Chapter 3, Problem 9RE

To determine

To find:The derivative of the given function.

Answer to Problem 9RE

The derivative function of $r=5{\theta }^2\sec \theta$ is $\frac{dr}{d\theta }=5\left(2\theta \sec \theta +\left(\sec \theta \tan \theta \right)\times {\theta }^2\right)$ .

Explanation of Solution

Given information: $r=5{\theta }^2\sec \theta$ .

Formula used:

• Product rule: $\left[u\left(x\right)\times v\left(x\right)\right]\prime =u\prime \left(x\right)\times v\left(x\right)+u\left(x\right)\times v\prime \left(x\right)$
• Constant out: $\left(a\times u\right)\prime =a\times f\prime$
• Derivative rule: $\left(x^n\right)\prime =n\left(x^{n-1}\right)$
• Derivative rule: $\left(a\right)\prime =0$
• Derivative rule: $\left(\sec \theta \right)\prime =\sec \theta \tan \theta$

Calculation:

  $\begin{array}{l}\frac{dr}{d\theta }=\frac{d}{d\theta }\left(5{\theta }^2\sec \theta \right)\\ \frac{dr}{d\theta }=5\frac{d}{d\theta }\left({\theta }^2\sec \theta \right)\\ \frac{dr}{d\theta }=5\left(\frac{d}{d\theta }{\theta }^2\times \sec \theta +\frac{d}{d\theta }\sec \theta \times {\theta }^2\right)\\ \frac{dr}{d\theta }=5\left(2\theta \sec \theta +\sec \theta \tan \theta \times {\theta }^2\right)\end{array}$

The derivative function of $r=5{\theta }^2\sec \theta$ is $\frac{dr}{d\theta }=5\left(2\theta \sec \theta +\left(\sec \theta \tan \theta \right)\times {\theta }^2\right)$ .