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A^(x^x) graph

  1. Jan 18, 2009 #1
    Hey guys,

    I was wondering about what a graph would look like where the power to a did not increase at a linear rate i.e

    a^(x^x)

    Is there such a recognised function as this? If so does it have any practical applications and what does the graph look like?

    Thanks,

    Oscar
     
  2. jcsd
  3. Jan 18, 2009 #2

    tiny-tim

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    Re: a^(x^x)???

    Hey 2Oscar! :smile:
    Never seen anything like it! :surprised

    I'd be very surprised if it does have any practical applications.

    Its graph would be like ax, only very much steeper. :smile:

    Why don't you try working out its derivative? :wink:
     
  4. Jan 18, 2009 #3
    Re: a^(x^x)???

    I'm unsure on how to differentiate a^x tbh... but i think i can do it for e^x...

    so y= e^(x^x)

    dy/dx = x2e^(x^x)?


    My graphical calculator goes wierd when i try to draw it lol... but i can see why it would be like a normal exponential graph just steeper.

    Thanks,

    Oscar
     
  5. Jan 18, 2009 #4

    tiny-tim

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    No …

    e^(xx) = e^(exlogx),

    so it's not x2, but d/dx(exlogx), = … ? :smile:

    (and a^(xx) = e^(xxloga) wink:)
     
  6. Jan 18, 2009 #5
    Re: a^(x^x)???

    sorry if im misunderstanding... is the log to the base e?

    If so then d/dx(e^xlogx) = (xlogx)e^(xlogx)?

    Or perhaps d/dx(e^xlogx) = e^x2?

    Sorry if i sound daft... not too familiar with this stuff :S


    Thanks,

    Oscar
     
  7. Jan 18, 2009 #6

    tiny-tim

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    No, use the chain rule …

    d/dx(exlogx) = exlogx d/dx(xlogx) :smile:

    (use the X2 tag just above the reply box :wink:)
     
  8. Jan 18, 2009 #7
    Re: a^(x^x)???

    so xlogx differentiates to 1+logx?

    so (1+logx)exlogx?
     
  9. Jan 18, 2009 #8

    tiny-tim

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    Yup! :biggrin:
     
  10. Jan 18, 2009 #9
    Re: a^(x^x)???

    Oh wow so the end differentiation is (1+logx)exlogxex^x?


    Thanks so much for your help :D

    Oscar
     
  11. Jan 18, 2009 #10

    tiny-tim

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    You can simplify it a bit more …

    (1+logx)xxex^x :wink:
     
  12. Jan 22, 2009 #11

    Mentallic

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    Re: a^(x^x)???

    Usually when you look into powers that increase at a rate rather than linear, you try quadratic, not exponential :tongue2:

    While I don't think it applies to anything, I'm more curious to dedicate my life studying:

    [tex]y=e^{x^{x^x}}[/tex]

    There is growing interest in the field of BS-mathematics :wink:

    If you haven't noticed yet, it grows pretty fast.

    [tex]x=1 ~ y=e[/tex]
    [tex]x=1.5 ~ y=8[/tex]
    [tex]x=2 ~ y=9,000,000[/tex]
    [tex]x=2.3 ~ y>googol[/tex]
     
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