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

Jul 8, 2017

$f ' \left(x\right) = - 2 {\left(2 x - 3\right)}^{2} \left(5 {x}^{2} + 3 x - 12\right)$

#### Explanation:

We have:

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

So using the product rule we have:

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

So using the chain rule we have:

$f ' \left(x\right) = \left\{4 - {x}^{2}\right\} \left\{3 {\left(2 x - 3\right)}^{2} \left(2\right)\right\} + \left\{- 2 x\right\} \left\{{\left(2 x - 3\right)}^{3}\right\}$
$\text{ } = 6 \left(4 - {x}^{2}\right) {\left(2 x - 3\right)}^{2} - 2 x {\left(2 x - 3\right)}^{3}$
$\text{ } = 2 {\left(2 x - 3\right)}^{2} \left\{3 \left(4 - {x}^{2}\right) - x \left(2 x - 3\right)\right\}$
$\text{ } = 2 {\left(2 x - 3\right)}^{2} \left(12 - 3 {x}^{2} - 2 {x}^{2} + 3 x\right)$
$\text{ } = 2 {\left(2 x - 3\right)}^{2} \left(- 12 - 5 {x}^{2} - 3 x\right)$
$\text{ } = - 2 {\left(2 x - 3\right)}^{2} \left(5 {x}^{2} + 3 x - 12\right)$