Question

Find $\frac{dy}{dx}$ of $y={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right),0<x<1$

Answer 1

$y={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$
Put $x=\tan \theta$
$\Rightarrow y={\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)$
$={\cos }^{-1}\cos \left(2\theta \right)=2\theta =2{\tan }^{-1}x$
$\therefore \frac{dy}{dx}=\frac{2}{1+x^2}$