求导：f(x)=(e^x+1)/e^x

f(x)=(e^x+1)/e^x = 1 + 1/e^x = 1 + e^(-x)f '(x) = [1 + e^(-x)] ' = [e^(-x)] ' =[e^(-x)]*(-x)'=[e^(-x)]*(-1)= -e^(-x)= -1/e^x,即f '(x) = -1/e^x