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

1 Answer
Jul 9, 2018

h''(x) = 9 sec(3x+1)[sec^2(3x+1) + tan^2(3x+1)]

Explanation:

Given: h(x) = sec(3x + 1)

Use the following derivative rules:

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

Product rule: (fg)' = f g' + g f'

Find the first derivative:

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

h'(u) = 3 sec u tan u

h'(x) = 3 sec (3x+1) tan (3x+1)

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''(x) = (3 sec(3x+1))(3 sec^2 (3x+1)) + ( tan (3x+1))(9 sec (3x+1) tan (3x+1))

h''(x) = 9 sec^3(3x+1) + 9tan^2(3x+1) sec(3x+1)

Factor:

h'(x) = 9 sec(3x+1) [sec^2(3x+1) + tan^2(3x+1)]