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Differentiation question

  1. May 6, 2005 #1
    how can this be differentiated,i need to know how.x^x^x^x^x^x....
    also ,my teacher posed this question to the whole class x^(x^2)^(x^3)^(x^4)................
  2. jcsd
  3. May 6, 2005 #2
    The greatest mathematician of all time would be able to solve this easily.
  4. May 6, 2005 #3


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    The first is really easy,if u consider logarithmic differentiation...

  5. May 6, 2005 #4
    Try to write : h(x)=x^x^x^x....

    Then clearly h(x)=x^h(x)....just differentiate and isolate h'(x)....

    Also x^(x^2)=x^(x.x)=(x^x)^x=x^x^x...or if you want : log(x^(x^2))=x^2*log(x)=x*log(x^x)=log((x^x)^x)=log(x^x^x)
  6. May 7, 2005 #5
    you are really funny juvenal.
    but then how about the other function, how can i get that differentiated?
  7. May 7, 2005 #6


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    Kleinwolf,i'm sure you're not familiar to the notation [tex] x^{x^{x^{...}}} [/tex]

    So basically i could write

    [tex] x^{x^{2}}=x^{x\cdot x}=\left(x^{x}\right)^{x}\neq x^{x^{x}}} [/tex]

  8. May 8, 2005 #7
    hey if its x^x^x^.......infinity
    y = x^x^x^x^......can be written as

    y = x^y ,
    then take log of both sides and then diffrentiate .
  9. May 8, 2005 #8
    Right Daniel, I messed up...

    I got disturbed by having no parenthesis, because [tex] x^{(x^2)}\neq (x^x)^2=x^{(2x)} [/tex].....

    Then I think u could use an iterative definition : [tex] f(x,y)=x^y [/tex]

    [tex] h_1(x)=x, h_2(x)=x^{(x^2)}, h_3(x)=x^{((x^2)^{(x^3)})}...[/tex]

    So that in fact : [tex] h_1(x)=f(x,1), h_2(x)=f(x,x^2), h_3(x)=f(x,f(x^2,x^3))...aso... [/tex]

    Now you "just" have to differentiate the imbrication of f(...f(...(x^{n-1},x^n)...)....this should be quite complicated.
    Last edited: May 8, 2005
  10. May 8, 2005 #9
    how about the other question ,no one is talking about that ,just the easy one.
    the most important is x^(x^2)^(x^3)....
  11. May 9, 2005 #10


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    For the first one I got the following answer, which I'm pretty sure is correct.
    y' = y^2 / ( x ( 1 - ln(x) y ) )

    I set it up as an implicit function, y(x) : (x^y - y)=0, and used
    y' = -(dQ/dx) / (dQ/dy), where Q(x,y) = x^y - y.
    Last edited: May 9, 2005
  12. May 9, 2005 #11


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    Recalling the ambiguity (that was clarified by Daniel) in the first expression is that second one meant to be evaluated from left to right or from right to left ?

    If it's evaluated left to right then it diverges for all x>1 and is one for 0<x<=1, so the derivative is not really very interesting as it's just zero in the region for which the function is meaningfully defined. If however it's evaluated right to left then hmmm, I'm not sure.
  13. May 10, 2005 #12
    That is the solution for the first function ,how about the other x^(x^2)^(x^3)^(x^4).....................
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