xy+sin(x+y)=1,两边求导数
y+xy'+cos(x+y)*(1+y')=0
xy'+cos(x+y)y'=-[cos(x+y)+y]
∴y'[cos(x+y)+x]=-[cos(x+y)+y]
∴y'=-[y+cos(x+y)]/[x+cos(x+y)]不好意思呀，回复晚了。这一步有点不明白：cos(x+y)*(1+y')。谢谢了sin(x+y) 是复合函数，求导时，先求外导，再乘上内导∴[sin(x+y)]'=cos(x+y)*(x+y)'=cos(x+y)*(x'+y')=cos(x+y)*(1+y')