函数y=1+xe^y 求二阶导数的问题

y=1+xe^y ==>y'=(1+xe^y )'==>y'=(xe^y)'==>y'=1*e^y+xe^y*y'==>y'(1-xe^y)=e^y==>y'=e^y/(1-xe^y)因为y=1+xe^y,则1-xe^y=2-y,得y'=e^y/(2-y)即dy/dx=e^y/(2-y)dy/dx=e^y/(2-y)==>d(dy/dx)/dx=d(e^y/(2-y))==>d(dy/dx)/dx=[e^y*dy*(2-y)-e^y*(-dy)]/(2-y)^2因为dy/dx=e^y/(2-y),则==>d(dy/dx)/dx=[e^2y+e^2y/(2-y)]/(2-y)^2==>d(dy/dx)/dx=e^2y[1+1/(2-y)]/(2-y)^2 求二阶导数是对一阶导数直接再次求导,可用d(dy/dx)/dx这个公式dx是微分变量