Derivative of $x^{x^{\cdot^{\cdot}}}$?

Question

The infinite tetration is defined as $$f(x)=x^{x^{\cdot^{\cdot}}}$$

This function is defined for $e^{-e} \leq x \leq e^{e-1}$.

(Wikipedia image)

Can one determine the derivative of this function?

Answer 1

Letting $h(x)$ be your infinite power tower, one can solve the functional equation $h(x)=x^{h(x)}$ in terms of the Lambert function $W(x)$, the inverse function of $x\exp\,x$. More specifically, we have

$$h(x)=\exp(-W(-\log\,x))$$

One can then apply the chain rule as usual. The formula

$$W^\prime(x)=\frac{\exp(-W(x))}{1+W(x)}$$

is easily derived through implicit differentiation of the relationship $W(x)\exp(W(x))=x$.

We thus have

$$h^\prime(x)=\frac{\exp(-2 W(-\log\,x))}{x (1+W(-\log\,x))}=\frac{h(x)^2}{x(1-h(x)\log\,x)}$$

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As lhf says, the functional equation for $h(x)$ can be differentiated implicitly, without needing to take the Lambert route:

$$\begin{align*} h^\prime(x)&=\frac{\mathrm d}{\mathrm dx}x^{h(x)}\\ h^\prime(x)&=x^{h(x)}\left(\frac{h(x)}{x}+h^\prime(x)\log\,x\right)\\ h^\prime(x)&=\frac{h(x)^2}{x}+h(x)h^\prime(x)\log\,x\\ h^\prime(x)-h(x)h^\prime(x)\log\,x&=\frac{h(x)^2}{x}\\ h^\prime(x)&=\frac{h(x)^2}{x(1-h(x)\log\,x)}\\ \end{align*}$$