Derivative of \({^{\infty}x}\)??? (Part 3 of my past 2 notes)

My last note went over differentiating ${^{n}x}$ and it turned out to be $x^{ {^{n-1}x} + {^{n-2}x} - 1}\overset{\small n-2}{\underset{\small k=0}{\huge \Beta }}{^{k}x}\ln(x)\beta + 1$ where $\overset{\small b}{\underset{\small k=a}{\huge \Beta }}f(x)\beta g(x) = \underbrace{f(x)(f(x)(\dots f(x)g(x))g(x))\dots)g(x)}_{b-a \text{ nested parentheses}}$ However, I feel that there is one more problem for this trilogy of notes to acknowledge: the derivative of ${^{\infty}x}$. We can utilise our generalisation from earlier to solve this: $\begin{aligned} \frac{d}{dx}({^{\infty}x}) &= x^{ {^{\infty-1}x} + {^{\infty-2}x} - 1}\overset{\small \infty-2}{\underset{\small k=0}{\huge \Beta }}{^{k}x}\ln(x)\beta + 1 \\ &= x^{ {^{\infty}x} + {^{\infty}x} - 1}\overset{\small \infty}{\underset{\small k=0}{\huge \Beta }}{^{k}x}\ln(x)\beta + 1 \\ &= x^{ 2({^{\infty}x}) - 1}\overset{\small \infty}{\underset{\small k=0}{\huge \Beta }}{^{k}x}\ln(x)\beta + 1 \end{aligned}$ But what is $\overset{\small \infty}{\underset{\small k=0}{\huge \Beta }}{^{k}x}\ln(x)\beta + 1$? It is the same as $\underbrace{ {^{\infty}x}\ln(x)({^{\infty-1}x}\ln(x)( \dots( \ln(x) +1)+1)\dots)+1}_{\text{infinite nested parentheses}}$ but can this be expressed more clearly? Maybe. Let us have another go at differentiating${^{\infty}x}$, but using a different method.

The equation $x^y = y$ is the same as ${^{\infty}x} = y$. This is because $y = \blue{x^y} = x^{\blue{x^y}} = x^{x^{\blue{x^y}}} = x^{x^{x^{x^{\dots}}}} = {^{\infty}x}$. We can use implicit differentiation here to find the derivative of ${^{\infty}x}$: $\begin{aligned} \frac{d}{dx}\left( y = x^y \right) \implies \frac{dy}{dx} = x^y \cdot \frac{d}{dx}(y\ln(x)) &= x^y \left( \ln(x)\frac{dy}{dx} + \frac{y}{x} \right)\\ &= x^y \ln(x)\frac{dy}{dx} + \frac{x^y y}{x} \end{aligned}$ $\implies \frac{dy}{dx} - x^y \ln(x)\frac{dy}{dx} = \frac{x^y y}{x}$ $\implies \frac{dy}{dx}(1-x^y\ln(x)) = x^{y-1} y$ $\implies \frac{dy}{dx} = \boxed{\frac{x^{y-1} y}{1-x^y\ln(x)}}$ Note that $y = x^y$ so we can also write this as $\frac{x^{y-1}x^y}{1-x^y\ln(x)} = x^{2y-1}\frac{1}{1-x^y\ln(x)}$ Also remember that $y = {^{\infty}x}$, so it can be written as $x^{2({^{\infty}x})-1}\frac{1}{1-{^{\infty}x}\ln(x)}$ This must be the same as our first attempt on the derivative of ${^{\infty}x}$, so we may be able to figure out what the infinite nested brackets could be: $\blue{x^{2({^{\infty}x})-1}}\frac{1}{1-{^{\infty}x}\ln(x)} = \blue{x^{ 2({^{\infty}x}) - 1}}\overset{\small \infty}{\underset{\small k=0}{\huge \Beta }}{^{k}x}\ln(x)\beta + 1$ $\implies \green{\boxed{\overset{\small \infty}{\underset{\small k=0}{\huge \Beta }}{^{k}x}\ln(x)\beta + 1 = \frac{1}{1-{^{\infty}x}\ln(x)}}}$ So the derivative of ${^{\infty}x}$ is $x^{2({^{\infty}x})-1}\cfrac{1}{1-{^{\infty}x}\ln(x)}$, but we also figured out that $\overset{\small \infty}{\underset{\small k=0}{\huge \Beta }}{^{k}x}\ln(x)\beta + 1 = \cfrac{1}{1-{^{\infty}x}\ln(x)}$ which is quite cool if i do say so myself.

I hope that you found this interesting!