# How do you use the product rule to differentiate  y= 4x(2x+3)^2?

Mar 4, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = 48 {x}^{2} + 96 x + 36$

#### Explanation:

Using the$\textcolor{b l u e}{\text{ Product rule }}$

If f(x) = g(x).h(x) then f'(x) = g(x)h'(x)+h(x)g'(x)

now g(x) = 4x $\Rightarrow g ' \left(x\right) = 4$

and $h \left(x\right) = {\left(2 x + 3\right)}^{2} \Rightarrow h ' \left(x\right) = 2 \left(2 x + 3\right) \frac{d}{\mathrm{dx}} \left(2 x + 3\right)$
Replace these results into f'(x)

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

$= 16 x \left(2 x + 3\right) + 4 {\left(2 x + 3\right)}^{2} = 32 {x}^{2} + 48 x + 16 {x}^{2} + 48 x + 36$

$= 48 {x}^{2} + 96 x + 36$

Mar 4, 2016

Let $f = 4 x , g = {\left(2 x + 3\right)}^{2} \to f ' = 4 , g ' = 2 \left(2 x + 3\right) \cdot 2$
$y ' = f g ' + g f ' = 16 x \left(2 x + 3\right) + 4 {\left(2 x + 3\right)}^{2} = 32 {x}^{2} + 48 x + 16 {x}^{2} + 48 x + 36 = 48 {x}^{2} + 96 x + 36$