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Math Help - Second Derivative

  1. #1
    Junior Member
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    Second Derivative

    Ok first post here so i apologize if i do something wrong.

    So i know to use implicit differentiation to get the first derivative
    5x^4/3x^2.
    But then i dont understand two things...

    1) Why do they multiply by xy/xy?
    2) And in the last step, what happened to the y' ? How did they get rid of it? I hope my question makes sense.

    Attached Thumbnails Attached Thumbnails Second Derivative-screen-shot-2012-10-09-8.49.24-pm.jpg  
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  2. #2
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    Re: Second Derivative

    Hey petenice.

    The reason why they multiply it is because they want to get it into the form of x^5/y^3 which is equal to 1 by the definition of the initial relationship so it cancels.

    For your second question, this is just the normal quotient rule where d/dx(u/v) = [vu' - uv']/v^2.

    u = 5y, v = 3x, u' = 5y', u' = 3 so [vu' - uv']/v^2 = [(3x)(5y') - (5y)(3)]/(9x^2) because y' = dy/dx and d/dx(3x) = 3.
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Re: Second Derivative

    Another approach:

    y=x^{\frac{5}{3}}

    \frac{dy}{dx}=\frac{5}{3}x^{\frac{2}{3}}

    \frac{d^2y}{dx^2}=\frac{10}{9}x^{-\frac{1}{3}}=\frac{10x^{\frac{5}{3}}}{9x^2}=\frac{  10y}{9x^2}
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  4. #4
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    Re: Second Derivative

    Im going to use this forum more often now. I've been staring at this question for hours and now I totally get it!

    Chiro, Mark,

    Thanks so much!
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