设z=z(x,y)是由方程x²+ y²+ z²=xyz 确定的函数,求dz

如题所述

F=x^2 +y^2 +z^2-xyz,

Fx=2x-yz, Fy=2y-xz, Fz=2z-xy,
∂z/∂x=-Fx/fz=-(2x-yz)/(2z-xy),
∂z/∂y=-Fy/fz=-(2y-xz)/(2z-xy),
dz=-(2x-yz)/(2z-xy)•dx-(2y-xz)/(2z-xy)•dy.
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