求二階導數""
y=sin(x-y)
ans=[-sin(x-y)]/[1+cos(x-y)]^3
要詳情步驟,,thz=]
F4,A-MATHS**
2007-03-19 4:59 am
回答 (2)
2007-03-19 5:22 am
參考: My Maths knowledge
2007-03-19 5:40 am
let z=x-y (0)
=> y=x-z (1)
=> y'=1-dz/dx (2)
=> y"=-d^2(z)/dx^2 (3)
y=sin(x-y)
=> x-z=sin(z)
=> 1-z'=cos(z).z'
=> 1=z'[1+cos(z)]
=> z'=[1+cos(z)]^(-1) (4)
=> z"=-1.[1+cos(z)]^(-2).[-sin(z).z']
from (4)
=> z"=[1+cos(z)]^(-2).[sin(z).[1+cos(z)]^(-1)]
=> z"=[1+cos(z)]^(-3).[sin(z)]
from (3) and (1)
=> -y"=[1+cos(x-y)]^(-3).[sin(x-y)]
=> y"=-sin(x-y)/[1+cos(x-y)]^3
=> y=x-z (1)
=> y'=1-dz/dx (2)
=> y"=-d^2(z)/dx^2 (3)
y=sin(x-y)
=> x-z=sin(z)
=> 1-z'=cos(z).z'
=> 1=z'[1+cos(z)]
=> z'=[1+cos(z)]^(-1) (4)
=> z"=-1.[1+cos(z)]^(-2).[-sin(z).z']
from (4)
=> z"=[1+cos(z)]^(-2).[sin(z).[1+cos(z)]^(-1)]
=> z"=[1+cos(z)]^(-3).[sin(z)]
from (3) and (1)
=> -y"=[1+cos(x-y)]^(-3).[sin(x-y)]
=> y"=-sin(x-y)/[1+cos(x-y)]^3
收錄日期: 2021-04-23 21:10:53
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070318000051KK05025