Answer
$$ k'(\theta )
=2\theta \sin\theta (\sin\theta+\theta \cos\theta ) .$$
Work Step by Step
Recall the product rule: $(uv)'=u'v+uv'$
Recall that $(x^n)'=nx^{n-1}$
Recall that $(\sin x)'=\cos x$.
Since $ k(\theta )= \theta^2 \sin^2 \theta $, then using the product rule, the derivative is giving by
$$ k'(\theta )= (\theta^2)' \sin^2 \theta+\theta^2( \sin^2 \theta)'\\
=2\theta \sin^2 \theta+2\theta^2 \sin \theta \cos\theta\\
=2\theta \sin\theta (\sin\theta+\theta \cos\theta ) .$$
