Question

微积分求解

Answer 1

f(x+y,x-y) = x^2.y +y^2
u=x+y (1)
v= x-y (2)
(1) +(2)
x= (1/2)(u+v)
(1)-(2)
y= (1/2)(u-v)

f(x+y,x-y) = x^2.y +y^2
f(u,v) = [(1/2)(u+v)]^2 . (1/2)(u-v) + [(1/2)(u-v)]^2
= (1/8) (u+v)^2.(u-v) + (1/4)(u-v)^2
= (1/8)(u-v) [ (u+v)^2 - 2(u-v) ]
f(x,y) =(1/8)(x-y) [ (x+y)^2 - 2(x-y) ]

追问

请问划线的地方不是加吗

追答

对
f(u,v) = [(1/2)(u+v)]^2 . (1/2)(u-v) + [(1/2)(u-v)]^2
= (1/8) (u+v)^2.(u-v) + (1/4)(u-v)^2
= (1/8)(u-v) [ (u+v)^2 + 2(u-v) ]
f(x,y) =(1/8)(x-y) [ (x+y)^2 + 2(x-y) ]

追问

谢谢