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Derviate x^x^x

  1. Oct 23, 2007 #1
    1. The problem statement, all variables and given/known data

    differentiate f(x)= x^x^x

    2. Relevant equations
    chain rule
    product rule

    3. The attempt at a solution

    x^x (lnx)

    i dont know what to do after this
  2. jcsd
  3. Oct 23, 2007 #2


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    A function raised to another function is an exponential:

    In general, [tex]f(x)^{g(x)} = \exp(\ln(f(x)^{g(x)}))=\exp(g(x)\ln(f(x)))[/tex]

    And you know how to differentiate an exponential.

    So, can you use what I wrote to write x^x^x as an exponential?
  4. Oct 23, 2007 #3


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    Would that be:
    1. [tex]x^{(x^{x})}=x^{x^{x}}[/tex]
    2. [tex](x^{x})^{x}=x^{x^{2}}[/tex]

    Learn to use parentheses..
  5. Oct 24, 2007 #4


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    Don't just leave x^x(ln x) by itself! If f= x^x^x, then ln(f)= x^x ln(x). Now DO IT AGAIN! ln(ln(f))= ln(x^x ln(x))= ln(x^x)+ ln(ln(x))= xln(x)+ ln(ln(x)).

    Use the chain rule to differentiate both ln(ln(f(x)) and ln(ln(x)).
  6. Oct 25, 2007 #5

    Gib Z

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    I sometimes get annoyed with exponential notation for exactly that reason. My opinion is if the exponent is any larger than 1 term, write it is terms of exp(...).
  7. Oct 26, 2007 #6
    Neglecting the given attempt, I put
    [tex] y = x^{x^{x}} [/tex]
    [tex] z = x^{x} [/tex]
    and develop as follows.
    \ln y = \ln x^{x^{x}}
    = z \ln x
    \frac{y'}{y} = z' \ln x + z \frac{1}{x}
    here I calculate the differentiation of [tex] z [/tex]
    [tex] z = x^{x} [/tex]
    [tex] \ln z = x \ln x [/tex]
    [tex] \frac{z'}{z} = \ln x + x \frac{1}{x} [/tex]
    z' = z \left( { \ln x + 1 } \right)
    = x^{x} \left( { \ln x + 1 } \right)
    y' = y \left( { x^{x} \left( { \ln x + 1 } \right) \ln x + x^{x} \frac{1}{x} } \right)
    = x^{x^{x}+x-1} \left( { x \left( { \ln x + 1 } \right) \ln x + 1 } \right)
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