Finding the derivative of $y^x = e^y$ I have a problem that I've been working on and I don't understand how to start:


  If $y = y(x)$ satisfies $y^x = e^y$, then when $(x, y) = (2\sqrt{e}, \sqrt{e})$, the derivative $\frac{dy}{dx} = \ ...$


I can't understand what $y = y(x)$ satisfies $y^x = e^y$ means. Any help or hints are greatly appreciated! Thanks!
Let us use implicit differentiation

$$ y^x = e^y \implies x \ln y = y \implies \ln y + x  \frac{y'}{y} = y'$$

$$ \implies \ln y = y'(1 - \frac{x}{y}) \implies \frac{dy}{dx} = \frac{\ln y }{1 - \frac{x}{y}}$$

$$ \therefore \; \text{at}\ \ (2 \sqrt{e}, \sqrt{e} ) ,\ \frac{dy}{dx} = \frac{\frac{1}{2}}{1 - 2} = - \frac{1}{2}$$
Rewrite it as $y^{1/y}=e^{1/x}$, look up Lambert W function and ask Wolfram...