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**