#### Answer

$$h'(\theta)=\frac{\theta(2\sin\theta+\theta\cos\theta)}{\cos^2(\theta^2\sin\theta)}$$

#### Work Step by Step

$$h'(\theta)=(\tan(\theta^2\sin\theta))'=\frac{1}{\cos^2(\theta^2\sin\theta)}(\theta^2\sin\theta)'=\frac{1}{\cos^2(\theta^2\sin\theta)}((\theta^2)'\sin\theta+\theta^2(\sin\theta)')=
\frac{1}{\cos^2(\theta^2\sin\theta)}(2\theta\sin\theta+\theta^2\cos\theta)=\frac{\theta(2\sin\theta+\theta\cos\theta)}{\cos^2(\theta^2\sin\theta)}$$