y=1+xe^y,求y'|x=0
1个回答

题目：y=1+xe^y 求d^2/dx^2

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

相关问题