Minimizing a minima.

Let $y = x^x$ . Then

$\begin{aligned} \frac {dy}{dx} & = \frac d{dx} x^x = \frac d{dx} e^{x\ln x} = (\ln x + 1) e^{x \ln x} = (\ln x +1) x^x \end{aligned}$

Equating $\dfrac {dy}{dx} = 0$ , since $x^x > 0$ , $\implies \ln x + 1 = 0$ or $x = \dfrac 1e$ .

And $\dfrac {d^2 y}{dx^2} = \dfrac {x^x}x + (\ln x+1)^2 x^x$ . $\implies \dfrac {d^2y}{dx^2} \bigg|_{x=\frac 1e} > 0$ . Therefore, $y = x^x$ is minimum when $x = \dfrac 1e$ and $\min (x^x) = \boxed{e^{-1/e}}$ .

Differentiate the given expression twice, and see that the critical point (i.e. exp(-1) ) is the point of minima..........Hence, we get the answer.........!!