Question 390517
Let {{{y = x^x}}}.  We have to restrict the domain of y to (0, {{{infinity}}}), otherwise we encounter a densely infinite number of problems, haha.(E.g., {{{(-1/4)^(-1/4) = 1/((-1/4)^(1/4))}}}, or {{{(-pi)^(-pi) = 1/((-pi)^pi)}}}).  Note also that y > 0 for all x in the domain.

That said, we can take ln of both sides of {{{y = x^x}}}:
ln y  = x lnx.
Differentiating both sides:
{{{(dy/dx)/y = lnx +1}}} <-----(A)
Set {{{dy/dx = 0}}}, so that the equation (A) becomes lnx = -1, or {{{x = e^(-1) = 1/e}}}.  Since there is only 1 critical value this is either an absolute min or absolute max.  We find out using the 2nd derivative test.

Differentiate (A) again:
{{{(y(d^2y/dx^2) - (dy/dx)^2)/y^2 = 1/x}}}.
When x = 1/e, the equation becomes {{{(d^2y/dx^2)/y = e}}}, or {{{d^2y/dx^2 = ye > 0}}}.  Thus the graph of the function y is concave up at x=1/e, and so there is an absolute minimum there for y.