Given $z^{z^x}=x$, estimate $\frac{dx}{dz}$ I'm reading Knoebel's Exponentials Reiterated, page 242 and have very hard time understanding the following:

If $z^{z^x}=x$, then $\frac{dx}{dz}\gt 0$ on the open line
$$L=\{<z,x>|0\lt z\lt e^{-e}\,\text{and}\, x=e^{-1}\}.$$


*

*This doesn't make sense to me. Because if $z^{z^x}=x$, then $x=e^{-1}$ implies $z=e^{-e}$, which is not in the open interval $\left(0,e^{-e}\right)$.


*I know that, if $j(z,x)=z^{z^x}-x$, then
$$dj(z,x)=z^{z^x}z^{x-1}(x\ln z+1)\,dz+\left(z^{z^x}z^x\ln ^2z-1\right)\,dx,$$
and there is a unique trajectory satisfying $j(z,x)=0$ when it is not true that $z=e^{-e}$ and $x=e^{-1}$ simultaneously by the Implicit function theorem,
but how could that be used to prove $\frac{dx}{dz}\gt 0$ on the line?
 A: Consider the implicit function
$$F(x,z)=z^{z^x}-x=0$$
$$\frac{\partial F(x,z)}{\partial x}=z^{z^x+x} \log ^2(z)-1$$
$$\frac{\partial F(x,z)}{\partial z}=z^{z^x} \left(z^{x-1}+x z^{x-1} \log (z)\right)$$
$$\frac {dx}{dz}=-\frac{\frac{\partial F(x,z)}{\partial z} } {\frac{\partial F(x,z)}{\partial x} }=-\frac{z^{z^x+x-1} (x \log (z)+1)}{z^{z^x+x} \log ^2(z)-1}$$
More pleasant would be the reverse way using
$$z=\left(\frac{x \log (x)}{W(x \log (x))}\right)^{\frac{1}{x}}$$ then logarithmic differentiation and chain rule with $t=x \log(x)$. We should end with
$$\frac {dx}{dz}=f(x)$$ which is positive for $0 \leq x \lt e$ with a vertical asymptote at $x=\frac 1e$. For $x>e$, the derivative is always negative.
Edit
For the calculation of the derivative, start with
$$z=\left(\frac{t}{W(t)}\right)^{\frac{1}{x}} \qquad t=x\log(x)$$
$$\log(z)={\frac{1}{x}} \log\left(\frac{t}{W(t)}\right)$$
$$\frac {z'} z=-\frac 1{x^2}\log\left(\frac{t}{W(t)}\right)+\frac 1x \frac d {dt}\Bigg[\log\left(\frac{t}{W(t)}\right) \Bigg]\frac {dt} {dx}$$
$$\frac d {dt}\Bigg[\log\left(\frac{t}{W(t)}\right) \Bigg]=\frac{W(t)}{t (1+W(t))}$$ All of the above gives
$$z'=\frac z {x^2 t (1+W(t)) }\Bigg[x W(t) t'-t (1+W(t)) \log \left(\frac{t}{W(t)}\right) \Bigg]$$
$$\frac {dx}{dz}=\frac 1 {z'}$$
A: *

*I think Knoebel meant $dx/dz>0$ where $z\in(0,e^{-e})$ and $x\in(0,e^{-1})$. Otherwise we would just be evaluating the derivative at one point, namely $(x,z)=(e^{-1},e^{-e})$.


*Note that $z^{z^x}=x$ is equivalent to $\ln\ln x=\ln\ln z+x\ln z$. Implicit differentiation gives $$\frac{dx}{dz}=\frac{x\ln x(1+x\ln z)}{z\ln z(1-x\ln x\ln z)}$$ after collecting terms. The derivative is positive if $\frac{1+x\ln z}{1-x\ln x\ln z}>0$ which is true following a case-by-case analysis of $1-x\ln x\ln z\in(-\infty,0)$, $(0,1/e)$, $(1/e,1)$ and $(1,\infty)$.
