等式两边分别对x求导,同时将y看做是由x确定的隐函数
1/√(x²+y²) * (√（x²+y²))' = 1/(1+ y²/x²) * (y/x)‘
1/√(x²+y²) * (1/2√(x²+y²))* (x²+y²)’ = 1/(1+ y²/x²) * ((y'x-y)/x²)
1/2(x²+y²) * (2x+2y*y') = 1/(1+ y²/x²) * ((y'x-y)/x²)
整理可得到
y‘=dy/dx
= (x+y)/(x-y)