# How do you find the derivative of y=2x(3x-1)(4-2x)?

May 27, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 12 {x}^{3} + 4 {x}^{2} + 28 x - 8$

#### Explanation:

$y = 2 x \left(3 x - 1\right) \left(4 - 2 x\right)$

$y = \left(6 {x}^{2} - 2 x\right) \left(4 - 2 x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(6 {x}^{2} - 2 x\right) \left(- 2 x\right) + \left(4 - 2 x\right) \left(12 x - 2\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 12 {x}^{3} + 4 {x}^{2} + 48 x - 8 - 24 x + 4 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 12 {x}^{3} + 4 {x}^{2} + 28 x - 8$

The product rule is given by

$y = u v$

$\frac{\mathrm{dy}}{\mathrm{dx}} = u v ' + v u '$