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设y=y(x)由方程e^xy+sin(xy)=y确定,求dy/dx.
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第1个回答 2022-09-16
e^(xy) + sin(xy) = y
(y+xy')e^(xy) + (y+xy')cos(xy) = y'
y'=(ye^(xy)+ycos(xy))/(1-xe^(xy)-xcos(xy))
相似回答
设y=y(x)由方程e^xy+sin(xy)=y确定,求dy
/
dx
.
答:
e^(xy)
+ sin(xy) = y (y+
xy'
)e^(xy)
+ (y+xy')cos(xy) = y'y'=(ye^(xy)+ycos(xy))/(1-xe^(xy)-xcos(xy))
这个
dy
/
dx
?
答:
e^(x+y)
·(1+y')-
sin(xy)
·(
y+
xy')=0 整理得到:[e^(x+y)-x·sin(xy)]·y'
=y
·sin(xy)-e^(x+y)∴y'=[y·sin(xy)-e^(x+y)]/[e^(x+y)-x·sin(xy)]
求下列
方程
所
确定
的隐函数的导数
dy
/
dx
答:
2
xy+
x^2 y'=cosy y' +2e^(2x) y'[x^2-cosy]=2e^(2x)-2xy y'=2[e^(2x)-xy]/(x^2-cosy)x
y=e^(x+y)
y+xy'=e^(x+y) (1+y') y'[1-e^(x+y)]=[e^(x+y)-y] y'=[e^(x+y)-y]/[1-e^(x+y)]
1、设函数
y=y(x)由方程e^
x-e^y=
sin(xy)
所
确定,求
(
dy
/
dx
)|x=0;2、设...
答:
1) x=0代入方程:1-e^y=0,得y(0)=0 两边对X求导:
e^x-y
'e^y=cos(xy)(
y+
xy')y'=[e^x-ycos(xy)]/[xcos
(xy)+
e^y]代入x=0
,y(
0)=0,故y'(0)=1 2)f'
(x)=
2x-1/x^2
求下列
由方程
所
确定
的隐函数
y=y(x)
的导数
dy
/
dx
e^
x-e^y-
sinx
y=0
答:
答:e^x-e^y-
sin(xy)=
0 两边对x求导:e^x -(e^
y)y
'-cos(xy)*(
y+
xy')=0 所以:[xcos(xy)+e^y]*y'=
e^x-y
cos(xy)所以:
dy
/
dx=y
'= [e^x-ycos(xy) ] / [ xcos(xy)+e^y ]
设函数
y=y(x)由方程e^
x-e^y=
sin(xy)
所
确定,求dy
/
dx
|x=0
答:
e^x-
e^y=sin(xy)
对x求导 e^x-e^y*y'=cos(xy)*(xy)' (xy)'=x'*y+x*y'
=y+
x*y' 所以e^x-e^y*y'=cos(xy)*(y+x*y') x=0,则1-e^y*y'=1*y 把x=0代入e^x-e^y=sin(xy) 1-e^y=0 所以y=0,即x=0时y=0 所以代入1-e^y*y'=1*y 1-y'...
y=x+e^x,求dy
/
dx= y
是
方程Sin(xy)=x+y
所
确定
的x函数,求dy/dx=?
答:
∵
y=x+e^
x∴dy/dx=d
(x+e^x)
/dx=1+e^x.解另一题:∵
sin(xy)=x+y=
=>d(sin(xy))=d
(x+y
)==>cos(xy)(xd
y+ydx)
=
dx+dy=
=>(xcos(xy)-1)dy=(1-ycos(xy))dx∴dy/dx=(1-ycos(xy))/(xcos(xy)-1)
设函数
y=y(x)由方程sin(x+
y
)+e
(x方
)=
0所
确定,
则
dy
/dc=
答:
∵d(2xy)=2xyln2?d
(xy)=
2xyln2?(
ydx+xdy)
d
(x+y)
=
dx+dy
∴2xyln2?(ydx+xdy)=dx+dy 又x=0时
,y=
1 ∴代入上式得:dy|x=0=(ln2-1
)dx
求
由方程
x
e^y+sin(xy)=
0所
确定
的隐函数的导数
dy
/
dx
答:
将原方程两边微分得d[xe^y
+sin(xy)
]=0→
e^ydx+
x
e^ydy
+cos(xy)(
ydx+xdy)=
0→移项 [x
e^y+
xcos(xy)]
dy=
-[e^y+ycos(xy)]dx整理→dy/dx=-[e^y+ycos(xy)]/[xe^y+xcos(xy)].这种方法是最快最不易出错的.
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