Question

find $\frac{dr}{d\theta}$. $r = 1-\theta^{3}sin\theta$ $\frac{dr}{d\theta}=square$

Explanation:

Step1: Apply sum - difference rule

The derivative of a sum/difference $u - v$ is $u'-v'$. Here $u = 1$ and $v=\theta^{3}\sin\theta$. The derivative of a constant $u = 1$ is $0$, so $\frac{dr}{d\theta}=0-\frac{d}{d\theta}(\theta^{3}\sin\theta)=-\frac{d}{d\theta}(\theta^{3}\sin\theta)$.

Step2: Apply product rule

The product rule states that if $y = uv$, then $y'=u'v + uv'$. For $u=\theta^{3}$ and $v = \sin\theta$, $u'=3\theta^{2}$ and $v'=\cos\theta$. So $\frac{d}{d\theta}(\theta^{3}\sin\theta)=3\theta^{2}\sin\theta+\theta^{3}\cos\theta$.

Step3: Get the final derivative

Since $\frac{dr}{d\theta}=-\frac{d}{d\theta}(\theta^{3}\sin\theta)$, then $\frac{dr}{d\theta}=- 3\theta^{2}\sin\theta-\theta^{3}\cos\theta$.

Answer:

$-3\theta^{2}\sin\theta-\theta^{3}\cos\theta$