Question

Differentiate the following w.r.to $x$

${\tan }^{-1}\sqrt{\frac{1-x}{1+x}}$

Answer 1

let $y={\tan }^{-1}\sqrt{\frac{1+x^2}{1+x^2}}$

let$x=\cos 2\theta ⟹dx=-2\sin 2\theta$ $\because \frac{d\theta }{dx}=\frac{-1}{\sqrt{1-x^2}}$

$1-x=1-\cos 2\theta$

$=2{\sin }^2\theta$

$1+x=1+\cos 2\theta$

$=2{\cos }^2\theta$

$\frac{1-x}{1+x}={\tan }^2\theta ⟹\sqrt{\frac{1+x^2}{1+x^2}}=\tan \theta$

$\because y={\tan }^{-1}\sqrt{\frac{1+x^2}{1+x^2}}$

$={\tan }^{-1}\left(\tan \theta \right)$

$=0$

$\frac{dy}{dx}=\frac{d\theta }{dx}=\frac{-1}{2\sqrt{1-x^2}}$