Derivative of $x^{x^{\cdot^{\cdot}}}$? 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?
 A: 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
$$h^\prime(x)=\frac{\exp(-2 W(-\log\,x))}{x (1+W(-\log\,x))}=\frac{h(x)^2}{x(1-h(x)\log\,x)}$$

As lhf says, the functional equation for $h(x)$ can be differentiated implicitly, without needing to take the Lambert route:
$$\begin{align*}
h^\prime(x)&=\frac{\mathrm d}{\mathrm dx}x^{h(x)}\\
h^\prime(x)&=x^{h(x)}\left(\frac{h(x)}{x}+h^\prime(x)\log\,x\right)\\
h^\prime(x)&=\frac{h(x)^2}{x}+h(x)h^\prime(x)\log\,x\\
h^\prime(x)-h(x)h^\prime(x)\log\,x&=\frac{h(x)^2}{x}\\
h^\prime(x)&=\frac{h(x)^2}{x(1-h(x)\log\,x)}\\
\end{align*}$$
A: its given that $y=f(x)=x^{x^{x^{x^\cdots}}}$
 I can write this function as,
$$y=x^{y}$$
Taking $log_e$ on both sides, we have,
$$\log y=y\log x$$
Now, differentiate w.r.to $x$ on both sides,
$$\frac 1 y y'=\frac y x + y'\log x $$
$$ \left(\frac 1 y-\log x\right)y'=\frac y x$$
$$y'=\frac {y^2}{x(1-x\log x)}$$
$$f'(x)=\frac {x^{x^{x^{x^\cdots}}}}{x(1-x\log x)}$$
