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Thread: Derivative of a fraction

  1. #1
    Aug 2009

    Derivative of a fraction

    I 've been trying to find the derivative of this function and it's a little confusing.

    $\displaystyle f(x)= \frac{3x^3-x^2+2}{x^\frac{1}{2}} $

    I started out by using the quotient rule getting:

    $\displaystyle f^l(x)= \frac{(9x^2-2x)(x^\frac{1}{2})-[(3x^3-x^2+2)(\frac{1}{2}x^\frac{-1}{2})]}{x^\frac{1}{4}} $

    Then multiplying this out and combining like terms:

    $\displaystyle f^l(x)= \frac{\frac{15}{2}x^\frac{5}{2}-\frac{3}{2}x^\frac{3}{2}+x^\frac{-1}{2}}{x^\frac{1}{4}} $

    The final solution is $\displaystyle f^l(x)= \frac{15}{2}x^\frac{3}{2}-\frac{3}{2}x^\frac{1}{2}-x^\frac{-3}{2} $

    I just cant seem to get from the last step to the solution. Can anyone help? Thanks
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  2. #2
    May 2009
    New Delhi
    $\displaystyle f^l(x)= \frac{(9x^2-2x)(x^\frac{1}{2})-[(3x^3-x^2+2)(\frac{1}{2}x^\frac{-1}{2})]}{(x^\frac{1}{2})^2} $
    $\displaystyle f^l(x)= \frac{(9x^2-2x)(x^\frac{1}{2})-[(3x^3-x^2+2)(\frac{1}{2}x^\frac{-1}{2})]}{x} $
    $\displaystyle f^l(x)= \frac{\frac{15}{2}x^\frac{5}{2}-\frac{3}{2}x^\frac{3}{2}+x^\frac{-1}{2}}{x} $
    $\displaystyle \\ f^l(x)= \frac{15}{2}x^\frac{3}{2}-\frac{3}{2}x^\frac{1}{2}-x^\frac{-3}{2} $
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