Question

find $\frac{dr}{d\theta}$. $r = 3-\theta^{4}sin\theta$. $\frac{dr}{d\theta}=square$

Explanation:

Step1: Differentiate constant term

The derivative of a constant 3 with respect to $\theta$ is 0. So $\frac{d}{d\theta}(3)=0$.

Step2: Apply product - rule

The product - rule states that if $u = \theta^{4}$ and $v=\sin\theta$, then $\frac{d(uv)}{d\theta}=u\frac{dv}{d\theta}+v\frac{du}{d\theta}$. Here, $\frac{du}{d\theta}=\frac{d(\theta^{4})}{d\theta}=4\theta^{3}$ and $\frac{dv}{d\theta}=\frac{d(\sin\theta)}{d\theta}=\cos\theta$. So $\frac{d(\theta^{4}\sin\theta)}{d\theta}=\theta^{4}\cos\theta + 4\theta^{3}\sin\theta$.

Step3: Differentiate r

Since $r = 3-\theta^{4}\sin\theta$, then $\frac{dr}{d\theta}=\frac{d(3)}{d\theta}-\frac{d(\theta^{4}\sin\theta)}{d\theta}$. Substituting the results from Step 1 and Step 2, we get $\frac{dr}{d\theta}=0-(\theta^{4}\cos\theta + 4\theta^{3}\sin\theta)=-\theta^{4}\cos\theta - 4\theta^{3}\sin\theta$.

Answer:

$-\theta^{4}\cos\theta - 4\theta^{3}\sin\theta$