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Derivative x^x^x

  1. Sep 9, 2003 #1
    i can solve x^x but adding this new x just confuses me any help will do, X^x^x
     
  2. jcsd
  3. Sep 9, 2003 #2
    Do you mean (x^x)^x or x^(x^x)? I'm assuming the former.
     
  4. Sep 9, 2003 #3
    sorry i meant x^(x^x)
     
  5. Sep 9, 2003 #4
    Glad you picked that one. Get something like ln(y) = xxln(x) and use the product rule. Don't forget you already know d(xx)/dx :wink:
     
  6. Sep 9, 2003 #5
    wow i never noticed that i was doing it a long way which would have come out wrong anyways. thanks for the help
     
  7. Feb 2, 2009 #6
    ok ok
     
    Last edited: Feb 2, 2009
  8. Feb 2, 2009 #7

    arildno

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    1. It is not at all imposible; note that this equals: [tex](x^{x})^{x}=x^{x^{2}}[/tex]
    Rewriting this as:
    [tex]x^{x^{2}}=e^{x^{2}\ln(x)}[/tex]
    We may readily differentate this by means of the chain rule, yielding the derivative:
    [tex]x^{x^{2}}(2x\ln(x)+x)[/tex]

    2. Please do not re-open nearly 6-year old threads.
     
  9. Feb 2, 2009 #8
    ok ok
     
    Last edited: Feb 2, 2009
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