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For the recent week I've been diving myself into tetration out of pure curiosity, and after watching videos about how to find the derivative of $^2x$ and $^3x$, I decided that I wanted to look into what a derivative rule for it would be. When I searched online I couldn't find anything that was satisfying to me, so I put some work in and through a couple iterations of solutions, I think I have a closed form derivative rule I am personally satisfied with. What I found was:

$$\frac{d}{dx}(^nx)=\frac{^nx}{x} \sum_{a=0}^{\lceil n \rceil -1} \ln(x)^a \prod_{b=0}^a \frac{\ln(^{n-b}x)}{\ln(x)}.$$

Firstly, I've been wanting to get on this website for a while, so I thought I would bring it here. Secondly, I wanted to know if this looks accurate to anyone here, as through my testing with it, it seems to be. And lastly, I'm just a Calc BC student, exploring into these further realms of mathematics myself because it interests me, so I don't know, given that this is accurate, if there is a more concise or elegant way to put this derivative rule together, so is there a way I could?

A note about this is that I have it working for $n$ being any nonnegative real number, and for that to work, I am going by the recursive definition for tetration of:

$$^nx:=\begin{cases} 1 & \text{if } n = 0\cr x^{\big(^{(n-1)}x\big)} & \text{if } n > 0, \end{cases}$$

which extends it so that $n$ can be any nonnegative real number, such that $^{2.47}x$ would be considered $x^{x^{x^{0.47}}}$.

Update:

As I know that the recursive definition for tetration is not exactly the definition for tetration across the reals, and that that is still a researched topic, I wanted to share my derivative rule which I made before this one that works for $n$ as any nonnegative integer. Here is what I have on that:

$$\frac{d}{dx}(^nx)=\frac{^nx}{x}\sum_{a=0}^{n-1}\ln(x)^a\prod_{b=0}^a{^{n-b-1}x}.$$

For this, I ask the same questions as I asked above. Thank you for your time reading this.

Brian359
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    A personal opinion (shared by others, I think): Don't lose too much time on Tetration: it is a curiosity but it has no interest either for pure nor for applied Mathematics. – Jean Marie Jan 25 '21 at 23:44
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    I verified your formula for $n \in \mathbb N$. Note that $\ln(^{n-b}x)/\ln x$ can be simplified to $^{n-1-b}x$. Your definition of tetration for real values of $n$ doesn't seem natural to me. See for example here. – posilon Jan 26 '21 at 00:05
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    @JeanMarie I can understand that, and thank you for the insight on that. This is something I just wanted to briefly try my hand at. – Brian359 Jan 26 '21 at 00:53
  • @posilon The only reason I had the $\ln(^{n-b}x)/\ln x$ was because I wasn't sure how a negative value for $n$ would work with tetration, so it was my way of getting around that happening. Would $^{n-1-(\lceil n \rceil -1)}x$ be equal to $\bmod n of 1$? Like would $^{.47-1}x=.47$? If so, then I would be able to replace$\ln(^{n-b}x)/\ln x$ with $^{n-1-b}x$. I had something similar to that anyway in a previous version of the rule I made for $n$ being any nonnegative integer. – Brian359 Jan 26 '21 at 01:07
  • @posilon Another note is that I pulled the recursive definition from the top of the same wiki page you shared, and in checking it visually myself, it does appear to show a smooth transition between nonnegative values of $N$ on the graph. I put together a demonstration of it using Desmos here. – Brian359 Jan 26 '21 at 01:14
  • What do you mean by smooth? The way you defined $^nx$, it is not differentiable w.r.t. $n$ (though it is piecewise differentiable). – posilon Jan 26 '21 at 01:38
  • @posilon What do you mean by differentiable with respect to n? n would always be a constant nonnegative real number, not a variable in these cases. Also when I said smooth I meant that when you changed n, the graph didn't suddenly jump to a new value when x>0, it smoothly transitions between values. Also in the link you sent, I believe this definition would fall under the first two requirements, but how would the third one ever be filled? Isn't it saying it would always have to be concave up? Or am I misinterpreting it? Because a function like $f(x)=^3x$ would have $f''(x)\le0$ for some $x>0$. – Brian359 Jan 26 '21 at 03:14
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    The Wiki link has $x$ as the height, while in your comment (and your question), you have it as the base. – Varun Vejalla Jan 26 '21 at 05:44
  • @Brian359 The expression $t(x,n) := {^nx}$ is a function of two variables, namely $x$ and $n$. If you plot $^nx$ for a fixed value of $n$ and then start changing that value, you won't notice any abrupt jumps happening. That's because $t(x,n)$ is continuous w.r.t. $n$. That is, for any fixed $x$, the function of one variable $g_x(n) := t(x,n)$ is continuous. But $g_x$ is not differentiable at the integer values of $n$ (unless $x = e$). – posilon Jan 26 '21 at 17:39
  • @posilon this makes sense, thank you for clearing this up. – Brian359 Jan 27 '21 at 17:32
  • @VarunVejalla I see where I went wrong with this now, thank you for the heads up with this – Brian359 Jan 27 '21 at 17:32

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