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Let $$f(x)=\lim\limits_{n\to\infty}\underbrace{x^{x^{x^{...}}}}_{n\text{ times}}$$ Is it possible to find $f'(x)$. If yes, please show all steps.

Conifold
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3 Answers3

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$\textbf{hint:}$ $$f(x) = x^{f(x)}$$ Then proceed

Chinny84
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  • Simply brilliant. – naslundx Jul 15 '14 at 22:17
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    I tried this. I've got f(x) in terms of Lambert W function, but I don't know W'(x). –  Jul 15 '14 at 22:18
  • Just take the derivative of both sides and solve for $f'(x)$ in terms of $f(x)$, @Mathematician171 – Thomas Andrews Jul 15 '14 at 22:19
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    This assumes that the limit exists and is differentiable, which isn't obvious at all. In fact for x>1 it obviously doesn't. Without specifying the domain an equally valid answer is $f'$ doesn't exist, or even $f$ doesn't exist for that matter. – Conifold Jul 15 '14 at 22:24
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    @conifold you are absolutely correct. If in fact the domain does not permit the approach above then I hold my hands up and take it down. But I am afraid that's a question for the original poster :). – Chinny84 Jul 15 '14 at 22:30
  • also $$\frac{{\rm d}W}{{\rm d}z}=\frac{W(z)}{z(1 + W(z))}\quad\text{for }z\not\in{0,-1/e}$$ if it's useful – John Fernley Jul 15 '14 at 22:36
  • @conifold: Without specifying a domain, the domain is whereever $f$ has a defined value. This appears to be the interval (0,1]. It looks like $f'$ is defined on (0,1). – Klaas van Aarsen Jul 15 '14 at 22:37
  • Make that the interval $[e^{-e}, e^{1/e}] = [0.0659, 1.4446]$ (after reading the Wolfram article). – Klaas van Aarsen Jul 15 '14 at 22:59
  • @IlikeSerena surprised that $e^{1/e}>1$ – John Fernley Jul 15 '14 at 23:04
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    @JohnFernley: Yep. I am surprised as well. Still trying to make sense of that. But trying it out in W|A confirms it. – Klaas van Aarsen Jul 15 '14 at 23:07
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    @ilikeserena thank you for providing the domain. You can edit my "hint" accordingly ;) if you want. Thanks – Chinny84 Jul 15 '14 at 23:13
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This function is known as infinite tetration. The sequence is undefined for $x\leq 0$ except at some rational points, so there is no point talking about the derivative there. Ditto for large $x$, where the sequence diverges to $\infty$.

Rather surprisingly for $x>0$ the limit only exists for $x\in[e^{-e},e^{1/e}]$ as shown by Euler, see here and here. On that interval $f(x)$ is the inverse function to $x^{1/x}$. This implies differentiability there, and that's where formal games with implicit differentiation actually work. At first glance it seems strange that the sequence converges even for some values $x>1$ since then $x^x>x$ and it is monotone increasing. However, as long as the value of $x$ is close enough to $1$ this is similar to $(1+\frac1n)^n$ that monotone increases but converges to $e$. The threshold value $e^{1/e}$ is the maximum of $x^{1/x}$, there is nothing to be inverse to for larger values.

Conifold
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WLOG, let $y = f(x)$. You know that $$y =x^y$$ So $$\ln(y) =y\ln(x)$$ $$\frac{dy}{dx}*\frac{1}{y} = \frac{dy}{dx}*\ln(x) + \frac{y}{x}$$ $$\frac{dy}{dx} = \frac{y^2}{x-xy\ln(x)}$$

Vishwa Iyer
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