Prove the following

Solution:

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

Putting $x^{2}=\cos 2 \theta \quad \therefore \theta=\frac{1}{2} \cos ^{-1} x^{2}$

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

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

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

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

$y=\tan ^{-1}\left[\frac{\frac{\cos \theta}{\cos \theta}+\frac{\sin \theta}{\cos \theta}}{\frac{\cos \theta}{\cos \theta}-\frac{\sin \theta}{\cos \theta}}\right]$

$y=\tan ^{-1}\left[\frac{1+\tan \theta}{1-\tan \theta}\right]$

$y=\tan ^{-1}\left[\frac{\tan \frac{\pi}{4}+\tan \theta}{1-\tan \frac{\pi}{4} \cdot \tan \theta}\right]$

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

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

Differentiating both sides w.r.t. $x$

$\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\pi}{4}\right)+\frac{1}{2} \frac{d}{d x}\left(\cos ^{-1} x^{2}\right)$

$=0+\frac{1}{2} \times \frac{-1}{\sqrt{1-x^{4}}} \cdot \frac{d}{d x}\left(x^{2}\right)=\frac{-1.2 x}{2 \sqrt{1-x^{4}}}=-\frac{x}{\sqrt{1-x^{4}}}$

Thus, $\quad \frac{d y}{d x}=-\frac{x}{\sqrt{1-x^{4}}}$