设z=f(2x-y,ysinx),其中f(u,v)具有连续的二阶偏导数,求?2z?x?y
展开全部
∵z=f(2x-y,ysinx)
∴
z=
f(2x-y,ysinx)
=f1′
(2x-y)+f2'
(ysinx)
=2f1′+ycosxf2'
=
(2f1′+ycosxf2')
=2
f1′+cosx
(yf2')
因为:
f1′=f11″
(2x-y)+f12″
(ysinx)
=-f11″+sinxf12″
(yf2')=f2'+y
f2'
=f2'+y[f21″
(2x-y)+f22″
(ysinx)]
=f2'+y[-f21″+sinxf22″]
=f2'-yf21″+ysinxf22″
所以:
=2
f1′+cosx
(yf2')
=2(-f11″+sinxf12″)+cosx(f2'-yf21″+ysinxf22″)
=-2f11″+2sinxf12″+cosxf2'-ycosf21″+ysinxcosxf22″
又因为函数f具有连续二阶导数,所以其二阶混合偏导数相等,即:
f12″=f21″
所以:
=-2f11″+2sinxf12″+cosxf2'-ycosf21″+ysinxcosxf22″
=-2f11″+(2sinx-ycosx)f12″+cosxf2'+ysinxcosxf22″
故
的值为:
-2f11″+(2sinx-ycosx)f12″+cosxf2'+ysinxcosxf22″
∴
? |
?x |
? |
?x |
=f1′
? |
?x |
? |
?x |
=2f1′+ycosxf2'
?2z |
?x?y |
? |
?y |
=2
? |
?y |
? |
?y |
因为:
? |
?y |
? |
?y |
? |
?y |
=-f11″+sinxf12″
? |
?y |
? |
?y |
=f2'+y[f21″
? |
?y |
? |
?y |
=f2'+y[-f21″+sinxf22″]
=f2'-yf21″+ysinxf22″
所以:
?2z |
?x?y |
? |
?y |
? |
?y |
=2(-f11″+sinxf12″)+cosx(f2'-yf21″+ysinxf22″)
=-2f11″+2sinxf12″+cosxf2'-ycosf21″+ysinxcosxf22″
又因为函数f具有连续二阶导数,所以其二阶混合偏导数相等,即:
f12″=f21″
所以:
?2z |
?x?y |
=-2f11″+(2sinx-ycosx)f12″+cosxf2'+ysinxcosxf22″
故
?2z |
?x?y |
-2f11″+(2sinx-ycosx)f12″+cosxf2'+ysinxcosxf22″
展开全部
zx=f1*2+f2
ycosx
=2f1+ycosxf2
zxy=-2f11+2sinxf12+cosxf2+ycosx(-f21+sinxf22)
=-2f11+2sinxf12+cosxf2-ycosxf21+ysinxcosxf22
ycosx
=2f1+ycosxf2
zxy=-2f11+2sinxf12+cosxf2+ycosx(-f21+sinxf22)
=-2f11+2sinxf12+cosxf2-ycosxf21+ysinxcosxf22
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询