# How do you integrate y=(2x^3)/(4-x) using the quotient rule?

Sep 16, 2016

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

#### Explanation:

The question should be asking how to differentiate the given function, and not how to integrate it.

We have: $y = \frac{2 {x}^{3}}{4 - x}$

This function can be differentiated using the "quotient rule":

$\implies y ' = \frac{\left(4 - x\right) \left(6 {x}^{2}\right) - \left(2 {x}^{3}\right) \left(- 1\right)}{{\left(4 - x\right)}^{2}}$

$\implies y ' = \frac{24 {x}^{2} - 6 {x}^{3} + 2 {x}^{3}}{{\left(4 - x\right)}^{2}}$

$\implies y ' = \frac{24 {x}^{2} - 4 {x}^{3}}{{\left(4 - x\right)}^{2}}$

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