# What is the second derivative of g(x) = sec(3x+1)?

Jul 9, 2018

$h ' ' \left(x\right) = 9 \sec \left(3 x + 1\right) \left[{\sec}^{2} \left(3 x + 1\right) + {\tan}^{2} \left(3 x + 1\right)\right]$

#### Explanation:

Given: $h \left(x\right) = \sec \left(3 x + 1\right)$

Use the following derivative rules:

(sec u)' = u' sec u tan u; " "(tan u)' = u' sec^2 u

Product rule: $\left(f g\right) ' = f g ' + g f '$

Find the first derivative:

Let u = 3x + 1; " "u ' = 3

$h ' \left(u\right) = 3 \sec u \tan u$

$h ' \left(x\right) = 3 \sec \left(3 x + 1\right) \tan \left(3 x + 1\right)$

Find the second derivative using the product rule:

Let f = 3 sec(3x+1); " "f' = 9 sec (3x+1) tan (3x+1)

Let g = tan (3x+1); " "g' = 3 sec^2 (3x+1)

$h ' ' \left(x\right) = \left(3 \sec \left(3 x + 1\right)\right) \left(3 {\sec}^{2} \left(3 x + 1\right)\right) + \left(\tan \left(3 x + 1\right)\right) \left(9 \sec \left(3 x + 1\right) \tan \left(3 x + 1\right)\right)$

$h ' ' \left(x\right) = 9 {\sec}^{3} \left(3 x + 1\right) + 9 {\tan}^{2} \left(3 x + 1\right) \sec \left(3 x + 1\right)$

Factor:

$h ' \left(x\right) = 9 \sec \left(3 x + 1\right) \left[{\sec}^{2} \left(3 x + 1\right) + {\tan}^{2} \left(3 x + 1\right)\right]$