Given function FX satisfies f (x + y) = f (x) + F (y) for any XY ∈ R Find the value of 1 f (0) 2 F (x) is an odd function

Given function FX satisfies f (x + y) = f (x) + F (y) for any XY ∈ R Find the value of 1 f (0) 2 F (x) is an odd function

(1) Let x = 0, y = 0, then f (0 + 0) = f (0) + F (0)
f(0)=2f(0),f(0)=0
(2) Let x = - y have f (x + y) = f (x) + F (y), that is, f (0) = f (x) + F (- x)
And f (0) = 0, so f (x) + F (- x) = 0, that is, f (x) is an odd function

If f (XY, X-Y) = x ^ 2 + y ^ 2, then FX (x, y) + FY (x, y)=

f(xy,x-y)=x^2+y^2=(x-y)^2+2xy
therefore
f(x,y)=2x+y^2
fx=2
fy=2y

Given z = e ^ XY, find the total differential when x = 1, y = 1, △ x = 0.15, △ y = 0.1

dz=d(e^xy)=e^ydx+e^xdy
Take it in and get e / 4
Note: DX and Dy can be replaced by △ X and △ y