二阶及更高阶偏导数

本文提供了一阶、二阶及更高阶偏导数的计算方法，并附有例子及其解答，包括详细的计算步骤。还通过例子验证了在一定连续性条件下混合偏导数相等的克莱劳定理。
二阶及更高阶偏导数本质上是对多元函数关于一个或多个变量多次求导所得的导数。

一阶偏导数

给定函数 \(f(x, y)\)，其一阶偏导数为：
相对于 \(x\) 的偏导数：\(\dfrac{\partial f}{\partial x}\)
相对于 \(y\) 的偏导数：\(\dfrac{\partial f}{\partial y}\)

二阶偏导数

在得到一阶偏导数后，可以再次求导得到二阶偏导数，具体如下：
相对于 \(x\) 的二阶导数：\(\dfrac{\partial^2 f}{\partial x^2}\)
相对于 \(y\) 的二阶导数：\(\dfrac{\partial^2 f}{\partial y^2}\)
先对 \(x\) 后对 \(y\) 的导数：\(\dfrac{\partial^2 f}{\partial y \partial x}\)
先对 \(y\) 后对 \(x\) 的导数：\(\dfrac{\partial^2 f}{\partial x \partial y}\)

更高阶偏导数

同样地，可以继续求导得到更高阶的偏导数。例如，函数 \(f\) 对 \(x\) 两次后对 \(y\) 一次的三阶偏导数表示为 \(\dfrac{\partial^3 f}{\partial y \partial x^2}\)。

例子及其解答

例1

计算函数 \[f(x, y) = x^2y + 3xy^2\] 的一阶和二阶偏导数。

例1解答与详细步骤

1. 一阶偏导数
a. 对 \(x\) 求导： \[ \dfrac{\partial f}{\partial x} = \dfrac{\partial}{\partial x}(x^2y) + \dfrac{\partial}{\partial x}(3xy^2) \\\\ = 2xy + 3y^2 \] b. 对 \(y\) 求导： \[ \dfrac{\partial f}{\partial y} = \dfrac{\partial}{\partial y}(x^2y) + \dfrac{\partial}{\partial y}(3xy^2) \\\\ = x^2 + 6xy \] 2. 二阶偏导数
a. 对 \(\dfrac{\partial f}{\partial x}\) 再对 \(x\) 求导： \[ \dfrac{\partial^2 f}{\partial x^2} = \dfrac{\partial}{\partial x} \left(\dfrac{\partial f}{\partial x} \right) \\\\ =\dfrac{\partial}{\partial x}(2xy + 3y^2) = 2y \] b. 对 \(\dfrac{\partial f}{\partial y}\) 再对 \(y\) 求导： \[ \dfrac{\partial^2 f}{\partial y^2} = \dfrac{\partial}{\partial y} \left(\dfrac{\partial f}{\partial y} \right) \\\\ = \dfrac{\partial}{\partial y}(x^2 + 6xy) = 6x \] c. 对 \(\dfrac{\partial f}{\partial x}\) 对 \(y\) 求导： \[ \dfrac{\partial^2 f}{\partial x \partial y} = \dfrac{\partial }{\partial y} \left(\dfrac{\partial f}{\partial x}\right) \\\\= \dfrac{\partial}{\partial y}(2xy + 3y^2) = 2x + 6y \] d. 对 \(\dfrac{\partial f}{\partial y}\) 对 \(x\) 求导（由于对称性，如果函数的混合偏导数是连续的，结果应与步骤3相同）： \[ \dfrac{\partial^2 f}{\partial y \partial x} = \dfrac{\partial }{\partial x} \left(\dfrac{\partial f}{\partial y}\right) \\\\ =\dfrac{\partial}{\partial x}(x^2 + 6xy) = 2x + 6y \] 注意 \(\dfrac{\partial^2 f}{\partial y \partial x} = \dfrac{\partial^2 f}{\partial x \partial y}\)，这说明了在一定连续性条件下克莱劳定理关于混合偏导数相等的结论。

例2

计算函数 \[g(x, y) = e^{xy} + \sin(x)y^2\] 的一阶和二阶偏导数。

例2解答与详细步骤

1. 一阶偏导数
a. 对 \(x\) 求导： \[ \dfrac{\partial g}{\partial x} = \dfrac{\partial}{\partial x}(e^{xy}) + \dfrac{\partial}{\partial x}(\sin(x)y^2) \\\\= ye^{xy} + \cos(x)y^2 \] b. 对 \(y\) 求导： \[ \dfrac{\partial g}{\partial y} = \dfrac{\partial}{\partial y}(e^{xy}) + \dfrac{\partial}{\partial y}(\sin(x)y^2) \\\\= xe^{xy} + 2y\sin(x) \] 2. 二阶偏导数
a. 对 \(\dfrac{\partial g}{\partial x}\) 再对 \(x\) 求导： \[ \dfrac{\partial^2 g}{\partial x^2} = \dfrac{\partial}{\partial x} \left(\dfrac{\partial g}{\partial x}\right) \\\\=\dfrac{\partial}{\partial x}(ye^{xy} + \cos(x)y^2) \\\\= y^2e^{xy} - y^2\sin(x) \] 此步骤涉及对每项分别应用微分的乘积法则。
b. 对 \(\dfrac{\partial g}{\partial y}\) 再对 \(y\) 求导： \[ \dfrac{\partial^2 g}{\partial y^2} = \dfrac{\partial}{\partial y} \left(\dfrac{\partial g}{\partial y}\right) \\\\ = \dfrac{\partial}{\partial y}(xe^{xy} + 2y\sin(x)) \\\\= x^2e^{xy} + 2\sin(x) \] 同样，应用乘积法则，此次重点关注指数项和正弦项如何随 \(y\) 的变化而变化。
c. 对 \(\dfrac{\partial g}{\partial x}\) 对 \(y\) 求导： \[ \dfrac{\partial^2 g}{\partial x \partial y} = \dfrac{\partial }{\partial y} \left(\dfrac{\partial g}{\partial x}\right) \\\\ = \dfrac{\partial}{\partial y}(ye^{xy} + \cos(x)y^2) \\\\= e^{xy} + xye^{xy} + 2y\cos(x) \] d. 对 \(\dfrac{\partial g}{\partial y}\) 对 \(x\) 求导（由于混合偏导数的连续性和对称性，结果与步骤3相同）： \[ \dfrac{\partial^2 g}{\partial y \partial x} = \dfrac{\partial }{\partial x} \left(\dfrac{\partial g}{\partial y}\right) \\\\ = \dfrac{\partial}{\partial x}(xe^{xy} + 2y\sin(x)) \\\\= e^{xy} + xye^{xy} + 2y\cos(x) \] \(\dfrac{\partial^2 g}{\partial y \partial x} = \dfrac{\partial^2 g}{\partial x \partial y}\) 的等式说明了在一定连续性条件下克莱劳定理关于混合偏导数相等的结论。

例3

计算函数 \( f(x, y) = x^3y^2 + x^2e^y \) 的一阶、二阶及两个三阶偏导数 \( \dfrac{\partial^3 f}{\partial y \partial x^2} \) 和 \( \dfrac{\partial^3 f}{\partial x \partial y \partial x} \)。

例3解答与详细步骤

1. 一阶偏导数
a. 对 \(x\) 求导： \[ \dfrac{\partial f}{\partial x} = \dfrac{\partial}{\partial x}(x^3y^2) + \dfrac{\partial}{\partial x}(x^2e^y) \\\\= 3x^2y^2 + 2xe^y \] b. 对 \(y\) 求导： \[ \dfrac{\partial f}{\partial y} = \dfrac{\partial}{\partial y}(x^3y^2) + \dfrac{\partial}{\partial y}(x^2e^y) \\\\= 2x^3y + x^2e^y \] 2. 二阶偏导数
a. 对 \(x\) 两次求导： \[ \dfrac{\partial^2 f}{\partial x^2} = \dfrac{\partial}{\partial x} \left(\dfrac{\partial f}{\partial x}\right) \\\\ = \dfrac{\partial}{\partial x}(3x^2y^2 + 2xe^y) \\\\= 6xy^2 + 2e^y \] b. 对 \(y\) 两次求导： \[ \dfrac{\partial^2 f}{\partial y^2} = \dfrac{\partial}{\partial y} \left(\dfrac{\partial f}{\partial y}\right) \\\\ = \dfrac{\partial}{\partial y}(2x^3y + x^2e^y) \\\\= 2x^3 + x^2e^y \] c. 先对 \(x\) 后对 \(y\) 求导： \[ \dfrac{\partial^2 f}{\partial y \partial x} = \dfrac{\partial}{\partial y}\left(\dfrac{\partial f}{\partial x}\right) \\\\ = \dfrac{\partial}{\partial y}(3x^2y^2 + 2xe^y) \\\\ = 6x^2y + 2xe^y \] d. 先对 \(y\) 后对 \(x\) 求导： \[ \dfrac{\partial^2 f}{\partial x \partial y} = \dfrac{\partial}{\partial x}\left(\dfrac{\partial f}{\partial y}\right)\\\\ = \dfrac{\partial}{\partial x}(2x^3y + x^2e^y) \\\\= 6x^2y + 2xe^y \] 在此例中，我们同样有 \(\dfrac{\partial^2 f}{\partial y \partial x} = \dfrac{\partial^2 f}{\partial x \partial y}\) 的等式，这说明了在一定连续性条件下克莱劳定理关于混合偏导数相等的结论。
3. 三阶偏导数
a. 先对 \(x\) 两次，再对 \(y\) 求导： \[ \dfrac{\partial^3 f}{\partial y \partial x^2} = \dfrac{\partial}{\partial y} \left(\dfrac{\partial^2 f}{\partial x^2} \right) \\\\ = \dfrac{\partial}{\partial y}(6xy^2 + 2e^y) = 12 x y + 2 e^y \] b. 先对 \(x\)，再对 \(y\)，最后对 \(x\) 求导： \[ \dfrac{\partial^3 f}{\partial x \partial y \partial x} = \dfrac{\partial}{\partial x} \left(\dfrac{\partial^2 f}{\partial x \partial y} \right) \\\\ = \dfrac{\partial}{\partial x}(6x^2y + 2xe^y) = 12xy + 2e^y \]

克莱劳定理

克莱劳定理，也称为混合偏导数相等定理，指出如果一个函数 \( f(x, y) \) 在包含点 \( (a, b) \) 的开区域内具有连续的二阶偏导数，那么在该点处混合偏导数的次序不会影响结果。换句话说： \[ \dfrac{{\partial^2 f}}{{\partial x \partial y}} = \dfrac{{\partial^2 f}}{{\partial y \partial x}} \] 以下是一个简单的例子，说明克莱劳定理：
考虑函数 \( f(x, y) = x^2y + y^3 \)。
首先，我们求混合偏导数：
1. 求 \( \dfrac{{\partial f}}{{\partial x}} \)： \[ \dfrac{{\partial f}}{{\partial x}} = 2xy \] 2. 求 \( \dfrac{{\partial f}}{{\partial y}} \)： \[ \dfrac{{\partial f}}{{\partial y}} = x^2 + 3y^2 \] 现在，我们求混合偏导数：
1. \( \dfrac{{\partial^2 f}}{{\partial x \partial y}} \)： \[ \dfrac{{\partial^2 f}}{{\partial x \partial y}} = \dfrac{{\partial}}{{\partial x}} \left( \dfrac{{\partial f}}{{\partial y}} \right) = \dfrac{{\partial}}{{\partial x}} (x^2 + 3y^2) = 2x \] 2. \( \dfrac{{\partial^2 f}}{{\partial y \partial x}} \)： \[ \dfrac{{\partial^2 f}}{{\partial y \partial x}} = \dfrac{{\partial}}{{\partial y}} \left( \dfrac{{\partial f}}{{\partial x}} \right) = \dfrac{{\partial}}{{\partial y}} (2xy) = 2x \] 根据克莱劳定理，由于混合偏导数是连续且相等的，我们有 \( \dfrac{{\partial^2 f}}{{\partial x \partial y}} = \dfrac{{\partial^2 f}}{{\partial y \partial x}} \)。