Slopey Slopes!

First, what is $f'(x)$ ? Let $y=f(x)$ .

$y=x^{x^{x^{x^{.^{.{^.}}}}}}=x^y=e^{y\ln{x}}$

$y'=e^{y\ln{x}}(\frac{d}{dx}y\ln{x})=x^y(\frac{y}{x}+y'\ln{x})=y(\frac{y}{x}+y'\ln{x})$

$y'(1-y\ln{x})=\frac{y^2}{x}$

$y'=\frac{y^2}{x-xy\ln{x}}=\frac{(x^{x^{x^{x^{.^{.{^.}}}}}})^2}{x-x(x^{x^{x^{x^{.^{.{^.}}}}}})\ln{x}}$

Next, what is $f(\sqrt{2})$ or $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{.^{.{^.}}}}}$ ?

Like before, $y=\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{.^{.{^.}}}}}$ can also be expressed as $y=\sqrt{2}^y$ . This gives candidates: $y=2,4$

Then, I want to show that $f(x)\leqslant e$ .

Raise both sides of $y=x^y$ to $\frac{1}{y}$ . We have: $x=y^\frac{1}{y}$ $\frac{dx}{dy}=y^\frac{1}{y}(\frac{1-\ln(y)}{y^2})$

$y^\frac{1}{y}$ and $\frac{1}{y^2}$ are both positive for $y>0$ , meaning the critical point is found where $1-\ln(y)=0$ : at $y=e$ and $x=e^\frac{1}{e}$ .

Consider the sequence: $x,x^x,x^{x^x},...,x^{x^{x^{x^{.^{.{^.}}}}}},...$ over the interval $1<x\leqslant e^\frac{1}{e}$ . Each term is greater than the last over this interval and it's upper bound is $f(x)$ . It will therefore converge to $f(x)$ from below. We know $y=e$ is a local maximum of $x$ , so greater $y$ 's would require the function to curve backwards, meaning: $1<x^{x^{x^{x^{.^{.{^.}}}}}}\leqslant e$ over the interval $1<x\leqslant e^\frac{1}{e}$ , which includes $\sqrt{2}$ .

Finally, $1<2<e<4$ . Since $4$ is not defined on the range of $f(x)$ : $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{.^{.{^.}}}}}=2$ Plugging in:

$f'(\sqrt{2})=\frac{(2)^2}{\sqrt{2}-\sqrt{2}(2)\ln{\sqrt{2}}}=\frac{2\sqrt{2}}{1-\ln{2}}$