differentiate

$\displaystyle y=\frac{x^2-3x+1}{x^3+2}$

my answer: $\displaystyle y'=\frac{-x^4+6x^3+4x-6-3x^2}{x^6+4x^3+4}$

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- Aug 27th 2009, 02:41 PMyoman360did i do this correctly?
differentiate

$\displaystyle y=\frac{x^2-3x+1}{x^3+2}$

my answer: $\displaystyle y'=\frac{-x^4+6x^3+4x-6-3x^2}{x^6+4x^3+4}$ - Aug 27th 2009, 02:43 PMJG89
It's fine.

- Aug 27th 2009, 04:26 PMyoman360