Answer
$g'(\theta)=-8\sin{16\theta}.$
Work Step by Step
$u=\cos{8\theta}$; $\dfrac{du}{d\theta}=-8\sin{8\theta}$
$\dfrac{d}{du}g(u)=2u$
$\dfrac{d}{d\theta}g(\theta)=\dfrac{d}{du}g(u)\times\dfrac{du}{d\theta}=(2cos{8\theta})(-8\sin{8\theta})$
$=-16\cos{8\theta}\sin{8\theta}$
$=-8\sin{16\theta}.$
