第二小题 答案在第二张图片上 要详细过程谢谢！
第1个回答 2019-10-15
y=tan(x+y)
两边对x求导
dy/dx=sec^2(x+y)*(1+dy/dx)
dy/dx=sec^2(x+y)+sec^2(x+y)*dy/dx
[sec^2(x+y)-1]*dy/dx=-sec^2(x+y)
tan^2(x+y)*dy/dx=-[tan^2(x+y)+1]
dy/dx=-1-cot^2(x+y)
两边再对x求导
d^2y/dx^2=-2cot(x+y)*[-csc^2(x+y)]*(1+dy/dx)
=2cot(x+y)*csc^2(x+y)*[-cot^2(x+y)]
=-2cot^3(x+y)*csc^2(x+y)本回答被网友采纳
两边对x求导
dy/dx=sec^2(x+y)*(1+dy/dx)
dy/dx=sec^2(x+y)+sec^2(x+y)*dy/dx
[sec^2(x+y)-1]*dy/dx=-sec^2(x+y)
tan^2(x+y)*dy/dx=-[tan^2(x+y)+1]
dy/dx=-1-cot^2(x+y)
两边再对x求导
d^2y/dx^2=-2cot(x+y)*[-csc^2(x+y)]*(1+dy/dx)
=2cot(x+y)*csc^2(x+y)*[-cot^2(x+y)]
=-2cot^3(x+y)*csc^2(x+y)本回答被网友采纳
第2个回答 2019-10-15
第3个回答 2019-10-15
(12)(2)y=tan(x+y),
y'=sec^(x+y)*(1+y'),
y'[sec^(x+y)-1]=-sec^(x+y),
y'=-1/sin^(x+y),
y''=2/[sin(x+y)]^3*cos(x+y)*(1+y')
=2cos(x+y)[sin^(x+y)+1]/[sin(x+y)]^5.
y'=sec^(x+y)*(1+y'),
y'[sec^(x+y)-1]=-sec^(x+y),
y'=-1/sin^(x+y),
y''=2/[sin(x+y)]^3*cos(x+y)*(1+y')
=2cos(x+y)[sin^(x+y)+1]/[sin(x+y)]^5.
第4个回答 2019-10-15
欢迎和我一起讨论数学，一起进步！
追问请问第三行是什么意思啊？是sec的公式吗？
我看懂了
