兩邊對x求導(dǎo)：
y'=[sec(x+y)]^2*(1+y')
即y'=(sec(x+y))^2/[1-(sec(x+y))^2]
化簡得：y'=-(sec(x+y))^2/(tan(x+y))^2=- [csc(x+y)]^2
兩邊再對x求導(dǎo)得：
y"=2csc(x+y)*csc(x+y)*ctg(x+y)*(1+y'),再代入y',得：
=2(csc(x+y))^2*ctg(x+y)*[1-(csc(x+y))^2]
=-2(csc(x+y))^2*(ctg(x+y))^2
=-2[csc(x+y)*ctg(x+y)]^2