Question

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

Answer 1

`(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