# What is the derivative of y= 2x^4 - 2x^3 - 8?

I found: $y ' = 8 {x}^{3} - 6 {x}^{2}$
You can use the fact that the derivative of $y = {x}^{n}$ is:
$y ' = n {x}^{n - 1}$ to get:
$y ' = 2 \cdot 4 {x}^{4 - 1} - 2 \cdot 3 {x}^{3 - 1} - 0 = 8 {x}^{3} - 6 {x}^{2}$