# What is the second derivative of f(x)= sec^2x?

Jan 28, 2016

$f ' ' \left(x\right) = 4 {\tan}^{2} x {\sec}^{2} x + 2 {\sec}^{4} x$

#### Explanation:

To find the first derivative, we will have to use the chain rule on the second power.

Use the rule that $\frac{d}{\mathrm{dx}} \left({u}^{2}\right) = 2 u \cdot u '$.

Thus, we see that

$f ' \left(x\right) = 2 \sec x \cdot \frac{d}{\mathrm{dx}} \left(\sec x\right)$

$f ' \left(x\right) = 2 \sec x \cdot \sec x \tan x$

$f ' \left(x\right) = 2 {\sec}^{2} x \tan x$

To find the second derivative, we will have to use the product rule.

$f ' ' \left(x\right) = 2 \tan x \frac{d}{\mathrm{dx}} \left({\sec}^{2} x\right) + 2 {\sec}^{2} x \frac{d}{\mathrm{dx}} \left(\tan x\right)$

Note that we already know that $\frac{d}{\mathrm{dx}} \left({\sec}^{2} x\right) = 2 {\sec}^{2} x \tan x$ and that $\frac{d}{\mathrm{dx}} \left(\tan x\right) = {\sec}^{2} x$.

This gives us

$f ' ' \left(x\right) = 2 \tan x \left(2 {\sec}^{2} x \tan x\right) + 2 {\sec}^{2} x \left({\sec}^{2} x\right)$

$f ' ' \left(x\right) = 4 {\tan}^{2} x {\sec}^{2} x + 2 {\sec}^{4} x$