Question

If $y={\tan }^{-1}\left[\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right]$, what would be $\frac{dy}{dx}$

• A. $\frac{-x}{\sqrt{1-x^4}}$
• B. $\frac{1}{\sqrt{1-x^4}}$
• C. $\frac{x}{\sqrt{1-x^4}}$
• D. none of these

Answer 1

Answer: (C)

$\frac{x}{\sqrt{1-x^4}}$

given that,

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

let$x^2=\sin 2\theta$

$2xdx=\cos 2\theta d\theta$

$\therefore \frac{d\theta }{dx}=\frac{x}{\sqrt{1-x^4}}$

$1+x^2=1+\sin 2\theta ={(\sin \theta +\cos \theta )}^2$

$1-x^2=1+\sin 2\theta ={(\sin \theta -\cos \theta )}^2$

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

$=\frac{\sin \theta +\cos \theta +\sin \theta -\cos \theta }{\sin \theta +\cos \theta +\cos \theta -\sin \theta }$

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

$\therefore \frac{dy}{dx}=\frac{d\theta }{dx}=\frac{x}{\sqrt{1-x^4}}$