How do you differentiate #g(x) = sqrt(2x^3-4)cos4x# using the product rule?

1 Answer
Oct 26, 2017

Answer:

#(3x^2cos(4x))/sqrt(2x^3-4) - 4(sqrt(2x^3-4)sin4x)#

Explanation:

Okay, the product rule pertains to the product of 2 functions.

#d/(dt)f(x)g(x) = f'(x)g(x) + f(x)g'(x)#

Here we have #f(x) = sqrt(2x^3 - 4)#
And #g(x) = cos4x#

We can attack the problem piece by piece. First, calculate #f'(x)#:

#d/dx(sqrt(2x^3-4)) = d/dx(2x^3-4)^(1/2)#

...this derivative is calculated via the chain rule:

#= 1/2(2x^3 - 4)^(-1/2) * 6x^2#

#f'(x) = (3x^2)/(sqrt(2x^3-4)#

and calculating g(x) also uses the chain rule, but is pretty simple:

#-4sin4x#

So #f'(x)g(x) + f(x)g'(x) = #

#(3x^2cos(4x))/sqrt(2x^3-4) - 4(sqrt(2x^3-4)sin4x)#

GOOD LUCK