Derivative of \(\tan ^{-1}\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right)\) w.r.t. \(\cos ^{-1} x^2\) is

Let \(y=\tan ^{-1}\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right)\) and \(\mathrm{z}=\cos ^{-1}\left(x^2\right)\)
Put \(x^2=\cos 2 \theta \Rightarrow \theta=\frac{1}{2} \cos ^{-1} x^2\)
\(\therefore y =\tan ^{-1}\left(\frac{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}\right) \)
\( \Rightarrow y =\tan ^{-1}\left(\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}\right) \)
\( \Rightarrow y =\tan ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right) \)
\( \Rightarrow y =\tan ^{-1}\left(\tan \left(\frac{\pi}{4}-\theta\right)\right) \)
\( \Rightarrow y =\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x^2 \)
\( \Rightarrow y =\frac{\pi}{4}-\frac{1}{2} \mathrm{z} \)
\( \therefore \quad \frac{\mathrm{dy}}{\mathrm{dz}} =-\frac{1}{2}\)