f(x)=logx,f'(x)=?

from the first principle:

f(x+h)-f(x)/h=log(x+h)-log(x)/h

=1/hlog[1+h/x]

=1/h.h/xlog[1+h/x]^x/h

=1/xlog[1+h/x]^x/h

put h/x=z,we get z0 as h0

therefore,log[1+h/x]^x/h=log(1+z)^1/z log e=1 as z0

hence f(x+h)-f(x)/h1/x as h0

f'(x)=1/x for each x>0

d/dx(logx)=1/x