Answer
$4 \cos (2x) \cos [1-\sin (2x)]-\sin [1-\sin 2x]$
Work Step by Step
Our aim is to solve the function $f(x)=\cos^2[1-\sin 2x]$
We differentiate both sides with respect to $x$.
$f^{\prime}(x)=\dfrac{d}{dx}[\cos^2[1-\sin 2x]]$
or, $=2 \cos [1-\sin 2x] [-\sin (1-\sin 2x)] \dfrac{d}{dx} [ 1-\sin 2x]$
or, $=2 \cos [1-\sin 2x] [-\sin (1-\sin 2x)] \times (-2 \cos 2x)$
or, $=4 \cos (2x) \cos [1-\sin (2x)]-\sin [1-\sin 2x]$
