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二阶偏导数

约 1252 字大约 4 分钟

2025-11-21

二阶偏导数（Second-order Partial Derivatives） 是多元函数微分学中的重要概念。对于多元函数 z=f(x,y)z = f(x,y)，其一阶偏导数 ∂z∂x\frac{\partial z}{\partial x} 和 ∂z∂y\frac{\partial z}{\partial y} 仍然是 xx 和 yy 的函数，我们可以对这些偏导数再求偏导，得到二阶偏导数。

基本定义

对于函数 z=f(x,y)z = f(x,y)，其二阶偏导数包括：

  1. 关于 xx 的二阶偏导数：

    ∂2z∂x2=∂∂x(∂z∂x)\frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x}\left(\frac{\partial z}{\partial x}\right)

  2. 关于 yy 的二阶偏导数：

    ∂2z∂y2=∂∂y(∂z∂y)\frac{\partial^2 z}{\partial y^2} = \frac{\partial}{\partial y}\left(\frac{\partial z}{\partial y}\right)

  3. 混合偏导数：

    ∂2z∂x∂y=∂∂x(∂z∂y)\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial x}\left(\frac{\partial z}{\partial y}\right)

    ∂2z∂y∂x=∂∂y(∂z∂x)\frac{\partial^2 z}{\partial y \partial x} = \frac{\partial}{\partial y}\left(\frac{\partial z}{\partial x}\right)

混合偏导数的连续性定理

如果函数 f(x,y)f(x,y) 的二阶混合偏导数 ∂2f∂x∂y\frac{\partial^2 f}{\partial x \partial y} 和 ∂2f∂y∂x\frac{\partial^2 f}{\partial y \partial x} 在某区域内连续，则在该区域内：

∂2f∂x∂y=∂2f∂y∂x\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}

这个性质在实际计算中非常有用，可以大大简化计算过程。

复合函数的二阶偏导数

对于复合函数 z=f(u(x,y),v(x,y))z = f(u(x,y), v(x,y))，计算二阶偏导数需要使用链式法则。设 u=u(x,y)u = u(x,y)，v=v(x,y)v = v(x,y)，则：

∂z∂x=∂f∂u⋅∂u∂x+∂f∂v⋅∂v∂x=f1⋅ux+f2⋅vx\begin{aligned} \frac{\partial z}{\partial x} &= \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x} \\ &= f_1 \cdot u_x + f_2 \cdot v_x \end{aligned}

其中 f1=∂f∂uf_1 = \frac{\partial f}{\partial u}，f2=∂f∂vf_2 = \frac{\partial f}{\partial v}。

例题 1

设 z=f(x2−y2,xy)z = f(x^2 - y^2, xy)，且 f(u,v)f(u,v) 有连续的二阶偏导数，求 ∂2z∂x∂y\frac{\partial^2 z}{\partial x \partial y}。

解：

设 u=x2−y2u = x^2 - y^2，v=xyv = xy，则 z=f(u,v)z = f(u,v)。

第一步：求一阶偏导数

∂z∂x=∂f∂u⋅∂u∂x+∂f∂v⋅∂v∂x=f1⋅2x+f2⋅y=2xf1+yf2∂z∂y=∂f∂u⋅∂u∂y+∂f∂v⋅∂v∂y=f1⋅(−2y)+f2⋅x=−2yf1+xf2\begin{aligned} \frac{\partial z}{\partial x} &= \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x} \\ &= f_1 \cdot 2x + f_2 \cdot y = 2xf_1 + yf_2 \\ \\ \frac{\partial z}{\partial y} &= \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial y} \\ &= f_1 \cdot (-2y) + f_2 \cdot x = -2yf_1 + xf_2 \end{aligned}

第二步：求混合偏导数 ∂2z∂x∂y\frac{\partial^2 z}{\partial x \partial y}

对 ∂z∂x=2xf1+yf2\frac{\partial z}{\partial x} = 2xf_1 + yf_2 关于 yy 求偏导：

∂2z∂x∂y=∂∂y(2xf1+yf2)=2x⋅∂f1∂y+f2+y⋅∂f2∂y\begin{aligned} \frac{\partial^2 z}{\partial x \partial y} &= \frac{\partial}{\partial y}(2xf_1 + yf_2) \\ &= 2x \cdot \frac{\partial f_1}{\partial y} + f_2 + y \cdot \frac{\partial f_2}{\partial y} \end{aligned}

其中：

∂f1∂y=∂f1∂u⋅∂u∂y+∂f1∂v⋅∂v∂y=f11⋅(−2y)+f12⋅x=−2yf11+xf12∂f2∂y=∂f2∂u⋅∂u∂y+∂f2∂v⋅∂v∂y=f21⋅(−2y)+f22⋅x=−2yf21+xf22\begin{aligned} \frac{\partial f_1}{\partial y} &= \frac{\partial f_1}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial f_1}{\partial v} \cdot \frac{\partial v}{\partial y} \\ &= f_{11} \cdot (-2y) + f_{12} \cdot x = -2yf_{11} + xf_{12} \\ \\ \frac{\partial f_2}{\partial y} &= \frac{\partial f_2}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial f_2}{\partial v} \cdot \frac{\partial v}{\partial y} \\ &= f_{21} \cdot (-2y) + f_{22} \cdot x = -2yf_{21} + xf_{22} \end{aligned}

由于 ff 有连续的二阶偏导数，所以 f12=f21f_{12} = f_{21}。

第三步：合并结果

∂2z∂x∂y=2x(−2yf11+xf12)+f2+y(−2yf12+xf22)=−4xyf11+2x2f12+f2−2y2f12+xyf22=f2−4xyf11+2(x2−y2)f12+xyf22\begin{aligned} \frac{\partial^2 z}{\partial x \partial y} &= 2x(-2yf_{11} + xf_{12}) + f_2 + y(-2yf_{12} + xf_{22}) \\ &= -4xyf_{11} + 2x^2f_{12} + f_2 - 2y^2f_{12} + xyf_{22} \\ &= f_2 - 4xyf_{11} + 2(x^2 - y^2)f_{12} + xyf_{22} \end{aligned}

提示

选择填空小技巧：

∂2z∂x∂y=(∂z∂x)y+∂z∂x⋅∂z∂y\frac{\partial^{2}z}{\partial x\partial y}=(\frac{\partial z}{\partial x})_{y}+\frac{\partial z}{\partial x}\cdot\frac{\partial z}{\partial y}

I=(2xf1+yf2)y+(2xf1+yf2)⋅(−2yf1+xf2)=f2−4xyf11+2(x2−y2)f12+xyf22\begin{aligned} I &= (2xf_{1}+yf_{2})_{y}+(2xf_{1}+yf_{2})\cdot(-2yf_{1}+xf_{2}) \\ &= f_2-4xyf_{11}+2(x^2-y^2)f_{12}+xyf_{22} \end{aligned}

例题 2

设 z=f(yx,xy)z = f\left(\frac{y}{x}, \frac{x}{y}\right)，其中 ff 有连续的二阶偏导数，求 ∂2z∂x2\frac{\partial^2 z}{\partial x^2}。

解：

设 u=yxu = \frac{y}{x}，v=xyv = \frac{x}{y}，则 z=f(u,v)z = f(u,v)。

第一步：求一阶偏导数

∂z∂x=∂f∂u⋅∂u∂x+∂f∂v⋅∂v∂x=f1⋅(−yx2)+f2⋅1y=−yx2f1+1yf2\begin{aligned} \frac{\partial z}{\partial x} &= \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x} \\ &= f_1 \cdot \left(-\frac{y}{x^2}\right) + f_2 \cdot \frac{1}{y} \\ &= -\frac{y}{x^2}f_1 + \frac{1}{y}f_2 \end{aligned}

第二步：求二阶偏导数

∂2z∂x2=∂∂x(−yx2f1+1yf2)=2yx3f1−yx2⋅∂f1∂x+1y⋅∂f2∂x\begin{aligned} \frac{\partial^2 z}{\partial x^2} &= \frac{\partial}{\partial x}\left(-\frac{y}{x^2}f_1 + \frac{1}{y}f_2\right) \\ &= \frac{2y}{x^3}f_1 - \frac{y}{x^2} \cdot \frac{\partial f_1}{\partial x} + \frac{1}{y} \cdot \frac{\partial f_2}{\partial x} \end{aligned}

其中：

∂f1∂x=∂f1∂u⋅∂u∂x+∂f1∂v⋅∂v∂x=f11⋅(−yx2)+f12⋅1y=−yx2f11+1yf12∂f2∂x=∂f2∂u⋅∂u∂x+∂f2∂v⋅∂v∂x=f21⋅(−yx2)+f22⋅1y=−yx2f21+1yf22\begin{aligned} \frac{\partial f_1}{\partial x} &= \frac{\partial f_1}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial f_1}{\partial v} \cdot \frac{\partial v}{\partial x} \\ &= f_{11} \cdot \left(-\frac{y}{x^2}\right) + f_{12} \cdot \frac{1}{y} \\ &= -\frac{y}{x^2}f_{11} + \frac{1}{y}f_{12} \\ \\ \frac{\partial f_2}{\partial x} &= \frac{\partial f_2}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial f_2}{\partial v} \cdot \frac{\partial v}{\partial x} \\ &= f_{21} \cdot \left(-\frac{y}{x^2}\right) + f_{22} \cdot \frac{1}{y} \\ &= -\frac{y}{x^2}f_{21} + \frac{1}{y}f_{22} \end{aligned}

第三步：合并结果

∂2z∂x2=2yx3f1−yx2(−yx2f11+1yf12)+1y(−yx2f21+1yf22)=2yx3f1+y2x4f11−1x2f12−1x2f21+1y2f22=2yx3f1+y2x4f11−2x2f12+1y2f22\begin{aligned} \frac{\partial^2 z}{\partial x^2} &= \frac{2y}{x^3}f_1 - \frac{y}{x^2}\left(-\frac{y}{x^2}f_{11} + \frac{1}{y}f_{12}\right) + \frac{1}{y}\left(-\frac{y}{x^2}f_{21} + \frac{1}{y}f_{22}\right) \\ &= \frac{2y}{x^3}f_1 + \frac{y^2}{x^4}f_{11} - \frac{1}{x^2}f_{12} - \frac{1}{x^2}f_{21} + \frac{1}{y^2}f_{22} \\ &= \frac{2y}{x^3}f_1 + \frac{y^2}{x^4}f_{11} - \frac{2}{x^2}f_{12} + \frac{1}{y^2}f_{22} \end{aligned}

（这里使用了 f12=f21f_{12} = f_{21}）