# How do you differentiate f(x)= 3x^2 (4x - 12)^2 using the product rule?

Nov 12, 2015

Taken you to a point where you should be able to complete the calculation.

#### Explanation:

Let $u = 3 {x}^{2} \text{ }$ then $\frac{\mathrm{du}}{\mathrm{dx}} = 6 x$ ..............(1)

Let $v = {\left(4 x - 12\right)}^{2} \text{ }$ then $\frac{\mathrm{dv}}{\mathrm{dx}} = 8 \left(4 x - 12\right) = 32 x - 96$.....(2)

Product $\left(v \frac{\mathrm{du}}{\mathrm{dx}} + u \frac{\mathrm{dv}}{\mathrm{dx}}\right) \text{ }$

$\textcolor{b l u e}{\text{~~~~~~~~~Foot Note showing How to obtain (2) ~~~~~~~~~~~~~~}}$

$\textcolor{b r o w n}{\text{Let } w = 4 x - 12}$

$\textcolor{b r o w n}{\frac{\mathrm{dw}}{\mathrm{dx}} = 4}$

$\textcolor{b r o w n}{\text{But "v=w^2 " so } \frac{\mathrm{dv}}{\mathrm{dw}} = 2 w}$

$\textcolor{b r o w n}{\text{But } \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{\mathrm{dv}}{\mathrm{dw}} \times \frac{\mathrm{dw}}{\mathrm{dx}}}$

color(brown)("(dv)/(dx)= 2w times 4 = 8w = 8(4x-12))

$\textcolor{b l u e}{\text{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}}$

By substitution we have:

$\left\{{\left(4 x - 12\right)}^{2} \times 6 x\right\} + \left\{3 {x}^{2} \times \left(32 x - 96\right)\right\}$

$\textcolor{b l u e}{\text{I have left the final calculation and simplification for you to complete}}$