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- info33
- ベストアンサー率50% (260/513)
回答No.2
z=f(x,y)=(x+y)/(x-y)= 1+2y/(x-y) = 2x/(x-y) -1 fx= -2y/(x-y)^2, fy=2x/(x-y)^2, fxy= -2/(x-y)^2 -4y/(x-y)^3= -2(x+y)/(x-y)^3, fyx=2/(x-y)^2 -4x/(x-y)^3= -2(x+y)/(x-y)^3 fxy=fyx (証明終)
- 178-tall
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回答No.1
f = f(x,y) = (x-y)/(x+y) について、 fxy = fyx が成立すること、かな? fx = 2y/(x+y)^2 fy = -2x/(x+y)^2 ↓ fxy = { 2(x+y)^2 - 4y(x+y) }/(x+y)^4 = 2(x-y)/(x+y)^3 fyx = {-2(x+y)^2 + 4y(x+y) }/(x+y)^4 = 2(x-y)/(x+y)^3