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Math Help - let y=x^x find dy/dx.... why is it hard ?

  1. #1
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    let y=x^x find dy/dx.... why is it hard ?

    This question is from a UNIVERSITY PAPER (just incase some of you ppl will ask why would you want to differentiate that) ....

    how would i go on about this... the answer is x^x(1+ln(x)) ???????


    Thank you....
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  2. #2
    Senior Member Twig's Avatar
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    Hi!

    Write x^{x} as  e^{x\cdot \ln(x)} .

    Now differentiate using normal rules, with u(x)=x\cdot \ln(x) .

    \frac{d}{dx}(e^{u(x)})=e^{u}\cdot u'(x) = e^{x\cdot \ln(x)}\cdot(1+\ln(x) =x^{x}\cdot(1+\ln(x))

    Note that I used the product rule on x\cdot \ln(x)
    Last edited by CaptainBlack; August 29th 2009 at 01:48 AM. Reason: improve the LaTeX
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  3. #3
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    Hello Khonics89!

    Quote Originally Posted by Khonics89 View Post
    This question is from a UNIVERSITY PAPER (just incase some of you ppl will ask why would you want to differentiate that) ....

    how would i go on about this... the answer is x^x(1+ln(x)) ???????


    Thank you....
    Hint:

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

    Do you know how to differentiate that?

    Yours
    Rapha


    Edit: Beaten to it by Twig.
    Last edited by CaptainBlack; August 29th 2009 at 01:47 AM. Reason: Improve the LaTeX
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  4. #4
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    ohh yes ofcourse hehe.. xD

    Thanks guys.. never knew we can write x^x as e^xln(x)....

    yeah all good thanks everyone
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  5. #5
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    logarithmic differentiation ...

    y = x^x

    \ln(y) = x \cdot \ln(x)

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

    y' = y[1 + \ln(x)] = x^x[1 + \ln(x)]
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