<s>[INST] How to simplify the trigonometric functions sine and cosine in a derivative? I am learning calculus and came across the following problem:

Find $\frac{dx}{d\theta}$ when $x=\sin^2\theta \cos^3\theta$

I solved this using the product rule.
$$u= \sin^2\theta \qquad v=\cos^3\theta$$
$$u'= \sin2\theta \qquad v'=-3\cos^2\theta \sin\theta$$
$$\begin{align}
\frac{dx}{d\theta}&= \sin^2\theta(-3\cos^2\theta \sin\theta) +\cos^3\theta(\sin2\theta) \\
&= -3\cos^2\theta \sin^3\theta + \cos^3\theta \sin2\theta
\end{align}$$

The problem is that this does not match any of the options the exercise provides, so how can I simplify it further? Are there common methods for simplification?

 [/INST] With
$u = \sin^2\theta, \; v = \cos^3 \theta, \tag 1$
the chain rule yields
$u'(\theta) = \dfrac{d\sin^2 \theta}{d\theta} = \dfrac{d\sin^2 \theta}{d\sin \theta}\dfrac{d\sin \theta}{d\theta} = 2(\sin \theta)(\cos\theta), \tag 2$
and
$v'(\theta) = \dfrac{d\cos^3\theta}{d\cos \theta} \dfrac{d\cos \theta}{d\theta} = 3(\cos^2 \theta)(-\sin \theta) = -3(\cos^2 \theta)(\sin \theta); \tag 3$
then
$\dfrac{d((\sin^2 \theta )(\cos^3 \theta)) }{d\theta} = \dfrac{d(uv)}{d\theta} = u'(\theta)v(\theta) + u(\theta)v'(\theta)$
$= (2(\sin \theta)(\cos \theta))(\cos^3 \theta) + (\sin^2 \theta)((-3\cos^2\theta)(\sin \theta))$
$= 2(\sin \theta)(\cos^4 \theta) -3(\sin^3\theta)(\cos^2 \theta)$
$= (\sin \theta)(\cos^2 \theta)(2\cos^2 \theta - 3\sin^2 \theta), \tag 4$
which agrees with the expression given by Blue in his/her comment to the question itself.
</s>