求由方程y=cot(x+y)所确定的隐函数y=y(x)的二阶导数

2024-11-28 05:43:41
推荐回答(2个)
回答1:

一阶导求完之后适当化简可以使二阶导的难度降低。

回答2:

y=cot(x+y)=cos(x+y)/sin(x+y)
y'=[-sin(x+y)·(1+y')·sin(x+y)-cos(x+y)·(1+y')·cos(x+y)]/sin²(x+y)
=-(1+y')/sin²(x+y)
y'[sin²(x+y)+1]=-1→y'=-1/[sin²(x+y)+1]
y''=[sin²(x+y)+1]'/[sin²(x+y)+1]²
=[sin2(x+y)·(1+y')]/[sin²(x+y)+1]²
=[sin2(x+y)·sin²(x+y)/[sin²(x+y)+1])]/[sin²(x+y)+1]²
=[sin2(x+y)·sin²(x+y)]/[sin²(x+y)+1]³