# Finding the derivative of the function

• May 24th 2011, 03:49 PM
purplec16
Finding the derivative of the function
$\displaystyle f(x)= \frac{{x}^{3 }-{4x}^{2 }+3 }{x }$

$\displaystyle f'(x)=\f{3x}^{2 }-8x +{x}^{ -2}$

I'm kinda confused on what happens to the x in the denominator i don't know if i should add it or multiply the numerator by it so that is the answer i felt most comfortable
• May 24th 2011, 03:55 PM
Mondreus
There are two ways to solve it. Either write it as you have and use the quotient rule. Or you can simply use $\displaystyle f(x)=\frac{x^3-4x^2+3}{x}=x^2-4x+\frac{3}{x}$
• May 24th 2011, 03:56 PM
purplec16
I really dont understand how you got that answer
• May 24th 2011, 03:57 PM
purplec16
oh ok cool thanks...i understand what you did you divided each term by x
• May 24th 2011, 03:59 PM
Mondreus
I just wrote f(x) in a different way using basic algebra. You should easily be able to differentiate the second form.
• May 24th 2011, 04:04 PM
purplec16
so is the answer $\displaystyle f'(x) = 2x-8x-{3x}^{-2 }$
• May 24th 2011, 04:12 PM
Mondreus
Almost, but the second term is not right.
• May 24th 2011, 04:25 PM
purplec16
$\displaystyle f'(x)=2x-8-{3x}^{-2 }$
No. $\displaystyle \frac{d}{dx}\left(-4x\right) \neq -8$