Respuesta :

sqdancefan
Define z = ln(x). Then
  y = z^z
  y' = z^z * (z')*(1 +ln(z))
Since
  z' = 1/x
You have
  y' = ln(x)^ln(x) * (1/x) * (1 +ln(ln(x))

Evaluated at x=e, this becomes
  y'(e) = 1^1 * (1/e) * (1 +ln(1))
  y'(e) = 1/e