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# 二阶偏导数测试

## 概括二阶导数

f, left parenthesis, x, comma, y, right parenthesis, equals, x, squared, y, cubed.

\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}

\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}

\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}

## 例 1: 树

：首先，求出两个偏导数：
\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}

\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}
$\begin{array}{ccccccc} &\large\sin(x)y^2 \\\\ &\small\dfrac{\partial}{\partial x}\large\swarrow\quad\searrow\small\dfrac{\partial}{\partial y} \\\\ &\large\cos(x)y^2\qquad\qquad\qquad 2\sin(x)y \\\\ \small\dfrac{\partial}{\partial x}\large\swarrow&\large\searrow\small\dfrac{\partial}{\partial y}\qquad\qquad\qquad\dfrac{\partial}{\partial x}\large\swarrow&\large\searrow\small\dfrac{\partial}{\partial y} \\\\ \large\sin(x)y^2&\large\maroonD{\underbrace{\blueE{2\cos(x)y\qquad2\cos(x)y}}_{\text{混合偏导数相等！}}}&\large2\sin(x) \end{array}$

## 二阶导数的对称

\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}

## 示例 2：更高阶的导数

: 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。

\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}

\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}

\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}