Question

z=y/√(x^2+y^2)的全微分

Answer 1

z'x=-xy/√(x^2+y^2)^3
Z'y=1/√(x^2+y^2)-y^2/√(x^2+y^2)^3
=x^2/√(x^2+y^2)^3
dz=z'x dx+z'ydy
=(-xy)/√(x^2+y^2)^3 dx + (x^2)/√(x^2+y^2)^3 dy