求由方程x+y-z=xe∧(x-y-z)所确定的隐函数z=z(x,y)的偏导数

Question

求由方程x+y-z=xe∧(x-y-z)所确定的隐函数z=z(x,y)的偏导数

Answer 1

求由方程x+y-z=xe∧(x-y-z)所确定的隐函数z=z(x,y)的偏导数
作函数F(x,y,z)=x+y-z-xe^(x-y-z)=0
则∂z/∂x=-(∂F/∂x)/(∂F/∂z)=-[1-e^(x-y-z)-xe^(x-y-z)]/[-1+xe^(x-y-z)]=[-(1+x)e^(x-y-z)]/[(1-xe^(x-y-z)]；
∂z/∂y=-(∂F/∂y)/(∂F/∂z)=-[1+xe^(x-y-z)]/[-1+xe^(x-y-z)]=[1+xe^(x-y-z)]/[1-xe^(x-y-z)].