$n^{th}$ derivative of a tetration function

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.

I found a method of doing the first derivative but it's relatively messy and makes you go all the way down the "n" chain as it were. Hopefully my ideas can give you some inspiration or spark a secondary idea.

I am using $^nx$ as the nth tetration of x.

What I did to reach my answer was start taking the derivatives of $^nx$ with increasing values of n using $e^{lnx}$. So for n=1 you obviously get $$\frac{d}{dx}(^1x) = \frac{d}{dx}(x) = 1$$

For n=2 you get $$\frac{d}{dx}(^2x) = \frac{d}{dx}(x^x)$$ $$=\frac{d}{dx}(e^{lnx^x})$$ $$=\frac{d}{dx}(e^{xlnx})$$ $$=e^{xlnx}\frac{d}{dx}(xlnx)$$ $$=(^2x)\Bigl(lnx\frac{d}{dx}(x)+x\frac{d}{dx}(lnx)\Bigl)$$ We already know the value of $\frac{d}{dx}(x)$ from the last problem, so we can plug it right in. $$=(^2x)\Bigl(lnx+\frac{x}{x}\Bigl)$$ $$=(^2x)(lnx+1)$$

For n=3 you get $$\frac{d}{dx}(^3x) = \frac{d}{dx}(x^{x^x})$$ $$=\frac{d}{dx}(e^{ln(x^{x^x})})$$ $$=\frac{d}{dx}(e^{(x^x)(lnx)})$$ $$=\frac{d}{dx}(e^{(e^{ln(x^x)})(lnx)})$$ $$=\frac{d}{dx}(e^{(e^{xln(x)})(lnx)})$$ $$=(^3x)\frac{d}{dx}(e^{xln(x)})(lnx)$$ $$=(^3x)\Bigl((lnx)\frac{d}{dx}(e^{xln(x)})+(e^{xln(x)})\frac{d}{dx}(lnx)\Bigl)$$ Note here that $e^{xln(x)}$ equals $^2x$. This means that we can substitute in values we already know, just like in the last problem, and a pattern starts to emerge. $$=(^3x)\Bigl((lnx)\frac{d}{dx}(^2x)+(^2x)\frac{d}{dx}(lnx)\Bigl)$$ $$=(^3x)\Bigl((lnx)\frac{d}{dx}(^2x)+\frac{^2x}{x}\Bigl)$$ We don't need to plug in the end result of $\frac{d}{dx}(e^{xln(x)})$ here, because the form the equation is in now will end up fitting our generalization later.

For now, let's check by plugging in n=4 $$\frac{d}{dx}(^4x) = \frac{d}{dx}(x^{x^{x^x}})$$ $$=\frac{d}{dx}(e^{ln(x^{x^{x^x}})})$$ $$.$$ $$.$$ $$.$$ $$=\frac{d}{dx}(e^{(e^{(e^{xln(x)})(lnx)}(lnx)})$$ $$=(^4x)\frac{d}{dx}(e^{(e^{xln(x)})(lnx)}(lnx)$$ $$=(^4x)\Bigl((lnx)\frac{d}{dx}(e^{(e^{xln(x)})(lnx)})+(e^{(e^{xln(x)}(lnx)})\frac{d}{dx}(lnx)\Bigl)$$ Here again, $e^{(e^{xln(x)})(lnx)}$ is the same as $^3x$. So if we rewrite our equation as $$(^4x)\Bigl((lnx)\frac{d}{dx}(^3x)+\frac{^3x}{x}\Bigl)$$ we can see that a general form of the first derivative can be written as $$\frac{d}{dx}(^nx) = (^nx)\Bigl((lnx)\frac{d}{dx}(^{n-1}x)+\frac{^{n-1}x}{x}\Bigl)$$

Obviously this has the problem of relying on the derivatives down the power tower, but I think this could have interesting applications. Hope these inane ramblings of a 17 year old help!

excellent question, and a good result. 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.

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.