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Here is my attempt:

Substituting y for infinite x powers:

$$x^{x^{x^{x^{.^{.^{.}}}}}}=y → x^y=y $$

Giving: $$x=y^{\frac{1}{y}}$$

Take natural logs & differentiate with respect to $y$: $$ln(x)=ln(y^\frac{1}{y}) → ln(x)=\frac{1}{y}ln(y^\frac{1}{y})$$ $$\frac{1}{x}\frac{dx}{dy}=-\frac{1}{y^2}ln(y)+\frac{1}{y^2}$$ $$\frac{dx}{dy}=x\left(\frac{1-ln(y)}{y^2}\right)$$

Sub. in $y^{\frac{1}{y}}$ for $x$: $$\frac{dx}{dy}=y^{\frac{1}{y}}\left(\frac{1-ln(y)}{y^2}\right)$$ $$\frac{dx}{dy}=y^{\frac{1}{y}-2}\left(1-ln(y)\right)$$

Inverse $\frac{dx}{dy}$ to get $\frac{dy}{dx}$: $$\frac{dy}{dx}=\left[y^{\frac{1}{y}-2}\left(1-ln(y)\right)\right]^{-1}$$ Therefore: $$\frac{dy}{dx}=\frac{y^{2-\frac{1}{y}}}{1-ln(y)}$$

Have I made a mistake anywhere? Have I made a false assumption? Please kindly provide some guidance, thanks.

Frank J
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1 Answers1

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I think that you could do it faster considering the implicit function $$F=\log(x)-\frac 1y \log(y)=0\implies F'_x=\frac{1}{x}\qquad F'_y=\frac{\log (y)}{y^2}-\frac{1}{y^2}$$ Now, from the implicit function theorem $$\frac{dy}{dx}=-\frac{F'_x}{F'_y}=\frac{y^2}{x-x \log (y)}$$ If now you replace $x$ by $y^y$, you effectively get $$\frac{dy}{dx}=\frac{y^{2-y}}{1-\log (y)}$$

Sooner or later, you will know that the solution of $x=y^{\frac{1}{y}}$ is explicitely given by $$y=-\frac{W(-\log (x))}{\log (x)}$$ where appears Lambert function and that $$\frac{dW(z)}{dz}=\frac{W(z)}{z (W(z)+1)}$$ Applying the chain rule, you would have $$\frac{dy}{dx}=\frac{W(-\log (x))^2}{x \log ^2(x) (1+W(-\log (x)))}$$

Claude Leibovici
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