这个高数题二阶导怎么求啊！困扰我一天了，跪求答案啊

Answer 1

这肯定是用隐函数求导方式求的
y=tan(x+y)
y'=sec²(x+y)*(x+y)'
=sec²(x+y)*(1+y')
=sec²(x+y)+y'sec²(x+y)
y'-y'sec²(x+y)=sec²(x+y)
y'=sec²(x+y)/[1-sec²(x+y)]
=sec²(x+y)/{-[sec²(x+y)-1]}
=sec²(x+y)/[-tan²(x+y)]
=-1/cos²(x+y)*cos²(x+y)/sin²(x+y)
=-csc²(x+y)

y''=-2csc(x+y)*[-csc(x+y)cot(x+y)]*(x+y)'
=2csc²(x+y)cot(x+y)*(1+y')
=2csc²(x+y)cot(x+y)*[1-csc²(x+y)]
=2csc²(x+y)cot(x+y)*{-1[csc²(x+y)-1]}
=-2csc²(x+y)cot(x+y)*[cot²(x+y)]
=-2csc²(x+y)cot³(x+y)

Answer 2

y=tan(x+y)
y'=sec(x+y)*(x+y)'
=sec(x+y)*(1+y')
=sec(x+y)+y'sec(x+y)
y'-y'sec(x+y)=sec(x+y)
y'=sec(x+y)/[1-sec(x+y)]
=sec(x+y)/{-[sec(x+y)-1]}
=sec(x+y)/[-tan(x+y)]
=-1/cos(x+y)*cos(x+y)/sin(x+y)
=-csc(x+y)

y''=-2csc(x+y)*[-csc(x+y)cot(x+y)]*(x+y)'
=2csc(x+y)cot(x+y)*(1+y')
=2csc(x+y)cot(x+y)*[1-csc(x+y)]
=2csc(x+y)cot(x+y)*{-1[csc(x+y)-1]}
=-2csc(x+y)cot(x+y)*[cot(x+y)]
=-2csc(x+y)cot(x+y)

追答

真麻烦！