# How do you differentiate f(x)= (2x+1)(2x-3)^3  using the product rule?

Jan 25, 2016

$f ' \left(x\right) = 16 x {\left(2 x - 3\right)}^{2}$

#### Explanation:

Given:

$f \left(x\right) = h \left(x\right) \cdot g \left(x\right)$

the Product Rule tells:

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

$h \left(x\right) = \left(2 x + 1\right) \implies h ' \left(x\right) = 2 + 0 = 2$
$g \left(x\right) = {\left(2 x - 3\right)}^{3}$

To find $g ' \left(x\right)$ you can use the Chain Rule:

$g \left(x\right) = {\left[i \left(x\right)\right]}^{n} \implies g ' \left(x\right) = n {\left[i \left(x\right)\right]}^{n - 1} \cdot i ' \left(x\right)$

$\therefore g ' \left(x\right) = 3 {\left(2 x - 3\right)}^{2} \cdot \left(2 + 0\right) = 2 \cdot 3 {\left(2 x - 3\right)}^{2} =$
$= 6 {\left(2 x - 3\right)}^{2}$

Thus:

$f ' \left(x\right) = 2 \cdot {\left(2 x - 3\right)}^{3} + 6 {\left(2 x - 3\right)}^{2} \cdot \left(2 x + 1\right) = 2 {\left(2 x - 3\right)}^{2} \left(2 x - 3 + 3 \left(2 x + 1\right)\right) =$
$= 2 {\left(2 x - 3\right)}^{2} \left(2 x \textcolor{g r e e n}{\cancel{- 3}} + 6 x \textcolor{g r e e n}{\cancel{+ 3}}\right) = 2 {\left(2 x - 3\right)}^{2} \left(8 x\right) =$
$= 16 x {\left(2 x - 3\right)}^{2}$