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Math Help - differentiating

  1. #1
    Member javax's Avatar
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    differentiating

    hello.
    I'm I bit confused on differentiating this function

    y = x^{x^x}
    Last edited by javax; June 29th 2008 at 08:16 AM.
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  2. #2
    Moo
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    Hello,

    Is it (x^x)^x or x^{x^x}?
    (see the latex code )

    In both cases, you will need that :

    x^x=e^{x \ln x}

    Using the chain rule :

    (x^x)'=(x \ln x)' e^{x \ln x}=(x \ln x)' x^x

    (x \ln x)'=1+\ln x (product rule)


    Therefore \boxed{(x^x)'=(1+\ln x)x^x}

    Does it help ?
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  3. #3
    Member javax's Avatar
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    y

    It was x^{x^x}. Yes Moo it helped. Thanks a lot!
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by javax View Post
    hello.
    I'm I bit confused on differentiating this function

    y = x^{x^x}
    y=x^{x^{x}}

    So \ln(y)=\ln(x)\cdot{x^x}

    Differentiating we get

    \frac{y'}{y}=\ln(x)\left(x^x\right)'+x^{x-1}

    Now as moo showed above

    \frac{y'}{y}=\ln(x)(1+\ln x)x^x+x^{x-1}

    Now seeing that y is our original equation and solving for y' we get

    y'=x^{x^{x}}\bigg[\ln(x)(\ln(x+1)x^x+x^{x-1}\bigg]
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