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Math Help - Tricky derivation...

  1. #1
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    Tricky derivation...

    Can someone help me to derivate x^x^x? I'm starting to belive it's not possible... : D heeelp! Thank you in advance!
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  2. #2
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    Quote Originally Posted by kukkeball View Post
    Can someone help me to derivate x^x^x? I'm starting to belive it's not possible... : D heeelp! Thank you in advance!
    Hello,

    you only have to change the x's into powers of base e:

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

    I presume that you know the chain rule and that you know how to derivate an exponentila function with base e.

    EB
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  3. #3
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    Quote Originally Posted by kukkeball View Post
    Can someone help me to derivate x^x^x? I'm starting to belive it's not possible... : D heeelp! Thank you in advance!
    Oh, it's perfectly possible, and fun too .

    I have class soon, but I will get you started.

    First, what is the derivative of x^x?

    We have y = x^x

    Note: x > 0

    ln(y) = ln(x^x)

    ln(y) = x*(ln(x))

    Differentiate both sides with respect to x;

    y'(1/y) = ln(x) + x*(1/x) = (ln(x) + 1)

    y' = (ln(x) + 1)*y

    But recall y = x^x

    y' = (ln(x) + 1)*x^x

    Now you can use this to find the derivative of x^x^x.
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  4. #4
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    Quote Originally Posted by earboth View Post
    Hello,

    you only have to change the x's into powers of base e:

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

    I presume that you know the chain rule and that you know how to derivate an exponentila function with base e.

    EB
    Perhaps I am missing something, but how is e^e^(x*ln(x)) = e^x^x the same as x^x^x. The derivatives are clearly different.
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  5. #5
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    Quote Originally Posted by AfterShock View Post
    Perhaps I am missing something, but how is e^e^(x*ln(x)) = e^x^x the same as x^x^x. The derivatives are clearly different.
    Hello, AfterShock,

    of course you are right: I messed up my notes. It should be:

    x^{x^x}=e^{\ln(x) \cdot e^{x*ln(x)}}

    As you can see, one factor ln(x) slipped away. But anyhow your method is much better as what I've planned to do.

    EB
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  6. #6
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    So what's the correct answer... i got something crazy... : F
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  7. #7
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    Hello, kukkeball!

    Here's another approach . . .


    Differentiate: . y \:=\:x^{x^x}

    Take logs: . \ln y \;=\;\ln\left(x^{x^x}\right) \;=\;x^x\ln x

    Take logs again: . \ln(\ln y)\;=\;\ln\left[x^x\ln x\right] \;=\;\ln\left(x^x\right) + \ln(\ln x) \;=\;x\ln x + \ln(\ln x)


    Differentiate implicitly (very carefully) . . .

    . . \frac{1}{\ln y}\cdot\frac{1}{y}\cdot\frac{dy}{dx}\;=\;x\cdot\fr  ac{1}{x} + \ln x + \frac{1}{\ln x}\cdot\frac{1}{x}

    We have: . \frac{1}{y\ln y}\cdot\frac{dy}{dx} \;=\;1 + \ln x + \frac{1}{x\ln x}

    Then: . \frac{dy}{dx}\;=\;y\ln y\left(1 + \ln x + \frac{1}{x\ln x}\right)

    Therefore: . \frac{dy}{dx}\;=\;x^{x^x}\ln\left(x^{x^x}\right)\l  eft(1 + \ln x + \frac{1}{x\ln x}\right)<br /> <br />
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