Question

differentiate with respect to x using chain rule​

Answer 1

Step-by-step explanation:

step by step solution is in the taken snapshot..

Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:log( \: x + \sqrt{ {x}^{2} - 1} \: )$

Let assume that

$\rm :\longmapsto\:y \: = \: log( \: x + \sqrt{ {x}^{2} - 1} \: )$

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

$\rm :\longmapsto\:\dfrac{d}{dx} y \: =\dfrac{d}{dx} \: log( \: x + \sqrt{ {x}^{2} - 1} \: )$

We know,

$\boxed{ \rm{\dfrac{d}{dx} \: logx \: = \: \frac{1}{x}}}$

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

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{x + \sqrt{ {x}^{2} - 1 } }\bigg(\dfrac{d}{dx} \: x + \dfrac{d}{dx}\sqrt{ {x}^{2} - 1} \: \bigg)$

We know,

$\boxed{ \rm{ \dfrac{d}{dx} \: x \: = \: 1}}$

and

$\boxed{ \rm{ \dfrac{d}{dx} \: \sqrt{x} \: = \: \frac{1}{2 \sqrt{x} } }}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{x + \sqrt{ {x}^{2} - 1} }\bigg(1 + \dfrac{1}{2 \sqrt{ {x}^{2} - 1}}\dfrac{d}{dx}( {x}^{2} - 1)\bigg)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{x + \sqrt{ {x}^{2} - 1} }\bigg(1 + \dfrac{1}{2 \sqrt{ {x}^{2} - 1}}(2x - 0)\bigg)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{x + \sqrt{ {x}^{2} - 1} }\bigg(1 + \dfrac{1}{2 \sqrt{ {x}^{2} - 1}}(2x)\bigg)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{x + \sqrt{ {x}^{2} - 1} }\bigg(1 + \dfrac{1}{ \sqrt{ {x}^{2} - 1}}(x)\bigg)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{x + \sqrt{ {x}^{2} - 1} }\bigg(1 + \dfrac{x}{ \sqrt{ {x}^{2} - 1}}\bigg)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{x + \sqrt{ {x}^{2} - 1} }\bigg(\dfrac{ \sqrt{ {x}^{2} - 1} + x}{ \sqrt{ {x}^{2} - 1}}\bigg)$

$\bf :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{ \sqrt{ {x}^{2} - 1} }$

Additional Information :-

$\boxed{ \rm{ \dfrac{d}{dx}x = 1}}$

$\boxed{ \rm{ \dfrac{d}{dx}k = 0}}$

$\boxed{ \rm{ \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} }}$

$\boxed{ \rm{ \dfrac{d}{dx}sinx = cosx}}$

$\boxed{ \rm{ \dfrac{d}{dx}cosx = - \: sinx}}$

$\boxed{ \rm{ \dfrac{d}{dx}cosecx = - \: cosecx \: cotx}}$

$\boxed{ \rm{ \dfrac{d}{dx}secx = \: secx \: tanx}}$

$\boxed{ \rm{ \dfrac{d}{dx}tanx = {sec}^{2}x}}$

$\boxed{ \rm{ \dfrac{d}{dx}cotx = - {cosec}^{2}x}}$