why is the derivative of f(x)= e^lnx
f'(x)=1?
if you look in your text, or in the properties of logs, you should find the following
a^loga k = k
so e^lnx = x
then f(x) = e^lnx
= x
therefore f'(x) = 1
f'(x)=1?
a^loga k = k
so e^lnx = x
then f(x) = e^lnx
= x
therefore f'(x) = 1