Question

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

Answer 1

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