If $f\left(x\right)=\sqrt{1-\sin 2x},$ ,then f '(x) is equal to

$f\left(x\right)=\sqrt{1-\sin 2x}=\sqrt{(\cos x-\sin x{)}^2}=|\cos x-\sin x|$

$=\left\{\begin{array}{cc}\cos x-\sin x,& \text{ for }0\le x\le \pi /4\\ -(\cos x-\sin x),& \text{ for }\pi /4<x\le \pi /2\end{array}\right.$

$f\text{'}\left(x\right)=\left\{\begin{array}{ll}-\sin x-\cos x& 0<x<\frac{\mathrm{\pi }}{4}\\ \sin x+\cos x& \frac{\mathrm{\pi }}{4}<x<\frac{\mathrm{\pi }}{2}\end{array}\right.$