求由方程xy-1+y^2=0所确定的隐函数的导数

2025-04-13 11:53:19
推荐回答(2个)
回答1:

y+xy'+2yy'=0
y'=-y/(x+2y)

回答2:

xy-1+y^2=0
→(1·y+xy')-0+2y·y'=0
∴y'=-y/(x+2y).