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

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Answer 1 · 2015-08-23T21:03:23

I found: $y'=8x^3-6x^2$

You can use the fact that the derivative of $y=x^n$ is:
$y'=nx^(n-1)$ to get:
$y'=2*4x^(4-1)-2*3x^(3-1)-0=8x^3-6x^2$

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