隐函数求导题目
答案:1 悬赏:40 手机版
解决时间 2021-07-26 05:57
- 提问者网友：wodetian
- 2021-07-25 22:56
隐函数求导题目
最佳答案
- 五星知识达人网友：洒脱疯子
- 2021-07-26 00:30
y=tan(x+y)
y'=[sec(x+y)]^2*(1+y')
则
y'=[sec(x+y)]^2/{1-[sec(x+y)]^2}=-[sec(x+y)]^2/tan(x+y)]^2=-1/[sin(x+y)]^2
则y''={-1/[sin(x+y)]^2}'={-2[sin(x+y)]cos(x+y)}*（1+y')1/[sin(x+y)]^4
则
y''={-2[sin(x+y)]cos(x+y)}（1-1/[sin(x+y)]^2)*（1+y')1/[sin(x+y)]^4
={-2[sin(x+y)]cos(x+y)}[cos(x+y)]^2*1/[sin(x+y)]^4
化简下去就OK了
y'=[sec(x+y)]^2*(1+y')
则
y'=[sec(x+y)]^2/{1-[sec(x+y)]^2}=-[sec(x+y)]^2/tan(x+y)]^2=-1/[sin(x+y)]^2
则y''={-1/[sin(x+y)]^2}'={-2[sin(x+y)]cos(x+y)}*（1+y')1/[sin(x+y)]^4
则
y''={-2[sin(x+y)]cos(x+y)}（1-1/[sin(x+y)]^2)*（1+y')1/[sin(x+y)]^4
={-2[sin(x+y)]cos(x+y)}[cos(x+y)]^2*1/[sin(x+y)]^4
化简下去就OK了
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