设z=f(2x-y)+g(x,xy),其中函数f(t)二阶可导,g(u,v)具有连续二阶偏导数,求?2z?x?y

2024-11-14 10:53:26
推荐回答(2个)
回答(1):

因为:z=f(2x-y)+g(x,xy)
所以:

?z
?x
=
?
?x
[f(2x-y)+g(x,xy)]
=
?
?x
f(2x-y)+
?
?x
g(x,xy)
=f′
?
?x
(2x-y)+g1
?
?x
(x)+g2
?
?x
(xy)
=2f′+g1′+yg2
?2z
?x?y
=
?
?y
(2f′+g1′+yg2′)
=2
?
?y
f′+
?
?y
g1′+
?
?y
(yg2′)
因为:
2
?
?y
f′=2f″
?
?y
(2x-y)=-2f″;
?
?y
g1′=g11
?
?y
(x)+g12
?
?y
(xy)=xg12″;
?
?y
(yg2′)=g2′+y
?
?y
g2
=g2′+yg21
?
?y
(x)+yg22
?
?y
(xy)
=g2′+xyg22
所以:
?2z
?x?y
=2
?
?y
f′+
?
?y
g1′+
?
?y
(yg2′)
=-2f″+xg12″+g2′+xyg22
?2z
?x?y
的值为:
-2f″+xg12″+g2′+xyg22

回答(2):