# Derivative of ${^{n}x}$?? (Continuation of my previous note)

In my last note, I went over the derivatives of $x^x$ and $x^{x^x}$. At the end, I went over a brief generalisation of differentiating ${^{n}x}$: $\frac{d}{dx}({^{n}x}) = {^{n}x} \cdot \frac{d}{dx}\left({^{n-1}x}\ln x\right)$ However, I am not satisfied with this because this is not fully calculated (you still need to differentiate ${^{n-1}x}\ln x$ and so on). I want to have a nice expression for this.

Before we start picking apart this problem, I want to define a new function (unless it isn't new) because it may or may not be useful later. Let $g(x) = f(x)(f(x)(f(x)(\dots(f(x)+t)+t)+t)\dots)+t$ where $f(x)$ is another function and $k$ is some constant. There could be $n$ many brackets here. For the purposes of this note, I want to notate this as $g(x) = \overset{\small n}{\underset{\small k=0}{\huge \Beta }}f(x) \beta + t$ Here, $\beta$ is where the nested brackets will go. Here is an example: $\overset{\small 2}{\underset{\small k=0}{\huge \Beta }}a \beta + b = a(a(a+b)+b)+b$ Here is another: $\overset{\small 6}{\underset{\small k=3}{\huge \Beta }}k\beta-t = 6(5(4(3-t)-t)-t)-t$

Hopefully you understand what this function does. Anyway, onto the main topic of this note!

Let us be reminded of our starter last note: the derivative of $x^x$. We found out that it was equal to $x^x(\ln x + 1)$ We also found out that the derivative of $x^{x^x}$ was $x^{x^x + x - 1}(x\ln^2(x) + x\ln x + 1)$ which can also be written as $x^{x^x+x-1}\left(x\ln\left(x\right)\left(\ln\left(x\right)+1\right)+1\right)$ For the sake of this note, I will not be calculating the derivative of $x^{x^{x^x}}$ by hand or any higher tetrations. If we look at the derivative of $x^{x^{x^x}}$, it turns out to be $x^{x^{x^x}+x^x-1}\left(x^x\ln\left(x\right)\left(x\ln\left(x\right)\left(\ln\left(x\right)+1\right)+1\right)+1\right)$ Maybe you start to see a pattern here. Just to check that there is one, we can observe the derivative of $x^{x^{x^{x^x}}}$: $x^{x^{x^{x^x}}+x^{x^x}-1}\left(x^{x^x}\ln\left(x\right)\left(x^x\ln\left(x\right)\left(x\ln\left(x\right)\left(\ln\left(x\right)+1\right)+1\right)+1\right)+1\right)$

That's a lot of $x$'s! However, in this giant mess, there is a pattern that begins to appear.

Can we try expressing these derivatives as the function that I established earlier? Let us do $x^{x^{x^x}}$: $x^{x^{x^x}+x^x-1}\left(x^x\ln\left(x\right)\left(x\ln\left(x\right)\left(\ln\left(x\right)+1\right)+1\right)+1\right) = x^{x^{x^x}+x^x-1}\overset{\small 2}{\underset{\small k=0}{\huge \Beta }}{^{k}x}\ln(x)\beta + 1$ That's nice! Let's try with $x^{x^{x^{x^x}}}$: $x^{x^{x^{x^x}}+x^{x^x}-1}\left(x^{x^x}\ln\left(x\right)\left(x^x\ln\left(x\right)\left(x\ln\left(x\right)\left(\ln\left(x\right)+1\right)+1\right)+1\right)+1\right) = x^{x^{x^{x^x}}+x^{x^x}-1}\overset{\small 3}{\underset{\small k=0}{\huge \Beta }}{^{k}x}\ln(x)\beta + 1$ Apart from the first bit it looks extremely similar to the previous one! To generalise the first bit, we can express it as $x^{ {^{n-1}x} + {^{n-2}x} - 1}$

So finally, we can generalise the derivative of ${^{n}x}$ as $x^{ {^{n-1}x} + {^{n-2}x} - 1}\overset{\small n-2}{\underset{\small k=0}{\huge \Beta }}{^{k}x}\ln(x)\beta + 1$ by using the new function.

I hope that this was not too confusing and that you may have learnt something!

(If you want the $\LaTeX$ for the nested bracket function, it is \overset{\small \text{some other number}}{\underset{\small k=\text{ a number}}{\huge \Beta }} ) $\overset{\small \text{some other number}}{\underset{\small k=\text{ a number}}{\huge \Beta }}$

Note by James Watson
1 month, 4 weeks ago

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