u=x/(x²+y²+z²)那么∂u/∂x=[(x²+y²+z²)-2x²] /(x²+y²+z²)²=(y²+z²-x²) /(x²+y²+z²)²而∂u/∂y=-2xy/(x²+y²+z²)²，∂u/∂z =-2xz/(x²+y²+z²)²所以得到全微分为du=(y²+z²-x²) /(x²+y²+z²)² dx -2xy/(x²+y²+z²)² dy -2xz/(x²+y²+z²)² dz
