h\rightarrow 0}\frac{\sin h}{h} \right )\left ( \lim_{h\rightarrow 0}\frac{1}{\cos x\cos \left ( x+h \right )} \right )-7\left ( \lim_{\frac{h}{2}\rightarrow 0}\frac{\sin \frac{h}{2}}{\frac{h}{2}} \right )$

$\displaystyle \left ( \lim_{h\rightarrow 0}\frac{\sin \left ( \frac{2x+h}{2} \right )}{\cos x\cos \left ( x+h \right )} \right )$
= 2.1.

$\displaystyle \frac{1}{\cos x\cos x}-7\cdot 1\left ( \frac{\sin x}{\cos x\cos x} \right )$
=

$\displaystyle 2\sec ^{2}x-7\sec x\tan x$

For some constant $a$ and $b$ , find the derivative of the following functions:
$(ax^{2} + b)^{2}$ .

Let

$\displaystyle f\left ( x \right )=\left ( ax^{2}+b \right )^{2}$

$\displaystyle \Rightarrow f\left ( x \right )=a^{2}x^{4}+2abx^{2}+b^{2}$

$\displaystyle \therefore f^{'}\left ( x \right )=\frac{d}{dx}\left ( a^{2} x^{4}+2abx^{2}\right )=a^{2}\frac{d}{dx}\left ( x^{4} \right )+2ab\frac{d}{dx}\left ( x^{2} \right )+\frac{d}{dx}\left ( b^{2} \right )$

$\displaystyle =a^{2}\left ( 4x^{3} \right )+2ab\left ( 2x \right )+b^{2}\left ( 0 \right )$

$=\displaystyle 4a^{2x^{3}}+4abx= \displaystyle 4ax\left ( ax^{2}+b \right )$

Find the derivatives of the following:
$\sec x$ .

Let

$f (x) = \displaystyle \sec x$
Thus using first principle,

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

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

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

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

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

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

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

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

$\displaystyle \frac{1}{\cos x}\cdot 1\cdot \frac{\sin x}{\cos x}= \sec x\tan x$

Find $\displaystyle \frac{dy}{dx}$ , if $\displaystyle y = \sin^{-1} x + \sin^{-1} \sqrt{1 - x^2} , - 1 \leq t \leq 1$

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

$\displaystyle \therefore \frac{dy}{dx} = \frac{d}{dx} [ \sin^{-1} x + \sin^{-1} \sqrt{1 - x^2} ]$

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{d}{dx} (\sin^{-1} x) + \frac{d}{dx} (\sin^{-1} \sqrt{1 - x^2} )$

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} + \frac{1}{\sqrt{1 - (\sqrt{1 - x^2})^2} } . \frac{d}{dx} (\sqrt{1 - x^2} )$

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} + \frac{1}{x} . \frac{1}{2 \sqrt{1 - x^2}} . \frac{d}{dx} (1 - x^2 )$

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

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} - \frac{1}{\sqrt{1 - x^2}}=0$

Find the derivatives of the following:
$3\cot x + 5 cosec x$ .

Let

$\displaystyle f\left ( x \right )=3\cot x+5\cos ec\,x$
Thus using first principle

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

$\displaystyle \lim_{h\rightarrow 0}\frac{3\cot \left ( x+h \right )+5\cos ec\left ( x+h \right )-3\cot x-5\cos ec\,x}{h}$
=

$\displaystyle 3\lim_{h\rightarrow 0}\frac{1}{h}\left [ \cot \left ( x+h \right ) -\cot x\right ]+5\lim_{h\rightarrow 0}\frac{1}{h}\left [ \cos ec\left ( x+h \right )-\cos ec\,x \right ]$ .....(1)
Now

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

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

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

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

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

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

$= -1\displaystyle \frac{1}{\sin x\cdot \sin \left ( x+0 \right )}=\frac{-1}{\sin ^{2}x}=-\cos ec^{2}x$ ...(2)

And

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

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

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

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

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

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

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

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

$\displaystyle -\cos ecx\,\cot \,x$ ...(3)
From (1), (2) and (3) we obtain

$\displaystyle f^{'}\left ( x \right )=-3\cos ec^{2}x-5\cos ec\,x\cot \,x$

Find $\dfrac{dy}{dx}$ if $x=a(\cos \theta + \theta \sin \theta)$ . and $y=a(\sin \theta - \theta \cos \theta)$ .

Given

$x=a(cos \theta+ \theta \sin \theta)$ and

$y=a(sin \theta- \theta \cos \theta)$

We need to find

$\dfrac{dy}{dx}$

$\dfrac{dy}{d \theta} = a(cos \theta-cos \theta +\theta sin \theta)=a( \theta sin \theta)$

Similarly , we get

$\dfrac{dx}{d \theta} = a(-sin \theta+sin \theta +\theta cos \theta)=a( \theta cos \theta)$

$\implies \dfrac{dy}{dx}=\dfrac{a(\theta sin \theta)}{a(\theta cos \theta)}=tan \theta$

Therefore

$\dfrac{dy}{dx}=tan \theta$

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

Given :

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

$(1)$

Now differentiate the given expression

$(1)$ with respective to

$x$ , we get

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

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

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

Show that when $n$ is infinite the limit of $nx^{n}$ tends to $0$ , when $x> 1$ .

Let

$x = \dfrac{1}{y}$ so that

$y > 1$

Also, let

$y^{n} = z$

$\Rightarrow n\log y = \log z$

Then,

$nx^{n} = \dfrac{n}{y^{n}} =\dfrac{1}{z}.\dfrac{\log z}{\log y} =\dfrac{1}{\log y} .\dfrac{\log z}{z}$

Now, when

$n$ is infinite,

$z$ is infinite and

$\dfrac{\log z}{z}=0$

Also,

$\log y$ is finite

Therefore,

$\displaystyle \lim_{n \rightarrow \infty} nx^{n} =0$

If $y=\sin ^{ -1 }{ (3x) } +\sec ^{ -1 }{ \left( \cfrac { 1 }{ 3x } \right) }$ , find $\cfrac { dy }{ dx }$

Given equation

$y=\sin ^{-1}(3x)+\sec ^{-1}\left(\dfrac 1{3x}\right)$

$\implies y=\sin ^{-1}(3x)+\cos ^{-1}(3x)$

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

$\implies y=\dfrac \pi 2$

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

If $x = a sin$ $\theta$ $+ b cos$ $\theta$ , $y= a cos$ $\theta$ $- b sin$ $\theta$ ,
then show that $(ax + ay)^2$ + $(bx - ay)^2$ = $(a^2 + b^2)^2$ .

Find the values of

$cos \theta$ and

$sin \theta$ .

$x=a\sin \theta + b \cos \theta$ ...(i)

$y = a\cos \theta - b \sin \theta$ ...(ii)

Multiplying eq.(i) by b and eq. (ii) by a

$bx = ab\sin \theta + b^2\cos \theta$

$ay = -ab\sin \theta + a^2 \cos \theta$

Adding resulting equations,

$\dfrac{bx+ay}{b^2+a^2} = \cos \theta$

similarly,

$\sin \theta = \dfrac{ax-by}{b^2+a^2}$

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

$(ax-by)^2 + (bx+ay)^2 = (a^2+b^2)^2$

Solving above equation will give:

$\Rightarrow$

$cos\theta = \dfrac {bx+ay}{b^{2} +a^{2}}$ and

$\Rightarrow$

$sin\theta=\dfrac{ax-by}{a^{2} + b^{2}}$

Squaring and adding both will give -

$\Rightarrow$

$(ax-by)^{2} + (bx+ay)^{2} = (a^{2} + b^{2})^{2}$

$\displaystyle \lim _{ x\rightarrow 1 }{ { \left( 2-x+a\left[ x-1 \right] +b\left[ 1+x \right] \right) } }$ =exists
So find $a,b$

$\displaystyle \lim _{ x\rightarrow 1^{ + } }{ \left( 2-x+a\left[ x-1 \right] +b\left[ 1+x \right] \right) }$

$=2-1+a\cdot 0+b(2)$

$x\rightarrow 1^{ + }$

$1>\left[ x-1 \right] >0$

$3>\left[ 1+x \right] >2$

$=2b+1$

$\displaystyle \lim _{ x\rightarrow 1^{ - } }{ \left( 2-x+a\left[ x-1 \right] +b\left[ 1+x \right] \right) }$

$=2-1+a(-1)+b(1)$

$x\rightarrow 1^{ - }$

$-1<\left[ x-1 \right] <0$

$2>\left[ 1+x \right] >1$

$=1-a+b$

Given limit exists at

$x=1$

$\Rightarrow$ LHL

$=$ RHL

$=f\left( 1 \right) =1$

$\Rightarrow 2b+1=1\Rightarrow b=0$

$1-a+b=1\Rightarrow a=0$

Find the following limit:
$\displaystyle \lim_{x \, \rightarrow \, -2 } \, \frac{x^4 \, + \, 5x^3 \, + \, 6x^2}{x^2 \, - \, 3x \, - \, 10}$

$\displaystyle \lim_{x\rightarrow - 2} \dfrac {x^{4} + 5x^{3} + 6x^{2}}{x^{2} - 3x - 10}$

$= \displaystyle \lim_{x\rightarrow - 2} \dfrac {x^{2}(x^{2} + 5x + 6)}{x^{2} - 3x - 10}$ By using factorization method

$= \displaystyle \lim_{x\rightarrow -2} \dfrac {x^{2} (x + 3)(x + 3)}{x^{2} - 5x + 2x - 10}$

$= \displaystyle \lim_{x \rightarrow - 2} \dfrac {x^{2}(x - 5)(x + 3)}{x(x - 5) + 2(x - 5)}$

$= \displaystyle \lim_{x\rightarrow -2} \dfrac {x^{2} (x + 3)(x + 2)}{(x + 2)(x - 5)}$

$= \displaystyle \lim_{x \rightarrow -2} \dfrac {x^{2} (x + 3)}{(x - 5)}$

$= \dfrac {4\times 1}{-2 - 5} = \dfrac {-4}{7}$ .

Calculate the following limits.
$\displaystyle \lim_{n \, \rightarrow \, \infty} \, \frac{n^2}{1 \, - \, 9n^2}, \, n \, \epsilon \, N.$

$\lim\limits_{n\to\infty}\dfrac{n^2}{1-9n^2}$

$=\lim\limits_{n\to\infty}\dfrac{1}{\dfrac{1}{n^2}-9}$

$=\dfrac{1}{\dfrac{1}{\infty}-9}$

$=-9$

Find the derivative of $\displaystyle\, y \, =\, (2 \, -\, x^2)\, \cos\, x \, +\, 2x \, \sin\, x$

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

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

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

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

$= x^{2} \sin x$ .

Find the following limit:
$\displaystyle \lim_{x \, \rightarrow \, 1 } \, \left [ \, \left (\frac{x^3 \, - \, 4x}{x^3 \, - \, 8} \right )^{-1} \, - \, \left (\frac{x \, + \, \sqrt{2x}}{x \, - \, 2} \, - \, \frac{\sqrt{2}}{\sqrt{x} \, - \, \sqrt{2}} \right )^{-1} \, \right]$

$\displaystyle \lim_{x\to 1}(\dfrac{x(x-2)(x+2)}{(x-2)(x^2+4x+4)})^{-1}-(\dfrac{\sqrt{x}(\sqrt{x}+\sqrt{2})}{(\sqrt{x}-\sqrt{2})(\sqrt{x}+\sqrt{2})}-\dfrac{\sqrt{2}}{\sqrt{x}-\sqrt{2}})^{-1}$

$=\displaystyle \lim_{x\to 1}(\dfrac{x(x+2)}{x^2+4x+4})^{-1}-1$

$=\displaystyle \lim_{x\to 1}(\dfrac{x}{x+2})^{-1}-1$

$=3-1=2$

Evaluate $\displaystyle \lim_{x\rightarrow 0} \dfrac {a^{\sin x} - 1}{\sin x}$ .

$\displaystyle \lim_{x\rightarrow 0} \dfrac {a^{\sin x} - 1}{\sin x}$ .

We have

$\displaystyle \lim_{x\rightarrow 0} \dfrac {a^{x} - 1}{x}={\log}_{e}{a}$ .

Using the above we have

$\displaystyle \lim_{x\rightarrow 0} \dfrac {a^{\sin x} - 1}{\sin x}={\log}_{e}{a}$ .

Find the following limit:
$\displaystyle \lim_{x \, \rightarrow \, \infty } \, \frac{x^2 \, + \, 3x \, - \, 4}{1 \, - \, 3x^2}.$

$\displaystyle \lim_{x\rightarrow \infty} \dfrac {x^{2} + 3x - 4}{1 - 5x^{2}}$

$= \displaystyle \lim_{x\rightarrow \infty} \dfrac {x^{2}\left (1 + \dfrac {3}{x} - \dfrac {4}{x^{2}}\right )}{x^{2} \left (\dfrac {1}{x^{2}} - 5\right )}$

$= \dfrac {1 + \dfrac {3}{\infty} - \dfrac {4}{\infty^{2}}}{\dfrac {1}{\infty^{2}} - 5} \left [\because \dfrac {1}{\infty} = 0\right ]$

$= \dfrac {1 + 0 - 0}{0 - 5} = \dfrac {-1}{5}$ .

Find the derivative of $\displaystyle\, y \, =\, \frac{1}{x} \, +\, \frac{1}{x^2} \, +\, \frac{3}{x^3}$

$y=\dfrac{1}{x}+\dfrac{1}{x^2}+\dfrac{3}{x^3}$

$\dfrac{dy}{dx}=-\dfrac{1}{x^2}-\dfrac{2}{x^3}-\dfrac{9}{x^4}$

Find the following limit.
$\displaystyle \lim_{x \, \rightarrow \, 1 } \, \frac{x^4 \, - \, 2x^2 \, + \, 1}{x^3 \, - \, 1}.$

$\displaystyle \lim_{x\rightarrow 1} \dfrac {x^{4} - 2x^{2} + 1}{x^{3} - 1}$

By using factorization method

$\displaystyle \lim_{x\rightarrow 1} \dfrac {(x^{2} - 1)^{2}}{(x - 1)(x^{2} + x + 1)}$ [here

$(x^{2} - 1)^{2} = x^{4} - 2x^{2} - 1\ x^{3} - 1^{3} = (x - 1)(x^{2} + x + 1)]$

$\displaystyle \lim_{x\rightarrow 1} \dfrac {[(x + 1)(x - 1)]^{2}}{(x - 1)(x^{2} + x + 1)}$

$\displaystyle \lim_{x\rightarrow 1} \dfrac {(x + 1)^{2} (x - 1)^{2}}{(x - 1)(x^{2} + x + 1)}$

$\displaystyle \lim_{x\rightarrow 1} \dfrac {(x + 1)^{2} (x - 1)}{x^{2} + x + 1}$

$= \dfrac {(1 + 1)^{2}(0)}{1 + 1 + 1} = 0$ .

Find the following limit.
$\displaystyle \lim_{x \, \rightarrow \, 0 } \, \frac{(1 \, + \, x) \, (x \, + \, 2x) \, (1 \, + \, 3x) \, - \, 1}{x}.$

$\displaystyle \lim_{x\rightarrow 0} \dfrac {(1 + x)(1 + 2x)(1 + 3x) - 1}{x}$

$= \displaystyle \lim_{x\rightarrow 0} \dfrac {(1 + 2x + x + 2x^{2})(1 + 3x) - 1}{x}$

$= \displaystyle \lim_{x\rightarrow 0} \dfrac {(1 + 3x + 2x^{2})(1 + 3x) - 1}{x}$

$= \displaystyle \lim_{x\rightarrow 0}\dfrac {(x + 3x + 2x^{2} + 3x + 9x^{2} + 6x^{3}) - 1}{x}$

$= \displaystyle \lim_{x \rightarrow 0} \dfrac {6x^{2} + 11x^{2} + 6x}{x}$

$\Rightarrow \displaystyle \dfrac {6x (6x^{2} + 11x + 6)}{x}$

$= 6$ .

Find the value of $x$ for which the derivative of the function $\displaystyle\, f(x) = 20\, \cos \, 3x + 12\, \cos\, 5x - 15\, \cos\, 4x$ is equal to zero?

$f(x) = 20\cos 3x + 12\cos 5x - 15\cos 4x$

$f'(x) = 20(-\sin 3x)(3) + 12(-\sin 5x)(5) + 15 \sin 4x (4)$

$= -60 \sin 3x - 60\sin 5x + 60\sin 4x$

$= 60 [\sin 3x - \sin 3x - \sin 5x] = 0$

$\Rightarrow \sin 4x = \sin 3x + \sin 5x$

$\Rightarrow \sin 4x = 2\sin \dfrac {8x}{2} \cos \dfrac {2x}{2}$

$\sin 4x = 2\sin 4x \cos x$

$\sin 4x (1 - 2\cos x) = 0$

$\therefore \sin 4x = 0$ and

$1 - 2\cos x = 0$

$x = \dfrac {\pi m}{4}$

and

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

$x = \pm \dfrac {\pi}{3}, \dfrac {-5\pi}{3}, \dfrac {7\pi}{4} ...$

$x = \pi \left [\dfrac {6n + 1}{3}\right ]$

where

$m, n\epsilon \mathbb {Z}$ (integers).

$\therefore x \, = \, {\dfrac{\pi m}{4}, \, \dfrac{\pi(6n \, + \, 1)}{3}, \, | \, m, \, n \, \epsilon Z}$

Find the slope of the tangent to the curve $y = x^{3} - x$ at $x = 2$ .

$y = x^{3} - x$ at

$x = 2$
Diff wrt

$x$

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

$= \dfrac {dy}{dx}|_{x = 2} = 3(2^{2}) - 1$

$= 12 - 1$

$\therefore$ Slope of tangent

$= 11$ .

Find the derivative of $\sin {x}$ with respect to $x$ from first principles.

First principle of differentiation :

$\dfrac {dy} {dx} =\lim_{\delta x\rightarrow 0} \dfrac {f(x+\delta x) - f(x)} {\delta x}$

Here

$f(x) =\sin x$

$\Rightarrow f(x+\delta x) =\sin(x+\delta x)$

$\Rightarrow f(x+\delta x) - f(x) =\sin(x+\delta x) - \sin x$

We know that

$\text {sin} C-\text {sin} D=2\text {cos} (\dfrac {C+D} {2})\text {sin} (\dfrac {C-D} {2})$

$\Rightarrow f(x+\delta x) - f(x) =2\cos(\dfrac {x+\delta x+x}{2})\sin(\dfrac {\delta x} {2})$

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

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

$\Rightarrow \dfrac {d(\sin x)} {dx} =\cos x$ as

$\lim_{x\rightarrow 0} \dfrac {\sin x} {x} =1$

Find the derivatives of the following functions.
(a) $\cot^3 \, x$ (b) $\sin \, \sqrt{x}$

$(a)\dfrac{d}{dx}\cot^3x$

Let,

$z=\cot{x}\implies \dfrac{dz}{dx}=-\text{cosec}^2x$

$=\dfrac{dz^3}{dx}=\dfrac{dz^3}{dz}\dfrac{dz}{dx}$

$=-3z^2\text{cosec}^2x$

$=-3\cot^2x\text{cosec}^2x$

$(b)\dfrac{d}{dx}\sin\sqrt{x}$

Let

$z=\sqrt{x}\implies \dfrac{dz}{dx}=\dfrac{1}{2\sqrt{x}}$

$=\dfrac{d\sin z}{dx}=\dfrac{d\sin z}{dz}\dfrac{dz}{dx}$

$=\cos z\dfrac{1}{2\sqrt{x}}$

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

Find the derivatives of the following functions at the indicated points.
$\displaystyle\, f(x) \, = \, sin \, 4x \, cos \, 4x, \, f' \, (\pi/3) \, = \, ?$

$f(x)=\sin4x\cos4x=\dfrac{2\sin4x\cos4x}{2}$

$\implies f(x)=\dfrac{1}{2}\sin8x$

$\implies f'(x)=\dfrac{1}{2}.8\cos8x=4\cos8x$

$\implies f'(\pi/3)=4\cos(8\pi/3)=4\cos(2\pi+2\pi/3)$

$\implies f'(\pi/3)=4\cos(2\pi/3)=4.\dfrac{-1}{2}$

$\implies f'(\pi/3)=-2$

Find the derivative of
$f(x)=(x^2-5)(x^3-2x+3)$

Given:

$f(x)=(x^2-5)(x^3-2x+3)$

Using product rule:

$f'(x)=2x(x^3-2x+3)+(x^2-5)(3x^2-2)=5x^4-21x^2+6x+10$ .

Find $\dfrac{dy}{dx}$ when $x$ and $y$ are connected by the relation given:
$\sin (xy)+\dfrac{x}{y}=x^2-y$

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

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

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

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

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

$(3x^4-x^3+4)^{5/2}$ differentiate w.r.t x.

Let

$y=(3x^{4}-x^{3}+4)^{\frac{5}{2}}$

Differentiate on both sides w.r.t x

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{5}{2}(3x^{4}-x^{3}+4)^{\frac{3}{2}}*\frac{\mathrm{d} }{\mathrm{d} x}(3x^{4}-x^{3}+4)$

$=\frac{5}{2}(3x^{4}-x^{3}+4)^{\frac{3}{2}}*(12x^{3}-3x^{2})$

$\therefore \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{15x^{2}}{2}(4x-1)(3x^{4}-x^{3}+4)^{\frac{3}{2}}$

Differentiate
$3x^{1/3} + \frac{6}{7} x^{7/6} +3x^{2/3} +C$

Let

$y=3{ x }^{ 1/3 }+\cfrac { 6 }{ 7 } { x }^{ 7/6 }+3{ x }^{ 2/3 }+C$

$\cfrac { d }{ dx } \left( { x }^{ n } \right) =n{ x }^{ n-1 }$

$\therefore \cfrac { dy }{ dx } =\cfrac { d }{ dx } \left( 3{ x }^{ 1/3 } \right) +\cfrac { d }{ dx } \left( \cfrac { 6 }{ 7 } { x }^{ 7/6 } \right) +\cfrac { d }{ dx } \left( 3{ x }^{ 2/3 } \right) +0$ (

$\because C$ is constant)

$=3.\cfrac { 1 }{ 3 } { x }^{ -2/3 }+\cfrac { 6 }{ 7 } .\cfrac { 7 }{ 6 } { x }^{ 1/6 }+3.\cfrac { 2 }{ 3 } { x }^{ -1/3 }$

$\therefore \cfrac { dy }{ dx } ={ x }^{ -2/3 }+{ x }^{ 1/6 }+2{ x }^{ -1/3 }$

solve: $x - \ sin x + C$

Let

$y=x-\sin { x } +C$

$\cfrac { dy }{ dx } =\cfrac { d }{ dx } \left( x-\sin { x } +C \right)$

$=1-\cos { x } +0$ (

$\because C$ is constant)

$\therefore \cfrac { dy }{ dx } =1-\cos { x }$

Differentiate
$\frac{2}{3} x^{3} + \frac{3}{2} x^{2} +C$

Let

$y=\cfrac { 2 }{ 3 } { x }^{ 3 }+\cfrac { 3 }{ 2 } { x }^{ 2 }+C$

$\cfrac { d }{ dx } \left( { x }^{ n } \right) =n{ x }^{ n-1 }$

$\therefore \cfrac { dy }{ dx } =\cfrac { d }{ dx } \left( \cfrac { 2 }{ 3 } { x }^{ 3 } \right) +\cfrac { d }{ dx } \left( \cfrac { 3 }{ 2 } { x }^{ 2 } \right) +0$ (

$\because C$ is constant )

$=\cfrac { 2 }{ 3 } .3{ x }^{ 2 }+\cfrac { 3 }{ 2 } .2x$

$\therefore \cfrac { dy }{ dx } =2{ x }^{ 2 }+3x$

Differentiate
$\sqrt{2x^{3/2} - 5x^{4/5}}$

Let

$y=\sqrt{2x^{\frac{3}{2}}-5x^{\frac{4}{5}}}$

Differentiate on both sides w.r.t x

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{2\sqrt{2x^{\frac{3}{2}}-5x^{\frac{4}{5}}}}\frac{\mathrm{d} }{\mathrm{d} x}(2x^{\frac{3}{2}}-5x^{\frac{4}{5}})$

$=\frac{2*\frac{3}{2}x^{\frac{1}{2}}-5*\frac{4}{5}x^{\frac{1}{5}}}{2\sqrt{2x^{\frac{3}{2}}-5x^{\frac{4}{5}}}}$

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{3x^{\frac{1}{2}}-4x^{\frac{-1}{5}}}{2\sqrt{2x^{\frac{3}{2}}-5x^{\frac{4}{5}}}}$

Differentiate
$\dfrac{ x^4 }{4} - \dfrac{x^{-3}}{3} - \dfrac{2}{x} +C$

Let

$y=\cfrac { { x }^{ 4 } }{ 4 } -\cfrac { { x }^{ -3 } }{ 3 } -\cfrac { 2 }{ x } +C$

$\cfrac { d }{ dx } \left( { x }^{ n } \right) =n{ x }^{ n-1 }$

$\therefore \cfrac { dy }{ dx } =\cfrac { d }{ dx } \left( \cfrac { { x }^{ 4 } }{ 4 } \right) -\cfrac { d }{ dx } \left( \cfrac { { x }^{ -3 } }{ 3 } \right) -\cfrac { d }{ dx } \left( \cfrac { 2 }{ x } \right) +0$ (

$\because C$ is constant)

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

$={ x }^{ 3 }+{ x }^{ -4 }+2{ x }^{ -2 }$

$\therefore \cfrac { dy }{ dx } ={ x }^{ 3 }+\cfrac { 1 }{ { x }^{ 4 } } +\cfrac { 2 }{ { x }^{ 2 } }$

(a) Differentiate $y = {\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$ with respect to $x,0<x<1$ ,
(b) Differentiate ${x^x} - {2^{\sin x}}$ with respect to $x$

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

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

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

$=\dfrac{-1}{\sqrt{1+{x}^{4}+2{x}^{2}-1-{x}^{4}+2{x}^{2}}}\left[\dfrac{2x\left(-1-{x}^{2}-1+{x}^{2}\right)}{1+{x}^{2}}\right]$

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

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

$(b)y={x}^{x}-{2}^{\sin{x}}$

Let

$u={x}^{x}$ and

$v={2}^{\sin{x}}$

$\Rightarrow\,\dfrac{dy}{dx}=\dfrac{du}{dx}-\dfrac{dv}{dx}$

We have

$u={x}^{x}$ and

$v={2}^{\sin{x}}$

$\Rightarrow\,\log{u}=\log{{x}^{x}}$ and

$\log{v}=\log{{2}^{\sin{x}}}$

$\Rightarrow\,\log{u}=x\log{x}$ and

$\log{v}=\sin{x}\log{2}=\log{2}\sin{x}$

$\Rightarrow\,\dfrac{1}{u}\dfrac{du}{dx}=x\times\dfrac{1}{x}+\log{x}$ and

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

$\Rightarrow\,\dfrac{1}{u}\dfrac{du}{dx}=1+\log{x}$ and

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

$\Rightarrow\,\dfrac{du}{dx}=u\left(1+\log{x}\right)$ and

$\dfrac{dv}{dx}=v\left(\log{2}\cos{x}\right)$

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

$\dfrac{dv}{dx}={2}^{\sin{x}}\left(\log{2}\cos{x}\right)$ where

$u={x}^{x}$ and

$v={2}^{\sin{x}}$

$\therefore\,\dfrac{dy}{dx}=\dfrac{du}{dx}-\dfrac{dv}{dx}$

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

Solve $\lim _{ x\rightarrow a }{ \{ \dfrac { { (x+2) }^{ 5/3 }-{ (a+2) }^{ 5/3 } }{ x-a } } \}$

Formula:

$\lim_{x\rightarrow a} {\{\dfrac{x^{n} - a^{n}}{x - a}}\} = a^{n-1}$

$\lim_{x\rightarrow a} {\{\dfrac{(x+2)^{\dfrac{5}{3}} - (a+2)^{\dfrac{5}{3}}}{x - a}}\}$

$= \lim_{x\rightarrow a} {\{\dfrac{(x+2)^{\dfrac{5}{3}} - (a+2)^{\dfrac{5}{3}}}{(x + 2) - (a + 2)}}\}$

$= (a+2)^{\dfrac{5}{3} - 1}$

$= (a+2)^{\dfrac{2}{3}}$ (Ans)

The value of $\lim\limits_{x\to 0}\left[\dfrac{\tan x}{x}+\dfrac{\sin x}{x}\right]$ is

We know that,

$\lim\limits_{x\to0}\dfrac{\tan x}{x}=1$ and

$\lim\limits_{x\to0}\dfrac{\sin x}{x}=1$

Hence,

$\lim\limits_{x\to 0}\left[\dfrac{\tan x}{x}+\dfrac{\sin x}{x}\right]=1+1=2$

$y={e}^{x}+{e}^{-x}$ prove that $\dfrac { dy }{ dx } =\sqrt { { y }^{ 2 }-4 }$

$\displaystyle \frac{dy}{dx}=e^x-e^{-x}$

and

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

$y^2-4=e^{2x}+e^{-2x}-2$

$\sqrt{y^2-4}=|e^x-e^{-x}|$

and we know that for

$\displaystyle x>0 \sqrt{y^2-4}=e^x-e^{-x}=\frac{dy}{dx}$

and for

$\displaystyle x<0 \sqrt{y^2-4}=-(e^x-e^{-x})=\frac{-dy}{dx}$

$\lim _{ n\rightarrow \infty }{ \dfrac { \cos { x } +\cos { 3x } +\cos { 5x } +\cos { \left( 2n-1 \right) } x }{ n } }$ where $x\neq k\pi ,k\epsilon$

$\displaystyle \lim{n\rightarrow \infty}\left[\dfrac{\cos(x)+\cos(3x)+\cos(5x)+\cos(2n-1)x}{n}\right]$

Apply the sequence theorem

$-1\le \cos (2n-1)\le 1$

$\displaystyle \lim{n\rightarrow \infty}\left(\dfrac{\cos(x)+\cos(3x)+\cos(5x)+(-1)x}{n})\le \displaystyle \lim{n\rightarrow \infty}(\dfrac{\cos(x)+\cos(3x)+\cos(5x)+\cos(2n-1)x}{n}\right)$

$\le \underset{n\rightarrow \infty}{lim}\left [\dfrac{\cos(x)+\cos(3x)+\cos(5x)+1x}{n}\right]=0$

Apply Infinity property :

$\displaystyle \lim{x\rightarrow \infty}(\dfrac{c}{x^a}=0)$

$\displaystyle \lim {n\rightarrow \infty}\left[\dfrac{\cos(x)+\cos(3x)+\cos(5x)+1.x}{n}\right]=0$

$\therefore$ By the sequence theorem

$\displaystyle \lim{n\rightarrow \infty}\left [\dfrac{\cos(x)+\cos(3x)+\cos(5x)+\cos(2n-1)x}{n}\right]$

$=0$

Differentiate with respect to $x$
$y=\sin{2x}-4{e}^{3}x$

Let $y = \sin 2x - 4{e^3}x$

Differentiating with respect to $x$ .

$\frac{{dy}}{{dx}} = 2\cos 2x - 4{e^3}$

If $y = \dfrac{{\sin \left( {x + 9} \right)}}{{\cos x}}$ , then $\dfrac{dy}{{dx}}$ at $x = 0$ , is

$y=\dfrac{\sin{\left(x+9\right)}}{\cos{x}}$

$y=\dfrac{\sin{x}\cos{9}+\cos{x}\sin{9}}{\cos{x}}$

$y=\cos{9}\tan{x}+\sin{9}$

$\Rightarrow\,\dfrac{dy}{dx}=\cos{9}{\sec}^{2}{x}$

$\Rightarrow\,\left[\dfrac{dy}{dx}\right]_{x=0}=\cos{9}{\sec}^{2}{0}=\cos{9}$

Solve $\lim_{x \rightarrow \infty}(\sqrt{x^{2} +ax + a^{2}} - \sqrt{x^{2} + a^{2}})$ =

$\lim_{x\rightarrow \infty} (\sqrt{x^2 + ax + a^2} - \sqrt{x^2 + a^2})$

$= \lim_{x\rightarrow \infty} \left(\dfrac{(\sqrt{x^2 + ax + a^2})^2 - (\sqrt{x^2 + a^2})^2}{\sqrt{x^2 + ax + a^2} + \sqrt{x^2 + a^2}}\right)$

$= \lim_{x\rightarrow \infty} \left(\dfrac{x^2 + ax + a^2 -x^2 - a^2}{\sqrt{x^2 + ax + a^2} + \sqrt{x^2 + a^2}}\right)$

$= \lim_{x\rightarrow \infty} \left(\dfrac{ax}{\sqrt{x^2 + ax + a^2} + \sqrt{x^2 + a^2}}\right)$

$= \lim_{x\rightarrow \infty} \left(\dfrac{ax}{x\sqrt{1 + \dfrac{a}{x} + \dfrac{a^2}{x^2}} + x\sqrt{1 + \dfrac{a^2}{x^2}}}\right)$

$= \lim_{x\rightarrow \infty} \left(\dfrac{a}{\sqrt{1 + \dfrac{a}{x} + \dfrac{a^2}{x^2}} + \sqrt{1 + \dfrac{a^2}{x^2}}}\right)$

$x\rightarrow \infty$ then

$\dfrac{1}{x} \rightarrow 0$

$= \lim_{\dfrac{1}{x}\rightarrow 0} \left(\dfrac{a}{\sqrt{1 + \dfrac{a}{x} + \dfrac{a^2}{x^2}} + \sqrt{1 + \dfrac{a^2}{x^2}}}\right)$

$= \dfrac{a}{\sqrt{1 + 0 + 0} + \sqrt{1 +0}}$

$= \dfrac{a}{\sqrt{2}}$

Solve: $\underset{x \rightarrow 0}{lim}\left(\dfrac{1}{x^2} - \cot^2 x\right)$ .

$\underset{x\rightarrow 0}{lt}\left(\dfrac{1}{x^2}-\cot^2x\right)$

$\Rightarrow \underset{x\rightarrow 0}{lt}\left(\dfrac{1}{\sin^2x}-\dfrac{\cos^2x}{\sin^2x}\right)$

$\because \underset{x\rightarrow 0}{lt}\sin x=x$

$\Rightarrow \underset{x\rightarrow 0}{lt}\dfrac{1-\cos^2x}{\sin^2x}=\underset{x\rightarrow 0}{lt}\dfrac{\not{\sin^2x}}{\not{\sin^2x}}=1$ .

Solve $x=a(\theta -\sin \theta),y=a(1+\cos \theta )$ find $\dfrac{dy}{dx}$ ?

$\because \quad x=a(\theta -\sin\theta )\\ \Rightarrow \dfrac { dx }{ d\theta } =a\left( 1-\cos\theta \right) \\ and,\quad y=a\left( 1+\cos\theta \right) \\ \Rightarrow \dfrac { dy }{ d\theta } =-a\sin\theta \\ \because \dfrac { dy }{ dx } =\dfrac { \dfrac { dy }{ d\theta } }{ \dfrac { dx }{ d\theta } } =\dfrac { -a\sin\theta }{ a\left( 1-\cos\theta \right) } =\dfrac { \sin\theta }{ \left( \cos\theta -1 \right) } \\$

Examine the graph of $y=f(x)$ as shown and evaluate the following limits
(i) $\lim_\limits{x \to1}f(x)$
(ii) $\lim_\limits{x \to2}f(x)$
(iii) $\lim_\limits{x \to3}f(x)$
(iv) $\lim_\limits{x \to199}f(x)$
(v) $\lim_\limits{x \to3^+}f(x)$

$\\(1.)LHL=2RHL=3\therefore\>limit\>doesn't\>exist\>\\(2.)LHL=3RHL=3\therefore\>limit=3\\(3.)LHL=3RHL=3\therefore\>limit=3\\(4.)doesn't\>exist\>at\>graph\>not\>available\>\\(5.)RHL=3$

If ${f}_{1}(x)=\cfrac{x}{2}+10\forall x\in R$ and ${f}_{n}(x)={f}_{1}({f}_{n-1}(x))\forall n\ge 2,n \in N$ , then evaluate $\lim _{ n\rightarrow \infty }{ { f }_{ n }(x) }$

$f_1(x)=\dfrac{x}{2}+10\implies f_1^{1}(x)=\dfrac{1}{2}$

$f_n(x)=f_1(f_{n-1}(x))\implies f^{1}_n(x)=f_1^{1}(f_{n-1}(x))f^{1}_{n-1}(x)=\dfrac{f^{1}_{n-1}(x)}{2}$

$\dfrac{f^{1}_n(x)}{f^{1}_{n-1}(x)}=\dfrac{1}{2}$

$\dfrac{f^{1}_{n}(x)}{f^{1}_{n-1}(x)}\dfrac{f^{1}_{n-1}(x)}{f^{1}_{n-2}(x)}\cdots\cdots \dfrac{f^{1}_2(x)}{f^{1}_{1}(x)}=\bigg(\dfrac{1}{2}\bigg)^{n-1}\implies f^{1}_{n}(x)=2^{-n}\implies f_{n}(x)=2^{-n}x+c$

$f_{1}(x)=\dfrac{x}{2}+c=\dfrac{x}{2}+10\implies c=10$

$f_{n}(x)=2^{-n}+10$

$\displaystyle\lim_{x\rightarrow \infty}f_{n}(x)=\displaystyle\lim_{x\rightarrow \infty}2^{-n}x+10=10$

$x=e^{\theta}(\sin\theta+\cos\theta)$ , $y=e^{\theta}(\sin\theta-\cos\theta)$
Fine $\dfrac{dy}{dx}$ .

Consider the following equation.

$x={{e}^{\theta }}(sin\theta +cos\theta ),y={{e}^{\theta }}(sin\theta -cos\theta )$

Differentiate $x\& y\$ w.r.t $\theta \$

$x\& y$

$\dfrac{dx}{d\theta }={{e}^{\theta }}(sin\theta +cos\theta )+{{e}^{\theta }}(cos\theta -sin\theta )\,\,\,\,\,\,\,\,.......\left( 1 \right)$

$\dfrac{dy}{d\theta }={{e}^{\theta }}(sin\theta -cos\theta )+{{e}^{\theta }}(cos\theta +sin\theta )\,\,\,\,\,\,\,\,.......\left( 2 \right)$

Divide by eq..(2) In eq. (1)

$\dfrac{dy}{dx}=\dfrac{{{e}^{\theta }}\left( sin\theta -cos\theta +cos\theta +sin\theta \right)}{{{e}^{\theta }}\left( sin\theta +cos\theta +cos\theta -sin\theta \right)}$

$=\dfrac{2\sin\theta }{2cos\theta }$

$=\tan\theta$

Hence, this is the correct answer.

Find the derivative of $f(x)$ from the first principles, where $f(x)$ is:
$x \sin x$

Given a function f(x)=x sin(x), and we have to find its derivative by the definition. Consider the expression (f(x+Delta x)-f(x))/(Delta x) and find its limit for Delta x->0:

(f(x+Delta x)-f(x))/(Delta x)=((x+Delta x)sin(x+Delta x)-x sin(x))/(Delta x)=

=(x(sin(x+Delta x)-sin(x))+Delta x sin(x+Delta x))/(Delta x)=

=x(sin(x+Delta x)-sin(x))/(Delta x)+ sin(x+Delta x).

The second summand has the limit sin(x), because sin(x) is continuous for any

The first summand has the limit x(sin(x))'=xcos(x) (by the definition of the derivative

Differentiate $\sqrt {\dfrac{{1 - \tan x}}{{1 + \tan x}}}$

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

$y=\sqrt{\dfrac{1+\tan x}{1-(1\times tan x)}}$

$y=\sqrt{\dfrac{\tan \dfrac{\pi}{4}+\tan x}{1-\tan \dfrac{\pi}{4}\tan x}}$

$y=\sqrt{\tan\left(\dfrac{\pi}{4}+x\right)}$

$\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{\tan \left(\dfrac{\pi}{4}+x\right)}}\cdot d\left(\tan \left(\dfrac{\pi}{4}+x\right)\right)$

$=\dfrac{\sec^2\left(\dfrac{\pi}{4}+x\right)}{2\sqrt{\tan \left(\dfrac{\pi}{4}+x\right)}}\cdot d\left(\dfrac{\pi}{4}+x\right)$

$\dfrac{dy}{dx}=\dfrac{\sec^2\left(\dfrac{\pi}{4}+x\right)}{2\sqrt{\tan\left(\dfrac{\pi}{4}+x\right)}}$ .

Find the derivative of $\sin \left( {x + 1} \right)$ , with respect to $x$ , from first principle.

Let

$f(x)=\sin (x+1)$

To find

$\dfrac{d}{dx}f(x)$ by

$1^{st}$ principle

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

$=\displaystyle\lim_{h\rightarrow 0}\dfrac{\sin (x+h+1)-\sin (x+1)}{h}$

$=\displaystyle\lim_{h\rightarrow 0}\dfrac{\cos (x+h+1)\cdot 1}{1}$

$\therefore f'(x)=\cos (x+1)$

$\therefore \dfrac{d}{dx}(\sin (x+1))=\cos (x+1)$ .

Solve: $\displaystyle\lim_{x\rightarrow \dfrac {3\pi}{4}} \dfrac {2 + \tan x}{5 + 4\tan x}$ .

1372805_1127042_ans_46323656b066407d81a8418650d6d866.jpg

$\displaystyle \lim_{x \to 0} \sin \dfrac{\pi}{x}$

1356111_1129755_ans_0434f5a3a0bd4f20bd0744280918e89a.jpg

Find
$\displaystyle \lim_{x \rightarrow \dfrac {5}{2}}[x]$

1951479_1145776_ans_791df99e088c4aeabad331164e49ba7b.jpg

$\lim_{x\rightarrow o}\frac{4sin^{2(x/_{2})cos^{2}(x/_{2})}}{sin^{2}(x/_{2})}=K$

We have,

$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{4{{\sin }^{2}}\dfrac{x}{2}{{\cos }^{2}}\dfrac{x}{2}}{{{\sin }^{2}}\dfrac{x}{2}}=k$

Then,

$\underset{x\to 0}{\mathop{\lim }}\,4{{\cos }^{2}}\dfrac{x}{2}=k$

Taking limit and we get,

$k=4$

Hence, this is the answer.

$\displaystyle\lim_{x\rightarrow\infty}\dfrac{\cos x+\sin^2x}{x+1}$ .

${ \lim }_{ x\to 8 }\, \, \dfrac { { \cos x+{ { \sin }^{ 2 } }x } }{ { x+1 } } \\ { \lim }_{ x\to 8 }\, \, \dfrac { { \cos x+\dfrac { { 1-\cos 2x } }{ 2 } } }{ { x+1 } } \\ multiply\, \, both\, \, Numerator\, \, and\, \, denominator\, \, by\, \, \dfrac { 1 }{ x } \\ { \lim }_{ x\to 8 }\, \, \dfrac { { \dfrac { { \cos x } }{ x } +\dfrac { 1 }{ { 2x } } -\dfrac { { \cos 2x } }{ { 2x } } } }{ { 1+\dfrac { 1 }{ x } } } \\ \left[ { \because { \lim }_{ x\to 8 }\, \, \dfrac { { \cos x } }{ x } =0 } \right] \\ \Rightarrow \dfrac { { 0+0-0 } }{ { 1+0 } } =0 \\$

$f(x)=\sec x-\cos x,x \epsilon (0,\pi/2)$
find f'(x)

Given that,

$f\left( x \right)=\sec x-\cos x\,\,\left( 0,\frac{\pi }{2} \right)$

Differentiate bathes sides with respect to $x$ ,

${{f}^{'}}\left( x \right)=\sec x.\tan x-\left( -\sin x \right)\,$

$=\sec x.\tan x+\sin x$

Hence this is the answer.

Which of the following limits vanish ?

Part (A)

Consider the given function:

$\underset{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\to \,\,1}{\mathop{f(x)=\lim }}\,\dfrac{{{\sin }^{-1}}x}{\tan \dfrac{\pi x}{2}}$

$=\dfrac{{{\sin }^{-1}}(-1)}{\tan \dfrac{\pi }{2}}$

$=\dfrac{-\dfrac{\pi }{2}}{\infty }$

$=0$

Hence this is the answer.

Part (B)

Consider the given function

$\underset{x\to 1}{\mathop{\lim }}\,\dfrac{{{e}^{\sin x}}}{{{e}^{x}}}$

$=\,\,\underset{x\to 1}{\mathop{\lim }}\,\dfrac{1+\sin x+\dfrac{\sin {{x}^{2}}}{2!}+\dfrac{\sin {{x}^{3}}}{3!}........\infty }{1+x+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}........\infty }$

$\underset{\,\,\,\,\,\,\,\,\,\,\,\,x\to 1}{\mathop{=\,\,\lim }}\,\dfrac{1+\sin x}{1+x}$

$=\,\,\dfrac{1+1}{1+1}\,\,\,$

$=\,\,\dfrac{2}{2}\,\,\,\,\,=0$

Hence this is the answer.

Solve: $\underset{x \rightarrow 2}{\lim} \dfrac{(1 - 3^x - 4^x + 12^x)}{\sqrt{(2 \cos x + 7)} - }$

$\displaystyle\lim_{x\rightarrow 2}\dfrac{(1-3^x-4^x+12^x)}{\sqrt{2\cos x+7}}$

$=\left(\displaystyle\lim_{x\rightarrow 2}1-\displaystyle\lim_{x\rightarrow 2}3^x-\displaystyle\lim_{x\rightarrow 2}4^x+\displaystyle\lim_{x\rightarrow 2}12^x\right)\times \dfrac{1}{\sqrt{2\displaystyle\lim_{x\rightarrow 2}\cos x-\displaystyle\lim_{x\rightarrow 2}7}}$

$=\dfrac{1-3^2-4^2+12^2}{\sqrt{2\cos 2-7}}$

$=\dfrac{1-9-16+144}{\sqrt{2\cos 2-7}}$

$=\dfrac{120}{\sqrt{2\cos 2-7}}$

$\therefore\displaystyle\lim_{x\rightarrow 2}\dfrac{1-3^x-4^x+12^x}{\sqrt{2\cos x+7}}=\dfrac{120}{\sqrt{2\cos 2-7}}$ .

Solve the equation:-
$\overset{lim}{n\rightarrow 20}\displaystyle\int_{r=0}^{n-1}\dfrac{n}{n^{1}+r^{2}}$

$\displaystyle \lim_{n\rightarrow 20} \displaystyle \int_r=0^{n-1} \dfrac {n}{n+r^2}$

$\Rightarrow \ \displaystyle \lim_{n\rightarrow 20} \displaystyle \int_r=0^{n-1} \dfrac {1}{1+(r/\sqrt n)^2}$

$r/\sqrt n =t$

$\dfrac {1}{\sqrt n}\times dr=dt$

$\Rightarrow \sqrt n \displaystyle \lim_{n\rightarrow } \displaystyle \int_0^{n-1} \dfrac {1}{1+t^2}$

$\Rightarrow \sqrt { n } \displaystyle \lim _{ n\rightarrow 20 }{ { _{ }^{ }{ \left[ \tan ^{ -1 }{ t } \right] } }_{ 0 }^{ \dfrac { n-1 }{ \sqrt { n } } } }$

$\Rightarrow \sqrt n \displaystyle \lim_{n\rightarrow 20} \left [\tan^{-1} \dfrac {n-1}{\sqrt n} -0 \right]$

$\Rightarrow \sqrt 20 \tan^{-1}\dfrac {19}{\sqrt 20}$

Find the limit :-
$\mathop {\lim }\limits_{x \to 1} \,\,\dfrac{{1 - \dfrac{1}{x}}}{{\sin \,\pi (x - 1)}}$

$\displaystyle \lim_{x\rightarrow 1}\dfrac {1-\dfrac {1}{x}}{\sin \pi (x-1)}$

Apply

$L$ Hospital rule

$\displaystyle \lim_{x\rightarrow 1}\dfrac {\dfrac {1}{x^2}}{\pi \cos \pi (x-1)}$

$=\dfrac {1}{\pi \times 1}$

$=\dfrac {1}{\pi }$

$\lim_{x\rightarrow o}cos\left ( \sqrt{1+x} \right )-cos\sqrt{x}$ .

We have,

$\underset{x\to 0}{\mathop{\lim }}\,\cos \left( \sqrt{1+x} \right)-\cos \sqrt{x}$

$\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,2sin\dfrac{\left( \sqrt{1+x}+\sqrt{x} \right)}{2}sin\dfrac{\left( \sqrt{x}-\sqrt{1+x} \right)}{2}\,\,\,\,\,\therefore \cos C-\cos D=2\sin \dfrac{C+D}{2}\sin \dfrac{D-C}{2}$

$\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,2sin\dfrac{\left( \sqrt{1+x}+\sqrt{x} \right)}{2}sin\dfrac{\left( \sqrt{x}-\sqrt{1+x} \right)}{2}\times \dfrac{\left( \sqrt{x}+\sqrt{1+x} \right)}{\left( \sqrt{x}+\sqrt{1+x} \right)}\,\,\,\,\,\,\text{on rationlize}$

$\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,2sin\dfrac{\left( \sqrt{1+x}+\sqrt{x} \right)}{2}sin\dfrac{\left( x-1-x \right)}{2\left( \sqrt{x}+\sqrt{1+x} \right)}$

Taking limit and we get,

$\Rightarrow 2\sin \dfrac{\left( \sqrt{1+0}+\sqrt{0} \right)}{2}\sin \dfrac{-1}{2\left( \sqrt{0}+\sqrt{1+0} \right)}$

$\Rightarrow 2\sin \dfrac{1}{2}\sin \dfrac{-1}{2}$

$\Rightarrow -2{{\sin }^{2}}\dfrac{1}{2}$

Hence, this is the answer

$\dfrac{sin2x}{a cos^{2}x+b sin ^{2}x}$

Let

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

On differentiating and we get,

$\dfrac{dy}{dx}=\dfrac{\left( a{{\cos }^{2}}x+b{{\sin }^{2}}x \right)\dfrac{d}{dx}\sin 2x-\sin 2x\dfrac{d}{dx}\left( a{{\cos }^{2}}x+b{{\sin }^{2}}x \right)}{{{\left( a{{\cos }^{2}}x+b{{\sin }^{2}}x \right)}^{2}}}$

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

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

Hence, this is the answer.

Solve: $\displaystyle \int^{1}_{-1} \frac{dx}{x^2 + 2x + 5}$

$\displaystyle\int^1_{-1}\dfrac{dx}{(x^2+2x+5)}$

$=\displaystyle\int^1_{-1}\dfrac{dx}{x^2+2x+4+1}$

$=\displaystyle\int^1_{-1}\dfrac{dx}{(x+2)^2+1^2}$

$=\tan^{-1}(x+2)\displaystyle\int^{1}_{-1}$

$=\tan^{-1}3-\tan^{-1}(1)$

$=\tan^{-1}3-\dfrac{\pi}{4}$

$\therefore \displaystyle\int^1_{-1}\dfrac{dx}{(x^2+2x+5)}=\tan^{-1}3-\dfrac{\pi}{4}$ .

Evaluate the following limits
$\mathop {\lim }\limits_{x \to 0} \left( {\frac{{{e^x} - \sin x}}{x}} \right)$

Solution

$\displaystyle \lim_{x\rightarrow 0}\left(\dfrac{e^x -\sin x}{x}\right)$

Value of numerator at

$x=0$

$e^o-\sin o=1-0=1$

value of denominator at

$x=0$

This is

$\dfrac{1}{ 0}$ is determinate form

So limit does not exist.

$y = \sin \left( {\pi /6{e^{xy}}} \right){\text{putting}}\;x = 0\;{\text{than}}\dfrac{{dy}}{{dx}}$

given

$\Rightarrow \ y=\sin{\left(\dfrac{\pi}{6{e}^{xy}}\right)}$ putting

$x=0$

$\Rightarrow \ y=\sin{\left(\dfrac{\pi}{6{e}^{0}}\right)}$

$\Rightarrow \ y=\sin{\left(\dfrac{\pi}{6}\right)}$

$\Rightarrow \ y=\dfrac{1}{2}$ then

$\dfrac{dy}{dx}$ of a constant is zero

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

Solve $\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x + 4}}{{x + 2}}} \right)^{x + 3}}$

1951561_1161698_ans_66fb607539f141779c08eed02618d8a8.jpg

Evaluate :
$\mathop {\lim }\limits_{x \to 0} \frac{{x - \sin x\cos x}}{{{x^3}}}.$

Solution -

$\lim_{x\rightarrow 0} \frac{x-sin\,xcos\,x}{x^{3}}$

$= \lim_{x\rightarrow 0} \frac{1-cos\,x}{x^{2}}$ because

$\lim_{x\rightarrow 0} \frac{sin\,x}{x} = 1$

Apply L Hospital Rule twice

$= \lim_{x\rightarrow 0} \frac{sin\,x}{2x}$

$= \lim_{x\rightarrow 0} \frac{cos\,x}{2}$

$= \frac{1}{2}$

1100719_1188309_ans_dacf33c3cd5a4ec594c5fc9f5783fb23.jpg

$y=6{{x}^{3}}+3{{x}^{2}}+4x+5$
Find the value of $\dfrac{dy}{dx}$ ?

Given that,

$y=6{{x}^{2}}+3{{x}^{2}}+4x+5$

On differentiating

$\dfrac{dy}{dx}=18{{x}^{2}}+6x+4$

Hence, this is the required solution

$\lim_{x\rightarrow 2}\frac{\sum 32x}{x^{3}-p}$

We have,

$\lim_{x\to\ 2}\dfrac{32x}{x^3-8}$

This is the

$\dfrac{0}{0}$ form.

So, apply L-hospital rule

$\lim_{x\to\ 2}\dfrac{32}{3x^2-0}$

$=\dfrac{32}{3\times 2^2}$

$=\dfrac{8}{3}$

Hence, this is the answer.

Solve
$\mathop {\lim }\limits_{x \to 0} \,\,\dfrac{{\tan 8x}}{{\sin 2x}}$

$\displaystyle \lim_{x\rightarrow 0}\dfrac {\tan 8x}{\sin 2x}$

$\displaystyle \lim_{x\rightarrow 0}\dfrac {\tan 8x}{8x}\times \dfrac {2x}{\sin 2x}\times \dfrac {8x}{2x}$

We know that,

$\displaystyle \lim_{x\rightarrow 0}\dfrac {\tan 8x}{8x}=1$

$\displaystyle \lim_{x\rightarrow 0}\dfrac {2x}{\sin 2x}=1$

$\therefore\displaystyle \lim_{x\rightarrow 0}\dfrac {\tan 8x}{8x}\times \dfrac {2x}{\sin 2x}\times \dfrac {8x}{2x}= \displaystyle \lim_{x\rightarrow 0}1\times 1\times \dfrac {8}{2}$

Hence,

$\displaystyle \lim_{x\rightarrow 0}\dfrac {\tan 8x}{\sin 2x}=4$

Solve the equation:-
$\lim_{{x\rightarrow \pi/2}}\ \ \dfrac{\tan\ 2x}{x-\pi/2}$

${\lim}_{x\rightarrow\,\frac{\pi}{2}}{\dfrac{\tan{2x}}{x-\dfrac{\pi}{2}}}$

$LHL={\lim}_{h\rightarrow\,{0}^{-}}{\dfrac{\tan{2\left(\dfrac{\pi}{2}-h\right)}}{\dfrac{\pi}{2}-h-\dfrac{\pi}{2}}}$

$={\lim}_{h\rightarrow\,{0}^{-}}{\dfrac{\tan{\left(\pi-2h\right)}}{-h}}$

$={\lim}_{h\rightarrow\,{0}^{-}}{-\dfrac{\tan{2h}}{-h}}$

$={\lim}_{h\rightarrow\,{0}^{-}}{-\dfrac{2\tan{2h}}{-2h}}=2$

$RHL={\lim}_{h\rightarrow\,{0}^{+}}{\dfrac{\tan{2\left(\dfrac{\pi}{2}+h\right)}}{\dfrac{\pi}{2}+h-\dfrac{\pi}{2}}}$

$={\lim}_{h\rightarrow\,{0}^{+}}{\dfrac{\tan{\left(\pi+2h\right)}}{h}}$

$={\lim}_{h\rightarrow\,{0}^{+}}{\dfrac{\tan{2h}}{h}}$

$={\lim}_{h\rightarrow\,{0}^{+}}{\dfrac{2\tan{2h}}{2h}}=2$

Thus,

${\lim}_{x\rightarrow\,\frac{\pi}{2}}{\dfrac{\tan{2x}}{x-\dfrac{\pi}{2}}}=2$

$\lim _ { x \rightarrow 0 } \dfrac { 1 - \cos ^ { 3 } x } { x \sin 2 x }$

$\displaystyle\lim_{x\rightarrow 0}\dfrac{1-\cos^3 x}{x\sin 2x}$

By L'Hospital's rule

$\displaystyle\lim_{x\rightarrow 0}\dfrac{f(x)}{g(x)}=\lim_{x\rightarrow 0}\dfrac{f'(x)}{g'(x)}$

$\Rightarrow$

$\displaystyle\lim_{x\rightarrow 0}\dfrac{0-3\cos^2 x(-\sin x)}{2x\cos 2x+\sin 2x}$ [ Taking derivative of numerator and denominator ]

$\Rightarrow$

$\displaystyle\lim_{x\rightarrow 0}\dfrac{3\cos^2 x \sin x}{2x(\cos^2 x-\sin^2 x)+2\sin x\cos x}$

$\Rightarrow$

$\dfrac{3}{2}\displaystyle\lim_{x\rightarrow 0}\dfrac{\cos^2 x \sin x}{x(\cos^2 x-\sin^2 x)+2\sin x\cos x}$

$\Rightarrow$

$\dfrac{3}{2}\displaystyle\lim_{x\rightarrow 0}\dfrac{\sin x}{x(1-\tan^2 x)+2\tan x}$ [ Dividing numerator and denominator by

$\cos^ 2x$ ]

$\Rightarrow$

$\dfrac{3}{2}\displaystyle\lim_{x\rightarrow 0}\dfrac{\cos x}{x(-2\tan x\sec^2 x)+(1-\tan^2 x).1+\sec^2 x}$ [ Taking derivative of numerator and denominator ]

$\Rightarrow$

$\dfrac{3}{2}\times\dfrac{1}{1+1}$

$\Rightarrow$

$\dfrac{3}{2}\times\dfrac{1}{2}$

$\Rightarrow$

$\dfrac{3}{4}$

$\displaystyle\lim_{x\rightarrow \frac{\pi}{2}}\dfrac{\tan 2x}{x-\dfrac{\pi}{2}}$ .

$\underset{x\rightarrow \pi /2}{\lim} \dfrac{\tan 2x}{x-\pi /2}$

$=\underset{x\rightarrow \pi /2}{\lim} \dfrac{-2 \tan 2(\pi /2-x)}{2(x-\pi /2)}$

$=\underset{x\rightarrow \pi /2}{\lim} \dfrac{2 \tan 2(\pi /2-x)}{2(\pi /2-x)}$

$=2 \left[\because \underset{x\rightarrow 0}{\lim}\dfrac{\tan x}{x}=1\right]$

Solve $\lim _ { x \rightarrow 0 } \dfrac { \sqrt { 2 + x } - \sqrt { 2 } } { x }$

$\\\lim_{x\rightarrow0}(\frac{\sqrt{2+x}-\sqrt{2}}{x})\times(\frac{\sqrt{2+x}+\sqrt{2}}{\sqrt{2+x}+\sqrt{2}})\\=\lim_{x\rightarrow0}(\frac{(2+x)-2}{x\left(\sqrt{2+x}+\sqrt{2}\right)})\\=\lim_{x\rightarrow0}(\frac{1}{\left(\sqrt{2+x}+\sqrt{2}\right)})\\=(\frac{1}{2\sqrt{2}})$

Find differentiation of $\sec ^ { - 1 } \tan x$ .

$\Rightarrow \sec^{-1} tan x$

$\because \dfrac{d}{dx}\sec^{-1}x = \dfrac{1}{x\sqrt{x^{2}-1}}$

$\dfrac{d}{dx} \sec^{-1}\tan x$

$= \dfrac{1}{\tan x\sqrt{\tan^{2}x-1}}.\sec^{2}x$

$= \dfrac{1}{\sin x\sqrt{\sin^{2}x-\cos^{3}x}}$

Find the diffrentiation of $xsinx$ .

$y = x \sin x$

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

Solve $\underset { x\rightarrow 1 }{ \lim } \dfrac { { x }^{ 3 }+{ 3x }^{ 2 }-6x+2 }{ { x }^{ 3 }+{ 3x }^{ 2 }-3x-1 }$

$\displaystyle \quad \lim _{ x\rightarrow 1 }{ \frac { { x }^{ 3 }+3{ x }^{ 2 }-6x+2 }{ { x }^{ 3 }+{ 3x }^{ 2 }-3x-1 } }$

$\displaystyle =\lim _{ x\rightarrow 1 }{ \frac { \left( x-1 \right) \left( { x }^{ 2 }+4x-2 \right) }{ \left( x-1 \right) \left( { x }^{ 2 }+4x+1 \right) } }$

$[\because a^3 +b^3=(a+b)(a^2-ab+b^2)]$

$\displaystyle =\lim _{ x\rightarrow 1 }{ \frac { { x }^{ 2 }+4x-2 }{ { x }^{ 2 }+4x+1 } }$

$\displaystyle =\frac { 3 }{ 6 }$

$\displaystyle =\frac { 1 }{ 2 }$

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

$y={\csc}^{2}{x}$ .......

$\left(1\right)$

Increase from

$y$ to

$y+\delta y$ correspondingly

$x$ to

$x+\delta x$ in the above equation

$\left(1\right)$

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

$\left(2\right)$

Eqn

$\left(2\right)$ -Eqn

$\left(1\right)$

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

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

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

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

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

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

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

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

$\sin{\left(-\theta\right)}=-\sin{\theta}$

$\Rightarrow {\lim}_{\delta{x}\rightarrow 0}\dfrac{\delta{y}}{\delta{x}}={\lim}_{\delta{x}\rightarrow 0}\dfrac{-sin{\left(2x+\delta{x}\right)}\dfrac{\sin{\delta{x}}}{\delta{x}}}{{\sin}^{2}{x}{\sin}^{2}{\left(x+\delta{x}\right)}}$ by applying limits to both sides

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

$\Rightarrow \dfrac{dy}{dx}=\dfrac{-sin{2x}}{{\sin}^{2}{x}{\sin}^{2}{x}}$ since

${\lim}_{\theta\rightarrow 0}\dfrac{\sin{\theta}}{\theta}=1$

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

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

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

$\therefore \dfrac{dy}{dx}=-2\cot{x}{\csc}^{2}{x}$

Let $y=$ $\sqrt { x } + 2 x ^ { \dfrac { 3 } { 4 } } + 3 x ^ { \dfrac { 5 } { 6 } } ( x > 0 )$ . Find the derivative of $y$ with respect to $x$ .

Given,

$y=$

$\sqrt { x } + 2 x ^ { \dfrac { 3 } { 4 } } + 3 x ^ { \dfrac { 5 } { 6 } }$

Now,

$\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x}}+2\times \dfrac{3}{4}\times x^{-\dfrac{1}{4}}+3\times \dfrac{5}{6}\times x^{-\dfrac{1}{6}}$

or,

$\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x}}+\dfrac{3}{2}x^{-\dfrac{1}{4}}+\dfrac{5}{2} x^{-\dfrac{1}{6}}$

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

$y={ c }^{ 2 }+\cfrac { c }{ x }$

$\cfrac{dy}{dx}=0+\cfrac{(-1)C}{{x}^{2}}$

$y'=\cfrac{-c}{{x}^{2}}$

Solve:
$\underset { x\rightarrow 0 }{ lim } \dfrac { sin3x }{ x }$

We have,

$\lim_{x\ \to\ 0}\dfrac{\sin 3x}{x}$

$\lim_{x\ \to\ 0}\dfrac{3\sin 3x}{3x}$

We know that

$\lim_{x\ \to\ 0}\dfrac{\sin x}{x}=1$

Therefore,

$=3\times 1$

$=3$

Hence, this is the answer.

Evaluate $\lim_{x\rightarrow 0}\dfrac { { x }^{ 2 }-\tan 2x }{ \tan x }$

$\lim _{ x\rightarrow 0 }{ (\dfrac { { x }^{ 2 }-tan2x }{ tanx } } )$

$=\lim _{ x\rightarrow 0 }{ (\dfrac { { x }^{ 2 } }{ tanx } -\dfrac { tan2x }{ tanx } ) }$

$=\lim _{ x\rightarrow 0 }{ \quad (\dfrac { { x }^{ 2 } }{ tanx } )- } \lim _{ x\rightarrow 0 }{ \quad (\dfrac { tan2x }{ tanx } ) }$

$=0-2=-2$ (Answer)

If $y=\tan x +\cot x$ find $\dfrac{dy}{dx}$

$y=\tan x+\cot x\\\dfrac{dy}{dx}=\dfrac d{dx}\tan x+\cot x\\\dfrac{dy}{dx}=\sec ^2x-\csc^2x$

Evaluate:
$\mathop {\lim }\limits_{x \to \tfrac{\pi }{3}} \dfrac{{\sin \left( {\frac{\pi }{3} - x} \right)}}{{2\cos x - 1}}$

$\displaystyle \lim_{x\rightarrow \dfrac {\pi}{3}} \dfrac {\sin \left (\dfrac {\pi}{3} - x\right )}{2\cos x - 1}$

Let

$x = \dfrac {\pi}{3} + t$

$\displaystyle \lim_{t\rightarrow 0} \dfrac {\sin \left (\dfrac {\pi}{3} - \dfrac {\pi}{3} - t\right )}{2\cos \left (\dfrac {\pi}{3} + t\right ) - 1}$

$=\displaystyle \lim_{t\rightarrow 0} \dfrac {-\sin t}{2[\cos (t + \dfrac {\pi}{3}] - 1}$

$=\displaystyle \lim_{t\rightarrow 0} \dfrac {-\sin t}{2\left [\dfrac {1}{2} \cos t - \dfrac {\sqrt {3}}{2} \sin t\right ] - 1}$

$=\displaystyle \lim_{t\rightarrow 0} \dfrac {-\sin t}{\cos t - \sqrt {3} \sin t - 1}$

$=\displaystyle \lim_{t \rightarrow 0} \dfrac {+\sin t}{\sqrt {3}\sin t + 1 - \cos t}$

$=\displaystyle \lim_{t\rightarrow 0} \dfrac {2\sin \dfrac {t}{2} \cos \dfrac {t}{2}}{\sqrt {3} \times 2\sin \dfrac {t}{2} \cos \dfrac {t}{2} + 2\sin^{2}\dfrac {t}{2}}$

$=\displaystyle \lim_{t \rightarrow 0} \dfrac {2\sin \dfrac {t}{2} (\cos \dfrac {t}{2})}{2\sin \dfrac {t}{2} \left (\sqrt {3} \cos \dfrac {t}{2} + \sin \dfrac {t}{2}\right )}$

$=\displaystyle \lim_{t\rightarrow 0} \dfrac {\cos \dfrac {t}{2}}{\sqrt {3} \cos \dfrac {t}{2} + \sin \dfrac {t}{2}}$

$\Rightarrow \dfrac {1}{\sqrt {3} + 0} = \dfrac {1}{\sqrt {3}}$ .

$\lim\limits_{x \rightarrow 0}\dfrac{\sin (\pi\, \cos^2\,x)}{x^2}$ =

We have,

$\lim\limits_{x\to\ 0}\dfrac{\sin (\pi \cos^2 x)}{x^2}$

$\lim\limits_{x\to\ 0}\dfrac{\sin (\pi (1-\sin^2 x))}{x^2}$

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

$\lim\limits_{x\to\ 0}\dfrac{\sin (\pi -\pi \sin^2 x))}{x^2}$

$\lim\limits_{x\to\ 0}\dfrac{\sin (\pi \sin^2 x)}{x^2}$

$(\because \sin (\pi-x)=\sin x)$

$\lim\limits_{x\to\ 0}\dfrac{\sin (\pi \sin^2 x)}{\pi \sin^2 x}\times \dfrac{\pi \sin^2 x}{x^2}$

We know that,

$\lim\limits_{x\to\ 0}\dfrac{\sin x}{x}=1$

Therefore,

$=1\times \pi \times 1$

$=\pi$ .

If $y=\cos^{-1}\left\{\dfrac{2x-3\sqrt{1-x^{2}}}{\sqrt{13}}\right\}$ , find $\dfrac{dy}{dx}$ .

Sol:

$y=\cos ^{ -1 }{ \left\{ \dfrac { 2x-3\sqrt { 1-{ x }^{ 2 } } }{ \sqrt { 13 } } \right\} }$ , To find

$\dfrac {dy}{dx}$

put

$x=\sin \theta$ , Hence

$\theta =\sin +x$

$y=\cos ^{ -1 }{ \left\{ \dfrac { 2\sin { \theta } -3\sqrt { 1-\sin ^{ 2 }{ \theta } } }{ \sqrt { 13 } } \right\} }$

$(1-\sin^{2}\theta =\cos^{2}\theta)(since \sin^{2} \theta +\cos \theta =1)$

$y=\cos ^{ -1 }{ \left\{ \dfrac { 2\sin { 6 } -3\sqrt { \cos ^{ 2 }{ 6 } } }{ \sqrt { 13 } } \right\} }$

$y=\cos ^{ -1 }{ \left\{ \dfrac { 2\sin { 6 } -3\cos { 6 } }{ \sqrt { 13 } } \right\} } \left( \cos ^{ 2 }{ 6 } =\cos { 6 } \right)$

$y=\cos ^{ -1 }{ \left\{ \dfrac { 2 }{ \sqrt { 13 } } \sin { 6 } -\dfrac { 2 }{ \sqrt { 13 } } \cos { \theta } \right\} }$

Now get

$\dfrac {2}{\sqrt {13}}=\cos \alpha , 2\dfrac {3}{\sqrt {13}}=\sin \alpha$

$\left[ \sin ^{ 2 }{ \alpha } +\cos ^{ 2 }{ \alpha } ={ \left( \dfrac { 2 }{ \sqrt { 13 } } \right) }^{ 2 }+{ { \left( \dfrac { 3 }{ \sqrt { 13 } } \right) } }^{ 2 }=\dfrac { 4 }{ 13 } +\dfrac { 3 }{ 13 } =\dfrac { 13 }{ 13 } =1 \right]$

$y=\cos ^{ -1 }{ \left\{ \cos { \alpha } \sin { 6 } -\sin { \alpha } \cos { \theta } \right\} }$

Now

$[\sin (\theta -\alpha) =\sin 6\cos \alpha -\cos 6 \sin \alpha ]$

$y=\cos ^{ -1 }{ \left\{ \sin { \left( \theta -\alpha \right) } \right\} } \left[ \cos { \left( \dfrac { \pi }{ 2 } \left( \theta -\alpha \right) \right) } =\sin { \left( \theta -\alpha \right) } \right]$

$y=\cos ^{ -1 }{ \left\{ \cos { \left( \frac { \pi }{ 2 } \left( \theta -\alpha \right) \right) } \right\} }$

$y=\pi /2 -(\theta -\alpha)\quad [\cos^{-1} (\cos x)=x]$

$y=\pi /2 -(\sin^{-1}x -\cos^{-1}\dfrac {2}{\sqrt {13}})$

Now differente wrt

$x$ we have

$\dfrac {dy}{dx}=0-\dfrac {1}{\sqrt {1-x^{2}}}-0 \quad \left (\cos^{_1}\dfrac {2}{\sqrt {13}}=\cos \tan \right)$

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

Evaluate: $\lim _{x\to 0}\:\dfrac{e^{\alpha x}-e^{\beta x}}{sin\:\alpha x-sin\:\beta x}$

$\lim _{x\to 0}\:\dfrac{e^{\alpha x}-e^{\beta x}}{sin\:\alpha x-sin\:\beta x}$

$={\lim}_{x\rightarrow 0}\dfrac{{e}^{\alpha x}\left(1-\dfrac{{e}^{\beta x}}{\alpha x}\right)}{2\cos{\left(\dfrac{\alpha x+\beta x}{2}\right)}\sin{\left(\dfrac{\alpha x-\beta x}{2}\right)}}$ .........

$\left(1\right)$ using transformation angle formula

$\sin{C}-\sin{D}=2\cos{\left(\dfrac{C+D}{2}\right)}\sin{\left(\dfrac{C-D}{2}\right)}$

Using the formulae,

${\lim}_{x\rightarrow 0}\dfrac{{e}^{x}-1}{x}=1$

${\lim}_{x\rightarrow 0}{\cos{\theta}}=1$ and

${\lim}_{\theta\rightarrow 0}{\dfrac{\sin{\theta}}{\theta}}=1$

we evaluate the above expression as

${\lim}_{x\rightarrow 0}{e}^{\alpha x}=1$

${\lim}_{x\rightarrow 0}\left(1-{e}^{\left(\beta-\alpha\right)x}\right)={\lim}_{x\rightarrow 0}\dfrac{\left(1-{e}^{\left(\beta-\alpha\right)x}\right)}{\left(\beta-\alpha\right)x}\times \left(\beta-\alpha\right)x=-1\times {\lim}_{x\rightarrow 0}\left(\beta-\alpha\right)x$

${\lim}_{x\rightarrow 0}\cos{\left(\dfrac{\alpha+\beta}{2}\right)x}=1$

${\lim}_{x\rightarrow 0}\sin{\left(\dfrac{\alpha-\beta}{2}\right)x}={\lim}_{x\rightarrow 0}\dfrac{\sin{\left(\dfrac{\alpha-\beta}{2}\right)x}}{\left(\dfrac{\alpha-\beta}{2}\right)x}\times \left(\dfrac{\alpha-\beta}{2}\right)x={\lim}_{x\rightarrow 0}{\left(\dfrac{\alpha-\beta}{2}\right)x}$

Substituting the above in

$\left(1\right)$ we get

$=\dfrac{-1\times{\lim}_{x\rightarrow 0}\left(\beta-\alpha\right)x }{2 \times 1\times {\lim}_{x\rightarrow 0}{\left(\dfrac{\alpha-\beta}{2}\right)x}}$ the term

$x$ gets cancelled in both the numerator and the denominator

$=\dfrac{-\beta+\alpha}{\alpha-\beta}=1$

Differentiate w.r.t $x$
$e ^ { x }\cos ^ { 3 } x\sin ^ { 2 } x$

$\dfrac{d}{dx}=u^1vw=u^1vw+uv^1w+uvw^1$

$e^x\cos ^3x\,\,\, \sin ^2x$

$u=e^x\,\,\,u^1=e^x$

$v=\cos^3x$

$v^1=3\cos^2x(-\sin x)=-3\cos ^2x\sin x$

$w=\sin ^2x$

$w^1=2\sin x\cos x=\sin 2x$

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

$e^x[\cos ^3x\sin ^2x-3\cos ^2x \sin x+\cos ^3x\sin 2x]$

$lim_{x \rightarrow 1}\dfrac{sin\, \pi\,x}{x-1}$ is

We have,

$\lim_{x\to\ 0}\dfrac{\sin \pi x}{x-1}$

This is the

$\dfrac{0}{0}$ form.

So, apply L-Hospital rule

$\lim_{x\to\ 0}\dfrac{\pi \cos \pi x}{1-0}$

$\lim_{x\to\ 0}{\pi \cos \pi x}$

$=\pi \cos 0$

$=\pi$

Hence, this is the answer.

Solve $\mathop {\lim }\limits_{x \to 2} \,f\left( x \right) = \mathop {\lim }\limits_{x \to 2} \,\dfrac{{\left( {{x^2} - 4} \right)}}{{\left( {x - 2} \right)}}$

$\underset{x \rightarrow 2}{\lim} \dfrac{(x^2 - 4)}{(x - 2)} = \underset{x \rightarrow 2}{\lim} \dfrac{(x - 2)(x + 2)}{(x -2)}$

$= \underset{x \rightarrow 2}{\lim} (x + 2) = 2 + 2$

$= 4$

If $\dfrac{3}{2}+y^{3}=3axy$ , then find $\dfrac {dy}{dx}$

$\cfrac{3}{2}+y^3=3axy$

On differentiating we get,

$3y^2\cfrac{dy}{dx}=3ay+3ax\cfrac{dy}{dx}$

$\implies \cfrac{dy}{dx}=\cfrac{3ay}{3y^2-3ax}$

$\implies \cfrac{dy}{dx}=\cfrac{ay}{y^2-ax}$

If $y = \log \left( {{e^{3x}}{{\left( {\frac{{x - 4}}{{x + 3}}} \right)}^{\frac{2}{3}}}} \right)$ , then find $\dfrac{{dy}}{{dx}}$

1351980_1263423_ans_bceff5db590141e19b59ed9f1e8e591f.PNG

Find $\dfrac{dy}{dx}$ , if $x=a(\theta-\sin \theta)$ and $y=a(1-\cos \theta)$ .

$x = a (\theta - \sin \theta)$ and

$y = a (1 - \cos \theta)$

Differentiating w.r.t to

$\theta$ , we get,

$\dfrac{dx}{d \theta} = a (1 - \cos \theta)$ and

$\dfrac{dy}{d \theta} = a \sin \theta$

$\therefore \dfrac{dy}{dx} = \dfrac{dy/d\theta}{dx / d \theta} = \dfrac{a \sin \theta}{a (1 - \cos \theta)} = \dfrac{2 \sin (\theta/2) \cos (\theta/2)}{2 \sin^2 (\theta/2)} = \cot (\dfrac{\theta}{2})$

Solve:
$\displaystyle \lim_{n\rightarrow \infty}{\dfrac{1.n+2(n-1)+3(n-2)+.....+n.1}{n^{3}}}$

We have,

${ \lim }_{ x\to \infty }\dfrac { { 1.n+2\left( { n-1 } \right) +3\left( { n-2 } \right) +....+n.1 } }{ { { n^{ 3 } } } } \\ { \lim }_{ x\to \infty }\dfrac { { \sum _{ r=1 }^{ n }{ r\left( { n-r+1 } \right) } } }{ { { n^{ 3 } } } }$

${ \lim }_{ x\to \infty }\dfrac { { n\times \dfrac { { n\left( { n+1 } \right) } }{ 2 } +\dfrac { { n\left( { n+1 } \right) } }{ 2 } -\dfrac { { n\left( { n+1 } \right) \left( { 2n+1 } \right) } }{ 6 } } }{ { { n^{ 3 } } } } \\ \lim_{x\to \infty}\dfrac { { \dfrac { { n\left( { n+1 } \right) } }{ 2 } \left( { n+1 } \right) -\dfrac { { n\left( { n+1 } \right) \left( { 2n+1 } \right) } }{ 6 } } }{ { { n^{ 3 } } } } \\ ={ \lim }_{ x\to \infty }\dfrac { { \dfrac { { { n^{ 3 } } } }{ 2 } -\dfrac { { { n^{ 3 } } } }{ 3 } +...... } }{ { { n^{ 3 } } } } =\dfrac { 1 }{ 2 } -\dfrac { 1 }{ 3 } \\ =\dfrac { 1 }{ 6 }$

Hence, this is the answer.

$lim_{x \rightarrow 3}\dfrac{x^5-243}{x^2-9}$

We have,

$\lim_{x\to\ 3}\dfrac{x^5-243}{x^2-9}$

This is

$\dfrac{0}{0}$ form.

So, apply L-hospital rule

$\lim_{x\to\ 3}\dfrac{5x^4-0}{2x-0}$

$\lim_{x\to\ 3}\dfrac{5x^4}{2x}$

$\lim_{x\to\ 3}\dfrac{5x^3}{2}$

$=\dfrac{5\times 3^3}{2}$

$=\dfrac{135}{2}$

Hence, this is the answer.

verify that the function y= $\sqrt{a^2 - x^2}, x\epsilon (-a,a_)$ is a solution of the diffrential equation x + y $\dfrac{dy}{dx}=0, (y \neq 0)$

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

$\Rightarrow x+y\dfrac{dy}{dx}=x-\dfrac{xy}{\sqrt{a^2-x^2}}=x-2$

$=0$ .

1167382_1267414_ans_a0069e38d49243fc80b31b917592283a.jpg

$\begin{matrix} lim \\x \xrightarrow\quad 1 \end{matrix}$ $\left(\dfrac{1}{x-1} - \dfrac{2}{x^2 - 1}\right)$

$\frac{1}{x-1}-\frac{2}{(x+1)(x-1)}=\frac{x+1-2}{(x+1)(x-1)}=\frac{x-1}{(x+1)(x-1)}=\frac{1}{x+1}$

$\lim_{x-1} \frac{1}{x+1}=\frac{1}{2}$

1181943_1294396_ans_b3c4ce3c44e94d14b0b7a1284d2bb33c.jpg

Solve:
$\displaystyle \lim _{ \theta \rightarrow \pi }{ \dfrac {\pi -\theta}{\sqrt {\cos \theta +1}} }$

We have,

$\underset{\theta \to \pi }{\mathop{\lim }}\,\dfrac{\pi -\theta }{\sqrt{\cos \theta +1}}$ $\left( \dfrac{0}{0}\,form \right)$

So,

$\Rightarrow \underset{\theta \to \pi }{\mathop{\lim }}\,\dfrac{\pi -\theta }{\sqrt{2{{\cos }^{2}}\dfrac{\theta }{2}-1+1}}$

$\Rightarrow \underset{\theta \to \pi }{\mathop{\lim }}\,\dfrac{\pi -\theta }{\sqrt{2}\cos \dfrac{\theta }{2}}$

Applying L’ Hospital rule and we get,

$\underset{\theta \to \pi }{\mathop{\lim }}\,\dfrac{0-1}{\sqrt{2}\left( -\sin \dfrac{\theta }{2} \right)\times \dfrac{1}{2}}$

Taking limit and we get,

$\dfrac{1}{\dfrac{\sqrt{2}}{2}\sin \dfrac{\pi }{2}}$

$=\dfrac{2}{\sqrt{2}}\,\,or\,\sqrt{2}$

Hence, this is the answer.

Solve :
$\displaystyle\lim _{ x\rightarrow 1 }{ \left[ \dfrac { 3 }{ { x }^{ 2 }+x-2 } -\dfrac { 4 }{ { x }^{ 2 }+2x-3 } \right] }$

We have,

${ \lim }_{ x\to 1 }\left[ { \dfrac { 3 }{ { { x^{ 2 } }+x-2 } } -\dfrac { 4 }{ { { x^{ 2 } }+2x-3 } } } \right] \\ ={ \lim }_{ x\to 1 }\left[ { \dfrac { 3 }{ { \left( { x+2 } \right) \left( { x-1 } \right) } } -\dfrac { 4 }{ { \left( { x-1 } \right) \left( { x+3 } \right) } } } \right] \\$

$={ \lim }_{ x\to 1 }\left[ { \dfrac { { 3\left( { x+3 } \right) -4\left( { x+2 } \right) } }{ { \left( { x-1 } \right) \left( { x+2 } \right) \left( { x+3 } \right) } } } \right]$

$={ \lim }_{ x\to 1 }\left[ { \dfrac { { 3x+9 -4x-8 } }{ { \left( { x-1 } \right) \left( { x+2 } \right) \left( { x+3 } \right) } } } \right]$

$\\ ={ \lim }_{ x\to 1 }\dfrac { { -x+1 } }{ { \left( { x-1 } \right) \left( { x+2 } \right) \left( { x+3 } \right) } } \\ ={ \lim }_{ x\to 1 }\dfrac { { -\left( { x-1 } \right) } }{ { \left( { x-1 } \right) \left( { x+2 } \right) \left( { x+3 } \right) } } \\ =\dfrac { { -1 } }{ { 3\times 4 } } =-\dfrac { { 1 } }{ { 12 } }$

Hence, this is the answer.

Solve:
$\displaystyle\lim _{ x\rightarrow 2 }{ \dfrac { { x }^{ 6 }+32 }{ x+2 } }$

$\displaystyle \lim_{x\rightarrow 2} \frac{x^{6}+32}{x+2}$

$= \frac{2^{6}+32}{2+2}$

$= \frac{64+32}{4} = \frac{96}{4}$

$= 24$

1209330_1284092_ans_b2399b1189d6468cb53f07ec6e324401.jpg

$\lim_{x\rightarrow\ 0}\dfrac{\cos\ x}{\pi-x}$

$\lim_{x\rightarrow 0} \frac{cos \, x}{\pi -x}$

Put x = 0

$\frac{cos \,0 }{\pi - 0}$

$[\because cos \, 0 = 1]$

$= \frac{1}{\pi }$

1211718_1285275_ans_033d107b8afe4aeb8b3e586a0dc1dd83.jpg

Evaluate: $\displaystyle\lim_{n\rightarrow \infty}(n-\sqrt{n^2+n})$ .

$\displaystyle\lim_{n\rightarrow \infty}\dfrac{(n-\sqrt{n^2+n})(n+\sqrt{n^2+n})}{(n+\sqrt{n^2+n})}$

$=\displaystyle\lim_{n\rightarrow \infty}\dfrac{n^2-n^2-n}{n+\sqrt{n^2+n}}$

$=\displaystyle\lim_{n\rightarrow \infty}\dfrac{-1}{1+\sqrt{1+1/n}}$

$=\dfrac{-1}{1+1}=\dfrac{-1}{2}$ .

1194389_1300249_ans_8485fed134aa402bbd2b3b2e9b53319d.jpg

$\underset { x\rightarrow 3 }{ lim } \left( [x-3] \right) +[3-x]-x$

$=\underset { x\rightarrow 3 }{ lim } \left( [x]-3+3+[-x]-x \right)$

$=\underset { x\rightarrow 3 }{ lim } \left( [x]+[-x]-x \right) =-1-3=-4$

Evaluate : $\underset { x\rightarrow 0 }{ lim } \frac { { 3 }^{ x }-1 }{ \sqrt { 1+x } -1 }$ .

$L=\lim _{ x\rightarrow 0 }{ \cfrac { { 3 }^{ x }-1 }{ \sqrt { 1+x } -1 } }$

Applying L' Hospital's rule

$\lim _{ x\rightarrow 0 }{ \cfrac { { 3 }^{ x }\log { 3 } }{ \cfrac { 1 }{ 2\sqrt { x+1 } } } } =\lim _{ x\rightarrow 0 }{ 2\sqrt { x+1 } .{ 3 }^{ x }.\log { 3 } } \\ =2\times \sqrt { 0+1 } \times { 3 }^{ 0 }.\log { 3 } =2\log { 3 }$

solve $\underset { x\rightarrow 0 }{ lim } \cfrac { { 5 }^{ x }-{ 4 }^{ x } }{ x }$

$\lim _{ x\rightarrow 0 }{ \cfrac { { 5 }^{ x }-{ 4 }^{ x } }{ x } } =\lim _{ x\rightarrow 0 }{ \cfrac { ({ 5 }^{ x }-1)-({ 4 }^{ x }-1) }{ x } } \\ =\lim _{ x\rightarrow 0 }{ \cfrac { { 5 }^{ x }-1 }{ x } } -\lim _{ x\rightarrow 0 }{ \cfrac { { 4 }^{ x }-1 }{ x } } =\log { 5 } -\log { 4 } =\log { \cfrac { 5 }{ 4 } }$

If $y=\dfrac{(ax+b)(cx+d)}{(ax-b)(cx-d)}, x \neq \dfrac{b}{a}, \dfrac{d}{c}$ then find $\dfrac{dy}{dx}$ .

$\begin{array}{l} We\, \, have\, \\ y=\dfrac { { \left( { ax+b } \right) \left( { cx+d } \right) } }{ { \left( { ax-b } \right) \left( { cx-d } \right) } } ,\, \, x\ne \dfrac { b }{ a } ,\, \dfrac { d }{ c } \\ y=\dfrac { { ac{ x^{ 2 } }+adx+bcx+bd } }{ { ac{ x^{ 2 } }-adx-bcx+bd } } \\ =\dfrac { { ac{ x^{ 2 } }+\left( { ad+bc } \right) x+bd } }{ { ac{ x^{ 2 } }-\left( { ad+bc } \right) x+bd } } \\ Now, \\ \dfrac { { dy } }{ { dx } } =\dfrac { { \left\{ { 2acx+\left( { ad+bc } \right) } \right\} \left\{ { ac{ x^{ 2 } }-\left( { ad+bc } \right) x+bd } \right\} -\left\{ { 2acx-\left( { ad+bc } \right) \left( { ac{ x^{ 2 } }+\left( { ad+bc } \right) x+bd } \right) } \right\} } }{ { { { \left( { ac{ x^{ 2 } }-\left( { ad+bc } \right) x+bd } \right) }^{ 2 } } } } \\ =\frac { { \left\{ { 2{ a^{ 2 } }{ c^{ 2 } }{ x^{ 3 } }-2ac\left( { ad+bc } \right) { x^{ 2 } }+2acbdx+\left( { ad+bc } \right) ac{ x^{ 2 } }-\left( { ad+bc } \right) bd } \right\} -\left\{ { \left\{ { 2{ a^{ 2 } }{ c^{ 2 } }{ x^{ 3 } }-2ac\left( { ad+bc } \right) { x^{ 2 } }+2acbdx-\left( { ad+bc } \right) ac{ x^{ 2 } }+{ { \left( { ad+bc } \right) }^{ 2 } }x-\left( { ad+bc } \right) bd } \right\} } \right\} } }{ { { { \left\{ { ac{ x^{ 2 } }-\left( { ad+bc } \right) x+\left( { bc } \right) } \right\} }^{ 2 } } } } \\ =\dfrac { { -2ac\left( { ad+bc } \right) { x^{ 2 } }-2\left( { ad+b{ c^{ 2 } }x } \right) +2\left( { ad+bc } \right) bd } }{ { { { \left\{ { ac{ x^{ 2 } }-\left( { ad+bc } \right) x+\left( { bd } \right) } \right\} }^{ 2 } } } } \\ =\dfrac { { -2\left( { ad+bc } \right) \left[ { ac{ x^{ 2 } }+\left( { ad+bc } \right) x+bd } \right] } }{ { { { \left[ { \left( { ax-b } \right) \left( { cx-d } \right) } \right] }^{ 2 } } } } \\ =\dfrac { { -2\left( { ad+bc } \right) \left[ { \left( { ax+b } \right) \left( { cx+d } \right) } \right] } }{ { { { \left[ { \left( { ax-b } \right) \left( { cx-d } \right) } \right] }^{ 2 } } } } \\ \therefore \dfrac { { dy } }{ { dx } } =\dfrac { { -2\left( { ad+bc } \right) y } }{ { \left( { ax-b } \right) \left( { cx-d } \right) } } \end{array}$

$\underset { x\rightarrow 1 }{ lim } \cfrac { (\log { (1+x) } -\log { 2 } )(3.{ 4 }^{ x-1 }-3x) }{ \{ { (7+x) }_{ 3 }^{ 1 })-{ (1+3x) }^{ \cfrac { 1 }{ 2 } })\} \sin { \pi x } }$

$L=\lim _{ x\rightarrow 1 }{ \cfrac { (\log { (1+x) } -\log { 2 } )({ 3.4 }^{ x-1 }-3x) }{ ({ (7+x) }^{ \cfrac { 1 }{ 3 } }-{ (1+3x) }^{ \cfrac { 1 }{ 2 } })\sin { \pi x } } } \\ L=\cfrac { \lim _{ x\rightarrow 1 }{ (\log { (1+x) } -\log { 2 } ) } .\lim _{ x\rightarrow 1 }{ (\cfrac { 3 }{ 4 } .{ 4 }^{ x }-3x) } }{ \lim _{ x\rightarrow 1 }{ [{ (7+x) }^{ \cfrac { 1 }{ 3 } }-{ (1+3x) }^{ \cfrac { 1 }{ 2 } }] } \lim _{ x\rightarrow 1 }{ \sin { \pi x } } }$

Applying L'Hospital's rule

$L=\cfrac { \lim _{ x\rightarrow 1 }{ (\cfrac { 1 }{ x+1 } )\times \lim _{ x\rightarrow 0 }{ (\cfrac { 3 }{ 4 } .{ 4 }^{ x }\log { 4 } -3) } } }{ \lim _{ x\rightarrow 1 }{ \pi \cos { \pi x } } \times \lim _{ x\rightarrow 1 }{ [\cfrac { 1 }{ 3 } { (7+x) }^{ \cfrac { -2 }{ 3 } }-\cfrac { 3 }{ 2 } { (1+3x) }^{ \cfrac { -1 }{ 2 } }] } } \\ =\cfrac { (\cfrac { 1 }{ 1+1 } )\times (\cfrac { 3 }{ 4 } \times { 4 }^{ 0 }\times \log { 4 } -3) }{ \pi \cos { 0 } \times [\cfrac { 1 }{ 3 } { (8) }^{ \cfrac { -2 }{ 3 } }-\cfrac { 3 }{ 2 } { (4) }^{ \cfrac { -1 }{ 2 } }] } \\ =\cfrac { \cfrac { 1 }{ 2 } \times (\cfrac { 3 }{ 4 } \log { 4 } -3) }{ \pi \times [\cfrac { 1 }{ 3 } \times \cfrac { 1 }{ 4 } -\cfrac { 3 }{ 2 } \times \cfrac { 1 }{ 2 } ] } \\ =\cfrac { \cfrac { 3 }{ 2 } (\cfrac { \log { 4 } }{ 4 } -1) }{ \pi \times (\cfrac { 1 }{ 12 } -\cfrac { 3 }{ 4 } ) } =\cfrac { 3 }{ 2 } \times \cfrac { (\cfrac { \log { 4 } }{ 4 } -1) }{ \pi \times (\cfrac { 1-9 }{ 12 } ) } \\ =\cfrac { 3\times 12 }{ 2\times \pi } (\cfrac { \log { 4 } }{ 4 } -1)\times \cfrac { 1 }{ -8 } \\ =\cfrac { -9 }{ 4\pi } (\cfrac { \log { 4 } }{ 4 } -1)$

$lim_{x\to \infty} (10e^{3x} +8)^{5/x}$

$\displaystyle\lim_{x\rightarrow \infty}(10 e^{3x}+8)^{5/x}$

$\Rightarrow \lim_{x\rightarrow \infty} exp(log ((10e^{3x}+8)^{5/x}))$

$=\displaystyle\lim_{x\rightarrow \infty} exp\left[\dfrac{s log(10 e^{3x}+8)}{x}\right]$

$\Rightarrow exp\left[\displaystyle\lim_{x\rightarrow \infty}\left(\dfrac{s log(10e^{3x}+8)}{x}\right)\right]$

$=exp\left[5 \displaystyle\lim_{x\rightarrow \infty}\left(\dfrac{log (10 e^{3x})+log \left(\dfrac{4e^{-3x}}{5}+1\right)}{x}\right)\right]$

but

$\displaystyle\lim_{x\rightarrow \infty} log\left(\dfrac{4e^{-3x}}{5}+1\right)=0$

$\Rightarrow exp\left[5\displaystyle\lim_{x\rightarrow \infty}\left(\dfrac{log(10 e^{3x})}{x}\right)\right]$

Applying L' Hospital's rule

$exp\left[5\displaystyle\lim_{x\rightarrow 0}\dfrac{d}{dx}\dfrac{log(10 e^{3x})}{\dfrac{dx}{dx}}\right]$

$=exp\left(5\displaystyle\lim_{x\rightarrow \infty}\dfrac{3}{1}\right)=exp(5\times 3)=e^{15}$ .

1214434_1317276_ans_e8269c7c50454ccf98645408d9bfb836.jpeg

$\underset { x\rightarrow 0 }{ lim } \dfrac { { sin(\pi cos }^{ 2 }x) }{ { x }^{ 2 } }$ is equal to

$\displaystyle\lim_{x\rightarrow 0}\dfrac{\sin (\pi-\cos^{2}x)}{x^{2}}=\lim_{x\rightarrow 0}\dfrac{\sin \left(\pi\left(1-\sin^{2}x\right)\right)}{x^{2}}$

$=\displaystyle\lim_{x\rightarrow 0}\dfrac{\sin \left(\pi-\pi\sin^{2}x\right)}{x^{2}}$

$=\displaystyle\lim_{x\rightarrow 0}\dfrac{\sin \left(\pi\sin^{2}x\right)}{x^{2}} \left[\because \sin (\pi-\theta)=\sin\theta\right]$

$=\displaystyle\lim_{x\rightarrow 0}\left\{\dfrac{\sin (\pi\sin^{2}x}{\pi\sin^{2}x}.\dfrac{\pi\sin^{2}x}{x^{2}}\right\}$

$=\left(\displaystyle\lim_{x\rightarrow 0}\dfrac{\sin (\pi\sin^{2}x}{\pi\sin^{2}x}\right)\left\{\displaystyle\lim_{x\rightarrow 0}\pi\left(\dfrac{\sin x}{x}\right)^{2}\right\}$

$=1.\pi.1\left[\because \displaystyle\lim_{x\rightarrow 0}\dfrac{\sin x}{x}=1\ and\ x\rightarrow 0\pi \sin^2 x\rightarrow 0\right]$

$=\pi$

Find $\dfrac{dy}{dx}$ , if $x^{y}+y^{x}=1$

Let

$b(x. y)=x^y+y^x-1$

$b'=0$

$b'xdx+b'ydy=0$

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

$\dfrac{dy}{dx}=\dfrac{-b'x}{b'y}=-\dfrac{yx^{y-1}+y^x10y}{xy^{x-1}+x^y10x}$ .

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If $y = \cos ^ { - 1 } \left( \dfrac { 5 \cos x - 12 \sin x } { 13 } \right) , x \in \left( 0 , \dfrac { \pi } { 2 } \right) ,$ then find the value of $dy/dx .$

$\begin{array}{l} y={ \cos ^{ -1 } }\left( { \frac { 5 }{ { 13 } } \cos x-\frac { { 12 } }{ { 13 } } \sin x } \right) \\ y={ \cos ^{ -1 } }\left( { \cos A\cos x-\sin A\sin x } \right) \\ y={ \cos ^{ -1 } }\left( { \cos \left( { A+x } \right) } \right) \\ where\, \, \, \cos A=\frac { 5 }{ { 13 } } \\ y=A+x \\ \frac { { dy } }{ { dx } } =1 \end{array}$

Solve: $\underset{x \rightarrow 0}{lim} \dfrac{(x + 1)^5 -1}{x}$

$We\, \, have \\$

$I={ \lim }_{ x\to 0 }\dfrac { { { { \left( { x+1 } \right) }^{ 5 } }-1 } }{ x } \\$

Using L-hospital rule

$I={ \lim }_{ x\to 0 }\dfrac { { { {5 \left( { x+1 } \right) }^{ 4 } }-0} }{ 1 } \\$

$= 5( 0+1 )^{ 4 } =5$

$\displaystyle x\xrightarrow { lim } 0\frac{log(2+x)-log(2-x)}{x}$

$\lim _{ x\rightarrow 0 }{ \cfrac { \log { \left( 2+x \right) } -\log { \left( 2-x \right) } }{ x } }$

Substituting

$x=0$ we get

$\cfrac { 0 }{ 0 }$ form

So we apply L-Hospital's rule

$=\lim _{ x\rightarrow 0 }{ \cfrac { \cfrac { 1 }{ x+2 } -\cfrac { 1 }{ 2-x } }{ 1 } }$

Substituting

$x=0$

$=\cfrac { 1 }{ 2 } -\cfrac { 1 }{ 2 } =0$

Evaluate $\displaystyle x\xrightarrow { lim }a\, \frac{{x}^{7}-{a}{7}}{x-a}$

$\lim _{ x\rightarrow a }{ \cfrac { { x }^{ 7 }-{ a }^{ 7 } }{ x-a } }$

On substituting

$x=a$ we get

$\cfrac { 0 }{ 0 }$ form

So we apply L-Hospital's rule

$\lim _{ x\rightarrow a }{ \cfrac { 6{ x }^{ 6 }-0 }{ 1 } } =6{ a }^{ 6 }$

Evaluate
$\lim_{x\rightarrow 0}\dfrac{cos(2x^3)-cos(5x^3)}{xsin^2(2x)tan^3(3x)}$

$\lim_{x\to 0}\cfrac{\cos(2x^3)-\cos(5x^3)}{x\sin^2(2x)\tan^3(3x)}$

$\lim_{x\to 0}\cfrac{2\sin((7/2)x^3)\sin(3x^3/2)}{x\sin^2(2x)\tan^3(3x)}$

$=\cfrac{2\times(7/2)x^3(3/2)x^3}{x.4x.27x^3}$

$=\cfrac{7}{18}$

Find $\dfrac{dy}{dx}$ for $y=x^{10}+10^x+10x+10$ .

Given,

$y=x^{10}+10^x+10x+10$ .

Now differentiating both sides with respect to

$x$ we get,

$\dfrac{dy}{dx}=10x^9+10^x.\log 10+10$ .

Solve:
$\displaystyle\lim _{ x\rightarrow 5 }{ \dfrac { x-5 }{ \sqrt { 6x-5 } -\sqrt { 4x+5 } } }$

Given,

$\lim _{x\rightarrow 5}\left(\dfrac{x-5}{\sqrt{6x-5}-\sqrt{4x+5}}\right)$

$=\lim _{x\rightarrow 5}\dfrac{\left(x-5\right)\left(\sqrt{6x-5}+\sqrt{4x+5}\right)}{\left(\sqrt{6x-5}-\sqrt{4x+5}\right)\left(\sqrt{6x-5}+\sqrt{4x+5}\right)}$

$\lim _{x\rightarrow 5}\dfrac{\left(x-5\right)\left(\sqrt{6x-5}+\sqrt{4x+5}\right)}{2\left(x-5\right)}$

$\lim _{x\rightarrow 5}\left(\dfrac{\sqrt{6x-5}+\sqrt{4x+5}}{2}\right)$

$=\dfrac{\sqrt{6\times 5-5}+\sqrt{4\times 5+5}}{2}$

$=\dfrac{2\sqrt{25}}{2}$

$=5$

Solve:
$\displaystyle \lim_{x\rightarrow 8}\dfrac {\sqrt {1+\sqrt {1+x}}-2}{x-8}$

$\lim_{x\to 8}\cfrac{\sqrt{1+\sqrt{1+x}}-2}{x-8}$

$=\lim_{x\to 8}\cfrac{1}{(\sqrt{x+1}+3)(\sqrt{1\sqrt{1+x}}+2)}$

$=\cfrac{1}{(3+3)(2+2)}$

$=\cfrac{1}{24}$

Evaluate : $\lim_{x-1}\frac{x^{7}-2x^{5}+1}{x^{3}-3x^{2}+2}$

$\displaystyle\lim_{x\rightarrow 1}\dfrac{x^{7}-2x^{5}+1}{x^{3}-3x^{2}+2}$

on differentiating both numertor & denomitator w.r.t x we get

$\Rightarrow \displaystyle\lim_{x\rightarrow 1} \dfrac{\dfrac{d}{dx}(x^{7}-2x^{5}+1)}{\dfrac{d}{dx}(x^{3}-3x^{2}+2)}$

$\Rightarrow \displaystyle\lim_{x\rightarrow 1}\dfrac{7x^{6}-10x^{4}}{3x^{2}-6x}$

on putting limit we get

$\Rightarrow \dfrac{7(1)^{6}-10(1)^{4}}{3(1)^{2}-6(1)}$

$\dfrac{7-1}{3-6}$

$=\dfrac{-3}{-3}$

$=1$

Hence

$\displaystyle\lim_{x\rightarrow 1} \dfrac{x^{7}-2x^{5}+1}{x^{3}-3x^{2}+2}$

$=1$

Hence

$\displaystyle\lim_{x\rightarrow 1} \dfrac{x^{7}-2x^{5}1}{x^{3}-3x^{2}+2}=1$ .

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if $\displaystyle \lim_{x\rightarrow 2}\dfrac{ x^5-32}{x^2-4}$

$lim_{x\rightarrow 2}\cfrac{x^5-32}{x^2-4}\\=lim_{x\rightarrow2}\cfrac{1}{x+2}\cfrac{x^5-2^5}{x-2}\\=\lim_{x\rightarrow 2}\cfrac{1}{x+2}lim_{x\rightarrow2}\cfrac{x^5-2^5}{x-2}\\=\cfrac{1}{2+2}5\times2^{5-1}=\cfrac{1}{4}\times5\times2^4\\=20$

Evaluate $\displaystyle \lim _{ x\rightarrow 0 } \dfrac{3\sin{x}-\sin{3x}}{x^{2}}$

$\underset{x\rightarrow 0}{Lt}\dfrac{3\sin x-\sin 3x}{x^2}$

$\underset{x\rightarrow 0}{Lt}\dfrac{3\sin x-3\sin x+4\sin^3x}{x^2}$

$[\sin 3x=3\sin x-4\sin^3x]$

$=\underset{x\rightarrow 0}{Lt}4\left(\dfrac{\sin x}{x}\right)^2\sin x$

$\underset{x\rightarrow 0}{Lt}\dfrac{\sin x}{x}=1$

$=\underset{x\rightarrow 0}{Lt}4\sin x=4\sin 0=0$ .

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$\text{Show that}$ $f\left(x\right)=\dfrac{x\tan{2x}}{\sin{3x}\sin{5x}}$ for $x\neq \,0, f(0)=\dfrac{2}{17} \, \text{is discontinuous at x=0}$ .

$f\left(0\right)=\dfrac{2}{17}$ is discontinuous at

$x=0$

$f\left(x\right)=\dfrac{x\tan{2x}}{\sin{3x}\sin{5x}}$ as

$x\rightarrow\,0$

$=\dfrac{x\dfrac{\tan{2x}}{2x}\times 2x}{\dfrac{\sin{3x}}{3x}\times 3x\dfrac{\sin{5x}}{5x}\times 5x}$

$=\dfrac{x\times 2x}{ 3x\times 5x}$

$=\dfrac{2}{15}\neq f\left(0\right)$

So,

$f\left(x\right)$ is discontinuous at

$x=0$

$\displaystyle \lim_{x\rightarrow 1}\left(\dfrac{1}{1-x}-\dfrac{3}{1-x^{3}}\right)$ is equal to

Consider

$\dfrac{1}{1-x}-\dfrac{3}{1-{x}^{3}}$

$=\dfrac{\left(1-{x}^{3}\right)-3\left(1-x\right)}{\left(1-x\right)\left(1-{x}^{3}\right)}$

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

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

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

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

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

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

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

Now

${\displaystyle\lim}_{x\rightarrow 1}\left(\dfrac{1}{1-x}-\dfrac{3}{1-{x}^{3}}\right)$

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

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

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

Find the slope of tangent to the curve $y=3x^{2}-6$ at the point on it whose x-coordinate is 2.

$y=3x^2-6$

Differentiating both sides, we get,

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

At x=2, slope of tangent =

$6(2)=12$

Therefore, required slope = 12

Differentiate $y=\sin{b{x}^{2}}$ w.r.t $x$

$y=\sin{\left(b{x}^{2}\right)}$

$\dfrac{dy}{dx}=\cos{\left(b{x}^{2}\right)}\dfrac{d}{dx}\left(b{x}^{2}\right)$

$\therefore \dfrac{dy}{dx}=2bx\cos{\left(b{x}^{2}\right)}$

Differentiate $y=\left(x+a\right)\left({x}^{2}+{a}^{2}\right)$ w.r.t $x$

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

$\dfrac{dy}{dx}=3{x}^{2}+{a}^{2}+2ax+0$

$\dfrac{dy}{dx}=3{x}^{2}+2ax+{a}^{2}$

Given:
$y = {\tan ^{ - 1}}\left( {\dfrac{{3x - {x^3}}}{{1 - 3{x^2}}}} \right), - \dfrac{1}{{\sqrt 3 }} < x < \dfrac{1}{{\sqrt 3 }}.$ Then find $\dfrac{dy}{dx}$ .

Given,

$y = {\tan ^{ - 1}}\left( {\dfrac{{3x - {x^3}}}{{1 - 3{x^2}}}} \right)$

or,

$y=3\tan^{-1}x$ [ Using direct formula]

Now differentiating both sides with respect to

$x$ we get,

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

Find $\dfrac{dy}{dx}$ $( x + y ) \frac { d y } { d x } = 1$

Consider given the deferential equation,

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

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

Hence, this is the answer.

If $y=\dfrac { { 2x }^{ 9 } }{ 3 } -\dfrac { 5 }{ 7 } { x }^{ 7 }+6{ x }^{ 3 }$ find $\dfrac { dy }{ dx }$ at x =1

Given:

$y=\dfrac{2{x}^{9}}{3}-\dfrac{5{x}^{7}}{7}+6{x}^{3}$

$\dfrac{dy}{dx}=\dfrac{2}{3}\dfrac{d}{dx}\left({x}^{9}\right)-\dfrac{5}{7}\dfrac{d}{dx}\left({x}^{7}\right)+6\dfrac{d}{dx}\left({x}^{3}\right)$

$=\dfrac{2}{3}\left(9{x}^{8}\right)-\dfrac{5}{7}\left(7{x}^{6}\right)+6\left(3{x}^{2}\right)$

$=6{x}^{8}-5{x}^{6}+18{x}^{2}$

$\therefore \dfrac{dy}{dx}=6{x}^{8}-5{x}^{6}+18{x}^{2}$ at

$x=1$

$=6\times {1}^{8}-5\times{1}^{6}+18\times{1}^{2}=6-5+18=19$

Differentiate : y = 5 sin x + 3 cos x

${\textbf{Step -1: Differentiating the given expression }}$

${\text{Given } y= 5} \sin x + 3\cos x$

${\text{We have to find }}\dfrac{{dy}}{{dx}}$

$\dfrac{d}{{dx}}[5\sin x + 3\cos x]$

$= 5\dfrac{d}{{dx}}\sin x + 3\dfrac{d}{{dx}}\cos x$

$= 5\cos x + 3( - \sin x)$

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

${\textbf{Hence, the answer is }} \bf {5 cos x - 3 sin x.}$

Find the derivative of ${\csc}^{2}{x}$ , by using first principle of derivatives.