Question

Differentiate with respect to x :

${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right),0<x<1$

Answer 1

Let $y={\cos }^{{}_1}\left(\frac{1-x^2}{1+x^2}\right)$, where $0<x<1$.

Putting $x=\tan \theta \Rightarrow \theta ={\tan }^{-1}x$,

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

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

$y=2\theta$

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

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