设z=e^xy,则dz|(1,1)=

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设z=e^xy,则dz|(1,1)=

设z=e^xy,则dz|(1,1)=
设z=e^xy,则dz|(1,1)=

设z=e^xy,则dz|(1,1)=

dZ = эZ/эx *dx + эz/эy*dy
= y*e^(xy)*cos(xy)*dx + e^(xy)*[-ysin(xy)]*dx
+ x*e^(xy)*cos(xy)*dy + e^(xy)*[-xsin(xy)*dy
dZ|(0,1) = 1*e^0*cos0*dx - e^0*y*sin0*dx +0*e^0*cos0*dy - e^0*0*sin0*dy
= dx
以上回答你满意么?


dZ = эZ/эx *dx + эz/эy*dy
= y*e^(xy)*cos(xy)*dx + e^(xy)*[-ysin(xy)]*dx
+ x*e^(xy)*cos(xy)*dy + e^(xy)*[-xsin(xy)*dy
dZ|(0,1) = 1*e^0*cos0*dx - e^0*y*sin0*dx +0*e^0*cos0*dy - e^0*0*sin0*dy
= dx

答案应该是2e。
dz=edx+edy。