用最基本的方法计算如下:令F(x,y,z)=e^z-xy²+sin(y+z)则 F'x=-y²,F'y=-2xy+cos(y+z),f'z=e^z+cos(y+z)于是Z'x=-(-y²)/[e^z+cos(y+z)],Z'y=-[-2xy+cos(y+z)]/[e^z+cos(y+z)],dz=Z'xdx+Z'ydy={y²dx+[2xy-cos(y+z)]dy}/[e^z+cos(y+z)]