设2sin(x+2y-3z)=x+2y-3x,证明:∂z/∂x+∂z/∂y=1.
设2sin(x+2y-3z)=x+2y-3x,证明:∂z/∂x+∂z/∂y=1.
【正确答案】:证明:令F(x,y,z)=2sin(x+2y-3z)-x-2y+3z,则 ∂F/∂x=2cos(x+2y-3z)-1,∂F/∂y=4cos(x+2y-3z)-2, ∂F/∂x=-6cos(z+2y-3z)+3, 所以 ∂F/∂x=-[(∂F/∂x)(∂F/∂z)]=-[2cos(x+2y-3z)-1]/[-6cos(x+2y-3z)+3]=1/3, ∂z/∂y=-[(∂F/∂y)(∂F/∂z)]=-[4cos(x+2y-3z)-2]/[-6cos(x+2y-3z)+3]=2/3 因此∂z/∂x+∂z/∂y=1.
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