# How do you differentiate r=5theta^2sectheta?

Jan 18, 2017

$\frac{\mathrm{dr}}{d \theta} = 5 \theta \sec \theta \left(\tan \theta + 2\right)$

#### Explanation:

differentiate using the $\textcolor{b l u e}{\text{product rule}}$

$\text{Given " r=f(theta).g(theta)" then}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{r ' \left(\theta\right) = f \left(\theta\right) g ' \left(\theta\right) + g \left(\theta\right) f ' \left(\theta\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}} \leftarrow \text{ product rule}$

The standard derivative of $\sec \left(\theta\right)$ which should be known is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\frac{d}{d \theta} \left(\sec \left(\theta\right)\right) = \sec \left(\theta\right) \tan \left(\theta\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{here } f \left(\theta\right) = 5 {\theta}^{2} \Rightarrow f ' \left(\theta\right) = 10 \theta$

$\text{and } g \left(\theta\right) = \sec \theta \Rightarrow g ' \left(\theta\right) = \sec \theta \tan \theta$

$\Rightarrow r ' \left(\theta\right) = 5 {\theta}^{2} \sec \theta \tan \theta + \sec \theta .10 \theta$

$= 5 \theta \sec \theta \left(\tan \theta + 2\right)$