Question

y=tan(x+y) 的微分dy 怎么求？

Answer 1

由y=tan(x+y)可得x=arctany-y，
∴dx=d（arctany-y）=[1／（1+y²）－1]dy≒[﹣y²／（1+y²)]dy，
∴dy=﹣[(1+y²)／y²]dx
另：lee77105 的回答略有失误，改正后最终答案也可转化为这种形式，如下：
dy/dx=(1+dy/dx)*[sec(x+y)]^2,
dy/dx=[sec(x+y)]^2/{1-*[sec(x+y)]^2},
则dy==[sec(x+y)]^2/{1-*[sec(x+y)]^2} dx
dy=[sec(x+y)]^2/{[1-sec(x+y)]^2} dx ，
∵sec²α=tan²α+1
∴[sec(x+y)]^2=tan²（x+y）+1=y²+1
代入可得dy=﹣[(1+y²)／y²]dx

Answer 2

y'=sec²(x+y)*(1+y')
y'[1-sec²(x+y)]=sec²(x+y)
y'=sec²(x+y)/[1-sec²(x+y)]
=-sec²(x+y)*cot²(x+y)
=-1/sin²(x+y)
即dy/dx=-1/sin²(x+y)
所以dy=-dx/sin²(x+y)

Answer 3

因为dy=(dy/dx)dx,
则只需求出dy/dx即可
这是一个隐函数求导问题：左右两边同时对x求导：则dy/dx=(1+dy/dx)*[sec(x+y)]^2,
dy/dx=[sec(x+y)]^2/{1+)*[sec(x+y)]^2},
则dy==[sec(x+y)]^2/{1+)*[sec(x+y)]^2} dx