求导：xy=x-e^xy,求dy/dx

答：xy=x-e^(xy)e^(xy)=x-xy=x(1-y)两边对x求导：(xy)' e^(xy)=1-y-xy'(y+xy')e^(xy)=1-y-xy'ye^(xy)+xy'e^(xy)+xy'=1-y[ 1+e^(xy) ] xy'=1-y-ye^(xy)y'=dy/dx= [ 1-y -ye^(xy) ] / [ x+xe^(xy) ]