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

Jan 31, 2016

$f ' \left(x\right) = 12 \left(2 x + 3\right) \left(2 x + 1\right)$

#### Explanation:

The product rule states that for a function $f \left(x\right) = g \left(x\right) h \left(x\right)$, the derivative of the function is

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

Applying this to $f \left(x\right) = 4 x {\left(2 x + 3\right)}^{2}$, we see that

$f ' \left(x\right) = \textcolor{red}{\frac{d}{\mathrm{dx}} \left[4 x\right]} {\left(2 x + 3\right)}^{2} + 4 x \textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left[{\left(2 x + 3\right)}^{2}\right]}$

We can find each of these derivatives separately and then plug them back in.

color(red)(d/dx[4x]=4

The following will require the chain rule.

color(blue)(d/dx[(2x+3)^2])=2(2x+3)^1d/dx[2x+3]color(blue)(=4(2x+3)

From these, we see that

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

Factor out a $\left(2 x + 3\right)$ term.

$f ' \left(x\right) = \left(2 x + 3\right) \left(4 \left(2 x + 3\right) + 16 x\right)$

Simplify.

$f ' \left(x\right) = 12 \left(2 x + 3\right) \left(2 x + 1\right)$

This is also equal to

$f ' \left(x\right) = 48 {x}^{2} + 96 x + 36$