Question

If $y={\tan }^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right);x^2\le 1$ then find $\frac{dy}{dx}$.

Answer 1

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

Put $x^2=\cos 2\theta \Rightarrow \theta =\frac{1}{2}{\cos }^{-1}{x}^2$--- (1)

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

$\Rightarrow {\tan }^{-1}\left(\frac{\sqrt{2}\cos \theta +\sqrt{2}\sin \theta }{\sqrt{2}\cos \theta -\sqrt{2}\sin \theta }\right)$ [ Since, $1+\cos 2\theta =2{\cos }^2\theta$ and $1-\cos 2\theta =2{\sin }^2\theta$]

$\Rightarrow {\tan }^{-1}\left(\frac{1+\tan \theta }{1-\tan \theta }\right)$

$\Rightarrow {\tan }^{-1}\left(\frac{\tan (\frac{\pi }{4}+\tan \theta )}{1-\tan \left(\frac{\pi }{4}\right)\tan \theta }\right)$

$\Rightarrow {\tan }^{-1}\tan (\frac{\pi }{4}+\theta )=\frac{\pi }{4}+\theta$

$\Rightarrow y=\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}{x}^2$

$\frac{dy}{dx}=0-\frac{1}{2}\times \frac{1}{\sqrt{1-x^4}}\times 2x=-\frac{x}{\sqrt{1-x^4}}$