第四章 多元函数微分学

第一节 基本概念机结论

定义1：（二元函数）

$$
\begin{align}
&z=f(x,y),(x,y)\in D\subset R^2\
\end{align}
$$

例题
$$
\begin{align}
&f(x,y)=\arcsin(2x)+\ln y+\frac{\sqrt{4x-y^2}}{\ln{(1-x2-y^2)}}\
&解：-1\leq 2x\leq 1,y0,1-x^2-y20,1-x^2-y2\neq1,4x-y^2\geq0\
&D={(x,y)|-\frac{1}{2}\leq x\leq\frac{1}{2},y0,x^2+y2<1,x\geq\frac{1}{4}y^2}\
\end{align}
$$

定义2：（二元函数的极限）

$$
\begin{align}
&\lim_{(x,y)\rightarrow(x_0,y_0)}f(x,y)=A或\lim_{x\rightarrow x_0,y\rightarrow y_0}f(x,y)=A\
\end{align}
$$

例题
$$
\begin{align}
&求极限\
&解法1；\
&\lim_{x\rightarrow 0,y\rightarrow 0}xy\frac{x^2-y2}{x^2+y2}\
&\xlongequal[x=r\cos\theta]{y=r\sin\theta}\lim_{r\rightarrow0}{r\cos\theta r\sin\theta\frac{r^{2\cos2\theta}{r}2}}=0\
&解法2：\
&0\leq|xy\frac{x^2-y2}{x^2+y2}|\leq{|xy|}\
&\lim_{x\rightarrow 0,y\rightarrow 0}{0}=0\leq\lim_{x\rightarrow 0,y\rightarrow 0}|xy\frac{x^2-y2}{x^2+y2}|\leq\lim_{x\rightarrow 0,y\rightarrow 0}|xy|=0\
&\Leftrightarrow\lim_{x\rightarrow 0,y\rightarrow 0}xy\frac{x^2-y2}{x^2+y2}=0\
&\
&验证极限不存在\
&(1)\lim_{x\rightarrow 0,y\rightarrow 0}\frac{xy}{x^2+y2}\
&(2)\lim_{x\rightarrow 0,y\rightarrow 0}\frac{x^3+y3}{x^2+y}\
&解：\
&(1)令y=kx\
&\lim_{x\rightarrow0,y=kx}\frac{xkx}{x^2+k2x^{2}=\frac{k}{1+k}2}\
&(2)令y=-x^2+x4\
&\lim_{x\rightarrow0,y=-x^2+x4}\frac{x^3+(x4-x²⁾3}{x^2+(-x2+x^{4)}=\lim_{x\rightarrow0}{[\frac{1}{x}}+\frac{(x}4-x²⁾2}{x^4}]=\infty\
\end{align}
$$
注解：
$$
\begin{align}
&一元函数{(x,f(x))|x\in D}\
&多元函数{(x,y,f(x,y))|(x,y)\in D}
\end{align}
$$

定义3（二院函数的连续性）

$$
\begin{align}
&f(x,y)在点P_0处连续:\lim_{x\rightarrow x_0,y\rightarrow y_0}f(x,y)=f(x_0,y_0)\
&注解1：z=f(x,y)于P_0点连续\Leftrightarrow\Delta z=f(x,y)-f(x_0,y_0)\rightarrow0(x\rightarrow x_0,y\rightarrow y_0)\
&注解2："二元初等函数"在其定义域内处处连续，\lim_{x\rightarrow 1,y\rightarrow2}\frac{x+y}{x-y}=-3\
\end{align}
$$

定理1

$$
\begin{align}
&有界闭区域D\subset{R}上的连续函数,必有界,且有最大值最小值 \
\end{align}
$$

定义4（偏导数）

$$
\begin{align}
&\frac{\partial{z}}{\partial x}|{(x_0,y_0)}=\lim\frac{\Delta Z_x}{\Delta x}=\lim_{\Delta x\rightarrow0}\frac{f(x_0+\Delta x,y_0)-f(x_0,y_0)}{\Delta x}\
&f_y'(x_0,y_0)=\lim_{\Delta y\rightarrow0}\frac{\Delta Z_y}{\Delta y}=\lim_{y\rightarrow y_0}\frac{f(x_0,y_)-f(x_0,y_0)}{y-y_0}
\end{align}
$$

例题
$$
\begin{align}
&(1)设f(x,y)=\begin{cases}&\frac{xy}{x^2+y2},(x,y)\neq(0,0)\&0,(x,y)=(0,0)\\end{cases},求f_x'(0,0)和f'y(0,0),但\limf(x,y)不存在\
&(2)求f(x,y)=\sqrt{x^2+y2}在(0,0)处的偏导数，并说明函数在此点的连续性\
\end{align}
$$

$$
\begin{align}
&（1）f_x'(0,0)=\lim_{x\rightarrow0}\frac{f(x,0)-f(0,0)}{x-0}=\lim_{x\rightarrow 0}{\frac{0-0}{x}}=0\
&同理得f_y'(0,0)=0
&\
\end{align}
$$

$$
\begin{align}
&\lim_{x\rightarrow 0^{\pm}}\frac{f(x,0)-f(0,0)}{x-0}=\lim_{x\rightarrow x^{{\pm}}\frac{\sqrt{x}2}-0}{x}=\pm 1\
&\Rightarrow f'_x(0,0),f'_y(0,0)不存在\
\end{align}
$$

定义5（全微分）

$$
\begin{align}
&若z=f(x,y),\Delta z=A\Delta x+B\Delta y+o(\rho)(\rho\rightarrow 0),\rho=\sqrt{\Delta x^2+\Delta y^2},则称z=f(x,y)在点P_0可微，\
&dz|{(x_0,y_0)}=df|=A\Delta x+B\Delta y
\end{align}
$$

注解：
$$
\begin{align}
&(1)若\exist A,B使得\lim_{\Delta x\rightarrow0,\Delta y\rightarrow 0}\frac{\Delta f-A\Delta x-B\Delta y}{\sqrt{\Delta x^2+\Delta y^2}}=0,则f于(0,0)可微\
&(2)若f于(x_0,y_0)可微,则\lim\frac{\Delta f-df}{\sqrt{\Delta x^2+\Delta y^2}}=0
\end{align}
$$

定理2

$$
\begin{align}
&若z=f(x,y)在点P_0(x_0,y_0)处可微,则偏导数存在，且dz|_{(x_0,y_0)}=f'_x(x_0,y_0)dx+f_y'(x_0,y_0)dy\
\end{align}
$$

定理3 几个命题之间的关系

$$
\begin{align}
&可微\Rightarrow\begin{cases}&连续\Rightarrow极限存在\&偏导数存在\&方向导数存在\\end{cases}\
\end{align}
$$

二元函数可微与偏导的联系 - blueflylabor - 博客园 (cnblogs.com)

例题

注解：
$$
\begin{align}
&可微充分条件\Leftrightarrow偏导函数在(x_0,y_0)处连续\Rightarrow 可微\
\end{align}
$$

第二节 多元函数微分法

初等函数的微分法

注解：
$$
\begin{align}
&(1)对x把y看做常数，对y把x看成常数\
&(2)u\begin{cases}&u_x'\begin{cases}&u''{xx}\&u''\\end{cases}\&u_y'\begin{cases}&u''{yx}\&u''\\end{cases}\\end{cases}
当偏导函数连续时,u''{xy}=u''
\end{align}
$$

例题
$$
\begin{align}
&1.设z=\arcsin\frac{x}{\sqrt{x^2+y2}}，求\frac{\partial^2z}{\partial x^{2},\frac{\partial}2z}{\partial x\partial y}\
&解\
&z'x=\frac{|y|}{x^2+y2}\
&z''=\frac{\partial}{\partial x}(\frac{|y|}{x^2+y2})=\begin{cases}&\frac{-2xy}{(x^2+y2)^{2},y0\&0,x\neq0,y=0\&\frac{2xy}{(x}2+y²⁾2},y0\end{cases}\
&z''_{xy}=\frac{\partial}{\partial y}(\frac{|y|}{x^2+y2})=\begin{cases}&\frac{x^2-y2}{(x^2+y2)^2},y0\&不存在,x\neq 0,y=0\&\frac{y^2-x2}{(x^2+y2)^2},y<0\\end{cases}\
\end{align}
$$

复合函数微分法

$$
\begin{align}
&\frac{\partial z}{\partial x}=-\frac{1}{x^2}f(xy)+\frac{1}{x}f'(xy)y+y\phi'(x+y)\
&\frac{\partial^2 z}{\partial x\partial y}=-\frac{1}{x^2}f'(xy)x+\frac{1}{x}[f''(xy)xy+f'(xy)]+\phi'(x+y)+y\phi''(x+y)\
\end{align}
$$
3.

$$
\begin{align}
&法1：\
&\frac{\partial f}{\partial x}=e^{-(xy)2}y-e^{-(x+y)2}\
&\frac{\partial^2 f}{\partial x\partial y}=\
&法2：\
&令u=x+y,v=xy,f(x,y)由\int_v^uedt与\begin{cases}&u=x+y\&v=xy\\end{cases}复合\
&\frac{\partial f}{\partial x}=e^{-(v)2}\frac{\partial u}{\partial x}-e^{-(u)2}\frac{\partial v}{\partial x}\
&\frac{\partial^2 f}{\partial x\partial y}=\
\end{align}
$$

多元隐函数的微分法

例题：

$$
\begin{align}
&法1:公式法\
&F(x,y,u)=u+e^u-xy,u'_x=\frac{\partial u}{\partial x}=-\frac{F'_x}{F'_x}=-\frac{-y}{1+e^u}\
&法2:\
&u'_x+e^{uu'_x=y,u_x'=\frac{y}{1+e}u},u'_y={\frac{x}{1+e^u}}\
&\frac{\partial ^2u}{\partial x\partial y}={\frac{1[1+e^u]-y*euu_y'}{[1+e^u]2}}\
\end{align}
$$

多元函数的极值与最值求法

无条件极值（二元）

$$
\begin{align}
&(1)定义\
&(2)必要条件\begin{cases}&z_x'(x_0,y_0)=0\&z_y'(x_0,y_0)=0\\end{cases}\
&(3)充分条件\Delta=AC-B^2\begin{cases}&0\&<0\\end{cases}\
&A=f_{xx}''(x_0,y_0),B=f_{xy}''(x_0,y_0),C=f''_{yy}(x_0,y_0)\
\end{align}
$$

有界闭区域

$$
\begin{align}
&有界闭区域D上的连续函数f(x,y)的最值：\
&如果函数f(x,y)在有界闭区域D\subset R^2上连续，则f(x,y)必在D上取得最大值和最小值\
\end{align}
$$

解题步骤

例题：