Question

find $\frac{dr}{d\theta}$. $r=(9 + sec\theta)sin\theta$ $\frac{dr}{d\theta}=square$

Explanation:

Step1: Expand the function

$r = 9\sin\theta+\sec\theta\sin\theta=9\sin\theta + \tan\theta$

Step2: Differentiate term - by - term

The derivative of $\sin\theta$ is $\cos\theta$ and the derivative of $\tan\theta$ is $\sec^{2}\theta$.
$\frac{dr}{d\theta}=\frac{d}{d\theta}(9\sin\theta)+\frac{d}{d\theta}(\tan\theta)$

Step3: Calculate the derivatives

$\frac{d}{d\theta}(9\sin\theta)=9\cos\theta$ and $\frac{d}{d\theta}(\tan\theta)=\sec^{2}\theta$
So, $\frac{dr}{d\theta}=9\cos\theta+\sec^{2}\theta$

Answer:

$9\cos\theta+\sec^{2}\theta$