Question

Differentiate the given function w.r.t. $x$.
$y=\sin \left({\tan }^{-1}{e}^{-x}\right)$

Answer 1

Let $\sin \left({\tan }^{-1}{e}^{-x}\right)$
Thus by chain rule,
$\frac{dy}{dx}=\frac{d}{dx}\left[\sin \right({\tan }^{-1}{e}^{-x}\left)\right]$
$=\cos \left({\tan }^{-1}{e}^{-x}\right).\frac{d}{dx}\left({\tan }^{-1}{e}^{-x}\right)$
$=\cos \left({\tan }^{-1}{e}^{-x}\right).\frac{1}{1+(e^{-x}{)}^2}\frac{d}{dx}\left(e^{-x}\right)$
$=\frac{\cos \left({\tan }^{-1}{e}^{-x}\right)}{1+e^{-2x}}.e^{-x}.\frac{d}{dx}(-x)$
$=\frac{e^{-x}\cos \left({\tan }^{-1}{e}^{-x}\right)}{1+e^{-2x}}.(-1)$
$=\frac{-e^{-x}\cos \left({\tan }^{-1}{e}^{-x}\right)}{1+e^{-2x}}$