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Math Help - Derivative help!

  1. #1
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    Derivative help!

    I'm having a hard time find 2nd derivative of the following function:

    \frac{p^3 }{ p^3+(1-p)^3}

    I have the first derivative:

    \frac{3p^2[p^3+(1-p)^3][-3p^2-3(1-p)^2](p^3)}{[(p^3+(1-p)^3]^2}

    2

    \frac{3p^5+3p^2(1-p)^3-3p^5+3p^3(1-p)^2}{[(p^3+(1-p)^3]^2}

    3

    \frac{3p^2(1-p)^3+3p^3(1-p)^2}{[p^3+(1-p)^3]^2}

    4

    \frac{3p^2(1-p)^3(1-p)+p}{[p^3+(1-p)^3]^2}

    5
    \frac{3p^2(1-p)^2}{p^3+(1-p)^3]^2}


    F

    3p^2(1-p)^2

    6p(1-p)^2+2(1-p)(-1)3p^2

    6p(1-p)^2-6(1-p)p^2

    6p(1-p)[(1-p)-p]

    6p(1-p)(1-2p)

    g

    2(p^3+(1-p)^3)[3p^2+3(1-p)^2(-1)]

     2[p^3+(1-p)^3][3p^2-3(1-p)^2]

    2[p^3+(1-p)^3[3p^2-3+6p-3p^2]

     2[p^3+(1-p)^3](-3+6p)

    6(p^3+(1-p)^3](2p-1)


    h2


    \frac{6p(1-p)(1-2p)[p^3+(1-p)^3]^2-[6(p^3+(1-p)^3](3p^2)(1-p)^2}{[p^3+(1-p)^3]^4}

    ...

    \frac{6(1-p)[p^3+1-p)^3[p^3+(1-p)^3](1-2p)-3p^2(1-p)}{p^3+[(1-p)^3]^4}

    ...

    \frac{6(1-p)[p(p^3+(1-p)^3](1-p)^3(1-2p)-3p^2(1-p)}{[p^3+(1-p)3]^3}

    (p^3+[1-p)^3]^2

    \frac{.61^3}{.61+(1-.61)^3}
    Last edited by minus; November 10th 2008 at 12:55 PM.
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  2. #2
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    Using MathCad, I get \frac{{12p^3  - 18p^2  + 6p}}{{27p^6  - 81p^5  + 108p^4  - 81p^3  + 36p^2  - 9p + 1}}
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  3. #3
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    [quote=Plato;212682]Using MathCad, I get \frac{{12p^3  - 18p^2  + 6p}}{{27p^6  - 81p^5  + 108p^4  - 81p^3  + 36p^2  - 9p + 1}}[/

    quote]

    Thanks for your help plato. Your results is for 2nd derivative right? If indeed it's 2nd derivative can you please double check my answer for first derivative.

    \frac{61^3}{39^3}

    3.826 + 1

    \frac{100}{4.826}

    (20.72)(3.826)
    Last edited by minus; November 11th 2008 at 06:23 AM.
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  4. #4
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    Quote Originally Posted by minus View Post
    Thanks for your help plato. Your results is for 2nd derivative right? If indeed it's 2nd derivative can you please double check my answer for first derivative.
    That is not my result; I used a computer algebra system to do it.
    I think doing this sort of a problem without a CAS is a pointless exercise!
    The first derivative is
    \frac{{3p^4  - 6p^3  + 3p^2 }}{{9p^4  - 18p^3  + 15p^2  - 6p + 1}}
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