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求由方程ysinx-cos(xy)=0所确定的隐函数y=y(x)的导数dy/dx
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推荐答案 2010-05-22
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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相似回答
由方程ysinx-cos(x
+
y)=0确定隐函数y(x)
,
求dy
|(0,π/2)
答:
两边求导:y'
sinx
+
ycos
x+sin(x+y)*(1+y'
)=0
令x=0,y=π/2:π/2+1+y'=0 y'=-(π/2+1
)dy=
-(π/2+1
)dx
由方程ysinx-cos(x
+
y)=0确定隐函数y(x)
,
求dy
|(0,π/2)
答:
两边求导:y'
sinx
+
ycos
x+sin(x+y)*(1+y'
)=0
令x=0,y=π/2:π/2+1+y'=0 y'=-(π/2+1
)dy=
-(π/2+1
)dx
由方程ysinx-cos(x-y)=0所确定的函数的导数dy
/
dx
答:
ysinx-cos(x-y)=0所确定的函数的导数dy
/dx是:y'= dy/
dx
= [
ycos
x + sin(x-y)]/[sin(x-y) - sinx]计算过程如下:方程两边同时求导,得到下面式子:y'sinx+ycosx+sin(x-y) (1-y') = 0 整理可得 y'[sinx -sin(x-y)] = -ycosx - sin(x-y)所以 y'=[ycosx + sin(x...
求下列
由方程所确定的隐函数y=y(x)的导数dy
/
dx
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-
ycos(xy)
所以:dy/
dx=y
'= [e^x-ycos(xy) ] / [ xcos(xy)+e^y ]
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