Answer
$\frac{x(-x^3-3x-2)}{(x^3-1)^2}$
Work Step by Step
$y=\frac{x^2+1}{x^3-1}$
Use the quotient rule.
$y'=\frac{(x^3-1)(2x)-(x^2+1)(3x^2)}{(x^3-1)^2}$
Distribute.
$y'=\frac{2x^4-2x-3x^4-3x^2}{(x^3-1)^2}$
Combine like terms.
$y'=\frac{-x^4-3x^2-2x}{(x^3-1)^2}$
Factor out the x.
$\frac{x(-x^3-3x-2)}{(x^3-1)^2}$
