# [SOLVED] Derivative

• Sep 15th 2008, 07:20 PM
john doe
[SOLVED] Derivative
f(x)= $
\frac {x^3-3x^2+4}{x^2}
$

well i used the power rule but came up with

then i used the quotient rule and came out with the

am i doing my algebra wrong is is there something im not seeing
• Sep 15th 2008, 07:24 PM
Jhevon
Quote:

Originally Posted by john doe
f(x)= x^3-3x^2+4/x^2

well i used the power rule but came up with

then i used the quotient rule and came out with the

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

why use the quotient rule (one thing you should have used though, was parentheses!)

$\frac {x^3 - 3x^2 + 4}{x^2} = x - 3 + 4x^{-2}$ .........(was that what you got?)

use the power rule
• Sep 15th 2008, 07:46 PM
john doe
o i see what you did now

you distributed the $x^-2$

$
\frac {x^3-8}{x^3}
$
• Sep 15th 2008, 07:48 PM
Jhevon
Quote:

Originally Posted by john doe
o i see what you did now

you distributed the $x^{-2}$

yes

(if your power has more than one term, use {}. you should have typed x^{-2} in your LaTeX code)

Quote:

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

$\frac {x^3 - 8}{x^3} = 1 - 8x^{-3}$

no more free rides :p you should be telling me this now
• Sep 15th 2008, 07:49 PM
javax
Like Jhevon said...much shorter way

$
(x-3+4x^{-2})' = 1-0+4(-2)x^{-3} = 1-\frac{8}{x^3} = \frac{x^3-8}{3}
$
• Sep 15th 2008, 08:50 PM
john 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, 08:52 PM
11rdc11
You could factor out the numerator if you wanted since it is the the form of $a^3-b^3$