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

Mar 11, 2018

$g ' \left(x\right) = 18 {x}^{2} - 8 x - 12$

#### Explanation:

We know,

The product rule says :

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

When $f \left(x\right)$ and $g \left(x\right)$ are two differentiable polynomials.

[Don't be confused with the question function $g \left(x\right)$ and the example here.]

So,

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

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

$= \left(\frac{d}{\mathrm{dx}} 3 x - \frac{d}{\mathrm{dx}} 2\right) \times \left(2 {x}^{2} - 4\right) + \left(3 x - 2\right) \times \left(\frac{d}{\mathrm{dx}} 2 {x}^{2} - \frac{d}{\mathrm{dx}} 4\right)$
[Subtraction Rule]

$= \left(3 - 0\right) \left(2 {x}^{2} - 4\right) + \left(3 x - 2\right) \left(2 \cdot 2 {x}^{2 - 1} - 0\right)$ [The derivative of a constant is always zero, and $\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$]

$= \left(6 {x}^{2} - 12\right) + 4 x \left(3 x - 2\right)$

$= 6 {x}^{2} - 12 + 12 {x}^{2} - 8 x$

$= 18 {x}^{2} - 8 x - 12$

So, Got The Answer, and I hope this will help.