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

Solution 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

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

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

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

Solution 2:

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,

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

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

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