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