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Thread: [SOLVED] Dirivative Help [Quotient Rule]

  1. #1
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    [SOLVED] Dirivative Help [Quotient Rule]

    $\displaystyle
    y=\frac{v^3-2v\sqrt{v}}{v}
    $

    i've tried various techniques and keep getting something like..
    $\displaystyle
    =\frac{(3v^2-3v\sqrt{v})-(v^3-2v\sqrt{v})}{v^2}
    $
    im not sure how to get the answer, but BoB says the answer is
    $\displaystyle
    y'=\frac{2v-1}{\sqrt{v}}
    $
    help? maybe you can tell what im doing wrong or what i still need to do
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  2. #2
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    Dec 2007
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    Lightbulb explanation of how i solved.

    sorry, i should of searched more before I posted. but this isn't really a quotient rule problem. it is initially set up like one, but sometimes it is easier to rewrite a quotient first to put it in a simpler form.
    see:
    $\displaystyle y=\frac{v^3-2v\sqrt{v}}{v}
    $
    $\displaystyle y=\frac{v^3-2v^\frac{3}{2}}{v}
    $

    divide out the $\displaystyle v$ to get:
    $\displaystyle y=v^2-2v^\frac{1}{2}$

    hope i helped someone learn something.
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