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

Jul 5, 2015

$y ' = 5 {x}^{4} + 8 x$

#### Explanation:

$y = \left({x}^{2}\right) \left({x}^{3} + 4\right)$

The product rule is:

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

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

$= 5 {x}^{4} + 8 x$

Or you could multiply out the brackets:

let $y = {x}^{5} + 4 {x}^{2}$

$y ' = 5 {x}^{4} + 8 x$