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

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

put $x^2=\cos 2\theta$

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

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

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

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

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

$\therefore y=\frac{\pi }{4}+\theta$

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

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

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