The minimum value of x^x (x > 0) is

Let $y = x^x \log y = x \log x$

$(\log y) = 1 + \log x$

and $(\log y) = x^{–1}$

minimum value of y or $\log y$

$(\log y) = 0\, 1 + \log x = 0$

$x = e^{–1}$

again $x = e^{–1}$

$(\log y ) = e > 0$ minimum value of at $x = e^{–1}$