0902偏导数-多元函数微分法及其应用

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一、偏导数的定义及其计算方法

1 偏导数的定义

定义 设函数$z=f(x,y)在点(x_0,y_0)的某一邻域内有定义，$当y固定在$y_0$,而x再$x_0处有增量\Delta x$时，相应的函数有增量

$f(x_0+\Delta x,y_0)-f(x_0,y_0)$

如果

$\underset{\Delta x\to 0}{\lim }\frac{f(x_0+\Delta x,y_0)-f(x_0,y_0)}{\Delta x}$

存在，那么称此极限为函数$z=f(x,y)在点(x_0,y_0)$处对x的偏导数，记作

$\frac{\partial z}{\partial x}{|}_{x=x_{0}y=y_0}$,或者$\frac{\partial f}{\partial x}{|}_{x=x_{0}y=y_0}$或者$z_x{|}_{x=x_{0}y=y_0}$或者$f_x(x_0,y_0)$

类似的，函数$z=f(x,y)在点(x_0,y_0)对y$的偏导数定义为

$\underset{\Delta y\to 0}{\lim }\frac{f(x_0,y_0+\Delta y)-f(x_0,y_0)}{\Delta y}$

记作

$\frac{\partial z}{\partial y}{|}_{x=x_{0}y=y_0}$,或者$\frac{\partial f}{\partial y}{|}_{x=x_{0}y=y_0}$或者$z_y{|}_{x=x_{0}y=y_0}$或者$f_y(x_0,y_0)$

如果函数$z=f(x,y)在区域D内每一点(x,y)$对x的偏导数都存在，那么这个偏导数就是x,y的函数，它就成为函数$z=f(x,y)对自变量x$的偏导函数，记作

$\frac{\partial z}{\partial x}$或者$\frac{\partial f}{\partial x}$或者$Z_x$或$f_x(x,y)$

类似的，可以定义函数$z=f(x,y)对自变量y$的偏导函数，记作

$\frac{\partial z}{\partial y}$或者$\frac{\partial f}{\partial y}$或者$Z_y$或$f_y(x,y)$

2 偏导数的求法

对于函数$z=f(x,y)求偏导\frac{\partial z}{\partial x},\frac{\partial z}{\partial y}$

• $\frac{\partial z}{\partial x}$: 把y看做常量，对x求导；
• $\frac{\partial z}{\partial y}$：把x看做常量，对y求导；

注：

1. 推广$u=f(x,y,z)$
   1. 定义：$u_x(x_0,y_0,z_0)=\underset{\Delta x\to 0}{\lim }\frac{u(x_0+\Delta x,y_0,z_0-u(x_0,y_0,z_0))}{\Delta x}$
   2. 先求$\frac{\partial u}{\partial x}$,在带入点$(x_0,y_0,z_0)$;$\frac{\partial u}{\partial x}$:把y，z看做常量，对x求导
2. $z=f(x,y)偏导数有2个\frac{\partial z}{\partial x}，\frac{\partial z}{\partial y}$；z=f(x,y,z)偏导数有3个$\frac{\partial u}{\partial x}$，$\frac{\partial u}{\partial y}$，$\frac{\partial u}{\partial z}$
3. $\frac{\partial u}{\partial x}$是一个整体，是一个符号，不能拆分；$\frac{dy}{dx}$是微分的商。

例1 求$z=x^2+3xy+y^2在(1,2)的偏导数$
$$\begin{gathered} 解： \\ \frac{\partial z}{\partial x}=2x+3y \\ {f}_x(1,2)=2+6=8 \\ \frac{\partial z}{\partial y}=3x+2y \\ {f}_y(1,2)=3+4=7 \end{gathered}$$
例2 求$z=x^2\sin (2y)$的偏导数
$$\begin{gathered} 解： \\ \frac{\partial z}{\partial x}=2x\sin (2y) \\ \frac{\partial z}{\partial y}=2x^2\cos (2y) \end{gathered}$$
例3 设$z=x^y(x>0,x\ne 1)$,求证$\frac{x}{y}\cdot \frac{\partial z}{\partial x}+\frac{1}{\ln x}\cdot \frac{\partial z}{\partial y}=2z$
$$\begin{gathered} 证明： \\ \frac{\partial z}{\partial x}=y\cdot x^{y-1} \\ \frac{\partial z}{\partial y}=\ln x\cdot x^6 \\ \therefore \frac{x}{y}\cdot \frac{\partial z}{\partial x}+\frac{1}{\ln x}\cdot \frac{\partial z}{\partial y}=\frac{x}{y}\cdot y\cdot x^{y-1}+\frac{1}{\ln x}\cdot \ln x\cdot x^y=2z \end{gathered}$$
例4 求$r=\sqrt{x^2+y^2+z^2}$的偏导数
$$\begin{gathered} 解： \\ \frac{\partial r}{\partial x}=\frac{1}{2}\cdot \frac{1}{\sqrt{x^2+y^2+z^2}}\cdot 2x=\frac{x}{\sqrt{x^2+y^2+z^2}}=\frac{x}{r} \\ \frac{\partial r}{\partial y}=\frac{y}{r} \\ \frac{\partial r}{\partial z}=\frac{z}{r} \end{gathered}$$

3 偏导数的几何意义

设$M_0(x_0,y_0,f(x_0,y_0))为z=f(x,y)$上的一点，过$M_0做平面y=y_0$则有曲线
$$\{\begin{array}{l}z=f(x,y)\\ y=y_0\end{array}$$
在平面$y=y_0上曲线z=f(x,y_0)$对其导数$\frac{d}{dx}f(x,y_0)即(f_x(x_0,y_0))切切线M_0{T}_x对x$轴的斜率。

同理$f_y(x_0,y_0)$代表曲线
$$\{\begin{array}{l}z=f(x,y)\\ x=x_0\end{array}$$
上过$M_0$处切线$M_0{T}_y对y$轴的斜率。

注：

1. 一元函数连续不一定可导，可导一定连续；多元连续函数不一定可导，多元可导函数不一定连续。

示例：
$$\begin{gathered} z=f(x,y)= \\ \{\begin{array}{l}\frac{xy}{x^2+y^2},x^2+y^2\ne 0\\ 0,x^2+y^2=0\end{array} \end{gathered}$$
$z=f(x,y)在(0,0)不连续$，通过定义看下在点$(0,0)$的偏导数
$$\begin{gathered} f_x(0,0)=\underset{\Delta x\to 0}{\lim }\frac{f(0+\Delta x,0)-f(0,0)}{\Delta x}=0 \\ {f}_y(0,0)=0 \end{gathered}$$

二、高阶偏导数

定义 $z=f(x,y)$，其偏导数为$\frac{\partial z}{\partial x}=f_x(x,y),\frac{\partial z}{\partial y}=f_y(x,y)$.若$f_x(x,y),f_y(x,y)$的偏导数也存在，则其偏导数就成为$z=f(x,y)$的二阶偏导数。按照对变量求导次序的不同有下列四个二阶偏导数：

$\frac{\partial }{\partial x}(\frac{\partial z}{\partial x})=\frac{{\partial }^{2}z}{\partial x^2}=f_{xx}(x,y),\frac{\partial }{\partial y}(\frac{\partial z}{\partial x})=\frac{{\partial }^{2}z}{\partial x\partial y}=f_{xy}(x,y)$

$\frac{\partial }{\partial x}(\frac{\partial z}{\partial y})=\frac{{\partial }^{2}z}{\partial y\partial x}=f_{yx}(x,y),\frac{\partial }{\partial y}(\frac{\partial z}{\partial y})=\frac{{\partial }^{2}z}{\partial y^2}=f_{yy}(x,y)$

其中第二、第三两个偏导数称为混合偏导数。同样可得三阶、四阶……以及n节偏导数。二阶及二阶以上偏导数统称为高阶偏导数。

例5： 设$z=x^3{y}^2-3x{y}^3-xy+1,求\frac{{\partial }^{2}z}{\partial x^2}、\frac{{\partial }^{2}z}{\partial y\partial x}、\frac{{\partial }^{2}z}{\partial x\partial y}、\frac{{\partial }^{2}z}{\partial y^2}、\frac{{\partial }^{3}z}{\partial x^3}$
$$\begin{gathered} 解： \\ \frac{\partial z}{\partial x}=3x^2{y}^2-3y^3-y,\frac{\partial z}{\partial y}=2x^{3}y-9x{y}^2-x \\ \frac{{\partial }^{2}z}{\partial x^2}=6x{y}^2 \\ \frac{{\partial }^{2}z}{\partial y\partial x}=6x^{2}y-9y^2-1 \\ \frac{{\partial }^{2}z}{\partial x\partial y}=6x^{2}y-9y^2-1 \\ \frac{{\partial }^{2}z}{\partial y^2}=2x^3-18xy \\ \frac{{\partial }^{3}z}{\partial x^3}=6y^2 \end{gathered}$$

定理 如果函数$z=f(x,y)$的两个混合偏导数$\frac{{\partial }^{2}z}{\partial y\partial x}及\frac{{\partial }^{2}z}{\partial x\partial y}$在区域D内连续，那么在该区域的两个混合偏导数必相等。

例6 验证函数$z=\ln \sqrt{x^2+y^2}$满足方程 $\frac{{\partial }^{2}z}{\partial x^2}+\frac{{\partial }^{2}z}{\partial 6^2}=0$
$$\begin{gathered} 证： \\ z=\ln \sqrt{x^2+y^2}=\frac{1}{2}\ln (x^2+y^2) \\ \frac{\partial z}{\partial x}=\frac{1}{2}\cdot \frac{1}{x^2+y^2}\cdot 2x=\frac{x}{x^2+y^2} \\ \frac{\partial z}{\partial y}=\frac{y}{x^2+y^2} \\ \frac{{\partial }^{2}z}{\partial x^2}=\frac{(x^2+y^2)-2x^2}{(x^2+y^2{)}^2}=\frac{y^2-x^2}{(x^2+y^2{)}^2} \\ \frac{{\partial }^{2}z}{\partial y^2}=\frac{x^2-y^2}{(x^2+y^2{)}^2} \\ \therefore \frac{{\partial }^{2}z}{\partial x^2}+\frac{{\partial }^{2}z}{\partial y^2}=\frac{y^2-x^2}{(x^2+y^2{)}^2}+\frac{x^2-y^2}{(x^2+y^2{)}^2}=0 \end{gathered}$$

例7 证明函数$u=\frac{1}{r}$满足方程$\frac{{\partial }^{2}z}{\partial x^2}+\frac{{\partial }^{2}z}{\partial y^2}+\frac{{\partial }^{2}z}{\partial z^2}=0$,其中$r=\sqrt{x^2+y^2+z^2}$
$$\begin{gathered} 证明: \\ \frac{\partial u}{\partial x}=\frac{\partial u}{\partial r}\cdot \frac{\partial r}{\partial x}=-\frac{1}{r^2}\cdot \frac{x}{r}=-\frac{x}{r^3} \\ \frac{{\partial }^{2}u}{\partial x^2}=-\frac{1}{r^3}+\frac{3x}{r^4}\cdot \frac{x}{r}=-\frac{1}{r^3}+\frac{3x^2}{r^5} \\ \because 函数关于自变量对称 \\ \therefore \frac{\partial u^2}{\partial y^2}=-\frac{1}{r^3}+\frac{3y^2}{r^5},\frac{\partial u^2}{\partial z^2}=-\frac{1}{r^3}+\frac{3z^2}{r^5} \\ \frac{\partial u^2}{\partial x^2}+\frac{\partial u^2}{\partial y^2}+\frac{\partial u^2}{\partial z^2}=-\frac{3}{r^3}+\frac{3(x^2+y^2+z^2)}{r^5}=0 \end{gathered}$$

结语

❓QQ:806797785

⭐️文档笔记地址：https://gitee.com/gaogzhen/math

参考：

[1]同济大学数学系.高等数学 第七版 下册[M].北京:高等教育出版社,2014.7.p65-71.

[2]同济七版《高等数学》全程教学视频[CP/OL].2020-04-16.p65.

[3]关于梯度下降算法的理解[CP/OL].

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