f(x+y,xy)=x^2+y^2求fxy(1,

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f(x+y,xy)=x^2+y^2求fxy(1,f(x+y,xy)=x^2+y^2求fxy(1,f(x+y,xy)=x^2+y^2求fxy(1,因为f(x+y,xy)=x^2+y^2=(x+y)^2-

f(x+y,xy)=x^2+y^2求fxy(1,
f(x+y,xy)=x^2+y^2
求fxy(1,

f(x+y,xy)=x^2+y^2求fxy(1,
因为f(x+y,xy)=x^2+y^2=(x+y)^2-2xy
所以f(x,y)=x^2-2y
现对x求导得到:
fx(x,y)=2x
再对y求导得到:
fxy(x,y)=0.
所以无论x,y为何值,fxy(x,y)=0.即fxy(1,1)=0.

因为f(x+y,xy)=x^2+y^2=(x+y)^2-2xy
所以f(x,y)=x^2-2y
所以fxy=0
fxy(1,1)=0