设函数z=x+y/x-y 则dz=?
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设函数z=x+y/x-y 则dz=?
![设函数z=x+y/x-y 则dz=?](/uploads/image/z/7628733-45-3.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0z%3Dx%2By%2Fx-y+%E5%88%99dz%3D%3F)
f(x)=z=x+y/x-y
dz=fxdx+fydy=[[(x-y)-(x+y)]/(x-y)^2 ]dx+[[(x-y)+(x+y)]/(x-y)^2]dy
=-2y/(x-y)^2 dx+2x/(x-y)^2 dy
=2/(x-y)^2 (xdy-ydx)
dz=fxdx+fydy=[[(x-y)-(x+y)]/(x-y)^2 ]dx+[[(x-y)+(x+y)]/(x-y)^2]dy
=-2y/(x-y)^2 dx+2x/(x-y)^2 dy
=2/(x-y)^2 (xdy-ydx)