z=y⼀√(x^2+y^2)的全微分

2025-01-07 03:42:08
推荐回答(1个)
回答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