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

Dec 5, 2015

$f ' \left(x\right) = 8 {x}^{3} - 6 {x}^{2}$

#### Explanation:

According to the product rule:

$f ' \left(x\right) = \left(2 x - 2\right) \frac{d}{\mathrm{dx}} \left[{x}^{3}\right] + {x}^{3} \frac{d}{\mathrm{dx}} \left[2 x + 2\right]$

Find each derivative separately.

$\frac{d}{\mathrm{dx}} \left[{x}^{3}\right] = 3 {x}^{2}$

$\frac{d}{\mathrm{dx}} \left[2 x - 2\right] = 2$

Plug back in.

$f ' \left(x\right) = 3 {x}^{2} \left(2 x - 2\right) + 2 {x}^{3}$

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

$f ' \left(x\right) = 8 {x}^{3} - 6 {x}^{2}$

Note that there's an easier way to do this, but it doesn't use prodyct rule:

Distribute the ${x}^{3}$ term first.

$f \left(x\right) = 2 {x}^{4} - 2 {x}^{3}$

$f ' \left(x\right) = 8 {x}^{3} - 6 {x}^{2}$