Question

设z=f（xy，x+y），且f有连续的二阶偏导数，求a^2z/axay

Answer 1

设u = xy，v = x + y
z = f(u，v)
∂z/∂x = ∂f/∂u · ∂u/∂x + ∂f/∂v · ∂v/∂x
= f₁ · ∂(xy)/∂x + f₂ · ∂(x + y)/∂x
= yf₁ + f₂

∂²z/∂x∂y = ∂/∂y (∂z/∂x) = ∂/∂y (yf₁ + f₂)
= ∂(yf₁)/∂y + ∂f₂/∂y
= f₁ · ∂y/∂y + y · ∂f₁/∂y + ∂f₂/∂y
= f₁ + y · (∂f₁/∂u · ∂u/∂y + ∂f₁/∂v · ∂v/∂y) + (∂f₂/∂u · ∂u/∂y + ∂f₂/∂v · ∂v/∂y)
= f₁ + y · [f₁₁ · ∂(xy)/∂y + f₁₂ · ∂(x + y)/∂y] + [f₂₁ · ∂(xy)/∂y + f₂₂ · ∂(x + y)/∂y]
= f₁ + y(xf₁₁ + f₁₂) + (xf₂₁ + f₂₂)
= f₁ + xyf₁₁ + yf₁₂ + xf₂₁ + f₂₂
记：
f₁ = ∂f/∂u，f₂ = ∂f/∂v
f₁₁ = ∂²f/∂u²，f₁₂ = ∂²f/∂u∂v
f₂₁ = ∂²f/∂v∂u，f₂₂ = ∂²f/∂v²

Answer 2

令u=xy，v=x+y
z=f(u,v)
az/ax=y(fu)+(fv)
a^2z/axay
=a(az/ax)/ay
=a(y(fu)+(fv))/ay
=(fu)+y(a(fu)/ay)+a(fv)/ay
=(fu)+y(a(fu)/au*au/ay+(a(fu)/av)*(av/ay))+((a(fv)/au)*(au/ay)+(a(fv)/av)*(av/ay))
=(fu)+(xy)*(fuu)+(x+y)*(fuv)+(fuv)