y=tan(x+y) 的微分dy 怎么求?
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y=tan(x+y) 的微分dy 怎么求?
y=tan(x+y) 的微分dy 怎么求?
y=tan(x+y) 的微分dy 怎么求?
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)
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
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)]...
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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
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