Answer
$$f'(\theta)=\frac{1}{2}\sin{2\theta}+\theta\cos{2\theta}$$
Work Step by Step
$$f(\theta)=\theta\cos\theta\sin\theta$$
$$f(\theta)=(\theta\cos\theta)\sin\theta$$
Apply the Product Rule with 2 elements $\theta\cos\theta$ and $\sin\theta$ first: $$f'(\theta)=(\theta\cos\theta)'\sin\theta+(\theta\cos\theta)(\sin\theta)'$$
$$f'(\theta)=(\theta\cos\theta)'\sin\theta+(\theta\cos\theta)\cos\theta$$
$$f'(\theta)=(\theta\cos\theta)'\sin\theta+\theta\cos^2\theta$$
Now apply the Product Rule for $(\theta\cos\theta)'$: $$f'(\theta)=[(\theta)'\cos\theta+\theta(\cos\theta)']\sin\theta+\theta\cos^2\theta$$
$$f'(\theta)=[\cos\theta-\theta\sin\theta]\sin\theta+\theta\cos^2\theta$$
$$f'(\theta)=\sin\theta\cos\theta-\theta\sin^2\theta+\theta\cos^2\theta$$
$$f'(\theta)=\frac{1}{2}\sin{2\theta}+\theta(\cos^2\theta-\sin^2\theta)$$
$$f'(\theta)=\frac{1}{2}\sin{2\theta}+\theta\cos{2\theta}$$
