## Calculus (3rd Edition)

$$\frac{d^3}{dx^3}\sin 2x = -8\cos 2x.$$
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, $$\frac{d^3}{dx^3}\sin 2x = -4\frac{d}{dx} \sin 2x= -8\cos 2x.$$