Differentiating an Infinite Tetration

$\begin{aligned} \color{#3D99F6}f(x) & = x^{x^{x^{\cdot^{\cdot^{\cdot}}}}} & \small \color{#3D99F6} \text{Let }y = f(x) \\ \color{#3D99F6} y & = x^y \\ \ln y & = y \ln x & \small \color{#3D99F6} \text{Differentiate both sides w.r.t. }x \\ \frac 1y \cdot \frac {dy}{dx} & = \ln x \cdot \frac {dy}{dx} + \frac yx \\ \frac {f'(x)}{f(x)} & = \ln x f'(x) + \frac {f(x)}x & \small \color{#3D99F6} \text{Put }x=\sqrt 2 \\ \frac {f'(\sqrt 2)}{\color{#3D99F6}f(\sqrt 2)} & = \ln \sqrt 2 f'(\sqrt 2) + \frac {\color{#3D99F6}f(\sqrt 2)}{\sqrt 2} & \small \color{#3D99F6} \text{Note that } f(\sqrt 2) = 2 \\ \frac {f'(\sqrt 2)}{\color{#3D99F6}2} & = \frac {\ln 2}2 \cdot f'(\sqrt 2) + \frac {\color{#3D99F6}2}{\sqrt 2} \\ \implies f'(\sqrt 2) & = \frac {2\sqrt 2}{1-\ln 2} \end{aligned}$

$\implies a+b+c+d = 2+2+1+2 = \boxed{7}$

Since $f(x) = \Large x^{x^{x^{.^{.^{.}}}}}$ , we can rewrite $f$ as $f(x) = x^{f(x)} \implies \ln{f(x)} = f(x)\ln x$ .

Differentiating both sides of this equation yields $\dfrac{f'(x)}{f(x)} = \dfrac{f(x)}{x} + f'(x) \ln x \implies f'(x) = \dfrac{(f(x))^2}{x(1-f(x)\ln x)}$ .

$f(\sqrt{2}) = \sqrt{2}^{f(\sqrt{2})} \implies \sqrt{2}^a = a \longrightarrow a \in \{2,4\}$ . Since $f$ is a continuous function on interval $[e^{-e},e^{1/e}]$ with a maximum at $(e^{1/e},e)$ , we see that $f(\sqrt{2}) = 2$ .

Therefore, evaluating $f'$ at $x = \sqrt{2}$ yields $\dfrac{2\sqrt{2}}{1-\ln 2}$ and thus, $a+b+c+d = 2+2+1+2 = \boxed{7}$ .

$\displaystyle \begin{aligned} y & = x^{x^{x^{\cdot^{\cdot^{\cdot}}}}} & \small \color{#3D99F6} \text{If } \lim_{n \ \to \ \infty} \underbrace{x^{x^{\cdot ^{\cdot ^{x}}}}}_{\text{n times}} \text{ exists, it is solution to } y = x^y \\ y & = x^{\color{#3D99F6}{y}} \\ \ln y & = y\ln x & \small \color{#3D99F6} \text{Differentiate both sides} \\ \frac{1}{y} \cdot \frac{dy}{dx} & = \frac{dy}{dx} \cdot \ln x + \frac yx & \small \color{#3D99F6} \text{Isolate } \frac{dy}{dx} \\ \frac{dy}{dx} \left(\frac 1y - \ln x \right) & = \frac yx \\ \implies \frac{dy}{dx} & = \frac{y}{x\left( \displaystyle \frac 1y - \ln x\right)} & \small \color{#3D99F6} y = x^{x^{x^{\cdot^{\cdot^{\cdot}}}}} \\ \implies \frac{dy}{dx} & = \frac{{\color{#3D99F6}{x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}}}}{x\Bigg( \displaystyle \frac {1}{{\color{#3D99F6}{x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}}}} - \ln x\Bigg)} & \small \color{#3D99F6} \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\cdot^{\cdot^{\cdot}}}}} = 2 \\ \implies \frac{dy}{dx} \bigg|_{x \ = \ \sqrt{2}} & = \frac{{\color{#3D99F6}{2}}}{\sqrt{2}\left(\frac {1}{{\color{#3D99F6}{2}}} - \ln \sqrt{2} \right)} \\ & = \frac{2\sqrt{2}}{1 - \ln 2} \end{aligned}$

$\displaystyle \implies a + b + c + d = \boxed{7}$