Question

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

Answer 1

`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)]`