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

1 Answer
Oct 26, 2017

(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