The derivative of $\sin ^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right)$ w.r.t. $\cos ^{-1} x$ is

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Put $x=\cos 	heta$, Then $	heta=\cos ^{-1} x$ $\begin{aligned}\left[\frac{\sqrt{1+x}+\sqrt{1-x}}{2}ight] &=\left[\frac{\sqrt{1+\cos 	heta}+\sqrt{1-\cos 	heta}}{2}ight]=\left[\frac{\sqrt{2 \cos ^{2} \frac{	heta}{2}}+\sqrt{2 \sin ^{2} \frac{	heta}{2}}}{2}ight] \ &=\frac{\sqrt{2} \cos \frac{	heta}{2}}{2}+\frac{\sqrt{2} \sin \frac{	heta}{2}}{2}=\frac{1}{\sqrt{2}} \cdot \frac{\cos 	heta}{2}+\frac{1}{\sqrt{2}} \frac{\sin 	heta}{2} \ &=\frac{\sin \pi}{4} \cdot \frac{\cos 	heta}{2}+\frac{\cos \pi}{4} \cdot \frac{\sin 	heta}{2}=\sin \left(\frac{\pi}{4}+\frac{	heta}{2}ight) \end{aligned}$ $	herefore y=\sin ^{-1}\left(\sin \left(\frac{\pi}{4}+\frac{	heta}{2}ight)ight)=\frac{\pi}{4}+\frac{	heta}{2}=\frac{\pi}{4}+\frac{\cos ^{-1} x}{2}$ $	herefore \frac{d y}{d x}=\frac{1}{2} 	imes \frac{-1}{\sqrt{1-x^{2}}}=\frac{-1}{2 \sqrt{1-x^{2}}}$

The derivative of $\sin ^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right)$ w.r.t. $\cos ^{-1} x$ is

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