# Question 16171

May 12, 2017

$f ' \left(x\right) = 2 {\left(2 x - 1\right)}^{2} \left({x}^{2} + 3\right) \left(7 {x}^{2} - 2 x + 9\right)$

#### Explanation:

We can use a combination of the chain rule and the product rule:

We have:

$f \left(x\right) = {\left(2 x - 1\right)}^{3} {\left({x}^{2} + 3\right)}^{2}$

We can use the chain rule on each individual function. Let

$\left\{\begin{matrix}u = {\left(2 x - 1\right)}^{3} \\ v = {\left({x}^{2} + 3\right)}^{2}\end{matrix}\right. \implies \left\{\begin{matrix}\frac{\mathrm{du}}{\mathrm{dx}} = 3 {\left(2 x - 1\right)}^{2} \left(2\right) & = 6 {\left(2 x - 1\right)}^{2} \\ \frac{\mathrm{dv}}{\mathrm{dx}} = 2 \left({x}^{2} + 3\right) \left(2 x\right) & = 4 x \left({x}^{2} + 3\right)\end{matrix}\right.$

And then:

$f \left(x\right) = u v$

And by the product rule we have:

$f ' \left(x\right) = \left(u\right) \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right) + \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \left(v\right)$
 " " = (2x-1)^3 4x(x^2+3) + 6(2x-1)^2(x^2+3)^2) #
$\text{ } = {\left(2 x - 1\right)}^{2} \left({x}^{2} + 3\right) \left\{4 x \left(2 x - 1\right) + 6 \left({x}^{2} + 3\right)\right\}$
$\text{ } = {\left(2 x - 1\right)}^{2} \left({x}^{2} + 3\right) \left(8 {x}^{2} - 4 x + 6 {x}^{2} + 18\right)$
$\text{ } = {\left(2 x - 1\right)}^{2} \left({x}^{2} + 3\right) \left(14 {x}^{2} - 4 x + 18\right)$
$\text{ } = 2 {\left(2 x - 1\right)}^{2} \left({x}^{2} + 3\right) \left(7 {x}^{2} - 2 x + 9\right)$