Question

微积分，多元函数的复合函数微分法问题，求解第(2)题，希望有讲解

Answer 1

(2) 记 u = xy^2, v = x+y, 则 z = f(u, v)
∂z/∂x = (∂f/∂u)(∂u/∂x) + (∂f/∂v)(∂v/∂x) = y^2∂f/∂u + ∂f/∂v
∂z/∂y = (∂f/∂u)(∂u/∂y) + (∂f/∂v)(∂v/∂y) = 2xy∂f/∂u + ∂f/∂v
∂^2z/∂x^2 = y^2[(∂^2f/∂u^2)(∂u/∂x) + (∂^2f/∂u∂v)(∂v/∂x)]
+ [(∂^2f/∂v∂u)(∂u/∂x) + (∂^2f/∂v^2)(∂v/∂x)]
= y^4∂^2f/∂u^2 + y^2∂^2f/∂u∂v + y^2∂^2f/∂v∂u + ∂^2f/∂v^2
= y^4∂^2f/∂u^2 + 2y^2∂^2f/∂u∂v + ∂^2f/∂v^2
∂^2z/∂x∂y = 2y∂f/∂u + y^2[(∂^2f/∂u^2)(∂u/∂y) + (∂^2f/∂u∂v)(∂v/∂y)]
+ [(∂^2f/∂v∂u)(∂u/∂y) + (∂^2f/∂v^2)(∂v/∂y)]
= 2y∂f/∂u + 2xy^3∂^2f/∂u^2+ y^2∂^2f/∂u∂v + 2xy∂^2f/∂v∂u + ∂^2f/∂v^2
= 2y∂f/∂u + 2xy^3∂^2f/∂u^2 + (y^2+2xy)∂^2f/∂u∂v + ∂^2f/∂v^2
∂^2z/∂y^2 = 2x∂f/∂u + 2xy[(∂^2f/∂u^2)(∂u/∂y) + (∂^2f/∂u∂v)(∂v/∂y)]
+ [(∂^2f/∂v∂u)(∂u/∂y) + (∂^2f/∂v^2)(∂v/∂y)]
= 2x∂f/∂u + 4x^2y^2∂^2f/∂u^2 + 2xy∂^2f/∂u∂v + 2xy∂^2f/∂v∂u + ∂^2f/∂v^2
= 2x∂f/∂u + 4x^2y^2∂^2f/∂u^2 + 4xy∂^2f/∂u∂v + ∂^2f/∂v^2