Answer
$$2\cos2x \cos^2x-\sin^2 2x$$
Work Step by Step
Recall the product rule: $(uv)'=u'v+uv'$
Recall that $(\sin x)'=\cos x$.
Recall that $(\cos x)'=-\sin x$.
Since $ y=\sin 2x \cos^2x $, then by the product rule, the derivative $ y'$ is given by
$$ y'=\cos 2x (2) \cos^2x+\sin 2x (2)\cos x (-\sin x)\\=2\cos2x \cos^2x-2\sin 2x \cos x \sin x.$$
Using the formula $\sin 2x=2\cos x \sin x$, we write:
$$2\cos2x \cos^2x-\sin^2 2x$$
