Answer
$y'=-{\frac{1}{2(1+x)\sqrt{\frac{1-x}{1+x}}}}$
Work Step by Step
$y=arctan(\sqrt{\frac{1-x}{1+x}})\\
y'=\frac{1}{1+(\sqrt{\frac{1-x}{1+x}})^2}\times{\frac{1}{2\sqrt{\frac{1-x}{1+x}}}}\times{-\frac{2}{(1+x)^2}}\\
y'=\frac{1+x}{2}\times{\frac{1}{2\sqrt{\frac{1-x}{1+x}}}}\times{-\frac{2}{(1+x)^2}}\\
y'=-{\frac{1}{2\sqrt{\frac{1-x}{1+x}}}}\times{\frac{1}{(1+x)}}\\
y'=-{\frac{1}{2(1+x)\sqrt{\frac{1-x}{1+x}}}}$
