∵x+y+z=0,x²+y²+z²=1
∴dx+dy+dz=0.(1)
xdx+ydy+zdz=0.(2)
故把(1)*(-y)+(2),得dx/dz=(y-z)/(x-y)
把(1)*(-x)+(2),得dy/dz=(x-z)/(y-x).