# How do you differentiate y=(x^5+1)(3x^2-x)?

Dec 7, 2016

The answer is $\frac{\mathrm{dy}}{\mathrm{dx}} = 21 {x}^{6} - 6 {x}^{5} + 6 x - 1$

#### Explanation:

We use $\left({x}^{n}\right) ' = n {x}^{n - 1}$

You apply the product role of differentiation

$\left(u v\right) ' = u ' v + u v '$

Here,

$u = \left({x}^{5} + 1\right)$, $\implies$, $u ' = 5 {x}^{4}$

$v = \left(3 {x}^{2} - x\right)$, $\implies$, $v ' = 6 x - 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 5 {x}^{4} \left(3 {x}^{2} - x\right) + \left({x}^{5} + 1\right) \left(6 x - 1\right)$

$= 15 {x}^{6} - 5 {x}^{5} + 6 {x}^{6} - {x}^{5} + 6 x - 1$

$= 21 {x}^{6} - 6 {x}^{5} + 6 x - 1$