The function f(x) = e^(ln((x)^x))
taking the log to the base e of both the sides we get:
ln f(x) = ln x^x
differentiate both the sides
f'(x)/f(x) = [ln x^x]'
=> f'(x)/f(x) = [x*ln x]'
=> f'(x)/ f(x) = x*(1/x) + ln x
=> f'(x) / f(x) = 1 + ln x
This proves that f'(x)/f(x) = 1 + ln x