设X^2+Y^2-XY=1,求dy/dx
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![设X^2+Y^2-XY=1,求dy/dx](/uploads/image/z/2651282-26-2.jpg?t=%E8%AE%BEX%5E2%2BY%5E2-XY%3D1%2C%E6%B1%82dy%2Fdx)
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设X^2+Y^2-XY=1,求dy/dx
设X^2+Y^2-XY=1,求dy/dx
设X^2+Y^2-XY=1,求dy/dx
x^2+y^2-xy=1
求导得:
2x+2yy'-2y-2xy'=0
y'(2y-2x)=2y-2x
y'=1
d(x^2+y^2-xy)=0
2xdx+2ydy-xdy-ydx=0
(2x-y)dx+(2y-x)dy=0
dy/dx=(2x-y)/(x-2y)
2x+2yy'-2y-2xy'=0 y'(2y-2x)=2y-2x y'=1 八字d(x^2+y^2-xy)=0 2xdx+2ydy-xdy-ydx=0 (2x-y)dx+(2y-x)dy=0 dy/dx=(
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