find f'(x) of f(x)=logax base a + lnax where a is constant.

The derivative of the given function
is:

f'(x) = [log a (x) + ln
(ax)]'

f'(x) = [log a (x)]' + [ ln
(ax)]'

We'll re-write the first term of the expression,
changing the "a" base into the natural base:

log a (x) = ln
x/ln a

We'll differentiate with respect to
x:

ln x/ln a = 1/x*ln a = 1/ln
(a^x)

[ ln (ax)]' = (ax)'/ax = a/ax =
1/x

The requested derivative of the function
is:f'(x)=1/ln (a^x) + 1/x