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多变量微积分

课程: 多变量微积分 > 单元 2

课程 1: 偏导数和梯度 (文章)
数学
多变量微积分
多元函数示例
偏导数和梯度 (文章)
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二阶偏导数测试

Google课堂
一个对二阶偏导数的简短回顾,  混合偏导数的对称, 和高阶偏导数.

背景：

概括二阶导数

思考一个有着二维输入值的函数，类似
f, left parenthesis, x, comma, y, right parenthesis, equals, x, squared, y, cubed.
它的偏导数 start fraction, \partial, f, divided by, \partial, x, end fraction 和 start fraction, \partial, f, divided by, \partial, y, end fraction 在同一二位输入值 left parenthesis, x, comma, y, right parenthesis 上推导：
∂f∂x=∂∂x(x2y3)=2xy3∂f∂y=∂∂y(x2y3)=3x2y2\begin{aligned} &\dfrac{\partial f}{\partial \blueE{x}} = \dfrac{\partial}{\partial \blueE{x}}(\blueE{x}^2 y^3) = 2\blueE{x} y^3 \\\\ &\dfrac{\partial f}{\partial \redE{y}} = \dfrac{\partial}{\partial \redE{y}}(x^2 \redE{y}^3) = 3x^2 \redE{y}^2 \end{aligned}
所以，我们也可以求偏导数的偏导数。
这些被称为二次偏导数，在单变量微积分中与一般二次导数表达方式 start fraction, d, squared, f, divided by, d, x, squared, end fraction 类似。
∂∂x(∂f∂x)=∂2f∂x2∂∂x(∂f∂y)=∂2f∂x∂y∂∂y(∂f∂x)=∂2f∂y∂x∂∂y(∂f∂y)=∂2f∂y2\begin{aligned} \dfrac{\partial}{\partial \blueE{x}} \left(\dfrac{\partial f}{\partial \blueE{x}} \right) &= \dfrac{\partial^2 f}{\partial \blueE{x}^2} \\\\ \dfrac{\partial}{\partial \blueE{x}} \left(\dfrac{\partial f}{\partial \redE{y}} \right) &= \dfrac{\partial^2 f}{\partial \blueE{x} \partial \redE{y}} \\\\ \dfrac{\partial}{\partial \redE{y}} \left(\dfrac{\partial f}{\partial \blueE{x}} \right) &= \dfrac{\partial^2 f}{\partial \redE{y} \partial \blueE{x}} \\\\ \dfrac{\partial}{\partial \redE{y}} \left(\dfrac{\partial f}{\partial \redE{y}} \right) &= \dfrac{\partial^2 f}{\partial \redE{y}^2} \end{aligned}
使用 f, start subscript, x, end subscript 来表达偏导数（在这里是关于x），你可能也会看到写成这样的二次偏导数：
(fx)x=fxx(fy)x=fyx(fx)y=fxy(fy)y=fyy\begin{aligned} (f_\blueE{x})_\blueE{x} &= f_{\blueE{x}\blueE{x}} \\\\ (f_\redE{y})_\blueE{x} &= f_{\redE{y}\blueE{x}} \\\\ (f_\blueE{x})_\redE{y} &= f_{\blueE{x}\redE{y}} \\\\ (f_\redE{y})_\redE{y} &= f_{\redE{y}\redE{y}} \end{aligned}
牵涉到多个不同的输入变量的二次偏导数，例如 f, start subscript, start color #bc2612, y, end color #bc2612, start color #0c7f99, x, end color #0c7f99, end subscript and f, start subscript, start color #0c7f99, x, end color #0c7f99, start color #bc2612, y, end color #bc2612, end subscript, 叫做”混合偏导数“。

例 1: 树

问题：求出f, left parenthesis, x, comma, y, right parenthesis, equals, sine, left parenthesis, x, right parenthesis, y, squared 的所有二次偏导数。
解：首先，求出两个偏导数：
∂∂x(sin⁡(x)y2)=cos⁡(x)y2∂∂y(sin⁡(x)y2)=2sin⁡(x)y\begin{aligned} \dfrac{\partial}{\partial \blueE{x}} (\sin(\blueE{x})y^2) &= \cos(\blueE{x})y^2 \\ \\ \dfrac{\partial}{\partial \redE{y}} (\sin(x)\redE{y}^2) &= 2\sin(x)\redE{y} \end{aligned}
然后将两个偏导数各自写出来：
∂∂x(∂∂x(sin⁡(x)y2))=∂∂x(cos⁡(x)y2)=−sin⁡(x)y2∂∂x(∂∂y(sin⁡(x)y2))=∂∂x(2sin⁡(x)y)=2cos⁡(x)y∂∂y(∂∂x(sin⁡(x)y2))=∂∂y(cos⁡(x)y2)=2cos⁡(x)y∂∂y(∂∂y(sin⁡(x)y2))=∂∂y(2sin⁡(x)y)=2sin⁡(x)\begin{aligned} \dfrac{\partial}{\partial \blueE{x}}\left( \dfrac{\partial}{\partial x}(\sin(x)y^2) \right) &= \dfrac{\partial}{\partial \blueE{x}}(\cos(\blueE{x})y^2) = -\sin(\blueE{x})y^2 \\\\ \dfrac{\partial}{\partial x}\left( \dfrac{\partial}{\partial y}(\sin(x)y^2) \right) &= \dfrac{\partial}{\partial \blueE{x}}(2\sin(\blueE{x})y) = 2\cos(\blueE{x})y \\\\ \dfrac{\partial}{\partial \redE{y}}\left( \dfrac{\partial}{\partial x}(\sin(x)y^2) \right) &= \dfrac{\partial}{\partial \redE{y}}(\cos(x)\redE{y}^2) = 2\cos(x)\redE{y} \\\\ \dfrac{\partial}{\partial \redE{y}}\left( \dfrac{\partial}{\partial y}(\sin(x)y^2) \right) &= \dfrac{\partial}{\partial \redE{y}}(2\sin(x)\redE{y}) = 2\sin(x) \end{aligned}
sin⁡(x)y2∂∂x↙↘∂∂ycos⁡(x)y22sin⁡(x)y∂∂x↙↘∂∂y∂∂x↙↘∂∂ysin⁡(x)y22cos⁡(x)y2cos⁡(x)y⏟混合偏导数相等！2sin⁡(x)\begin{array}{ccccccc} &\large\sin(x)y^2 \\\\ &\small\dfrac{\partial}{\partial x}\large\swarrow\quad\searrow\small\dfrac{\partial}{\partial y} \\\\ &\large\cos(x)y^2\qquad\qquad\qquad 2\sin(x)y \\\\ \small\dfrac{\partial}{\partial x}\large\swarrow&\large\searrow\small\dfrac{\partial}{\partial y}\qquad\qquad\qquad\dfrac{\partial}{\partial x}\large\swarrow&\large\searrow\small\dfrac{\partial}{\partial y} \\\\ \large\sin(x)y^2&\large\maroonD{\underbrace{\blueE{2\cos(x)y\qquad2\cos(x)y}}_{\text{混合偏导数相等！}}}&\large2\sin(x) \end{array}

二阶导数的对称

注意，在以上的例题中，两个混合偏导数 start fraction, \partial, squared, f, divided by, \partial, x, \partial, y, end fraction 和 start fraction, \partial, squared, f, divided by, \partial, y, \partial, x, end fraction 是一样的。这不是巧合；这几乎在你遇到的所有练习题里都会出现。例如，看看在一般多项式 x, start superscript, n, end superscript, y, start superscript, k, end superscript 上会发生什么：
∂∂x(∂∂y(xnyk))=∂∂x(kxnyk−1)=nkxn−1yk−1∂∂y(∂∂x(xnyk))=∂∂y(nxn−1yk)=nkxn−1yk−1\begin{aligned} \dfrac{\partial}{\partial \blueE{x}}\left( \dfrac{\partial}{\partial \redD{y}}(\blueE{x}^n \redD{y}^k) \right) &= \dfrac{\partial}{\partial \blueE{x}}(k \blueE{x}^n \redD{y}^{k-1}) = nk \blueE{x}^{n-1} \redD{y}^{k-1} \\\\ \dfrac{\partial}{\partial \redD{y}}\left( \dfrac{\partial}{\partial \blueE{x}}(\blueE{x}^n \redD{y}^k) \right) &= \dfrac{\partial}{\partial \redD{y}}(n \blueE{x}^{n-1} \redD{y}^k) = nk \blueE{x}^{n-1} \redD{y}^{k-1} \end{aligned}
通常来说，二次导数的对称不是一直一样的。根据施瓦兹定理或克莱罗定理，如果二次偏导数在一个点周围是连续的，二次导数的对称性会一直在这个点上。为了真正了解这个精华，我们需要一些实际分析。
你应该时刻记住有例外的存在，但是二次导数的对称性对于所有你将看到的”一般“函数都有用。

示例 2：更高阶的导数

为什么在二次偏导数就停下了？我们应该求五个关于不同输入值变量的偏导数。
问题: 如果 f, left parenthesis, x, comma, y, comma, z, right parenthesis, equals, sine, left parenthesis, x, y, right parenthesis, e, start superscript, x, plus, z, end superscript, f, start subscript, z, y, z, y, x, end subscript 是多少？
解: f, start subscript, z, y, z, y, x, end subscript 的表达方式是 left parenthesis, left parenthesis, left parenthesis, left parenthesis, f, start subscript, z, end subscript, right parenthesis, start subscript, y, end subscript, right parenthesis, start subscript, z, end subscript, right parenthesis, start subscript, y, end subscript, right parenthesis, start subscript, x, end subscript 的简写，所以我们关于 z 求导，然后关于 y, 然后 z, 然后 y, 然后 x. 简洁的说，我们从左到右读。
值得一提的是在其他表达方式中有不同的阶。
∂∂x∂∂y∂∂z∂∂y∂f∂z=∂5f∂x⏟5th∂y⏟4th∂z⏟3rd∂y⏟2nd∂z⏟1st\begin{aligned} \quad \dfrac{\partial}{\partial x} \dfrac{\partial}{\partial y} \dfrac{\partial}{\partial z} \dfrac{\partial}{\partial y} \dfrac{\partial f}{\partial z} = \dfrac{\partial^5 f}{ \underbrace{\partial x}_{5^{th}} \underbrace{\partial y}_{4^{th}} \underbrace{\partial z}_{3^{rd}} \underbrace{\partial y}_{2^{nd}} \underbrace{\partial z}_{1^{st}} } \end{aligned}
所以分母从左到右的顺序决定了求导的阶。
无论如何，回到我们的问题。这是你应该挽起袖子前进的任务之一，但是为了方便，让我们用有颜色的 start color #11accd, x, end color #11accd, comma, start color #e84d39, y, end color #e84d39, comma, start color #0d923f, z, end color #0d923f 来追踪它们在哪里。
f(x,y,z)=sin⁡(xy)ex+zfz(x,y,z)=∂f∂z(sin⁡(xy)ex+z)=sin⁡(xy)ex+zfzy(x,y,z)=∂f∂y(sin⁡(xy)ex+z)=cos⁡(xy)xex+zfzyz(x,y,z)=∂f∂z(cos⁡(xy)xex+z)=cos⁡(xy)xex+zfzyzy(x,y,z)=∂f∂y(cos⁡(xy)xex+z)=−sin⁡(xy)x2ex+zfzyzyx(x,y,z)=∂f∂x(−sin⁡(xy)x2ex+z)=−cos⁡(xy)y⏟∂∂x(−sin⁡(xy))x2ex+z=−sin⁡(xy)2x⏟∂∂xx2ex+z=−sin⁡(xy)x2ex+z⏟∂∂xex+z\begin{aligned} \quad f(\blueD{x}, \redD{y}, \greenE{z}) &= \sin(\blueD{x}\redD{y})e^{\blueD{x} + \greenE{z}} \\\\ f_{\greenE{z}}(\blueD{x}, \redD{y}, \greenE{z}) &= \dfrac{\partial f}{\partial \greenE{z}} \left( \sin(\blueD{x}\redD{y})e^{\blueD{x} + \greenE{z}} \right) \\\\ &= \sin(\blueD{x}\redD{y})e^{\blueD{x} + \greenE{z}} \\\\ f_{\greenE{z}\redD{y}}(\blueD{x}, \redD{y}, \greenE{z}) &= \dfrac{\partial f}{\partial \redD{y}} \left( \sin(\blueD{x}\redD{y})e^{\blueD{x} + \greenE{z}} \right) \\\\ &= \cos(\blueD{x}\redD{y})\blueD{x}e^{\blueD{x} + \greenE{z}} \\\\ f_{\greenE{z}\redD{y}\greenE{z}}(\blueD{x}, \redD{y}, \greenE{z}) &= \dfrac{\partial f}{\partial \greenE{z}} \left( \cos(\blueD{x}\redD{y})\blueD{x}e^{\blueD{x} + \greenE{z}} \right) \\\\ &= \cos(\blueD{x}\redD{y})\blueD{x}e^{\blueD{x} + \greenE{z}} \\\\ f_{\greenE{z}\redD{y}\greenE{z}\redD{y}}(\blueD{x}, \redD{y}, \greenE{z}) &= \dfrac{\partial f}{\partial \redD{y}} \left( \cos(\blueD{x}\redD{y})\blueD{x}e^{\blueD{x} + \greenE{z}} \right) \\\\ &= -\sin(\blueD{x}\redD{y})\blueD{x}^2 e^{\blueD{x} + \greenE{z}} \\\\ f_{\greenE{z}\redD{y}\greenE{z}\redD{y}\blueD{x}}(\blueD{x}, \redD{y}, \greenE{z}) &= \dfrac{\partial f}{\partial \blueD{x}} \left( -\sin(\blueD{x}\redD{y})\blueD{x}^2 e^{\blueD{x} + \greenE{z}} \right) \\\\ &= \underbrace{-\cos(\blueD{x}\redD{y})\redD{y}}_{ \small\dfrac{\partial}{\partial x} (-\sin(\blueD{x}\redD{y})) }\blueD{x}^2 e^{\blueD{x} + \greenE{z}} \\\\ &\phantom{=}-\sin(\blueD{x}\redD{y}) \underbrace{2\blueD{x}}_{ \small\dfrac{\partial}{\partial x}\blueD{x}^2 }e^{\blueD{x} + \greenE{z}} \\\\ &\phantom{=}-\sin(\blueD{x}\redD{y})\blueD{x}^2 \underbrace{e^{\blueD{x} + \greenE{z}}}_{ \small\dfrac{\partial}{\partial x} e^{\blueD{x} + \greenE{z}} } \end{aligned}
最后一步是使用延申乘积法则。
=ddx(f(x)g(x)h(x))=f′(x)g(x)h(x)+f(x)g′(x)h(x)+f(x)g(x)h′(x)\begin{aligned} &\phantom{=} \dfrac{d}{dx}\Big(f(x)g(x)h(x)\Big) \\\\ &= f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x) \end{aligned}
天呐！这真是一个冗长乏味的例题。但是如果你能坚持跟下来，求多个偏导数对你来说将不成问题。它是比任何别的更需要记忆的东西之一。

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