If \(f(x)=\sin ^{-1}\left(\sqrt{\frac{1-x}{2}}\right)\), then \(f^{\prime}(x)=\)

(B)
\(\frac{d}{d x}\left[\sin ^{-1}\left(\sqrt{\frac{1-x}{2}}\right)\right]=\frac{1}{\sqrt{1-\left(\sqrt{\frac{1-x}{2}}\right)^{2}}} \cdot \frac{d}{d x}\left(\sqrt{\frac{1-x}{2}}\right)\)
\(=\frac{1}{\sqrt{1-\frac{1-x}{2}}} \times \frac{1}{2 \sqrt{\frac{1-x}{2}}} \times \frac{d}{d x}\left(\frac{1-x}{2}\right)=\frac{1}{\sqrt{\frac{2-1+x}{2}}} \times \frac{1}{\sqrt{2} \cdot \sqrt{1-x}} \times \frac{-1}{2}\)
\(=\frac{\sqrt{2}}{\sqrt{1+x}} \times \frac{1}{\sqrt{2} \cdot \sqrt{1-x}} \times \frac{-1}{2}=\frac{-1}{2 \sqrt{1-x^{2}}}\)
This problem can also be solved as follows
\(\begin{array}{l}
f(x)=\sin ^{-1}\left(\sqrt{\frac{1-x}{2}}\right) \\
\text { Put } x=\cos \theta \Rightarrow \sqrt{\frac{1-x}{2}}=\sqrt{\frac{1-\cos \theta}{2}}=\sqrt{\frac{2 \sin ^{2} \frac{\theta}{2}}{2}}=\sin \frac{\theta}{2} \\
\therefore f(x)=\sin ^{-1}\left(\sin \frac{\theta}{2}\right)=\frac{\theta}{2}=\frac{\cos ^{-1} x}{2} \\
\therefore f^{\prime}(x)=\frac{-1}{2 \sqrt{1-x^{2}}}
\end{array}\)