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I Why is [x^x^x]' NOT (xln(x)^2+ln(x)+xln(x)+1)*x^(x+x^x)?

  1. Jul 14, 2017 #1
    Mathamatica, Symbolab, manual differentiation all return diffrent forumals, I cannot see wolfram's intermediary steps, symbolab's intermediary steps seem like they're treating a function like a constant and I can't trust my work when everything else disagrees, but i don't see where my logic would falter as long as x^x^x can be represented by f(x) = x^x, g(x) = x^x, x^x^x = f(g(x))

    $$\frac{dy}{dx} of y = e^x^x $$ is

    y = e^x^x

    y' = q(g(x))' = q'(g(x))*g'

    g(x) = x^x, q(x) = e^x
    g'(x) = (ln(x) + 1)*x^x, q'(x) = e^x

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

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

    The above is correct, dy/dx x^x was obtained via:
    g(x) = x^x
    ln(g) = xln(x)
    ln(g)' = (xln(x))'
    ln(g)' = 1 * ln(x) + x * 1/x
    g'/g = 1 * ln(x) + x * 1/x
    g'/g = ln(x) + 1
    g' = (ln(x) + 1)*x^x

    But when i try continue for e^x^x^x via chain rule
    t=x^x^x = f(g(x))
    f=x^x g = x^x
    f' = (ln(x) + 1)*x^x, g' = (ln(x) + 1)*x^x

    (f(g(x)))' = f'(g(x))*g'(x) = (ln(x) + 1)(ln(x^x) + 1)*x^x^x * x^x = (xln(x)^2 + ln(x) + xln(x) + 1) * x^(x + x^x)

    dy/dx x^x^x = (xln(x)^2 + ln(x) + xln(x) + 1) * x^(x + x^x)

    I obtain the incorrect answer, as wolfram says dy/dx of x^x^x = e^x^x^x * (xln(x)^2 + xln(x) + 1) * x^(x + x^x -1), but I do not have the professional version of wolfram, thus I cannot see the intermediary steps, and cannot find this information anywhere else,

    Any help appreciated.
     
    Last edited: Jul 14, 2017
  2. jcsd
  3. Jul 14, 2017 #2

    jbriggs444

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    Science Advisor

    If f(x) = g(x) = ##x^x## then f(g(x)) = ##g(x)^{g(x)}## = ##(x^x)^{(x^x)}##
     
  4. Jul 14, 2017 #3
    ah, damn, should have noticed that. Thanks so much.
     
  5. Jul 14, 2017 #4
    How would i make it into a chain rule form? x^x^x not (x^x)^(x^x)?
     
  6. Jul 14, 2017 #5

    Charles Link

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    Homework Helper

    For ## dy/dx ## where ## y=x^{(x^x)} ##, I do not get the answer you quoted from Wolfram, but the calculation should be straightforward. Write ## y=x^{(x^x)} ## and take the natural twice on both sides of the equation. Then simply apply the chain rule and implicit differentiation of both sides and solve for ## dy/dx ##. ## \\ ## In fact I am in agreement with what you quoted from Wolfram other than the factor ## e^{(x^{(x^x)})} ## which I have as simply being 1.
     
  7. Jul 14, 2017 #6
    I checked with a knowingly working tetration differentiation formula, although berkeman pointed out , my mistake was that what I typed as x^x^x is actually (x^x)^(x^x). Thanks though.
     
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