Question

Differentiate $y={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ with respect to $x,0<x<1$,

Answer 1

$y={\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)$

Let $x=\tan \theta$

$y={\cos }^{-1}\left(\cos 2\theta \right)$

$=2\theta$

$=2{\tan }^{-1}x$

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