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

Oct 26, 2017

$\frac{3 {x}^{2} \cos \left(4 x\right)}{\sqrt{2 {x}^{3} - 4}} - 4 \left(\sqrt{2 {x}^{3} - 4} \sin 4 x\right)$

#### Explanation:

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

$\frac{d}{\mathrm{dt}} f \left(x\right) g \left(x\right) = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

Here we have $f \left(x\right) = \sqrt{2 {x}^{3} - 4}$
And $g \left(x\right) = \cos 4 x$

We can attack the problem piece by piece. First, calculate $f ' \left(x\right)$:

$\frac{d}{\mathrm{dx}} \left(\sqrt{2 {x}^{3} - 4}\right) = \frac{d}{\mathrm{dx}} {\left(2 {x}^{3} - 4\right)}^{\frac{1}{2}}$

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

$= \frac{1}{2} {\left(2 {x}^{3} - 4\right)}^{- \frac{1}{2}} \cdot 6 {x}^{2}$

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

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

$- 4 \sin 4 x$

So $f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right) =$

$\frac{3 {x}^{2} \cos \left(4 x\right)}{\sqrt{2 {x}^{3} - 4}} - 4 \left(\sqrt{2 {x}^{3} - 4} \sin 4 x\right)$

GOOD LUCK