设隐函数y(x)由方程y=sin(x+y)确定,求y"

Question

如题所述

Answer 1

y'=cos(x+y)(1+y') ...(1)
y'=cos(x+y)+cos(x+y)y'
y'=cos(x+y)/[1-cos(x+y)]
对(1)再对x求导，则y''=-sin(x+y)(1+y')(1+y') +cos(x+y)*y''
即[cos(x+y)-1]y''=sin(x+y)(1+y')²
∴【cos(x+y)-1]y''=sin(x+y)/[1-cos(x+y)]²
∴y''=-sin(x+y)/[1-cos(x+y)]³本回答被提问者采纳

Answer 2

两边同时求导
dy/dx=cos(x+y)*(1+dy/dx)
解出dy/dx=cos(x+y)/(1-cos(x+y))
再次两边求导,设右式分子为u，分母为v
y"=(u/v)'=[u'*v-u*v']/v^2

Answer 3

把Y看作是X的函数，Y当复合函数来求导。
Y''=sin(x+y)/[1-cos(x+y)]^3追问

有过程吗?

Answer 4

y'=cos(x+y) (1+y')
y'=cos(x+y)/[1-cos(x+y)]
y"=[-sin(x+y)(1+y')(1-cos(x+y)-cos(x+y)sin(x+y)(1+y')]/[1-cos(x+y)]^2
=-(1+y')sin(x+y)/[1-cos(x+y)]^2
=-sin(x+y)/[1-cos(x+y)]^3