Verify the relation f'(x)/f(x)=1+lnx, if f(x)=e^(ln((x)^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.