find $\frac{dr}{d\theta}$. r = 1 - $\theta^{2}$ sin $\theta$ $\frac{dr}{d\theta}=$

Explanation:

Step1: Differentiate constant term

The derivative of a constant ($1$) is $0$. The derivative of $r$ with respect to $\theta$ starts with the derivative of the constant part and the non - constant part. The derivative of the constant $1$ is $0$, so we focus on differentiating $-\theta^{2}\sin\theta$. $\frac{d}{d\theta}(1) = 0$

Step2: Apply product rule

The product rule states that if $u = -\theta^{2}$ and $v=\sin\theta$, then $\frac{d(uv)}{d\theta}=u\frac{dv}{d\theta}+v\frac{du}{d\theta}$. First, find $\frac{du}{d\theta}$ and $\frac{dv}{d\theta}$. $\frac{du}{d\theta}=\frac{d(-\theta^{2})}{d\theta}=- 2\theta$ and $\frac{dv}{d\theta}=\frac{d(\sin\theta)}{d\theta}=\cos\theta$. Then $u\frac{dv}{d\theta}+v\frac{du}{d\theta}=-\theta^{2}\cos\theta+\sin\theta(-2\theta)$.

Step3: Combine results

Combining the derivative of the constant part and the result from the product - rule application, we get: $\frac{dr}{d\theta}=0-\theta^{2}\cos\theta - 2\theta\sin\theta=-\theta^{2}\cos\theta-2\theta\sin\theta$

Answer:

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