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

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

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Let $y=	an ^{-1}\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}ight)$ and $\mathrm{z}=\cos ^{-1}\left(x^2ight)$ Put $x^2=\cos 2 	heta \Rightarrow 	heta=\frac{1}{2} \cos ^{-1} x^2$ $\begin{aligned} 	herefore \quad y & =	an ^{-1}\left(\frac{\sqrt{1+\cos 2 	heta}-\sqrt{1-\cos 2 	heta}}{\sqrt{1+\cos 2 	heta}+\sqrt{1-\cos 2 	heta}}ight) \ \Rightarrow y & =	an ^{-1}\left(\frac{\cos 	heta-\sin 	heta}{\cos 	heta+\sin 	heta}ight) \ \Rightarrow y & =	an ^{-1}\left(\frac{1-	an 	heta}{1+	an 	heta}ight) \ \Rightarrow y & =	an ^{-1}\left(	an \left(\frac{\pi}{4}-	hetaight)ight) \ \Rightarrow y & =\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x^2 \ \Rightarrow y & =\frac{\pi}{4}-\frac{1}{2} \mathrm{z} \ 	herefore \quad \frac{\mathrm{dy}}{\mathrm{dz}} & =-\frac{1}{2} \end{aligned}$

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.…

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