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Thread: implicit differentiation

  1. #1
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    implicit differentiation

    Find $\displaystyle \frac{d^2y}{dx^2}$ by implicit differentiation: $\displaystyle 2x^3-3y^2=8$

    So far I get
    $\displaystyle \frac{dy}{dx}=\frac{x^2}{y}$
    $\displaystyle \frac{d^2y}{dx^2}=\frac{2y}{x}$ <----- incorrect


    correct answer= $\displaystyle \frac{2xy^2-x^4}{y^3}$

    How do i find the correct answer?
    Last edited by yoman360; Sep 12th 2009 at 01:49 PM.
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  2. #2
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    Quote Originally Posted by yoman360 View Post
    Find $\displaystyle \frac{d^2y}{dx^2}$ by implicit differentiation: $\displaystyle 2x^3-3y^2=8$
    show some effort ...

    what did you get for $\displaystyle \frac{dy}{dx}$ ?
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  3. #3
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    I get $\displaystyle \frac{dy}{dx}=\frac{x^2}{y}$

    i keep getting the same answer

    this one is difficult
    Last edited by mr fantastic; Sep 12th 2009 at 03:10 PM. Reason: Merged posts
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  4. #4
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    $\displaystyle 2x^3-3y^2=8$

    $\displaystyle 6x^2 - 6y \cdot y' = 0$

    $\displaystyle y' = \frac{x^2}{y}$

    $\displaystyle y'' = \frac{2xy - x^2y'}{y^2}$

    $\displaystyle y'' = \frac{2xy - \frac{x^4}{y}}{y^2}$

    $\displaystyle y'' = \frac{2xy^2 - x^4}{y^3}
    $
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  5. #5
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    Quote Originally Posted by skeeter View Post
    $\displaystyle 2x^3-3y^2=8$

    $\displaystyle 6x^2 - 6y \cdot y' = 0$

    $\displaystyle y' = \frac{x^2}{y}$

    $\displaystyle y'' = \frac{2xy - x^2y'}{y^2}$

    $\displaystyle y'' = \frac{2xy - \frac{x^4}{y}}{y^2}$

    $\displaystyle y'' = \frac{2xy^2 - x^4}{y^3}
    $
    oh i see now I forgot to use the quotient rule
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