y'=cos(x+y)(1+y')...(1)

　　y'=cos(x+y)+cos(x+y)y'

　　y'=cos(x+y)/[1-cos(x+y)]

　　对(1)再对x求导,则y''=-sin(x+y)(1+y')(1+y')+cos(x+y)*y''

　　即[cos(x+y)-1]y''=sin(x+y)(1+y')²

　　∴【cos(x+y)-1]y''=sin(x+y)/[1-cos(x+y)]²

　　∴y''=-sin(x+y)/[1-cos(x+y)]³