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Find the derivative of $f(x)$: $$f(x)=x^{x^{\dots}{^{x}}}$$

Let $n$ be the number of overall $x's$ in $f(x)$. So for $n=1$, $f(x)=x$. I then tried to determine a pattern by solving for the derivative from $n=1$ to $n=5$. Here's what I got:

\begin{align} n = 2 \Longrightarrow f(x) &= x^x \\ f'(x) &= \frac{d}{dx}\left(e^{x\ln \left(x\right)}\right) \\ &= e^{x\ln \left(x\right)}\frac{d}{dx}\left(x\ln \left(x\right)\right) \\ &= e^{x\ln \left(x\right)}\left(\ln \left(x\right)+1\right) \\ &= x^x\left(\ln \left(x\right)+1\right) \end{align}

\begin{align} n = 3 \Longrightarrow f(x) &= x^{x^{x}} \\ f'(x) &= \frac{d}{dx}\left(e^{x^x\ln \left(x\right)}\right) \\ &= e^{x^x\ln \left(x\right)}\frac{d}{dx}\left(x^x\ln \left(x\right)\right) \\ &= e^{x^x\ln \left(x\right)}\left(x^x\ln \left(x\right)\left(\ln \left(x\right)+1\right)+x^{x-1}\right) \\ &= x^{x^x}\left(x^x\ln \left(x\right)\left(\ln \left(x\right)+1\right)+x^{x-1}\right) \end{align}

\begin{align} n = 5 \Longrightarrow f(x) &= x^{x^{x^{x^{x}}}} \\ f'(x) &= ... \\ &= x^{x^{x^{x^x}}}\left(x^{x^{x^x}}\ln \left(x\right)\left(x^{x^x}\ln \left(x\right)\left(x^x\ln \left(x\right)\left(\ln \left(x\right)+1\right)+x^{x-1}\right)+x^{x^x-1}\right)+x^{x^{x^x}-1}\right) \end{align}

However, I am not sure if I see a pattern here that can help solve the question.

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    $\begingroup$ Hint: $y=x^y$ where $y=f(x)$ $\endgroup$ – DARK Jan 20 at 11:07
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    $\begingroup$ Are you asking about $f'(x)$ for $f(x)$ being a finite tower of $x$s of height $n$, or about the limit when $n\to\infty$? $\endgroup$ – Christoph Jan 20 at 11:14
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All the other answers are for the case where there are infinite x's. However, there are finite ($n$) x's in this problem!

Let's first denote the function as $f_n(x)=x^{x^{\cdots^x}}$ with $n$ x's in the exponent. For example, $f_1(x)=x^x, f_0(x)=x.$

Take the logarithm of both sides, we get $\ln f_n(x)=f_{n-1}(x)\ln x$. Taking the derivative gives $$\frac{f'_n(x)}{f_n(x)}=f'_{n-1}(x)\ln x+\frac{f_{n-1}(x)}{x}$$

$$f'_n(x)=f_n(x) f'_{n-1}(x)\ln x+\frac{f_n(x) f_{n-1}(x)}{x}$$

That's just a basic recurrence relation. We know that $$a_n=Ca_{n-1}+D\implies a_n=c C^{n-1} + \frac{D(C^n-1)}{C-1}$$ where $c$ is an arbitrary constant.

Similarly, we may get $$f'_n(x)=c(f_n(x)\ln x)^{n-1}+\frac{f_n(x)f_{n-1}(x)}{x(f_n(x)\ln x-1)} ((f_n(x)\ln x)^n-1)$$

To get the value of the parameter $c$, let $n=1$, then $f'_1(x)=c+x^x$. Hence, $c=x^x \ln x$. Substituting in, we get the final answer (in a rather cumbersome form) $$f'_n(x)=x^x \ln x(f_n(x)\ln x)^{n-1}+\frac{f_n(x)f_{n-1}(x)}{x(f_n(x)\ln x-1)} ((f_n(x)\ln x)^n-1)$$

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Let $f_1(x)=x$ and $f_n(x)=x^{f_{n-1}(x)}=x^{x^{\ldots^{x}}}$ ($n$ $x's$ on the exponent). Then $\ln(f_n(x))=f_{n-1}(x)\cdot\ln(x)$ so: $$\frac{f'_n(x)}{f_{n}(x)}=f'_{n-1}(x)\cdot\ln(x)+\frac{f_{n-1}(x)}{x}\implies f'_{n}(x)=f_{n}(x)\cdot\left(f'_{n-1}(x)\cdot\ln(x)+\frac{f_{n-1}(x)}{x}\right).$$

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$$ f(x) = x^{f(x)}\\ \ln(f(x))=f(x) \ln(x)\\ \frac{1}{f(x)}f'(x) = \frac{f(x)}{x} + \ln(x)f'(x),\\ f'(x)=\left(\frac{1}{f(x)}-\ln(x)\right)^{-1}\frac{f(x)}{x},\\ f'(x)=\left(\frac{f(x)}{1-\ln(x)f(x)}\right)\frac{f(x)}{x}. $$

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    $\begingroup$ This answer refers to $x^{x^{x^{\cdots}}}$ instead of a finite tower $x^{x^{\cdots^x}}$. $\endgroup$ – Christoph Jan 20 at 11:17
  • $\begingroup$ Although this doesn't strictly answer the question, I think it's a very nice answer. +1 $\endgroup$ – A-Level Student Jan 20 at 14:50
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$$x^{x^{x^\cdots}}= y \implies x^y = y \implies1 = yx^{-y} = ye^{-y\log (x)}$$

which means

$$y = \frac{-W(-\log (x))}{\log (x)}$$

where $W$ is the principal branch of the Lambert-W function. Now, the problem of the successive derivatives is "much" simpler (have a look here for the derivatives of $W(t)$).

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    $\begingroup$ This answer refers to $x^{x^{x^{\cdots}}}$ instead of a finite tower $x^{x^{\cdots^x}}$. $\endgroup$ – Christoph Jan 20 at 11:17

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