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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)

Infinite Tetration

Can one determine the derivative of this function?

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If you're talking about the infinite power tower, then you shouldn't be ending the tower with an $x$. – J. M. Apr 29 '12 at 17:55
@J.M. Very true. Edited. – Argon Apr 29 '12 at 18:01 – Arjang Jan 14 '15 at 17:33
up vote 12 down vote accepted

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


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


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

We thus have

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

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

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

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There is no need to bring $W$ into this. You can just differentiate $h(x)=x^{h(x)}$ and solve for $h'$. – lhf Apr 30 '12 at 0:45
@lhf: huh, I really was getting lazy there. Thanks for the note; I edited my answer. – J. M. Apr 30 '12 at 0:57

its given that $y=f(x)=x^{x^{x^{x^\cdots}}}$ I can write this function as,


Taking $log_e$ on both sides, we have,

$$\log y=y\log x$$

Now, differentiate $x$ on both sides,

$$\frac 1 y y'=\frac y x + y'\log x $$

$$ \left(\frac 1 y-\log x\right)y'=\frac y x$$ $$y'=\frac {y^2}{x(1-x\log x)}$$

$$f'(x)=\frac {x^{x^{x^{x^\cdots}}}}{x(1-x\log x)}$$

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