Question

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

Answer 1

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