Question

Differentiate using chain rule:$y=\left[{cos}^2\right({tan}^{-1}\left(sin\right({cot}^{-1}x\left)\right)\left)\right]$

Answer 1

Let $y={\cos }^2\left({\tan }^{-1}\left(\sin \left({\cot }^{-1}x\right)\right)\right)$

Let $u={\cot }^{-1}x\Rightarrow \frac{du}{dx}=\frac{-1}{1+x^2}$

Let $v=\sin u\Rightarrow \frac{dv}{du}=\cos u$

$\Rightarrow y={\cos }^2\left({\tan }^{-1}v\right)$

Let $w={\tan }^{-1}v\Rightarrow \frac{dw}{dv}=\frac{1}{1+v^2}$

$\Rightarrow y={\cos }^{2}w\Rightarrow \frac{dy}{dw}=-2\cos w\sin w$

Using chain rule of differentiation, we have

$\frac{dy}{dx}=\frac{dy}{dw}\times \frac{dw}{dv}\times \frac{dv}{du}\times \frac{du}{dx}$

$=-2\cos w\sin w\times \frac{1}{1+v^2}\times \cos u\times \frac{-1}{1+x^2}$

$=-2\cos \left({\tan }^{-1}v\right)\sin \left({\tan }^{-1}v\right)\times \frac{1}{1+v^2}\times \cos u\times \frac{-1}{1+x^2}$ where $w={\tan }^{-1}v$

$=-2\cos \left({\tan }^{-1}\left(\sin u\right)\right)\sin \left({\tan }^{-1}\left(\sin u\right)\right)\times \frac{1}{1+{\sin }^{2}u}\times \cos u\times \frac{-1}{1+x^2}$ where $v=\sin u$

$=-2\cos \left({\tan }^{-1}\left(\sin \left({\cot }^{-1}x\right)\right)\right)\sin \left({\tan }^{-1}\left(\sin \left({\cot }^{-1}x\right)\right)\right)\times \frac{1}{1+{\sin }^2\left({\cot }^{-1}x\right)}\times \cos \left({\cot }^{-1}x\right)\times \frac{-1}{1+x^2}$ where $u={\cot }^{-1}x$

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