Question

If $y=\sin \left\{{\tan }^{-1}\left[\frac{1-x}{1+x}\right]\right\}$ prove that $\frac{dy}{dx}=\frac{-x}{\sqrt{1-x^2}}$

Answer 1

Given that,

$y=\sin \left\{{\tan }^{-1}\left[\frac{1-x}{1+x}\right]\right\}$

$y=\sin \left[{\tan }^{-1}{\tan }^{-1}x\right]$

$y=\sin \left[\frac{\pi }{4}-{\tan }^{-1}x\right]$

Differentiation to $x$

$\frac{dy}{dx}=\cos \left(\frac{\pi }{4}-{\tan }^{-1}x\right)\times 0\left(0-\frac{1}{1+x^2}\right)$

$\frac{dy}{dx}=\frac{-\cos \left(\frac{\pi }{4}-{\tan }^{-1}x\right)}{1+x^2}$

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