The derivative of \(\tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)\) is

\(y=\tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right) \text { let } x=\cos 2 \theta \)
\( \Rightarrow \theta=\frac{1}{2} \cos ^{-1} x \)
\( \Rightarrow y=\tan ^{-1}\left(\frac{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}\right) \)
\( \Rightarrow y=\tan ^{-1}\left(\frac{\sqrt{2} \cos \theta-\sqrt{2} \sin \theta}{\sqrt{2} \cos \theta+\sqrt{2} \sin \theta}\right)=\tan ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right) \)
\( \Rightarrow y=\tan ^{-1} \tan \left(\frac{\pi}{4}-\theta\right)=\frac{\pi}{4}-\theta=\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x \)
\( \Rightarrow \frac{d y}{d x}=0-\frac{1}{2} \times \frac{-1}{\sqrt{1-x^2}}=\frac{1}{2 \sqrt{1-x^2}}\)