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

##### 1 Answer

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

#### Explanation:

given function:

$f \left(x\right) = \setminus \sec x$

Differentiating w.r.t. $x$ as follows

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

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

Again, differentiating $f ' \left(x\right)$ w.r.t. $x$, we get

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

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

$= \setminus \sec x {\sec}^{2} x + \setminus \tan x \setminus \sec x \setminus \tan x$

$= {\sec}^{3} x + \setminus \sec x \setminus {\tan}^{2} x$

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