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

Nov 19, 2015

Use both the Product and the Chain Rules to get
$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = {\left({x}^{3} - 3\right)}^{3} \cdot \left(24 {x}^{5} + 6 {x}^{4} - 18 x\right)$

#### Explanation:

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

Let $f \left(x\right) = p \left(x\right) \cdot q \left(x\right)$
where $p \left(x\right) = 2 {x}^{3}$ and $q \left(x\right) = {\left({x}^{3} - 3\right)}^{4}$

Further let $q \left(x\right) = r \left(s \left(x\right)\right)$
where $r \left(x\right) = {x}^{4}$ and $s \left(x\right) = {x}^{3} - 3$

$\frac{\mathrm{dp} \left(x\right)}{\mathrm{dx}} = \frac{d 2 {x}^{3}}{\mathrm{dx}} = 6 x$

$\frac{\mathrm{ds} \left(x\right)}{\mathrm{dx}} = \frac{d {x}^{3} - 3}{\mathrm{dx}} = 3 {x}^{2}$

$\frac{\mathrm{dr} \left(x\right)}{\mathrm{dx}} = \frac{d {x}^{4}}{\mathrm{dx}} = 4 {x}^{3}$
$\Rightarrow \frac{d r \left(s \left(x\right)\right)}{\mathrm{ds} \left(x\right)} = 4 {\left({x}^{3} - 3\right)}^{3}$

The Chain Rule tells us that
$\frac{\mathrm{dq} \left(x\right)}{\mathrm{dx}} = \frac{d r \left(s \left(x\right)\right)}{\mathrm{dx}} = \frac{d r \left(s \left(x\right)\right)}{d s \left(x\right)} \cdot \frac{d s \left(x\right)}{\mathrm{dx}}$
$\textcolor{w h i t e}{\text{XXX}} = 4 {\left({x}^{3} - 3\right)}^{3} \cdot 3 {x}^{2} = 12 {x}^{2} {\left({x}^{3} - 3\right)}^{3}$

The Product Rule tells us that
$\textcolor{w h i t e}{\text{XXX}} \frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{d \left(p \left(x\right) \cdot q \left(x\right)\right)}{\mathrm{dx}} = q \left(x\right) \cdot \frac{\mathrm{dp} \left(x\right)}{\mathrm{dx}} + p \left(x\right) \cdot \frac{\mathrm{dq} \left(x\right)}{\mathrm{dx}}$

$\textcolor{w h i t e}{\text{XXXXXX}} = {\left({x}^{3} - 3\right)}^{4} \cdot 6 x + 2 {x}^{3} \cdot 12 {x}^{2} {\left({x}^{3} - 3\right)}^{3}$

$\textcolor{w h i t e}{\text{XXXXXX}} = {\left({x}^{3} - 3\right)}^{3} \cdot \left[\left({x}^{3} - 3\right) \cdot 6 x + 2 {x}^{3} \cdot 12 {x}^{2}\right]$

$\textcolor{w h i t e}{\text{XXXXXX}} = {\left({x}^{3} - 3\right)}^{3} \left(24 {x}^{5} + 6 {x}^{4} - 18 x\right)$