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

(B)
Put \(x=\cos \theta\), Then \(\theta=\cos ^{-1} x\)
\(\left[\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right] =\left[\frac{\sqrt{1+\cos \theta}+\sqrt{1-\cos \theta}}{2}\right]=\left[\frac{\sqrt{2 \cos ^{2} \frac{\theta}{2}}+\sqrt{2 \sin ^{2} \frac{\theta}{2}}}{2}\right] \)
\( =\frac{\sqrt{2} \cos \frac{\theta}{2}}{2}+\frac{\sqrt{2} \sin \frac{\theta}{2}}{2}=\frac{1}{\sqrt{2}} \cdot \frac{\cos \theta}{2}+\frac{1}{\sqrt{2}} \frac{\sin \theta}{2} \)
\( =\frac{\sin \pi}{4} \cdot \frac{\cos \theta}{2}+\frac{\cos \pi}{4} \cdot \frac{\sin \theta}{2}=\sin \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\)
\(\therefore y=\sin ^{-1}\left(\sin \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right)=\frac{\pi}{4}+\frac{\theta}{2}=\frac{\pi}{4}+\frac{\cos ^{-1} x}{2}\) \(\therefore \frac{d y}{d x}=\frac{1}{2} \times \frac{-1}{\sqrt{1-x^{2}}}=\frac{-1}{2 \sqrt{1-x^{2}}}\)