Answer
$$\frac{d^3}{dx^3}\sin 2x = -8\cos 2x.$$
Work Step by Step
Recall that $(\sin x)'=\cos x$.
Recall that $(\cos x)'=-\sin x$.
We have
$$\frac{d}{dx} \sin 2x=2\cos 2x $$
and
$$\frac{d^2}{dx^2}\sin 2x =1\frac{d}{dx} \cos 2x=-4\sin 2x.$$
Hence, $$\frac{d^3}{dx^3}\sin 2x = -4\frac{d}{dx} \sin 2x= -8\cos 2x.$$
