Answer
$$dy = \sin 2xdx$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\sin ^2}x \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{{\sin }^2}x} \right] \cr
& f'\left( x \right) = 2\sin x\frac{d}{{dx}}\left[ {\sin x} \right] \cr
& f'\left( x \right) = 2\sin x\cos x \cr
& or \cr
& f'\left( x \right) = \sin 2x \cr
& {\text{Write in the form }}dy = f'\left( x \right)dx \cr
& dy = \sin 2xdx \cr} $$