# How do you calculate the derivative of r= 2thetasqrt(sec theta)?

Jul 9, 2015

Here we have $r \left(\theta\right) = 2 \theta \sqrt{\sec} \theta$

$\left(\frac{\mathrm{dr}}{d \theta}\right) \left[f \left(\theta\right) g \left(\theta\right)\right] = f \left(\theta\right) g ' \left(\theta\right) + g \left(\theta\right) f ' \left(\theta\right)$

$= \left(2 \theta\right) \left(\frac{1}{2 \sqrt{\sec} \theta} \cdot \sec \theta \tan \theta\right) + \left(\sqrt{\sec} \theta\right) \left(2\right)$

$= \theta \sqrt{\sec} \theta \tan \theta + 2 \sqrt{\sec} \theta$

$= 2 \sqrt{\sec} \theta + \theta \sqrt{\sec} \theta \sin \frac{\theta}{\cos} \theta$

$= \textcolor{b l u e}{2 \sqrt{\sec} \theta + \theta \sin \theta {\sec}^{\text{3/2}} \theta}$