If $y={\mathrm{Tan}}^{-1}\left[\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right]$ for $0<\left|x\right|<1\text{ find }\frac{dy}{dx}$

Given $y={\mathrm{Tan}}^{-1}\left[\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right]$

put $x^2=\cos 2\theta$

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

$$\begin{gathered} y={\mathrm{Tan}}^{-1}\left[\frac{\sqrt{1+\cos 2\theta }+\sqrt{1-\cos 2\theta }}{\sqrt{1+\cos 2\theta }-\sqrt{1-\cos 2\theta }}\right]\begin{array}{r} \\ \left(\because 1+\cos 2\theta =2{\cos }^2\theta \\ 1-\cos 2\theta =2{\sin }^2\theta \right) \\ \end{array} \end{gathered}$$

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

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

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

diff ‘$y$’ w.r.t. ‘$x$’ on B.S

$\frac{dy}{dx}=\frac{1}{2}\frac{-1}{\sqrt{1-x^4}}\left(2x\right)=\frac{-x}{\sqrt{1-x^4}}\left(\because \frac{d}{dx}\left({\cos }^{-1}\left(x\right)\right)=\frac{-1}{\sqrt{1-x^2}}\right)$