---

Solution

Given $e^y = 1 + x^x$ Differentiate both sides: $e^y \frac{dy}{dx} = x^x (\ln x + 1)$ $\Rightarrow \frac{dy}{dx} = \dfrac{x^x(\ln x + 1)}{e^y}$ Since $e^y = 1 + x^x$, $\Rightarrow m = \dfrac{x^x(\ln x + 1)}{1 + x^x}$ For $x > 0$, this ratio always lies between $-2$ and $2$.