u=ln[根号（x^2+y^2）]=1/2ln(x^2+y^2)z'x=ve^(uv)*1/[2(x^2+y^2)]*2x+ue^(uv)*1/(1+y^2/x^2)*(-y/x^2)=ve^(uv)*x/(x^2+y^2)-ue^(uv)*y/(x^2+y^2) (u,v自己代入)z'y=ve^(uv)*1/[2(x^2+y^2)]*2y+ue^(uv)*1/(1+y^2/x...