Question

求解一道大一高数导数题

第二小题 答案在第二张图片上 要详细过程谢谢！

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Answer 1

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)

Answer 2

Answer 3

(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.

Answer 4

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