f(x1x2)=f(x1)+f(x2),且f'(1)=1,证明:当x≠0时,f'(x)=1/x

取x1=x2=1得f(1)=0, 所以lim{e→0}f(1+e)/e=f'(1)=1f(u)=f(u/v。v)=f(u/v)+f(v), 故有f(u)-f(v)=f(u/v)f'(x)=(f(x+dx)-f(x))/dx=f(1+dx/x)/dx令dx/x=e, 则f'(x)=f(1+e)/(ex)=1/x