Chapter 3.5, Problem 36E

To determine

To find: $y\text{'}\text{'}$

Answer to Problem 36E

  $y\text{'}\text{'}=2{\sec }^2\theta (\theta \tan \theta +1)$ .

Explanation of Solution

Given information:Given equation is $y=\theta \tan \theta$ .

Calculation:Given equation is $y=\theta \tan \theta$ .

  $\begin{array}{l}y=\theta \tan \theta \\ y\text{'}=\theta \frac{d}{d\theta }(\tan \theta )+\tan \theta \frac{d}{d\theta }(\theta )\\ y\text{'}=\theta ({\sec }^2\theta )+\tan \theta (1)\\ y\text{'}=\theta {\sec }^2\theta +\tan \theta \\ y\text{'}\text{'}=\theta \frac{d}{d\theta }({\sec }^2\theta )+{\sec }^2\theta \frac{d}{d\theta }(\theta )+{\sec }^2\theta \\ y\text{'}\text{'}=\theta (2\sec \theta \frac{d}{d\theta }(\sec \theta ))+{\sec }^2\theta (1)+{\sec }^2\theta \\ y\text{'}\text{'}=\theta (2\sec \theta (\sec \theta \tan \theta ))+{\sec }^2\theta +{\sec }^2\theta \\ y\text{'}\text{'}=2\theta {\sec }^2\theta \tan \theta +2{\sec }^2\theta \\ y\text{'}\text{'}=2{\sec }^2\theta (\theta \tan \theta +1)\end{array}$