e^(x+2y-z)=1+xy^(2/3)z 且z=f(x,y) 则dz(1,0)=?我做的和答案有出入,

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e^(x+2y-z)=1+xy^(2/3)z 且z=f(x,y) 则dz(1,0)=?我做的和答案有出入,

答:
两边求微分得:
(dx+2dy-dz)e^(x+2y-z)=y^(2/3)zdx+2xy^(-1/3)z/3dy+xy^(2/3)dz
dz=[(1-y^(2/3)z)dx+(2-2xy^(-1/3)z/3)dy]/[1+xy^(2/3)]
当x=1,y=0时,e^(1-z)=1,解得z=1
代入x=1,y=0,z=1得:
dz=dx+2dy