过程：y'=tan'(x+y)=(1+tg^2(x+y))(1+y')推导出y'=-ctg^2(x+y)-1再求导y''=2ctg(x+y)(1+ctg^2(x+y))(1+y')代入y'y''=-2ctg^3(x+y)-2ctg^5(x+y)手算,没验算过,大致的思路应该如此
