Question

求方程2xz－2xyz+ln(xyz)=0所确定的函数z=f(x,y)的全微分

Answer 1

dz=(∂z/∂x)dx+(∂z/∂y)dy.
2xz-2xyz+ln(xyz)=0,对x求偏导数,则2z+2x(∂z/∂x)-2yz-2xy(∂z/∂x)+[1/(xyz)][yz+xy(∂z/∂x)]=0,
整理,[2z-2yz+(1/x)]+[2x-2xy+(1/z)](∂z/∂x)=0,所以∂z/∂x=[-2z+2yz-(1/x)]/[2x-2xy+(1/z)].
2xz-2xyz+ln(xyz)=0,对y求偏导数,则2x(∂z/∂y)-2xz-2xy(∂z/∂y)+[1/(xyz)][xz+xy(∂z/∂y)]=0,
整理,[-2xz+(1/y)]+[2x-2xy+(1/z)](∂z/∂y),所以∂z/∂y=[2xz-(1/y)]/[2x-2xy+(1/z)].
故dz={[-2z+2yz-(1/x)]dx+[2xz-(1/y)]dy}/[2x-2xy+(1/z)]