设z=ycos(x>2-y2),证明y2∂z/∂x+xy(∂z/∂y)=xz.
【正确答案】:证明:∂z/∂x=[ycos(x2-y2)]x′=-ysin(x2-y2)•(x2-y2)x′,=-2xysin(x2-y2), ∂z/∂y=[ycos(x2-y2)]y′, =cos(x2-y2)-ysin(x2-y2)(x2-y2)y′ =cos(x2-y2)+2y2•sin(x2-y2), 左端=-2xy3•sin(x2-y2)+xycos(x2-y2)+2xy3•sin(x2-y2) =x•ycos(x2-y2)=x•z =右端 即等式成立.