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Differentiating x^x

  1. Aug 3, 2006 #1
    Is it possible to differentiate x^x? I’ve tried with the use of logs, but it doesn’t seem to work…….any help would be kindly appreciated :smile:
  2. jcsd
  3. Aug 3, 2006 #2
    you ahve to say what yiou ahve done first ie workings and the such like
  4. Aug 3, 2006 #3

    when you differentiate e, it stays the same, but you multiply by the inner derrivative:

    Last edited: Aug 3, 2006
  5. Aug 3, 2006 #4


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    First why is a question about the derivative in "Precalculus"? It sounds pretty "calculus" to me!

    Logs is, of course, the way to go. If y= xx, then log y= x log x. Differentiate both sides:
    [tex]\frac{d log y}{dx}= \frac{1}{y}\frac{dy}{dx}[/tex]
    of course. And
    [tex]\frac{d(x log x)}{dx}= log x+ \frac{x}{x}= log x+ 1[/tex]

    Here's an interesting point: to find the derivative of [itex]f(x)^{g(x)}[/itex] there are two obvious mistakes you could make:

    1) Treat the exponent, g(x), as if it were a constant and write
    [tex]\frac{f(x)^{g(x)}}{dx}= g(x)f(x)^{g(x)-1}\frac{df}{dx}[/tex]

    2) Treat the base, f(x), as if it were a constant and write

    Because those are mistakes, neither is, of course, correct. The correct derivative of [itex]f(x)^{g(x)}[/itex] is the sum of those!
  6. Aug 3, 2006 #5
    he did say, he tried logs...
  7. Aug 3, 2006 #6
    Just because he tried logs doesn't mean he did it right. HallsofIvy way is the correct way to do it. Log both sides then differentiate both.
  8. Aug 3, 2006 #7
    yeah, you're right, we should have lead him to the solution and not just give it out like that...

    anyway, doing it my way or differentiationg both sides is really the same thing... i didn't need to look at both sides because i just used the equation [tex]a^b=e^{bln(a)}[/tex]
    so y is still y... so if i differentiate both sides i get:
  9. Aug 4, 2006 #8
    Instead of starting a new thread I thought I'd "borrow" this one:

    I am trying to differentiate the function [tex]y=(x^x)^x[/tex]

    Would this working be correct?

    [tex]y = (x^x)^x = x^f^(^x^) [/tex]

    [tex] \rightarrow ln(y) = f(x)ln(x) [/tex]

    [tex]\frac{dy}{dx}\frac{1}{y} = (f(x))(\frac{dln(x)}{dx}) + (ln(x))(\frac{df(x)}{dy})[/tex]

    [tex]= \frac{f(x)}{x} + f'(x) lnx[/tex]

    [tex]\rightarrow \frac{dy}{dx} = y((\frac{f(x)}{x}) + f'(x)ln(x))[/tex]

    But since:

    [tex] f(x) = x^x[/tex]

    [tex]f'(x) = x^x(ln(x) + 1)

    (I've spared the working for that since it's covered up there)

    So does that make:

    [tex] \frac {d}{dx} (x^x)^x = (x^x)^x ( \frac{x^x}{x} + (ln(x)+1)x^x)


    Or and I missing something along the way? This is my first time using the product rule so I want to make sure I'm doing it correctly..


    Last edited: Aug 4, 2006
  10. Aug 4, 2006 #9


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    Nope, this line is wrong. Instead, it should read:
    [tex]y = {(x ^ x)} ^ x = f(x) ^ x[/tex], where f(x) = xx. :)
    Ok, let's take log of both sides:
    [tex]\ln y = x \ln f(x)[/tex]
    Differentiate both sides with respect to x, we have:
    [tex]\frac{y'_x}{y} = \ln f(x) + x \frac{f'(x)}{f(x)} \Rightarrow y'_x = y \left( \ln f(x) + x \frac{f'(x)}{f(x)} \right) = {(x ^ x)} ^ x \left( \ln f(x) + x \frac{f'(x)}{f(x)} \right)[/tex], right?
    And from the above posts, we have:
    f'(x) = (xx)' = (xx) (ln(x) + 1). So plug that into the expression above, we have:
    [tex]y'_x = {(x ^ x)} ^ x \left( \ln f(x) + x \frac{x ^ x (\ln (x) + 1)}{x ^ x} \right) = {(x ^ x)} ^ x \left( \ln (x ^ x) + x (\ln (x) + 1)} \right)[/tex]
    [tex]= {(x ^ x)} ^ x \left( \ln (x ^ x) + \ln (x ^ x) + x)} \right) = {(x ^ x)} ^ x \left( 2 \ln (x ^ x) + x)}[/tex]
    Ok. Can you get this? :)
    Last edited: Aug 4, 2006
  11. Aug 4, 2006 #10


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    Lewis, your way would be correct if you wanted to differentiate [itex]y=x^{(x^x)}[/itex]. But as you have it written now, [itex]y=(x^x)^x=x^{(x^2)}[/itex].
  12. Aug 5, 2006 #11
    That's what I was trying to write! I can't get the hang of tex at all... I editted it a few times but couldn't get it to look like: [tex]y=x^{(x^x)}[/tex]

    Thanks though at least I know I'm on the right path (I should have explained what I was trying to do a little better in hindsight..)

    I'll try and tackle the:


    Problem now and see if I can get Vietdao's solution. :)

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