## Calculus 10th Edition

$h'(x)=2\cos{4x}.$
You can use the product rule to differentiate but it would be much easier to use the identity $(2\sin{x}\cos{x}=\sin{2x})$ to rewrite $h(x)$ as $\dfrac{\sin{4x}}{2}$ $u=4x$; $\dfrac{du}{dx}=4$ $\dfrac{d}{du}h(u)=\dfrac{\cos{u}}{2}$ $\dfrac{d}{dx}h(x)=\dfrac{d}{du}h(u)\times\dfrac{du}{dx}=2\cos{4x}.$