1. The problem is to find the derivative $\frac{du}{d\theta}$ given $u = \sin^2 \theta$.\n\n2. Recall the chain rule for derivatives: if $u = f(g(\theta))$, then $\frac{du}{d\theta} = f'(g(\theta)) \cdot g'(\theta)$.\n\n3. Here, $u = (\sin \theta)^2$, so $f(x) = x^2$ and $g(\theta) = \sin \theta$.\n\n4. Differentiate $f(x) = x^2$ with respect to $x$: $f'(x) = 2x$.\n\n5. Differentiate $g(\theta) = \sin \theta$ with respect to $\theta$: $g'(\theta) = \cos \theta$.\n\n6. Apply the chain rule: $$\frac{du}{d\theta} = 2 \sin \theta \cdot \cos \theta.$$\n\n7. Note that $2 \sin \theta \cos \theta = \sin 2\theta$ by the double-angle identity for sine.\n\n8. Therefore, $$\frac{du}{d\theta} = \sin 2\theta.$$\n\nThis matches the given expression and confirms the derivative.