Answer
\[y'=\frac{-1}{2\sqrt{1-x^2}}\]
Work Step by Step
\[y=arc\tan\sqrt{\frac{1-x}{1+x}}\]
Put $x=\cos 2z\;\;\;...(1)$
\[arc\tan\sqrt{\frac{1-x}{1+x}}=arc\tan\sqrt{\frac{1-\cos 2z}{1+\cos 2z}}\]
\[arc\tan\sqrt{\frac{1-x}{1+x}}=arc\tan\sqrt{\frac{2\sin^2 z}{2\cos^2 z}}\]
\[arc\tan\sqrt{\frac{1-x}{1+x}}=arc\tan\sqrt{\tan z}\]
\[arc\tan\sqrt{\frac{1-x}{1+x}}=z\]
Using (1)
\[arc\tan\sqrt{\frac{1-x}{1+x}}=\frac{1}{2}\cos^{-1}x\]
\[\Rightarrow y=\frac{1}{2}\cos^{-1}x\;\;\;...(2)\]
Differentiate (2) with respect to $x$
\[y'=\frac{1}{2}\left(\frac{-1}{\sqrt{1-x^2}}\right)\]
Hence \[y'=\frac{-1}{2\sqrt{1-x^2}}\]
