f(xy,x+y)=(x+y)²-xy,求af(x,y)/ax=?

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f(xy,x+y)=(x+y)²-xy,求af(x,y)/ax=?f(xy,x+y)=(x+y)²-xy,求af(x,y)/ax=?f(xy,x+y)=(x+y)²-xy

f(xy,x+y)=(x+y)²-xy,求af(x,y)/ax=?
f(xy,x+y)=(x+y)²-xy,求af(x,y)/ax=?

f(xy,x+y)=(x+y)²-xy,求af(x,y)/ax=?
f(xy,x+y)=(x+y)²-xy
所以
f(x,y)=y²-x
所以
af(x,y)/ax=-1