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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.

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marked as duplicate by Argon, Michael Albanese, rogerl, apnorton, Live Forever Jul 16 '14 at 1:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

What is the domain of your function? – Crostul Jul 15 '14 at 22:07
A recurrence is easy. Writing out the "unwound form" in LaTex, not so much. – André Nicolas Jul 15 '14 at 22:10 see the great section on the infinite tower, halfway down the page – John Fernley Jul 15 '14 at 22:52
Possible duplicate: – Argon Jul 16 '14 at 0:19

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

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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). – Mathematician171 Jul 15 '14 at 22:18
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
@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
@JohnFernley: Yep. I am surprised as well. Still trying to make sense of that. But trying it out in W|A confirms it. – I like Serena Jul 15 '14 at 23:07
@ilikeserena thank you for providing the domain. You can edit my "hint" accordingly ;) if you want. Thanks – Chinny84 Jul 15 '14 at 23:13

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.

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It appears the upper limit is $e^{1/e} > 1$. I don't quite understand it though. – I like Serena Jul 15 '14 at 23:09
I personally think this is a better answer than mine when I look at the overall picture of the problem. +1 – Chinny84 Jul 15 '14 at 23:10
@Chinny84: Oh, your hint is still quite valid. It appears there is just more to it. – I like Serena Jul 15 '14 at 23:12

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

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