设z=f(2x-y,ysinx),其中f(u,v)具有二阶偏导数,求z先对x再对y的二阶偏导。
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∵z=f(2x-y,ysinx)
∴
?
?x
z=
?
?x
f(2x-y,ysinx)
=f1′
?
?x
(2x-y)+f2'
?
?x
(ysinx)
=2f1′+ycosxf2'
?2z
?x?y
=
?
?y
(2f1′+ycosxf2')
=2
?
?y
f1′+cosx
?
?y
(yf2')
因为:
?
?y
f1′=f11″
?
?y
(2x-y)+f12″
?
?y
(ysinx)
=-f11″+sinxf12″
?
?y
(yf2')=f2'+y
?
?y
f2'
=f2'+y[f21″
?
?y
(2x-y)+f22″
?
?y
(ysinx)]
=f2'+y[-f21″+sinxf22″]
=f2'-yf21″+ysinxf22″
所以:
?2z
?x?y
=2
?
?y
f1′+cosx
?
?y
(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:
?2z
?x?y
=-2f11″+2sinxf12″+cosxf2'-ycosf21″+ysinxcosxf22″
=-2f11″+(2sinx-ycosx)f12″+cosxf2'+ysinxcosxf22″
故
?2z
?x?y
的值为:
-2f11″+(2sinx-ycosx)f12″+cosxf2'+ysinxcosxf22″
∴
?
?x
z=
?
?x
f(2x-y,ysinx)
=f1′
?
?x
(2x-y)+f2'
?
?x
(ysinx)
=2f1′+ycosxf2'
?2z
?x?y
=
?
?y
(2f1′+ycosxf2')
=2
?
?y
f1′+cosx
?
?y
(yf2')
因为:
?
?y
f1′=f11″
?
?y
(2x-y)+f12″
?
?y
(ysinx)
=-f11″+sinxf12″
?
?y
(yf2')=f2'+y
?
?y
f2'
=f2'+y[f21″
?
?y
(2x-y)+f22″
?
?y
(ysinx)]
=f2'+y[-f21″+sinxf22″]
=f2'-yf21″+ysinxf22″
所以:
?2z
?x?y
=2
?
?y
f1′+cosx
?
?y
(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:
?2z
?x?y
=-2f11″+2sinxf12″+cosxf2'-ycosf21″+ysinxcosxf22″
=-2f11″+(2sinx-ycosx)f12″+cosxf2'+ysinxcosxf22″
故
?2z
?x?y
的值为:
-2f11″+(2sinx-ycosx)f12″+cosxf2'+ysinxcosxf22″
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