# How do you differentiate f(x)=x(3x-9)^3 using the chain rule?

Feb 25, 2016

$f ' \left(x\right) = 3 {\left(3 x - 9\right)}^{2} \left(4 x - 3\right)$

#### Explanation:

The $\textcolor{b l u e}{\text{ Product rule " "will have to be used as well as the }}$
$\textcolor{b l u e}{\text{ chain rule }}$

There is a product of 2 functions here , x and ${\left(3 x - 9\right)}^{3}$

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)

using the$\textcolor{b l u e}{\text{ chain rule }}$

$\frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = f ' \left(g \left(x\right)\right) . g ' \left(x\right)$

hence: $f ' \left(x\right) = x \frac{d}{\mathrm{dx}} {\left(3 x - 9\right)}^{3} + {\left(3 x - 9\right)}^{3} \frac{d}{\mathrm{dx}} \left(x\right)$

$= x \left[3 {\left(3 x - 9\right)}^{2} \frac{d}{\mathrm{dx}} \left(3 x - 9\right)\right] + {\left(3 x - 9\right)}^{3} .1$

$= x \left[3 {\left(3 x - 9\right)}^{2} .3\right] + {\left(3 x - 9\right)}^{3}$

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

( can be simplified by 'taking out' common factor )

$= {\left(3 x - 9\right)}^{2} \left[9 x + 3 x - 9\right] = {\left(3 x - 9\right)}^{2} . 3 \left(4 x - 3\right)$