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**

To verify the given identity, we'll have to differentiate the function f(x):

f'(x) =e^[ln((x)^x)]*[ln((x)^x)]'

[ln((x)^x)]' = [x*ln(x)]'

We'll use the product rule:

(u*v)' = u'*v + u*v'

[x*ln(x)]' = x'*ln x + x*[ln(x)]'

[x*ln(x)]' = ln x + x/x

[x*ln(x)]' = ln x + 1

So, the derivative of the function f(x) is:

f'(x) = e^[ln((x)^x)]*(ln x + 1)

But e^[ln((x)^x)] = f(x).

f'(x) = f(x)*(ln x + 1)

We'll divide both sides by f(x) and we'll get:

f'(x)/ f(x) = (ln x + 1) q.e.d.

**We notice that the identity f'(x)/f(x)=1+lnx is verified.**