设函数z=z(x,y)由方程x∧2z∧3+2y∧2z∧2-x∧2+y∧3=0所确定,求dz
发布网友
共1个回答
热心网友
两边对x求偏导： 2xz^3+x^2*3z^2 Z'x+2y^2*2z*Z'x-2x=0,得：Z'x=(2x-2xz^3)/(3x^2z^2+4y^2z)
两边对y求偏导： x^2*3z^2*Z'y+4yz^2+2y^2*2z*Z'y+3y^2=0,得：Z'y=-(3y^2-4yz^2)/(3x^2z^2+4y^2z)
dz=Z'xdx+Z'ydy=(2x-2xz^3)/(3x^2z^2+4y^2z)dx-(3y^2-4yz^2)/(3x^2z^2+4y^2z)dy
