设2sin(x+2y-3z)=x+2y-3z,证明∂z/∂x+∂z/∂y=1.
设F(x,y,z)=2sin(x+2y-3z)-x-2y+3z,ze: Fx=2cos(x+2y-3z)-1,Fy=4cos(x+2y-3z)-2, Fz=-6cos(x+2y-3z)+3 ∴∂z/∂x=-(Fx/Fz)=[1-2cos(x+2y-3z)]/[3-6cos(x+2y-3z)]=1/3 ∂z/∂y=-(Fy/Fz)=[2-4cos(x+2y-3z)]/[3-6cos(x+2y-3z)]=2/3 ∴∂z/∂x+∂z/∂y=1