General Rule for Differentiation of Tetrations

Question

I'll start at the beginning. Initially, this sort of began as just what is $\frac{d}{dx}$[$x^x$] the answer being $x^x$+ln(x)$x^x$. This wasn't difficult to achieve, just some chain rule and product rule. I then took $\frac{d}{dx}$[$x^{x^x}$], this took some more chain and product rule but one should end up with an answer of ($x^{x^x}$$x^x$$)\frac{1}{x}$+$x^{x^x}$$x^x$ln$(x)$+$x^{x^x}$$x^x$ln$^2(x)$.

Evidently, I now could just take the $\frac{d}{dx}$[$x^{x^{x^x}}$], but I saw the pattern with the other derivatives and thought that there might be a general formula or rule for the derivatives of tetration. After a search online, and on the exchange, it led to no nothing. So I decided to differentiate the 4th tetration of x, which is $x^{x^{x^x}}$.

For future reference the 4 tetration of x, I'll call $x^{x|4}$. Additionally, I'll call $x^{x^{x^x}}$*$x^{x^x}$ -> $x^{x|4,3}$ just to make my life easier typing this.

The $\frac{d}{dx}$[$x^{x|4}$], is equal to ($x^{x|4,3}$+$x^{x|4,2}$ln$(x)$)$\frac{1}{x}$+$x^{x|4,2}$*ln$^2(x)$+$x^{x|4,2}$*ln$^3(x)$

From here I noticed a pattern, to where I could accurately predict the $\frac{d}{dx}$[$x^{x|n}$] (The derivative of the nth tetration of x)

The general pattern is:

$\frac{d}{dx}$[$x^{x|2}$] = $x^{x|2}$+$x^{x|2}$ln$(x)$

$\frac{d}{dx}$[$x^{x|3}$] = $\frac{x^{x|3,2}}{x}$+ $x^{x|3,2}$ln$(x)$+$x^{x|3,2}$ln$^2(x)$

$\frac{d}{dx}$[$x^{x|4}$] = $\frac{x^{x|4,3}}{x}$+ $\frac{x^{x|4,2}ln(x)}{x}$+$x^{x|4,2}ln^2(x)$+$x^{x|4,2}$ln$^3(x)$

Some rules of the pattern are as follows:

All terms except for the last two will be over x.

The maxima, which is the highest tetration, will always be the number of tetrations in the original tetration that is being differentiated.

The minima, the lowest tetration, will be the maxima-1 then count down to 2, where it will remain until the addition ends.

The exponent on the logarithm starts at zero and counts up until it reaches the maxima-1.

Note: Maxima and minima aren’t accepted terms so use your own as you wish.

For example:

The first segment of the $\frac{d}{dx}$[$x^{x|4}$]. $\frac{x^{x|4,3}}{x}$ notice how it is over x, the maxima is 4, the minima is 4-1, and there is no logarithm because ln$^0(x)$=1. Then one would add, $\frac{x^{x|4,2}ln(x)}{x}$ this segment is also over x and has a maxima of 4 and a minima of one less than the previous. Additionally, this segment has an ln$(x)$, this is because ln$^1(x)$ = ln$(1)$, no surprises.

This pattern continues bound by these rules, so now the $\frac{d}{dx}$[$x^{x|10}$]. However, I wasn’t satisfied with this answer; because I wanted a general formula.

I had my friend, who’s a lot better at math than me, assist me in finding a general formula.

He eventually devised $\frac{d}{dx}[x^{x|n}]$:

$$\sum_{k=0}^{n-2}\frac{x^{x|n,n-k}ln^{k-1}(x)}{x}+x^{x|2}ln^{n-2}+x^{x|2}ln^{n-1}, (n\ge2)$$

We've gone through the mathematical rigor, and so far this formula has seemed to uphold. The main reason I’m writing this is that I have a few questions about this formula and the general idea of there being a general formula.

Can this formula be disproved?

Is there a nicer-looking formula?

Is there a formula with proper notation, not my made-up one?

Can a formula be made so that n is a fraction or even a complex number?

Answer 1

Some possibly relevant published items are given below, in order of likely relevance to you. Although this is not really an answer, I thought that because these references are probably not very well known, maybe I should avoid burying them in a comment that isn't google-searchable (something I've done far too often in MSE).

[1] $\;$Exponential derivative [Solution to Problem #653], Mathematics Magazine 40 #5 (November 1967), pp. 283-284.

Three published solutions to a problem proposed in the March 1967 issue that asks for the derivative of $\;{}^{n+1} x\;$ (this notation is not used).

[2] $\;$S. B. Kulkarni, Solution to Problem #3977, School Science and Mathematics 84 #7 (November 1984), pp. 629-230.

Gives a recursion formula for the $n$-th derivative of $x^x$ -- see my answer to The $n$'th derivative of $x^x$.

[3] $\;$Jekuthiel Ginsburg, Iterated exponentials, Scripta Mathematica 11 (1945), pp. 340-353.

Combinatorial relationships for the coefficients of the Maclaurin expansions of $e_2(x) = e^{e^x},$ $e_3(x) = e^{e^{e^x}}, \,\ldots, \; e_n(x), \; \ldots$

Answer 2

The rigorous way to approach this is to use recursive relations. One can consider the sequence of functions $f$ such that $f_0(x)=x$ and $f_{n+1}(x)=x^{f_n(x)},$ and in this manner, $f_n(x)=\,^{n+1}x,$ where $^nx$ is the $n$th tetration of $x.$ We have that $f_0'(x)=1$ and $$f_{n+1}'(x)=x^{f_n(x)}\left[\ln(x)f_n'(x)+\frac{f_n(x)}{x}\right]=f_{n+1}(x)\left[\ln(x)f_n'(x)+\frac{f_n(x)}{x}\right],$$ which implies $$\frac{f_{n+1}'(x)}{f_{n+1}(x)}=e^{\ln[f_n(x)]}\left[\ln(x)\frac{f_n'(x)}{f_n(x)}+\frac1{x}\right].$$ This suggests defining $$g_n(x):=\ln[f_n(x)],$$ so that the equation can be rewritten $$g_{n+1}'(x)=e^{g_n(x)}\left[\ln(x)g_n'(x)+\frac1{x}\right].$$ As such, $$g_{n+2}'(x)=e^{g_{n+1}(x)}\left[\ln(x)g_{n+1}'(x)+\frac1{x}\right]=e^{g_{n+1}(x)}\ln(x)g_{n+1}'(x)+e^{g_{n+1}(x)}\frac1{x}$$ $$=e^{g_{n+1}(x)}\ln(x)\left[e^{g_n(x)}\ln(x)g_n'(x)+e^{g_n(x)}\frac1{x}\right]+e^{g_{n+1}(x)}\frac1{x}$$ $$=e^{g_{n+1}(x)}e^{g_n(x)}\ln(x)^2g_n'(x)+\frac{e^{g_{n+1}(x)}e^{g_n(x)}\ln(x)+e^{g_{n+1}(x)}}{x}.$$ Thus $$g_{n+3}'(x)=e^{g_{n+2}(x)}e^{g_{n+1}(x)}\ln(x)^2g_{n+1}'(x)+\frac{e^{g_{n+2}(x)}e^{g_{n+1}(x)}\ln(x)+e^{g_{n+2}(x)}}{x}$$ $$=e^{g_{n+2}(x)}e^{g_{n+1}(x)}\ln(x)^2\left[e^{g_n(x)}\ln(x)g_n'(x)+e^{g_n(x)}\frac1{x}\right]+\frac{e^{g_{n+2}(x)}e^{g_{n+1}(x)}\ln(x)+e^{g_{n+2}(x)}}{x}$$ $$=e^{g_{n+2}(x)}e^{g_{n+1}(x)}e^{g_n(x)}\ln(x)^3g_n'(x)+\frac1{x}\left[e^{g_{n+2}(x)}e^{g_{n+1}(x)}e^{g_n(x)}\ln(x)^2+e^{g_{n+2}(x)}e^{g_{n+1}(x)}\ln(x)+e^{g_{n+2}(x)}\right]$$ $$=\prod_{m=0}^{3-1}e^{g_{n+m}(x)}\ln(x)^3g_n'(x)+\frac1{x}\prod_{m=0}^{3-1}e^{g_{n+m}(x)}\left[\frac{\ln(x)^2}{1}+\frac{\ln(x)^1}{e^{g_n(x)}}+\frac{\ln(x)^0}{e^{g_n(x)}e^{g_{n+1}(x)}}\right]$$ $$=\prod_{m=0}^{3-1}e^{g_{n+m}(x)}\ln(x)^3g_n'(x)$$ $$+\frac{\ln(x)^2}{x}\prod_{m=0}^{3-1}e^{g_{n+m}(x)}\left[\frac1{\ln(x)^0\prod_{p=0}^{0-1}e^{g_{n+p}(x)}}+\frac1{\ln(x)^1\prod_{p=0}^{1-1}e^{g_{n+p}(x)}}+\frac1{\ln(x)^2\prod_{p=0}^{2-1}e^{g_{n+p}(x)}}\right]$$ $$=\prod_{m=0}^{3-1}e^{g_{n+m}(x)}\ln(x)^3g_n'(x)+\frac{\ln(x)^2}{x}\prod_{m=0}^{3-1}e^{g_{n+m}(x)}\sum_{m=0}^{3-1}\frac1{\ln(x)^m\prod_{p=0}^{m-1}e^{g_{n+p}(x)}}.$$ Continuing the pattern, one finds that $$g_{n+q}'(x)=\prod_{m=0}^{q-1}e^{g_{n+m}(x)}\ln(x)^qg_n'(x)+\frac{\ln(x)^{q-1}}{x}\prod_{m=0}^{q-1}e^{g_{n+m}(x)}\sum_{m=0}^{q-1}\frac1{\ln(x)^m\prod_{p=0}^{m-1}e^{g_{n+p}(x)}}$$ $$=\ln(x)^{q-1}\prod_{m=0}^{q-1}e^{g_{n+m}(x)}\left[\ln(x)g_n'(x)+\frac1{x}\sum_{m=0}^{q-1}\frac1{\ln(x)^m\prod_{p=0}^{m-1}e^{g_{n+p}(x)}}\right],$$ hence $$g_q'(x)=\ln(x)^{q-1}\prod_{m=0}^{q-1}e^{g_m(x)}\left[\ln(x)g_0'(x)+\frac1{x}\sum_{m=0}^{q-1}\frac1{\ln(x)^m\prod_{p=0}^{m-1}e^{g_p(x)}}\right]$$ $$=\ln(x)^{q-1}\prod_{m=0}^{q-1}e^{g_m(x)}\left[\frac{\ln(x)}{x}+\frac1{x}\sum_{m=0}^{q-1}\frac1{\ln(x)^m\prod_{p=0}^{m-1}e^{g_p(x)}}\right]$$ $$=\frac1{x}\left[\ln(x)^q\prod_{m=0}^{q-1}e^{g_m(x)}+\ln(x)^{q-1}\prod_{m=0}^{q-1}e^{g_m(x)}\sum_{m=0}^{q-1}\frac1{\ln(x)^m\prod_{p=0}^{m-1}e^{g_p(x)}}\right].$$ Noting that $$f_n(x)=e^{g_n(x)},$$ this means that $$\frac{f_q'(x)}{f_q(x)}=\frac1{x}\left[\ln(x)^q\prod_{m=0}^{q-1}f_m(x)+\ln(x)^{q-1}\prod_{m=0}^{q-1}f_m(x)\sum_{m=0}^{q-1}\frac1{\ln(x)^m\prod_{p=0}^{m-1}f_p(x)}\right],$$ which is equivalent to $$f_q'(x)=\frac1{x}\left[\ln(x)^q\prod_{m=0}^qf_m(x)+\ln(x)^{q-1}\prod_{m=0}^qf_m(x)\sum_{m=0}^{q-1}\frac1{\ln(x)^m\prod_{p=0}^{m-1}f_p(x)}\right]$$ $$=\frac{\ln(x)^q}{x}\prod_{m=0}^qf_m(x)\left[1+\sum_{m=0}^{q-1}\frac1{\ln(x)^{m+1}\prod_{p=0}^{m-1}f_p(x)}\right].$$ Finally, notice that $$\prod_{m=0}^{n+1}h_m=h_{n+1}\prod_{m=0}^nh_m$$ implies $$\prod_{m=0}^{-1}h_m=h_{-1}\prod_{m=0}^{-2}h_m=1,$$ which is equivalent to $$\prod_{m=0}^{-2}h_m=\frac1{h_{-1}}.$$ Givne the recursion for $f$ that we gave, we can conclude $f_{-1}(x)=1,$ so $$\prod_{p=0}^{-2}f_p(x)=1,$$ and since $$1=\ln(x)^0=\ln(x)^{(-1)+1},$$ it means $$1=\frac1{\ln(x)^{(-1)+1}\prod_{p=0}^{(-1)-1}f_p(x)},$$ hence $$f_q'(x)=\frac{\ln(x)^q}{x}\prod_{m=0}^qf_m(x)\sum_{m=-1}^{q-1}\frac1{\ln(x)^{m+1}\prod_{p=0}^{m-1}f_p(x)}=\frac{\ln(x)^q}{x}\prod_{m=0}^qf_m(x)\sum_{m=0}^q\frac1{\ln(x)^m\prod_{p=0}^{m-2}f_p(x)}.$$ Since $f_n(x)=\,^{n+1}x,$ it follows that $$\frac{\mathrm{d}}{\mathrm{d}x}(^nx)=\frac{\ln(x)^{n-1}}{x}\prod_{m=0}^{n-1}\,^{m+1}x\sum_{m=0}^{n-1}\frac1{\ln(x)^m\prod_{p=0}^{m-2}\,^{p+1}x}$$ $$=\frac{\ln(x)^{n-1}}{x}\prod_{m=1}^n\,^mx\sum_{m=0}^{n-1}\frac1{\ln(x)^m\prod_{p=1}^{m-1}\,^px}$$ It turns out this formula can be proven correct via induction. If I understand the notation in your formula correctly, then I think that means your formula is equivalent to this formula here, and so is also correct.

Answer 3

Using W|A and analyzing the pattern (I've been too lazy to derive the derivatives myself) one can state the following recursive pattern.

• First: define notation $x_0 = 1$, $x_1=x$, $x_2=x^{x_1}$,$\cdots,$ $x_{k+1}=x^{x_k}$ , and moreover $x_{-1}= 0 \qquad (=\log_x(1))$

• Second:

  • $\log(x)'=\frac1x \qquad $, note: we're not deriving $\log_x(x)$ !
  • $ x_0'=0 \quad$, $\quad x_1'=1$
  • $x_2' = x_2 \cdot(x_1' \log(x)+x_1 \log(x)')$
  • $x_3' = x_3 \cdot(x_2' \log(x)+x_2 \log(x)')$
  • $ \vdots \quad = \qquad \vdots $

$\qquad \qquad $ (Note, that $x_1'$ can easily & consistently be expressed in the same style as the higher $x_k$ iteratives)
I think, this is a nice scheme for programming purposes.

Moreover, I had such scheme for higher derivatives, but don't have it at hand and think it is not too difficult to extrapolate this yourself.