# How do you differentiate y=(8x^3-6x^2-4)/(2x^2)?

Sep 26, 2016

$y ' = 4 + \frac{4}{x} ^ 3$

#### Explanation:

re- write $y = \frac{8 {x}^{3} - 6 {x}^{2} - 4}{2 {x}^{2}}$ as

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

and apply the power rule $\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$

to get

$y ' = 4 + \frac{4}{x} ^ 3$