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

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Answer 1 · 2018-07-09T04:31:44

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

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)]#