1. The problem asks to find $\frac{dx}{d\theta}$ when $x = \cos \theta \sin \theta$. 2. We use the product rule for differentiation: if $x = u(\theta) v(\theta)$, then $\frac{dx}{d\theta} = u'(\theta) v(\theta) + u(\theta) v'(\theta)$. 3. Here, let $u(\theta) = \cos \theta$ and $v(\theta) = \sin \theta$. 4. Differentiate each: $u'(\theta) = -\sin \theta$ and $v'(\theta) = \cos \theta$. 5. Apply the product rule: $$\frac{dx}{d\theta} = (-\sin \theta)(\sin \theta) + (\cos \theta)(\cos \theta)$$ 6. Simplify: $$\frac{dx}{d\theta} = -\sin^2 \theta + \cos^2 \theta$$ 7. Recall the trigonometric identity $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$. 8. Therefore, $$\frac{dx}{d\theta} = \cos 2\theta$$ 9. So the derivative of $x = \cos \theta \sin \theta$ with respect to $\theta$ is $\cos 2\theta$.