Answer
$H'(\theta)=\sin \theta+\theta \cos \theta$
$H''(\theta)= 2\cos \theta-\theta \sin \theta$
Work Step by Step
$H'(\theta)= \frac{d}{d\theta}(\theta\sin \theta)= 1\times\sin \theta+\theta\times\cos\theta$
$=\sin \theta+\theta \cos \theta$
$H''(\theta)=\frac{d}{d\theta}(\sin\theta)+\frac{d}{d\theta}(\theta\cos\theta)$
$\cos \theta+ 1\times\cos \theta+ \theta\times-\sin\theta$
$=2\cos \theta-\theta \sin \theta$