Partial Differentiation 3

To:laisai yu
∂u/∂x=2x, 但 ∂x/∂u不是1/(2x)

(x+y)^2=u+2xy=(vxy)^2, 則(vxy- 1/v)^2=u+1/v^2
vxy= 1/v +/- √(u+ 1/v^2) (負不合,因x,y>0)
故xy= [1+√(uv^2+1)]/v^2
x+y=[1+√(uv^2+1)]/v
(x-y)^2=u-4xy=u- 4√[1+(uv^2+1)]/v^2

(i) lnz=w+x+y
(1/z)(∂z/∂x)=(∂w/∂u)*(∂u/∂x)+(∂w/∂v)*(∂v/∂x)+1
=(∂w/∂u)*(2x)+(∂w/∂v)*(-1/x^2)+1 ----(1)
(1/z)(∂z/∂y)=(∂w/∂u)*(∂u/∂y)+(∂w/∂v)*(∂v/∂y)+1
=(∂w/∂u)*(2y)+(∂w/∂v)*(-1/y^2)+1 ----(2)
(ii)
y*(1)-x*(2)得
[y(∂z/∂x)-x(∂z/∂y)]/z=(∂w/∂u)*0-(∂w/∂v)*(y/x^2-x/y^2)+y-x
(y-z)=(∂w/∂v)*(x^3-y^3)/(x^2y^2)+y-x
則 (∂w/∂v)(x^3-y^3)=0
又x, y不相等, 故∂w/∂v=0