f(x)= $\displaystyle

\frac {x^3-3x^2+4}{x^2}

$

well i used the power rule but came up with

the wrong answer

then i used the quotient rule and came out with the

wrong answer

am i doing my algebra wrong is is there something im not seeing

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- Sep 15th 2008, 06:20 PMjohn doe[SOLVED] Derivative
f(x)= $\displaystyle

\frac {x^3-3x^2+4}{x^2}

$

well i used the power rule but came up with

the wrong answer

then i used the quotient rule and came out with the

wrong answer

am i doing my algebra wrong is is there something im not seeing - Sep 15th 2008, 06:24 PMJhevon
- Sep 15th 2008, 06:46 PMjohn doe
o i see what you did now

you distributed the $\displaystyle x^-2 $

$\displaystyle

\frac {x^3-8}{x^3}

$ - Sep 15th 2008, 06:48 PMJhevon
- Sep 15th 2008, 06:49 PMjavax
Like Jhevon said...much shorter way

$\displaystyle

(x-3+4x^{-2})' = 1-0+4(-2)x^{-3} = 1-\frac{8}{x^3} = \frac{x^3-8}{3}

$ - Sep 15th 2008, 07:50 PMjohn doe
I think you made a mistake

and i did solve it that way

at first i tryed to do all this other stuff though - Sep 15th 2008, 07:52 PM11rdc11
You could factor out the numerator if you wanted since it is the the form of $\displaystyle a^3-b^3$