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For the people who doesn't know what tetration is:-

$^na=\begin{cases}1&\text{if}\,n=0\\a^{a^{.^{.^{.^{a}}}}}&\text{if}\,n\in\mathbb{N}\end{cases}$

There are $n$ a's in that power tower.

And also see the wikipedia page on tetration.

We all know about the power rule for differentiation-

$\frac{dx^n}{dx}=nx^{n-1}$

So I started thinking about a formula for the tetration rule for differentiation

And here's what I did:-

I didn't really used any clever tricks from advanced mathematics, I just tried to find formulas like a layperson.

Recurrence formula:-

We knew that-

$^nx=x^{(^{n-1}x)}$ $=e^{^{n-1}x\ln(x)}$

We can use this property of tetration to get a recurrence relation, derivative of the $^nx$ based on the derivative of $^{n-1}x$.

$\frac{d}{dx}(^nx)$ $=\frac{d(e^{^{n-1}x\ln(x)})}{dx}$

$$\fbox{$\frac{d}{dx}(^nx)$ $={^nx}(\frac{^{n-1}x}{x}+\ln(x)\frac{d}{dx}(^{n-1}x))$}$$

Explicit formula:-

Instead using a clever way to find an explicit formula, I just expanded out the expression and found a pattern.

$\frac{d}{dx}(^nx)$

$=\frac{{^nx}{^{n-1}x}}{x}+{^nx}\ln(x)\frac{d}{dx}(^{n-1}x)$

$=\frac{{^nx}{^{n-1}x}}{x}+\ln(x)\frac{{^nx}{^{n-1}x}{^{n-2}x}}{x}+\ln^2(x){^nx}{^{n-1}x}\frac{d}{dx}(^{n-2}x)$

$=\frac{{^nx}{^{n-1}x}}{x}$ $+\ln(x)\frac{{^nx}{^{n-1}x}{^{n-2}x}}{x}$ $+\ln^2(x)\frac{{^nx}{^{n-1}x}{^{n-2}x}{^{n-3}x}}{x}$ $+\ln^3(x){^nx}{^{n-1}x}{^{n-2}x}\frac{d}{dx}(^{n-3}x)$

$\vdots$

Well obviously we can't continue this process forever, so how many should we do this process?

At the $1st$ step in the formula the term inside the derivative was $^{n-1}x$

So in the $n'th$ step, the term inside the derivative will be $^{n-n}x={^0x}=1$, so it's derivative will be $0$. So the whole last term will vanish.

So instead of repeating this process $n$ times, we have to repeat it $n-1$ times.

Finally the formula is

$\frac{d}{dx}(^nx)$

$=\ln^0(x)\frac{{^nx}{^{n-1}x}}{x}+\ln(x)\frac{{^nx}{^{n-1}x}{^{n-2}x}}{x}+\cdots+\ln^{n-1}(x)\frac{{^nx}\cdots{^2x}}{x}$

$$\fbox{$\frac{d}{dx}(^nx)=\frac{1}{x}\sum_{k=0}^{n-1}(\ln^k(x)\prod_{j=0}^{k-1}{^{n-j}x})$}$$

Questions:-

I have just two questions.

$(1)$ Is my derivation correct?

$(2)$ I derived the formulas in the stupid way. So can we derive these formulas in the clever way? And what are those formulas?

Rounak Sarkar
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    From a quick read-through, it seems like this should be the correct formula if $^n x$ is actually differentiable. However, the chain rule only applies if you already know both of the original functions are differentiable. I think it's likely that $^n x$ is differentiable, but if we're being really rigorous we need to do a bit more work to show that there even is a derivative. – memerson Apr 05 '21 at 08:34
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    As far as whether or not there's a better way to go about this, I don't think there is. It seems like the best approach is to really leverage that recurrence relation like you did and then find the formula inductively/recursively. – memerson Apr 05 '21 at 08:36
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    @memerson. Do we really need a proof? Can this be considered as a proof:- $x$ and $e^z$ is differentiable. So $e^{x\ln(x)}=x^x$ is also differentiable. Which means $e^{x^x\ln(x)}=x^{x^x}$ is also differentiable. And we can continue this $n$ times. So $^nx$ will obviously be differentiable. – Rounak Sarkar Apr 05 '21 at 08:59
  • For $x>0$ , this should work. – Peter Apr 05 '21 at 10:56

1 Answers1

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The mathematical problem itself has other solutions here: $n^{th}$ derivative of a tetration function and here: Derivation of tetration by iteration.

Tetration with fixed height $n$ is an elementary function (differential algebra). The elementary functions are analytic almost everywhere. They are closed regarding differentiation. Therefore differentiability is given almost everywhere here.

(1)
Your differential equations seem to be correct. But your explicit formula seems to be wrong. It already doesn't work for $n=1$: I get $(^1x)′=\frac{1}{x}$, but it should be $1$. I found a correction of your equation: $$\frac{d}{dx}\left(^nx\right)=\frac{1}{x}\sum_{k=0}^{n-1}\left(\ln^k(x)\prod_{j=n-k-1}^n{}^jx\right)\tag{1}$$ So we found a further formula.

(2)
You derived the formula by guessing. Verification by a proof is still missing.
There are several ways to get identities and generating functions as explicit or recurrence equations, e.g. series methods combined with summation methods. Applying the chain rule of differentiation could also be possible.
Applying the chain rule of differentiation to the exp-ln-form of $^nx$ yields equation (1) in a systematic way without guessing.
All derivatives of an elementary function are elementary functions. Each elementary function can be represented by applying finite numbers of only elementary functions. Therefore a generating function in this form should be possible, and your formula has this form.

IV_
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