Question

differentiate with respect to x using chain rule​

Answer 1

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

Given function is

$\rm :\longmapsto\:log \: sin( \sqrt{cosx} \: )$

Let we assume that

$\rm :\longmapsto\:y \: = \: log \: sin( \sqrt{cosx} \: )$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y \: = \dfrac{d}{dx} \: log \: sin( \sqrt{cosx} \: )$

We know,

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

Using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{sin \sqrt{cosx} }\dfrac{d}{dx}sin (\sqrt{cosx} )$

We know,

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

Using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{sin \sqrt{cosx} }\cos (\sqrt{cosx} )\dfrac{d}{dx} \sqrt{cosx}$

We know,

$\boxed{ \rm{ \frac{cosx}{sinx} = cotx}}$

and

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

Using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = cot( \sqrt{cosx}) \times \dfrac{1}{2 \sqrt{cosx} } \dfrac{d}{dx}cosx$

We know,

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

So, using this result, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{cot( \sqrt{cosx})}{2 \sqrt{cosx} } ( - sinx)$

$\bf :\longmapsto\:\dfrac{dy}{dx} = \dfrac{ - \: sinx \: cot( \sqrt{cosx})}{2 \sqrt{cosx} }$

Additional Information :-

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

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

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

$\boxed{ \rm{ \dfrac{d}{dx} {e}^{x} = {e}^{x}}}$

$\boxed{ \rm{ \dfrac{d}{dx} {a}^{x} = {a}^{x} \: loga}}$

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

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

Answer 2

Answer:

The above answer is right