cos(xy)=x－y所确定的隐函数y=y(x)的导数dy/dx
答案:2 悬赏:40 手机版
解决时间 2021-12-26 14:12
- 提问者网友：辞取
- 2021-12-25 22:26
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最佳答案
- 五星知识达人网友：詩光轨車
- 2021-12-25 23:35
cos(xy)=x-y,隐函数,两边求导
-sin(xy)*(xy)'=1-y'
-sin(xy)*(y+xy')=1-y'
-ysin(xy)-xcos(xy)*y'=1-y'
y'[1-xsin(xy)]=1+ysin(xy)
y'=[1+ysin(xy)]/[1-xsin(xy)]
也可用设二元函数f(x,y)=cos(xy)-x+y
用隐函数求导法:f'x(x,y)+f,y(x,y)*y'=0
f'x(x,y)=-sin(xy)*(xy)'-1
=-ysin(xy)-1
f'y(x,y)=-sin(xy)*(xy)'+1
=-xsin(xy)+1
∴[-ysin(xy)-1]+[-xsin(xy)+1]*y'=0
y'=-[ysin(xy)-1]/[-xsin(xy)+1]
y'=[1+ysin(xy)]/[1-xsin(xy)]
-sin(xy)*(xy)'=1-y'
-sin(xy)*(y+xy')=1-y'
-ysin(xy)-xcos(xy)*y'=1-y'
y'[1-xsin(xy)]=1+ysin(xy)
y'=[1+ysin(xy)]/[1-xsin(xy)]
也可用设二元函数f(x,y)=cos(xy)-x+y
用隐函数求导法:f'x(x,y)+f,y(x,y)*y'=0
f'x(x,y)=-sin(xy)*(xy)'-1
=-ysin(xy)-1
f'y(x,y)=-sin(xy)*(xy)'+1
=-xsin(xy)+1
∴[-ysin(xy)-1]+[-xsin(xy)+1]*y'=0
y'=-[ysin(xy)-1]/[-xsin(xy)+1]
y'=[1+ysin(xy)]/[1-xsin(xy)]
全部回答
- 1楼网友：猎心人
- 2021-12-26 00:10
ysinx=cos(xy)
两边分别求导
y'sinx+ycosx=-sin(xy)(y+xy')
y'=-y(sin(xy)+cosx)/(sinx+xsin(xy))
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