y=sin(x+y)求y''(隐式函数二阶导)
答案:2 悬赏:30 手机版
解决时间 2021-01-03 21:56
- 提问者网友：放下
- 2021-01-03 18:06
y=sin(x+y)求y''(隐式函数二阶导)
最佳答案
- 五星知识达人网友：人類模型
- 2021-01-03 18:32
y'=[sin(x+y)]'(x+y)'=(1+y')cos(x+y)=cos(x+y)+y'cos(x+y)
y'=cos(x+y)/[(1-cos(x+y)]
y''=[cos(x+y]'(x+y)'+y''cos(x+y)+y'[cos(x+y)]'
=-(1+y')sin(x+y)+y''cos(x+y)-y'(1+y')sin(x+y)
y''[cos(x+y)-1]=(1+y')^2sin(x+y)
={1+cos(x+y)/[(1-cos(x+y)]}^2sin(x+y)
=sin(x+y)/[cos(x+y)-1]^2
y''=sin(x+y)/[cos(x+y)-1]^3
y'=cos(x+y)/[(1-cos(x+y)]
y''=[cos(x+y]'(x+y)'+y''cos(x+y)+y'[cos(x+y)]'
=-(1+y')sin(x+y)+y''cos(x+y)-y'(1+y')sin(x+y)
y''[cos(x+y)-1]=(1+y')^2sin(x+y)
={1+cos(x+y)/[(1-cos(x+y)]}^2sin(x+y)
=sin(x+y)/[cos(x+y)-1]^2
y''=sin(x+y)/[cos(x+y)-1]^3
全部回答
- 1楼网友：渡鹤影
- 2021-01-03 18:53
手写过称,还可以采用隐函数求导公式,都差不多吧,不难就是计算过程复杂一点.仔细就是了!
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