# How do you find the derivative of 3x^2 (4x - 12)^2 + x^3 (2) (4x - 12) (4)?

Aug 23, 2015

Rewrite a bit and then you have a choice.

#### Explanation:

$3 {x}^{2} {\left(4 x - 12\right)}^{2} + {x}^{3} \left(2\right) \left(4 x - 12\right) \left(4\right) = 3 {x}^{2} {\left(4 x - 12\right)}^{2} + 8 {x}^{3} \left(4 x - 12\right)$

Now we could use the product rule to differentiate each of the two terms.

I prefer to factor first.

$3 {x}^{2} {\left(4 x - 12\right)}^{2} + 8 {x}^{3} \left(4 x - 12\right) = {x}^{2} \left(4 x - 12\right) \left[3 \left(4 x - 12\right) + 8 x\right]$

$= {x}^{2} \left(4 x - 12\right) \left(20 x - 36\right)$

$= 16 {x}^{2} \left(x - 3\right) \left(5 x - 9\right)$

Now we can either distrubute the $16 {x}^{2}$ to one factor and use the usual product rule, or we can use the three factor product rule:

$\frac{d}{\mathrm{dx}} \left(u v w\right) = u ' v w + u v ' w + u v w '$

So we get:

$32 x \left(x - 3\right) \left(5 x - 9\right) + 16 {x}^{2} \left(1\right) \left(5 x - 9\right) + 16 {x}^{2} \left(x - 3\right) \left(5\right)$

This can also be factored and simplified if you wish.