y=1+xe^y的导数~
问题描述：
y=1+xe^y的导数~
答
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==>y'(1-xe^y)=e^y这一步什么意思？？可以解析下妈？？(*^__^*) 嘻嘻……