由隐函数xz=ln(z+y^2)确定了z=z(x,y) 求全微分dz

设F(x)=xz-ln(z+y²)
那么əz/əx=-F′x/F′z=-z/[x-1/(z+y²)]=[z(z+y²)]/(1-xz-xy²)
əz/əy=-F′y/F′z=2y/(xz+xy²-1)
dz=(əz/əx)dx+(əz/əy)dy
={[z(z+y²)]/(1-xz-xy²)}dx+[2y/(xz+xy²-1)]dy