x^x find minimum

Let f(x) = xx

lnf(x) = xlnx

Differentiate both sides with respect to x

1/f(x) f'(x) = x(1/x) + lnx

f'(x) = xx(1 + lnx)

Set f'(x) = 0

xx(1 + lnx) = 0

1 + lnx = 0, since xx =/= 0

-lnx = 1

ln(1/x) = 1

1/x = e

x = 1/e

When x < 1/e, f'(x) < 0

When x > 1/e, f'(x) > 0

So, the f(x) attains the minimum when x = 1/e

The minimum value

= (1/e)^(1/e)