I stumbled upon this very peculiar function last summer, namely: $f(x)=x^{x^{x^{...^{x}}}}$, where there is a number $n$ of $x$'s in the exponent, I tried to find the derivative for the function and I was successful, it turned out not to be the most elegant formula but it worked. (Firstly, I invented a new notation, namely, a function such as $f(x)$ we can write it as the following: $f(x) =x^{\langle x \vert n\rangle}$ where $x$ is the exponent that is getting "powered" up $n$ times.) The formula I obtained by pattern matching was: $$f^{\prime}(x)=x^{\langle x \vert n\rangle +\langle x \vert n-1\rangle -1}\left[1+\prod_{i=0}^{n-2}x^{\langle x \vert i\rangle}\cdot \ln(x)^n+\sum_{j=1}^{n-1}\prod_{k=n-1-j}^{n-2}x^{\langle x \vert k\rangle}\cdot \ln(x)^j\right]\tag{$n\geqslant 2$}.$$ I know this looks like a mad mess and I am aware that people like this have done it more elegantely, but now for the question. This is only the first derivative of the function, is there a way, or rather is there a general derivative i.e a $n^{th}$ derivative of this function?
Update: December 23th
I have tried to approach the problem myself since I asked the question and I have not gotten to a stage to say if it is impossible or possible to do, however I think I am on the right track. At first, I thought of distributing the factor $x^{\langle x \vert n \rangle +\langle x \vert n-1 \rangle -1}$ to all the terms in the parentheses, but I quickly realized I had to deal with at least derivatives of triple products. Now I have come to realize that the easiest way is to differentiate the function just as it is and get a normal product and thus I must use the following formula: $$(f \cdot g)^{(n)}=\sum_{k=0}^{n}{n\choose k}f^{(k)}\cdot g^{(n-k)}$$ where $f =x^{\langle x \vert n \rangle +\langle x \vert n-1 \rangle -1}$ and $g=1+\prod_{i=0}^{n-2}x^{\langle x \vert i\rangle}\cdot \ln(x)^n+\sum_{j=1}^{n-1}\prod_{k=n-1-j}^{n-2}x^{\langle x \vert k\rangle}\cdot \ln(x)^j$. Since $k$ and $n-k$ are arbitrary numbers this leads us to find the general derivative for $f$ and $g$, this is where I am right now. (I do realize that I am trying to find the $n^{th}$ derivative of the first derivative but that is easily fixed later). Please come with suggestions on how to tackle this problem.
Update: December 24th
I have made progress with the help of Maple 17, namely, I have found a repeating pattern in at least a part of the general derivative, but there is still a part of it I cannot yet explain. Nonetheless, I present to you the part of the general derivative I have found: $$D_x^{\xi}f(x) = x^{\langle x \vert n\rangle +\langle x \vert n-1\rangle -\xi} \Big[(-1)^{\xi}\cdot\xi! +O(x)\Big]$$
I renamed the degree of the derivative as $\xi$ since $n$ is taken for the number of $x$s. The $O(x)$ is the (perhaps) series which I am currently working on finding, I do think I am on the right track though. The approach above with the product rule turned out to be less successful.
Excellent question, and a good result. I am also impressed that you have developed your own notation. That is often a very effective way of getting to grips with a problem, especially one that has not yet become popular. I think there is a trend, however.
The notation tends to evolve with use. The notation here involves a redundancy which one can ill-afford in a subject already pushing at against conceptual boundaries. if you study your remarkable formula for the derivative, you will see that all the references to tetration involve the incomplete symbol: $x^{<x\mid...}$
In this usage, the initial exponent symbol $x$ is redundant, and complicates the expression. Thus the evolutionary pressure of being concise will force the rejection of this appendage, and one may use the symbol $\langle x \vert n \rangle$ by itself. this is conveniently defined by (if I have understood correctly): $$ \langle x \vert 0 \rangle = 1 \\ \langle x \vert n+1 \rangle = x^{\langle x \vert n \rangle} $$ For the purpose of differentiation, the logarithm is useful i.e. since $$ \ln \langle x \vert n+1 \rangle = \langle x \vert n \rangle \ln \;x $$ we obtain : $$\frac{\langle x \vert n+1 \rangle'} {\langle x \vert n+1 \rangle} = \frac{ \langle x \vert n \rangle }{x} \left( \frac{\langle x \vert n \rangle'}{\langle x \vert n \rangle}x\ln \;x + 1 \right) $$perhaps as might be expected, the logarithmic derivative $\frac{f'}f$ looms large here, and it is hardly surprising to see the "entropy" function also make an appearance.
We may abbreviate the form considerably if we define: $$T^n(x) = \frac{\langle x \vert n \rangle'}{\langle x \vert n \rangle} $$ So that we have a form fairly well-suited to recursive evaluation : $$T^{n+1}(x) = \frac{ \langle x \vert n \rangle }{x} \left( x \ln\ x\; T^n(x)+ 1 \right) $$ Congratulations on your achievement! I hope these casual remarks will be of some use or interest.