Question

Solve : $y={\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right),0<x<1$

Answer 1

$y={\sin }^{-1}\left(\frac{1-a^2}{1+x^2}\right)$

Putting $x=\tan \theta$

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

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

$y={\sin }^{-1}\left(\sin (\frac{\pi }{2}-2\theta )\right)$

$y=\frac{\pi }{2}-2\theta$

$y=\frac{\pi }{2}-2{\tan }^{-1}x$

Differentiating both sides w.r.t. $x$

$\frac{dy}{dx}=\frac{d(\frac{\pi }{2}-2{\tan }^{-1}x)}{dx}$

$=0-\frac{2d\left({\tan }^{-1}x\right)}{dx}$

$=-2\left(\frac{1}{1+x^2}\right)$

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