Question

find $\frac{dr}{d\theta}$. $r = 6\thetacos\theta+2sin\theta$ $\frac{dr}{d\theta}=square$

Explanation:

Step1: Apply product - rule on $6\theta\cos\theta$

The product - rule states that if $y = uv$, where $u$ and $v$ are functions of $\theta$, then $\frac{dy}{d\theta}=u\frac{dv}{d\theta}+v\frac{du}{d\theta}$. For $u = 6\theta$ and $v=\cos\theta$, $\frac{d(6\theta\cos\theta)}{d\theta}=6\theta(-\sin\theta)+6\cos\theta=- 6\theta\sin\theta + 6\cos\theta$.

Step2: Differentiate $2\sin\theta$

The derivative of $\sin\theta$ with respect to $\theta$ is $\cos\theta$. So, $\frac{d(2\sin\theta)}{d\theta}=2\cos\theta$.

Step3: Find $\frac{dr}{d\theta}$

$\frac{dr}{d\theta}=\frac{d(6\theta\cos\theta + 2\sin\theta)}{d\theta}=\frac{d(6\theta\cos\theta)}{d\theta}+\frac{d(2\sin\theta)}{d\theta}$. Substitute the results from Step1 and Step2: $\frac{dr}{d\theta}=-6\theta\sin\theta + 6\cos\theta+2\cos\theta=-6\theta\sin\theta + 8\cos\theta$.

Answer:

$-6\theta\sin\theta + 8\cos\theta$