设z=ycos(x>²-y²)，证明y²∂z/∂x+xy(∂z/∂y)=xz．

设z=ycos(x>²-y²)，证明y²∂z/∂x+xy(∂z/∂y)=xz．

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．