求下列函数的全微分:
(1)z=xy+x/y;
(2)z=In(1+x2+y2);
(3)z=yx
(4)u=xyz
(1)zx=y+1/y,zy=x-x/y2 ∴dz=(y+1/y)dx+(x-x/y2)dy (2)zx=2x/(1+x2+y2),zy=2y/(1+x2+y2) ∴dz=2(xddx+ydy)/(1+x2+y2) (3)zx=yxlny,zy=yx•(x/y) ∴dz=yx[lnydx+(x/y)dy] (4)ux=yzxyz-1,uy=zxyz,uz=yxyzlnx ∴du=yzxyz-1dx+zxyzlnxdy+yxyzlnxdz