$x$ th root of $x$

From $y = \sqrt[x]x \implies \dfrac {dy}{dx} = \left(\dfrac {1-\ln x}{x^2}\right) x^\frac 1x$ $\implies \dfrac {dy}{dx} = 0$ , when $\ln x = 1$ or $x = e$ . Note that $\dfrac {d^2y}{dx^2} = \left(\dfrac {1-\ln x}{x^2}\right)^2 x^\frac 2x + \left(\dfrac {2\ln x - 3}{x^3}\right) x^\frac 1x$ $\implies \dfrac {d^2y}{dx^2} \bigg|_{x=e} < 0$ . Therefore, $y$ is maximum when $x=e \approx \boxed{2.718}$ .