Chapter 8 多元函数微分学¶

1. 多元函数的极限与连续性¶

1.3 二元函数的极限与连续¶

极限¶

设$n$元函数$f(P)$在$U^0(P_0,\delta_0)$内有定义，若存在一常数$A,\forall \varepsilon>0,\exists\delta>0(\delta\le\delta_0)$，当$0<\rho(P,P_0)<\delta$，都有$|f(P)-A|<\varepsilon$成立，则称$A$为$n$元函数$f(P)$当$P\to P_0$时的极限，记作 $$ \lim_{P\to P_0} f(x)=A 或 f(P)\to A(P\to P_0) $$ 注意：这个极限与点$P(x,y)$趋于点$P_0(x_0,y_0)$的方式无关，只要$P$距离$P_0$充分接近，点$P$趋于$P_0$点的方式可有无数种.

多元函数极限的归结原则¶

若$f(P)$在$U^0(P_0,\delta_0)$内有定义，则 $$ \lim_{P\to P_0}f(P)=A\iff 对于U^0(P_0,\delta_0)的任何子集E,\lim_{\stackrel{P\to P_0}{(P\in E)}}f(P)=A $$ 同样可用归结原则，若发现点$P$按两个特殊的路径趋于点$P_0$时，$f(P)$极限存在但不相等，则$f(P)$在该点$P_0$极限不存在. 这是判断多元函数极限不存在的重要方法之一.

题型：证明极限：用定义；求极限：夹逼定理

累次极限与二重极限¶

累次极限$\lim_{x\to x_0}\lim_{y\to y_0}f(x,y),\lim_{y\to y_0}\lim_{x\to x_0}f(x,y)$本质上是二次求一元函数的极限，累次极限存在性与二重极限存在性没有一定联系.

定理：若累次极限$\lim_{x\to x_0}\lim_{y\to y_0}f(x,y),\lim_{y\to y_0}\lim_{x\to x_0}f(x,y)$与二重极限都存在，则三者相等.

推论：若$\lim_{x\to x_0}\lim_{y\to y_0}f(x,y),\lim_{y\to y_0}\lim_{x\to x_0}f(x,y)$存在但不相等，则二重极限不存在.

连续¶

定义：设$f(P)=f(x,y)$在点$P_0(x_0,y_0)$的某邻域$U(P_0)$内有定义，且$\pmb{\lim_{(x,y)\to(x_0,y_0)}f(x,y)=f(x_0,y_0)}$，则称函数$f(P)=f(x,y)$在点$P_0(x_0,y_0)$处连续，记 $$ \Delta z=f(x,y)-f(x_0,y_0)=f(x_0+\Delta x_,y_0+\Delta y)-f(x_0.y_0) $$ 称为函数（值）的全增量，则连读定义可写为$\lim_{\Delta x\to0,\Delta y\to0}\Delta z=0$.

记$\Delta_xz=f(x,y_0)-f(x_0,y_0)=f(x_0+\Delta x,y_0)-f(x_0,y_0)$称为函数（值）对$x$的偏增量.

记$\Delta_yz=f(x_0,y)-f(x_0,y_0)=f(x_0,y_0+\Delta y)-f(x_0,y_0)$称为函数（值）对$y$的偏增量.

若$f(P)$在点$P(x_0,y_0)$不连续，称$P_0$是$f(x,y)$的间断点，若$f(x,y)$在某区域$G$上每一点都连续，则称$f(x,y)$在区域$G$上连续. 若$f(x,y)$在闭域$G$上每一内点都连续，并在$G$的边界点$P_0(x_0,y_0)$处成立$\pmb{\lim_{P\to P_0,P\in G}f(P)=f(P_0)}$，则称$f(P)=f(x,y)$在闭域$G$上连续. 闭域上连续的二元函数的图形称为连续曲面.

注：初等多元函数在他们的定义域内都是连续的.

定理：设$f(P)=f(x,y)$在平面上一个有界闭区域$G$上连续，则

（1）$f(P)$必在$G$上取到最大值，最小值及其中间的一切值.

（2）$f(P)$在$G$上一致连续，即$\pmb{\forall\varepsilon>0,\exists\delta>0\ \ (\delta\le\delta_0)}$，当$\pmb{0<\rho(P_1,P_2)<\delta}$时，都有$\pmb{|f(P_1)-f(P_2)|<\varepsilon}$.

计算二元（或多元）函数的极限常用方法：

（1）利用不等式，使用两边夹逼定理.

（2）变量替换化为已知极限，或化为一元函数极限，利用极坐标.

（3）利用初等函数的连续性，利用极限的四则运算性质.

（4）利用初等变形，特别是指数形式常可先求其对数的极限.

（5）若事先能看出极限值，可用$\varepsilon-\delta$方法进行证明.

证明二元（或多元）函数极限不存在：

（1）证明径向路径的极限与辐角（或斜率）有关.

（2）证明某个特殊路径的极限不存在.

（3）证明两个特殊路径的极限存在但不相等.

（4）若二元函数在该点某空心邻域里连续，而两个累次极限存在但不相等，则该点极限不存在.

2 偏导数与全微分¶

2.1 偏导数¶

偏导数的定义¶

定义：设函数$z=f(x,y)$在点$P_0(x_0,y_0)$的某邻域内有定义，若极限 $$ \lim_{\Delta x\to0}\frac{\Delta_xz}{\Delta x}=\lim_{\Delta x\to0}\frac{f(x_0+\Delta x,y_0)-f(x_0,y_0)}{\Delta x}=\lim_{x\to x_0}\frac{f(x,y_0)-f(x_0,y_0)}{x-x_0} $$ 存在，则称该极限值为函数$z=f(x,y)$在点$P_0(x_0,y_0)$关于$x$的偏导数，记为 $$ f'x(x_0,y_0) 或 \frac{\partial z}{\partial x}\bigg| 或 z'}x=x_0\y=y_0\end{split}x| $$ 否则称}} 或 \lim_{\Delta y\to0}\frac{f(x,y+\Delta y)-f(x,y)}{\Delta y$z=f(x,y)$在点$P_0(x_0,y_0)$处对$x$的偏导数不存在.

若对于某一区域$G$上的每一点$(x,y)$，极限$\lim_{\Delta x\to0}\frac{\Delta_xz}{\Delta x}=\lim_{\Delta x\to0}\frac{f(x+\Delta x,y)-f(x,y)}{\Delta x}$都存在，它是$x,y$的函数，称为函数$z=f(x,y)$在$G$上关于$x$的偏导函数，简称偏导数，记作 $$ f'_x(x,y)=\frac{\partial }{\partial x}f(x,y)=z'_x=\frac{\partial z}{\partial x} $$ 同理可以定义函数对$y$的偏导数 $$ f'_y(x,y)=\frac{\partial }{\partial y}f(x,y)=z'_y=\frac{\partial z}{\partial y} $$

偏导数的计算¶

求$f'_x(x_0,y_0)$有三种方法：

（1）按定义.

（2）求导函数$\frac{d}{dx}f(x,y_0)$，然后把$x=x_0$代入.

（3）求偏导函数$f'_x(x,y)$，然后把$x=x_0,y=y_0$代入.

例1：$f(x,y)=x+(y-1)\arcsin{\sqrt{\frac{x}{y}}}$，求$f'_x(\frac{1}{2},1),f'_y(\frac{1}{2},1)$.

解：

（1） $$ \begin{align} f'x(\frac12,1)&=\lim=1\ f'}\frac{f(x,1)-f(\frac12,1)}{x-\frac12y(\frac12,1)&=\lim=\frac\pi4 \end{align} $$ （2） $$ \begin{align} f'}\frac{f(\frac12,y)-f(\frac12,1)}{y-1}=\lim_{y\to1}\frac{\frac12+(y-1)\arcsin{\sqrt\frac{1}{2y}}-\frac12}{y-1}=\lim_{y\to1}\arcsin{\sqrt\frac1{2y}x(\frac12,1)&=\frac{d}{dx}f(x,1)\bigg|=x'|{x=\frac12}=1 \f'_y(\frac12,1)&=\frac{d}{dy}f(\frac12,y)\bigg|=\frac\pi4 \end{align} $$ （3） $$ \begin{align} f'_x(x,y)&=1+\frac12(y-1)\frac{1}{\sqrt{1-\frac xy}}\cdot\frac{1}{\sqrt{\frac xy}}\cdot\frac{1}{y}\Longrightarrow f'_x(\frac12,1)=1 \f'_y(x,y)&=\arcsin{\sqrt{\frac xy}}+\frac12(y-1)\frac{1}{\sqrt{1-\frac xy}}\cdot\frac{1}{\sqrt{\frac xy}}\cdot\frac{-x}{y^2}\Longrightarrow f'_y(\frac12,1)=\arcsin{\frac{\sqrt2}{2}}=\frac\pi4 \end{align} $$}=\bigg[\frac12+(y-1)\arcsin{\sqrt{\frac{1}{2y}}}\bigg]'\Bigg|_{y=1

偏导数的几何意义¶

偏导数$f'_x(x_0,y_0)$的几何意义是表示$\begin{cases}z=f(x,y)\\y=y_0\end{cases}$ 在$M_0$处的切线对$Ox$轴的斜率.

偏导数$f'_y(x_0,y_0)$的几何意义是表示$\begin{cases}z=f(x,y)\\x=x_0\end{cases}$ 在$M_0$处的切线对$Oy$轴的斜率.

例1：讨论函数$f(x,y)=\begin{cases}\frac{2xy}{x^2+y^2},&x^2+y^2\ne0\\0,&x^2+y^2=0\end{cases}$ 在$(0,0)$点的偏导数是否存在，函数$f$是否连续？

解：

按定义求得 $$ f'x(0,0)=0,f'_y(0,0)=0\Longrightarrow函数f在点(0,0)处的偏导数都存在 $$ 但是 $$ \lim\Longrightarrow函数f在(0,0)处不连续 $$ }}f(x,y)=\lim_{\substack{x\to0\y=kx}}\frac{2k}{1+k^2注：多元函数中连续与可导无必然联系.

高阶偏导数¶

与一元函数的高阶导数一样，可以定义多元函数的高阶偏导数.

若函数$z=f(x,y)$的偏导数$\frac{\partial z}{\partial x},\frac{\partial z}{\partial y}$存在，称为一阶偏导数，它们仍是$x,y$的函数，如果它们对$x,y$的偏导仍存在，得到 $$ \frac{\partial }{\partial x}(\frac{\partial z}{\partial x}),\frac{\partial }{\partial y}(\frac{\partial z}{\partial x}),\frac{\partial }{\partial x}(\frac{\partial z}{\partial y}),\frac{\partial }{\partial y}(\frac{\partial z}{\partial y}) $$ 称为$z=f(x,y)$的二阶偏导数，记作 $$ \begin{align} z_{xx}''&=\frac{\partial ^2z}{\partial x^2}=\frac{\partial }{\partial x}(\frac{\partial z}{\partial x})=f''{xx}(x,y)\ z)=f''}''&=\frac{\partial ^2z}{\partial y^2}=\frac{\partial }{\partial y}(\frac{\partial z}{\partial y{yy}(x,y)\ z)=f''}''&=\frac{\partial ^2z}{\partial x\partial y}=\frac{\partial }{\partial y}(\frac{\partial z}{\partial x{xy}(x,y)\ z(x,y)\ \end{align} $$ 其中}''&=\frac{\partial ^2z}{\partial y\partial x}=\frac{\partial }{\partial x}(\frac{\partial z}{\partial y})=f''_{yx$f_{xy}''(x,y)$和$f_{yx}''(x,y)$称为二阶混合偏导数. 二阶及二阶以上的偏导数，统称为高阶偏导数.

定理：若函数$z=f(x,y)$的二阶偏导（函数）$f_{xy}''(x,y),f_{yx}''(x,y)$都在$(x_0,y_0)$处连续，则$f_{xy}''(x_0,y_0)=f_{yx}''(x_0,y_0)$.

2.2 全微分¶

全微分的概念¶

定义：若二元函数$z=f(x,y)$在点$(x,y)$处的全增量$\Delta z=f(x+\Delta x,y+\Delta y)-f(x,y)$可以表示为 $$ \Delta z=A\Delta x+B\Delta y+o(\rho) (\rho=\sqrt{\Delta x^2+\Delta y^2}\to0) $$ 其中，$A,B$与变量的增量$\Delta x,\Delta y$无关，而仅与$x,y$有关，则称函数$z=f(x,y)$在点$(x,y)$处可微. 其中$\Delta z$的线性主部$A\Delta x+B\Delta y$称为函数$z=f(x,y)$在点$(x,y)$处的全微分，记作$dz$，即 $$ dz=A\Delta x+B\Delta y $$ 函数$z=f(x,y)$在点$P(x_0,y_0)$处可微的充要条件 $$ \lim_{\substack{\Delta x\to0\\Delta y\to0}}\frac{\Delta z-\big[f'_x(x,y)\Delta x+f'_y(x,y)\Delta y\big]}{\sqrt{\Delta x^2+\Delta y^2}}=0 $$

可微的必要条件、充分条件¶

定理1：若$z=f(x,y)$在点$(x,y)$处可微，则$z=f(x,y)$在点$(x,y)$处连续，反之不成立.

定理2：若$z=f(x,y)$在点$(x,y)$处可微，则$z=f(x,y)$在点$(x,y)$的两个偏导数$f'_x(x,y),f'_y(x,y)$都存在，且$A=f'_x(x,y),B=f'_y(x,y)$.

例1：研究$f(x,y)=\begin{cases}\frac{2xy}{x^2+y^2},&x^2+y^2\ne0\\0,&x^2+y^2=0\end{cases}$ 在原点的可微性.

验证多元函数不可微有下述方法：

（1）若$f(x,y)$在点$(x,y)$处不连续，则$f$在点$(x,y)$处不可微.

（2）若$f(x,y)$在点$(x,y)$处至少有一个偏导数不存在，则$f$在点$(x,y)$处不可微.

（3）若$f(x,y)$在点$(x,y)$处两个偏导数都存在，但$\lim_{\substack{\Delta x\to0\\\Delta y\to0}}\frac{\Delta z-dz}{\rho}$极限不存在或极限存在但不为零，则$f$在点$(x,y)$处不可微.

定理3：若函数$z=f(x,y)$的偏导数$f_x'(x,y),f_y'(x,y)$在点$(x_0,y_0)$处连续，则函数$z=f(x,y)$在点$(x_0,y_0)$处可微.

全微分四则运算与计算¶

设$u,v$都是多元函数，且具有连续的偏导数，则

(1) $d(u\pm v)=du\pm dv$.

(2) $d(uv)=vdu+udv$；特别地$d(cu)=cdu$（$c$是常数）.

(3) $d(\frac{u}{v})=\frac{vdu-udv}{v^2}(v\ne0)$.

下面对（2）做出证明

设$u=u(x,y),v=v(x,y)$，有条件知，$uv$可微，且 $$ \begin{align} d(uv)&=\frac{\partial }{\partial x}(uv)dx+\frac{\partial }{\partial y}(uv)dy=(\frac{\partial u}{\partial x}v+u\frac{\partial v}{\partial x})dx+(\frac{\partial u}{\partial y}v+u\frac{\partial v}{\partial y})dy\ &=(\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy)v+(\frac{\partial v}{\partial x}dx+\frac{\partial v}{\partial y}dy)u=vdu+udv \end{align} $$

全微分在近似计算和误差估计中的应用¶

由全微分的定义，当$|\Delta x|,|\Delta y|$很小时，有$\Delta z\approx dz=f_x'(x,y)\Delta x+f_y'(x,y)\Delta y$. 得 $$ f(x+\Delta x,y+\Delta y)-f(x,y)\approx f_x'(x,y)\Delta x+f_y'(x,y)\Delta y $$ 设函数$z=f(x,y)$，若测得$x$的近似值为$x_0$，$y$的近似值为$y_0$，用测量的近似值$x_0,y_0$分别代替$x,y$来计算函数值$z$，就会引起绝对误差 $$ |\Delta z|=|f(x,y)-f(x_0,y_0)|\approx|dz|=|f_x'(x_0,y_0)\Delta x+f_y'(x_0,y_0)\Delta y|\ \le|f_x'(x_0,y_0)||\Delta x|+|f_y'(x_0,y_0)||\Delta y|\le|f_x'(x_0,y_0)|\delta_1 +|f_y'(x_0,y_0)|\delta_2 $$ 因此 $$ \bigg|\frac{\Delta z}{z_0}\bigg|\approx\bigg|\frac{dz}{z_0}\bigg|\le\Bigg|\frac{f_x'(x_0,y_0)}{f(x_0,y_0)}\Bigg|\delta_1+\Bigg|\frac{f_y'(x_0,y_0)}{f(x_0,y_0)}\Bigg|\delta_2 $$

例1：计算$1.007^{2.98}$.

解：设$f(x,y)=x^y\Longrightarrow f_x'(x,y)=yx^{y-1},f_y'(x,y)=x^y\ln{x}$，有 $$ \begin{align} 1.007^{2.98}&=f(1.007,2.98)=f(1+0.007,3-0.02)\ &\approx f(1,3)+f_x'(1,3)\times0.007+f_y'(1,3)\times(-0.02)\ &=1+3\times0.007+1^3\times\ln{1}\times(-0.02)=1.021 \end{align} $$

计算原函数¶

设二元函数$u(x,y)$存在二阶连续偏导，且 $$ du(x,y)=P(x,y)dx+Q(x,y)dy $$ 则$P(x,y)=\frac{\partial u}{\partial x},Q(x,y)=\frac{\partial u}{\partial y}$，得 $$ \frac{\partial P}{\partial y}=\frac{\partial ^2u}{\partial x\partial y}=\frac{\partial ^2u}{\partial y\partial x}=\frac{\partial Q}{\partial x} $$ 反之，若$P(x,y),Q(x,y)$在区域D上存在一阶连续偏导，且$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$，则存在二元函数$u(x,y)$满足 $$ du(x,y)=P(x,y)dx+Q(x,y)dy $$

例1：当$y>0$时，$u(x,y)$可微，且 $$ du(x,y)=(\frac{y}{x^2+y2}+xy^{2)dx+(-\frac{x}{x}2+y^2}+x2y+2y)dy $$ 求$u(x,y)$.

解：$du(x,y)\stackrel{\triangle}{=}P(x,y)dx+Q(x,y)dy$，则 $$ \begin{align} &P(x,y)=\frac{\partial u}{\partial x}=\frac{y}{x^2+y2}+xy^2\ &Q(x,y)=\frac{\partial u}{\partial y}=-\frac{x}{x^2+y2}+x^2y+2y\ &\Longrightarrow u(x,y)=\int(\frac{y}{x^2+y2}+xy^{2)dx=\arctan{\frac{x}{y}}+\frac{1}{2}x}2y^2+\psi(y)\ &\Longrightarrow\frac{\partial u}{\partial y}=\frac1{1+(\frac xy)^2}\cdot(-\frac x{y^2})+x2y+\psi'(y)=-\frac{x}{x^2+y2}+x^2y+2y\ &\Longrightarrow\psi'(y)=2y\Longrightarrow\psi(y)=y^2+C\ &\Longrightarrow u(x,y)=\arctan{\frac xy}+\frac12x^2y2+y^2+C \end{align} $$

3 复合函数微分法¶

3.1 复合函数的偏导公式¶

若$z=f(u,v)$定义在$uv$平面$\Omega_{uv}$上，函数$u=\varphi(x,y),v=\psi(x,y)$定义在$xy$平面$\Omega_{xy}$上 $$ \big{(u,v):u=\varphi(x,y),v=\psi(x,y),(x,y\in\Omega_{xy})\big}\subset\Omega_{uv} $$ 于是$z$是以$f$为外函数，$\varphi,\psi$为内函数的复合函数$z=f[\varphi(x,y),\psi(x,y)]$，$x,y$为自变量，$u,v$为中间变量.

定理：若函数$u=\varphi(x,y),v=\psi(x,y)$在点$(x,y)$处偏导数都存在，$z=f(u,v)$在点$(u,v)=(\varphi(x,y),\psi(x,y))$处可微，则复合函数$z=f[\varphi(x,y),\psi(x,y)]$在点$(x,y)$处的偏导数存在，且 $$ \frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\cdot\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\cdot\frac{\partial v}{\partial x}\ \frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}\cdot\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\cdot\frac{\partial v}{\partial y} $$ 注：设$z=f(x,u,v),u=u(x),v=v(x)$，这里$z$是通过三个中间变量的$x$的一元复合函数，则 $$ \frac{dz}{dx}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial u}\cdot\frac{du}{dx}+\frac{\partial f}{\partial v}\cdot\frac{dv}{dx} (全导数) $$

例1：设函数$W=F(x,y,z),z=f(x,y),y=\varphi(x)$，其中$F,f$具有连续的偏导数，$\varphi$可导，求$\large{\frac{dW}{dx}}$.

解： $$ \frac{dW}{dx}=F_x'+F_y'\varphi'+F_z'\big[f_x'+f_y'\cdot\varphi'\big] $$ 注：此处$W$是$x$的一元复合函数，因此$\large{\frac{dW}{dx}=\frac{\partial W}{\partial x}}$.

例2：设$z=f(x,y,u),u=u(x,y)$，请问：$\Large{\frac{\partial z}{\partial x}=\frac{\large\partial f}{\partial x}}$？

解： $$ \frac{\partial z}{\partial x}=\frac{\partial f}{\partial x}\cdot\frac{dx}{dx}+\frac{\partial f}{\partial y}\cdot\frac{\partial y}{\partial x}+\frac{\partial f}{\partial u}\cdot\frac{\partial u}{\partial x}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial u}\cdot\frac{\partial u}{\partial x} $$ 注：$\large{\frac{\partial z}{\partial x}}$是复合函数，$z=f[x,y,u(x,y)]$对自变量$x$的偏导，而$\large{\frac{\partial f}{\partial x}}$是外函数，$f(x,y,u)$对中间变量$x$的偏导，这时$y,u$都看成常数.

3.2 复合函数的全微分¶

一阶微分不变性¶

若$z=z(x,y)$，$x,y$是自变量，且$z=z(x,y)$可微，则其全微分为 $$ dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy $$ 设$z=z(x,y),x=x(s,t),y=y(s,t)$都具有连续的偏导数，则复合函数$z=z(x(s,t),y(s,t))$可微，且 $$ dz=\frac{\partial z}{\partial s}ds+\frac{\partial z}{\partial t}dt $$ 证明： $$ \begin{cases} \large\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\cdot\frac{\partial y}{\partial s}\ \large\frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\cdot\frac{\partial y}{\partial t} \end{cases} $$

\[ \begin{align} \Longrightarrow \frac{\partial z}{\partial s}ds+\frac{\partial z}{\partial t}dt&=(\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\cdot\frac{\partial y}{\partial s})ds+(\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\cdot\frac{\partial y}{\partial t})dt\\ &=\frac{\partial z}{\partial x}(\frac{\partial x}{\partial s}ds+\frac{\partial x}{\partial t}dt)+\frac{\partial z}{\partial y}(\frac{\partial y}{\partial s}ds+\frac{\partial y}{\partial t}dt)=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy \end{align} \]

上述表明虽然$x,y$不是自变量，但全微分的形式与当$x,y$是自变量时是一样的，这就是全微分的一阶微分形式不变性.

换句话说，若$z=z(u,v)$，且$dz=\varphi(u,v)du+\psi(u,v)dv$，则 $$ \frac{\partial z}{\partial u}=\varphi(u,v),\frac{\partial z}{\partial v}=\psi(u,v) $$ 注：

（1）一元、多元函数都不具有高阶微分不变性.

（2）当$x,y$是自变量时，$dx,dy$各自取值，

当$x,y$是中间变量时，它们的取值由$s,t,ds,dt$确定.

4 隐函数的偏导数¶

4.1 隐函数的偏导数¶

隐函数存在定理¶

定理1：若函数$F(x,y,z)$满足下列条件

（1）函数$F$在以$P_0(x_0,y_0,z_0)$为内点的某一区域$D\subset R^3$上连续.

（2）$F(x_0,y_0,z_0)=0$.（初始条件）

（3）在$D$内存在连续的偏导数$F'_x(x,y,z),F_y'(x,y,z),F_z'(x,y,z)$.

（4）$F_z'(x_0,y_0,z_0)\ne0$.

则在点$P_0$的某邻域$U(P_0)$内唯一确定一个具有连续导数的函数$z=f(x,y)$，且满足$z_0=f(x_0,y_0),F(x,y,f(x,y))\equiv0$，且有$\large\frac{\partial z}{\partial x}=-\frac{F_x'(x,y,z)}{F_z'(x,y,z)},\frac{\partial z}{\partial y}=-\frac{F_y'(x,y,z)}{F_z'(x,y,z)}$.

说明：对方程$F(x,y,z(x,y))\equiv0$两边关于$x,y$求偏导得 $$ F_x'\cdot1+F_z'\cdot\frac{\partial z}{\partial x}=0\Longrightarrow\frac{\partial z}{\partial x}=-\frac{F_x'}{F_z'}\ F_y'\cdot1+F_z'\cdot\frac{\partial z}{\partial y}=0\Longrightarrow\frac{\partial z}{\partial y}=-\frac{F_y'}{F_z'} $$ 注：$\large\frac{\partial z}{\partial x}=-\frac{F_x'}{F_z'},\frac{\partial z}{\partial y}=-\frac{F_y'}{F_z'}$仅适用于$F(x,y,z)=0$，对于一般形式不适用，一般形式只能用复合函数求偏导或一阶微分不变性来求.

4.2 隐函数组偏导数¶

隐函数组存在定理¶

定理：设函数$F(x,y,u,v),G(x,y,u,v)$在点$P_0(x_0,y_0,u_0,v_0)$的某邻域内满足

（1）$F(x_0,y_0,u_0,v_0)=G(x_0,y_0,u_0,v_0)=0$.（初始条件）

（2）$F,G$存在一阶连续偏导数$F_x',F_y',F_u',F_v',G_x',G_y',G_u',G_v'$.

（3）在点$P_0$处行列式$J\stackrel{\small\triangle}{=}\frac{\partial (F,G)}{\partial (u,v)}=\left|\begin{array}{c} F_u' &F_v'\\G_u' &G_v'\end{array}\right|\ne0$.（Jacobi行列式）

则方程组$\begin{cases}F(x,y,u,v)=0\\G(x,y,u,v)=0\end{cases}$ 在点$P_0(x_0,y_0,u_0,v_0)$的某邻域内唯一确定一组函数组$\begin{cases}u=u(x,y)\\v=v(x,y)\end{cases}$满足

（1）$u_0=u(x_0,y_0),v_0=v(x_0,y_0)$.

（2）$F(x,y,u(x,y),v(x,y))=0,G(x,y,u(x,y),v(x,y))=0$.

（3）$u=u(x,y),v=v(x,y)$具有一阶连续偏导数： $$ \frac{\partial u}{\partial x}=-\frac{1}{J}\cdot\frac{\partial (F,G)}{\partial (x,v)}=-\frac{\left|\begin{array}{c}F_x' &F_v'\G_x' &G_v'\end{array}\right|}{\left|\begin{array}{c} F_u' &F_v'\G_u' &G_v'\end{array}\right|}, \frac{\partial v}{\partial x}=-\frac{1}{J}\cdot\frac{\partial (F,G)}{\partial (u,x)}=-\frac{\left|\begin{array}{c}F_u' &F_x'\G_u' &G_x'\end{array}\right|}{\left|\begin{array}{c} F_u' &F_v'\G_u' &G_v'\end{array}\right|}\ \frac{\partial u}{\partial y}=-\frac{1}{J}\cdot\frac{\partial (F,G)}{\partial (y,v)}=-\frac{\left|\begin{array}{c}F_y' &F_v'\G_y' &G_v'\end{array}\right|}{\left|\begin{array}{c} F_u' &F_v'\G_u' &G_v'\end{array}\right|}, \frac{\partial v}{\partial y}=-\frac{1}{J}\cdot\frac{\partial (F,G)}{\partial (u,y)}=-\frac{\left|\begin{array}{c}F_u' &F_y'\G_u' &G_y'\end{array}\right|}{\left|\begin{array}{c} F_u' &F_v'\G_u' &G_v'\end{array}\right|} $$

以下仅有关$\large\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial v}{\partial x},\frac{\partial v}{\partial y}$计算公式推导： $$ \begin{cases} F(x,y,u,v)=0\ &\Longrightarrow\ G(x,y,u,v)=0 \end{cases} \begin{cases} \begin{align} F_x'+F_u'\cdot\frac{\partial u}{\partial x}+F_v'\cdot\frac{\partial v}{\partial x}&=0\ G_x'+G_u'\cdot\frac{\partial u}{\partial x}+G_v'\cdot\frac{\partial v}{\partial x}&=0 \end{align} \end{cases} $$

\[ \Longrightarrow \begin{cases} \Large{\frac{\partial u}{\partial x}}=-\LARGE\frac{\frac{\partial (F,G)}{\partial (x,v)}}{\frac{\partial (F,G)}{\partial (u,v)}}\\ \Large{\frac{\partial v}{\partial x}}=-\LARGE\frac{\frac{\partial (F,G)}{\partial (u,x)}}{\frac{\partial (F,G)}{\partial (u,v)}}\\ \end{cases} \]

解得$\large\frac{\partial u}{\partial x},\frac{\partial v}{\partial x}$，同理可得$\large\frac{\partial u}{\partial y},\frac{\partial v}{\partial y}$.

注：在实际计算时，一般用公式直接求.

例1：设$u=f(x,y,z),\psi(x^2,e^y,z)=0,y=\sin{x}$，其中$f,\psi$具有连续的偏导数且$\psi_3'\ne0$，求$\large\frac{du}{dx}$.

解：

法一

\[ \begin{align} u=f(x,y,z)&\Longrightarrow \frac{du}{dx}=f_x'+f_y'\frac{dy}{dx}+f_z'\frac{dz}{dx}&(1)\\ y=\sin x&\Longrightarrow \frac{dy}{dx}=\cos x &(2)\\ \psi(x^2,e^y,z)=0&\Longrightarrow\psi_1'\cdot2x+\psi_2'\cdot e^y\cdot\frac{dy}{dx}+\psi_3'\cdot\frac{dz}{dx}=0\\ &\Longrightarrow\frac{dz}{dx}=-\frac1{\psi_3'}(2x\psi_1'+e^y\cos x\psi_2')&(3) \end{align} \]

把$(2),(3)$代入$(1)$： $$ \frac{du}{dx}=f_x'+f_y'\cos x-f_z'\frac1{\psi_3'}(2x\psi_1'+e^y\cos x\psi_2') $$

法二 一阶微分不变性+微分四则运算

\[ \begin{align} u=f(x,y,z)&\Longrightarrow du=f_x'dx+f_y'dy+f_z'dz&(4)\\ \psi(x^2,e^y,z)=0&\Longrightarrow \psi_1'd(x^2)+\psi_2'd(e^y)+\psi_3'dz=0&(5)\\ &\Longrightarrow dz=-\frac1{\psi_3'}(2x\psi_1'+e^y\cos x\psi_2')dx&(6) \end{align} \]

把$(5),(6)$代入$(4)$： $$ du=[f_x'+f_y'\cos x-f_z'\frac1{\psi_3'}(2x\psi_1'+e^y\cos x\psi_2')]dx\ \Longrightarrow\frac{du}{dx}=f_x'+f_y'\cos x-f_z'\frac1{\psi_3'}(2x\psi_1'+e^y\cos x\psi_2') $$

4.3 反函数组的偏导数¶

\[ {\begin{cases} x=x(u,v)\\ y=y(u,v) \end{cases} \iff \begin{cases} u=u(x,y)\\ v=v(x,y) \end{cases} \ \ \ \ \ \ (反函数组)}\\ \begin{cases} F(x,y,u,v){\stackrel{\small\triangle}{=}} x-x(u,v)\\ G(x,y,u,v){\stackrel{\small\triangle}{=}} y-y(u,v) \end{cases} \Longrightarrow \begin{cases} F_u'=-x_u';F_v'=-x_v'\\ G_u'=-y_u';G_v'=-y_v' \end{cases} \Longrightarrow \frac{\partial (x,y)}{\partial (u,v)}=\frac{\partial (F,G)}{\partial (u,v)} \]

隐函数组存在定理$\large\Longrightarrow\frac{\partial (F,G)}{\partial (u,v)}\ne0\Longrightarrow\frac{\partial (x,y)}{\partial (u,v)}\ne0\Longrightarrow$反函数组的存在性. $$ {\begin{cases} x=x(u,v)\ y=y(u,v) \end{cases} \iff \begin{cases} x-x(u,v)\equiv0\ y-y(u,v)\equiv0 \end{cases} \iff \begin{cases} x-x[u(x,y),v(x,y)]\equiv0\ y-y[u(x,y),v(x,y)]\equiv0 \end{cases}}\ {\Longrightarrow \begin{cases} 1-x_u'u_x'-x_v'v_x'=0\ 0-y_u'u_x'-y_v'v_x'=0\ 0-x_u'u_y'-x_v'v_y'=0\ 1-y_u'u_y'-y_v'v_y'=0 \end{cases} \Longrightarrow \begin{cases} x_u'u_x'+x_v'v_x'=1\ y_u'u_x'+y_v'v_x'=0\ x_u'u_y'+x_v'v_y'=0\ y_u'u_y'+y_v'v_y'=1 \end{cases}}\ \frac{\partial (u,v)}{\partial (x,y)}\cdot\frac{\partial (x,y)}{\partial (u,v)}=\left|\begin{array}{c}u_x' &v_x'\u_y' &v_y'\end{array}\right|\cdot\left|\begin{array}{c}x_u' &x_y'\y_u' &y_v'\end{array}\right|=\left|\begin{array}{c}x_u'u_x'+x_v'v_x' &y_u'u_x'+y_v'v_x'\x_u'u_y'+x_v'v_y' &y_u'u_y'+y_v'v_y'\end{array}\right|=\left|\begin{array}{c}1 &0\0 &1\end{array}\right|=1 $$ 注：这结果与一元函数的反函数的导数公式${\large\frac{dy}{dx}\cdot\frac{dx}{dy}}=1$一致.

5 场的方向导数和梯度¶

5.1 场的概念¶

场在数学上是指一个向量到另一个向量或数的映射.

5.2 场的方向导数¶

定义：设三元函数$u$在$P_0(x_0,y_0,z_0)$的某邻域$u(P_0)\subset R^3$内有定义，$l$为从点$P_0$出发的射线，$P(x,y,z)$为$l$上且含于$u(P_0)$内的任一点，以$\rho$表示$P$与$P_0$两点的距离，若极限 $$ \lim_{\rho\to0}\frac{u(P)-u(P_0)}{\rho}=\lim_{\rho\to0}\frac{\Delta_{\vec i}u}{\rho} $$ 存在，则称此极限为函数$u$在点$P_0$沿方向$\vec l$的方向导数，记作${\large\frac{\partial u}{\partial \vec l}}\Big|_{P_0}$.

定理：若函数$u$在点$P_0(x_0,y_0,z_0)$处可微，则$u$在点$P_0$处沿任一方向$\vec l$的方向导数都存在，且 $$ \frac{\partial u}{\partial l}\Bigg|{P_0}=\frac{\partial u}{\partial x}\Bigg|\Bigg|}\cos\alpha+\frac{\partial u}{\partial y{P_0}\cos\beta+\frac{\partial u}{\partial z}\Bigg|\cos\gamma $$ 其中，$\vec l=(\cos\alpha,\cos\beta,\cos\gamma)$.

例1：讨论$f(x,y)=\sqrt{x^2+y^2}$在$(0,0)$处的任何方向$\vec l$的方向导数以及两个偏导数的存在性.

解：

\[ \frac{\partial f}{\partial \vec l}\Bigg|_{(0,0)}=\lim_{\rho\to0}\frac{f(\Delta x,\Delta y)-f(0,0)}\rho=\lim_{\rho\to0}\frac{\sqrt{(\Delta x)^2+(\Delta y)^2}}{\sqrt{(\Delta x)^2+(\Delta y)^2}}=1 \]

方向导数存在. $$ \frac{\partial f}{\partial x}\Bigg|{(0,0)}=\lim\ \frac{\partial f}{\partial y}\Bigg|}\frac{f(\Delta x,0)-f(0,0)}{\Delta x}=\lim_{\Delta x\to 0}\frac{\sqrt{(\Delta x)^2}}{\Delta x{(0,0)}=\lim $$ 偏导数}\frac{f(0,\Delta y)-f(0,0)}{\Delta y}=\lim_{\Delta y\to 0}\frac{\sqrt{(\Delta y)^2}}{\Delta y不存在.

例2：讨论$f(x,y)=\begin{cases}\large\frac{xy}{x^2+y^2},&x^2+y^2\ne0\\0,&x^2+y^2=0\end{cases}$ 在$(0,0)$处的方向导数以及两个偏导数的存在性.

解：

\[ \frac{\partial f}{\partial \vec l}\Bigg|_{(0,0)}=\lim_{\rho\to0}\frac{f(\Delta x,\Delta y)-f(0,0)}\rho=\lim_{\rho\to0}\frac{\large\frac{\Delta x\Delta y}{\rho^2}}\rho=\lim_{\rho\to0}\frac{\Delta x}\rho\cdot\frac{\Delta y}\rho\cdot\frac1\rho \]

方向导数不存在. $$ \frac{\partial f}{\partial x}\Bigg|{(0,0)}=\lim=0\ \frac{\partial f}{\partial y}\Bigg|}\frac{f(\Delta x,0)-f(0,0)}{\Delta x}=\lim_{\Delta x\to 0}\frac{0}{\Delta x{(0,0)}=\lim=0 $$ 偏导数}\frac{f(0,\Delta y)-f(0,0)}{\Delta y}=\lim_{\Delta y\to 0}\frac{0}{\Delta y存在.

结论：偏导数存在$\not\Leftrightarrow$方向导数存在.

5.3 梯度¶

定理：若函数$u$在点$P(x,y,z)$处可微，则$u$在点$P$处沿任一方向$\vec l=(\cos\alpha,\cos\beta,\cos\gamma)$的方向导数都存在，且 $$ \frac{\partial u}{\partial \vec l}=\frac{\partial u}{\partial x}\cdot\cos\alpha+\frac{\partial u}{\partial y}\cdot\cos\beta+\frac{\partial u}{\partial z}\cdot\cos\gamma=(\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial u}{\partial z})\cdot(\cos\alpha,\cos\beta,\cos\gamma) $$

定义：矢量${\large\frac{\partial u}{\partial x}}\vec i+{\large\frac{\partial u}{\partial y}}\vec j+{\large\frac{\partial u}{\partial z}}\vec k=({\large\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial u}{\partial z}})$称为函数$u(P)$在点$P$处的梯度，记为$\pmb{grad}\ u$，即 $$ \pmb {grad} u\stackrel{\small\triangle}{=}\frac{\partial u}{\partial x}\vec i+\frac{\partial u}{\partial y}\vec j+\frac{\partial u}{\partial z}\vec k=(\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial u}{\partial z}) $$ 注：梯度这个矢量实质上是和坐标系的选择无关的.

梯度的运算法则：设$u,v$可微，$\alpha,\beta$为常数，则

（1）$\pmb{grad}\ (\alpha u+\beta v)=\alpha\pmb{grad}\ u+\beta\pmb{grad}\ v$.

（2）$\pmb{grad}\ (uv)=v\pmb{grad}\ u+u\pmb{grad}\ v$.

（3）$\pmb{grad}\ f(u)=f'(u)\pmb{grad}\ u$.

6 多元函数的极值及应用¶

6.1 多元函数的泰勒公式¶

定理（泰勒定理）：若函数$f$在点$P_0(x_0,y_0)$某邻域$U(P_0)$内有直到$\pmb{n+1}$阶的连续偏导数，则对$U(P_0)$内任一点$(x_0+h,y_0+k)$，存在相应$\theta\in(0,1)$，使得 $$ \begin{align} &f(x_0+h,y_0+k)=f(x_0,y_0)+\Big(h\frac{\partial }{\partial x}+k\frac\partial {\partial y}\Big)f(x_0,y_0)\ &+\frac1{2!}\Big(h\frac{\partial }{\partial x}+k\frac\partial {\partial y}\Big)^2f(x_0,y_0)+···+\frac1{n!}\Big(h\frac{\partial }{\partial x}+k\frac\partial {\partial y}\Big)^nf(x_0,y_0)\ &+\frac1{(n+1)!}\Big(h\frac{\partial }{\partial x}+k\frac\partial {\partial y}\Big)^{n+1}f(x_0+\theta h,y_0+\theta k) \end{align} $$ 上式称为二元函数$f$在点$P_0$的$\pmb n$阶泰勒公式.

注：

（1）拉格朗日余项 $$ R_n\stackrel{\small\triangle}{=}\frac1{(n+1)!}\Big(h\frac{\partial }{\partial x}+k\frac\partial {\partial y}\Big)^{n+1}f(x_0+\theta h,y_0+\theta k),(0<\theta<1) $$ $R_n$称为泰勒公式的拉格朗日余项.

（2）当$(x_0,y_0)=(0,0)$时，则称为二元函数的麦克劳林公式.

（3）二元函数的中值公式 $$ \begin{align} f(x_0+h,y_0+k)-f(x_0,y_0)&=\Big(h\frac{\partial }{\partial x}+k\frac\partial {\partial y}\Big)f(x_0+\theta h,y_0+\theta k)\ &=f_x'(x_0+\theta h,y_0+\theta k)h+f_y'(x_0+\theta h,y_0+\theta k)k \end{align} $$ 这就是二元函数的拉格朗日中值公式.

（4）推论：设$f(x,y)$在区域$G$上具有连续的一阶偏导.

（i）若$f_x'(x,y)\equiv0,(x,y)\in G$，则$f(x,y)$在$G$上仅是$\pmb y$的函数.

（ii）若$f_y'(x,y)\equiv0,(x,y)\in G$，则$f(x,y)$在$G$上仅是$\pmb x$的函数.

（iii）若$f_x'(x,y)\equiv0,f_y'(x,y)\equiv0,(x,y)\in G$，则$f(x,y)$在$G$上是常值函数.

二元函数的中值定理：设函数$f(x,y)$在凸开域$D$上连续，在$D$的所有内点均可微，则对$D$内任意两点$P(x_0,y_0),Q(x_0+h,y_0+h),\exists \theta\in(0,1)$，使得 $$ f(x_0+h,y_0+k)-f(x_0,y_0)=f_x'(x_0+\theta h,y_0+\theta k)h+f_y'(x_0+\theta h,y_0+\theta k)k $$

例1：计算$\lim_{\substack{x\to0\\y\to0}}\frac{\ln{x^2+e^{y^2}}}{x^2+y^2}$.

解：

\[ f(x,y)\stackrel{\small\triangle}{=}\ln(x^2+e^{y^2})\Longrightarrow \begin{cases} f_x'(x,y)=\large\frac{2x}{x^2+e^{y^2}}\\ f_y'(x,y)=\large\frac{2ye^{y^2}}{x^2+e^{y^2}} \end{cases}\\ \Longrightarrow \begin{cases} f_{xx}''(x,y)=\large\frac{2e^{y^2}-2x^2}{(x^2+e^{y^2})^2}\\ f_{xy}''(x,y)=-\large\frac{4xye^{y^2}}{(x^2+e^{y^2})^2}\\ f_{yy}''(x,y)=\large\frac{(2+4y^2)e^{y^2}(x^2+e^{y^2})-4y^2e^{y^2}}{(x^2+e^{y^2})^2} \end{cases} \Longrightarrow \begin{cases} f_x'(0,0)=f_y'(0,0)=0\\ f_{xy}''(0,0)=0\\ f_{xx}''(0,0)=f_{yy}''(0,0)=2 \end{cases} \]

由泰勒公式得 $$ f(x,y)=(x^2+y2)+o(\rho^{2) (\rho\stackrel{\small\triangle}{=}\sqrt{x}2+y^{2}\to0)\Longrightarrow\lim_{\substack{x\to0\y\to0}}\frac{\ln(x}2+e^{y2})}{x^2+y2}=1 $$

6.2 多元函数的极值¶

多元函数的极值概念¶

极值定义：设函数$z=f(x,y)$在点$P_0(x_0,y_0)$的某邻域$U(P_0)$内有定义，$\forall P(x,y)\in U(P_0)$，都有$f(P)\le f(P_0)$或$f(P)\ge f(P_0)$，则称函数$f$在点$P_0$取到极大（极小）值，点$P_0$称为$f$的极大（极小）值点.

极值的必要条件：若函数$f$在点$P_0(x_0,y_0)$存在偏导数且在$P_0$取极值，则有 $$ f_x'(x_0,y_0)=0,f_y'(x_0,y_0)=0 $$ 反之，若函数$f$在点$P_0$满足上式，则称点$P_0$为$f$的稳定点或驻点.

注：

（1）上述定理指出，若$f$存在偏导数，则其极值点必是稳定点.

（2）稳定点不一定是极值点.

（3）函数$f(x,y)$在点$(x_0,y_0)$取极值，$f(x,y)$的偏导数只有两种情形：

（i）$f_x'(x_0,y_0),f_y'(x_0,y_0)$都存在，则$f_x'(x_0,y_0)=0,f_y'(x_0,y_0)=0$，即点$P_0(x_0,y_0)$为稳定点.

（ii）$f_x'(x_0,y_0),f_y'(x_0,y_0)$至少有一个不存在.

极值的充分条件：设函数$z=f(x,y)$在点$(x_0,y_0)$的某邻域$U(P_0)$连续，且有一阶与二阶连续偏导数，若$f_x'(x_0,y_0)=0,f_y'(x_0,y_0)=0$，设$A=f_{xx}''(x_0,y_0),B=f_{xy}''(x_0,y_0),C=f_{yy}''(x_0,y_0)$，则

（1）当$B^2-AC<0$时，$f(x_0,y_0)$一定为极值，并且

当$A(或C)>0$时，$f(x_0,y_0)$为极小值；

当$A(或C)<0$时，$f(x_0,y_0)$为极大值.

（2）当$B^2-AC>0$时，$f(x_0,y_0)$不是极值.

（3）当$B^2-AC=0$时，无法判断，要进一步研究.

多元函数的最大值、最小值¶

$f(P)$在有界闭区域$G$上连续$\Longrightarrow f(P)$在$G$上一定能取到最大（小）值.

$\iff \exists P_1,P_2\in G,f(P_1)=m,f(P_2)=M,\forall P\in G$，有$m\le f(P)\le M$.

最值在

边界点取到（边界函数值最大值与最小值点）.

内部取到：一定是极值点（稳定点或偏导数不存在点）.

条件极值¶

条件极值的极值点的搜索范围受到条件的限制.

拉格朗日乘数法（拉格朗日乘子法）：

条件极值问题：$\max(\min)u=f(x,y,z)&(目标函数)\\s.t.\ G(x,y,z)=0&(约束函数)$.

设$f$和$G$具有连续的偏导数且${\large\frac{\partial G}{\partial z}}\ne0$，隐函数存在定理得方程$G(x,y,z)=0$确定隐函数$z=z(x,y)$.

将$z=z(x,y)$代入目标函数$f$ $$ \begin{align} &条件极值问题\ u=&f(x,y,z)极值\ s.t. & G(x,y,z)=0 \end{align} \Longrightarrow \begin{aligned} &无条件极值问题\ u=&f(x,y,z(x,y))极值 \end{aligned} $$ 设$(x_0,y_0)$为$u=f(x,y,z(x,y))$的极值点$(z_0=z(x_0,y_0))$，满足二元函数的极值必要条件 $$ \begin{cases} {\large\frac{\partial u}{\partial x}}=0\ {\large\frac{\partial u}{\partial y}}=0 \end{cases} \Longrightarrow \begin{cases} {\large\frac{\partial u}{\partial x}}={\large f_x'+f_z'\frac{\partial z}{\partial x}}={\large f_x'-\frac{G_x'}{G_z'}f_z'}=0\ {\large\frac{\partial u}{\partial y}}={\large f_y'+f_z'\frac{\partial z}{\partial y}}={\large f_y'-\frac{G_y'}{G_z'}f_z'}=0 \end{cases} $$ $\iff$所求问题的解$(x_0,y_0,z_0)$必须满足关系式 $$ \frac{f_x'(x_0,y_0,z_0)}{G_x'(x_0,y_0,z_0)}=\frac{f_y'(x_0,y_0,z_0)}{G_y'(x_0,y_0,z_0)}=\frac{f_z'(x_0,y_0,z_0)}{G_z'(x_0,y_0,z_0)}\stackrel{\small\triangle}{=}-\lambda $$ 综上所述，$(x_0,y_0,z_0)$是以下方程组的解 $$ \begin{cases} f_x'(x,y,z)+\lambda G_x'(x,y,z)=0\ f_y'(x,y,z)+\lambda G_y'(x,y,z)=0\ f_z'(x,y,z)+\lambda G_z'(x,y,z)=0\ G(x,y,z)=0 \end{cases} $$ 若函数$L(x,y,z,\lambda)\stackrel{\small\triangle}{=}f(x,y,z)+\lambda G(x,y,z)$存在极值点$(x^*,y^*,z^*,\lambda^*)$，则满足极值的必要条件 $$ \begin{cases} L_x'(x,y,z,\lambda)=f_x'(x,y,z)+\lambda G_x'(x,y,z)=0\ L_y'(x,y,z,\lambda)=f_y'(x,y,z)+\lambda G_y'(x,y,z)=0\ L_z'(x,y,z,\lambda)=f_z'(x,y,z)+\lambda G_z'(x,y,z)=0\ L_\lambda'(x,y,z,\lambda)=G(x,y,z)=0 \end{cases} $$ $\Longrightarrow(x^*,y^*,z^*)=(x_0,y_0,z_0)$.

注：

（1）最后一定要根据实际情况验证怀疑点是否是极值点（最值点）.

（2）上述函数$L(x,y,z,\lambda)$称为拉格朗日函数.

推广到多变量与多约束条件的情形：一般条件极值问题 $$ \max(\min) y=f(x_1,x_2,···,x_n)\ s.t. \psi_k(x_1,x_2,···,x_n)=0,k=1,2,···,m(m<n) $$ 此时拉格朗日函数为$L(x_1，x_2,···,x_n,\lambda_1,\lambda_2,···,\lambda_m)=f(x_1,x_2,···,x_n)+\Sigma_{k=1}^{m}\lambda_k\psi_k(x_1,x_2,···,x_n)$

其中$\lambda_1,\lambda_2,···\lambda_m$为拉格朗日乘数.

7 偏导数在几何上的应用¶

7.1 矢量函数的微分法¶

矢量函数的概念¶

如果把动点$P$与原点$O$连结起来就得到以原点为起点，动点为终点的矢量：$\vec r=\vec {OP}$（矢径），其坐标为$(x(t),y(t),z(t))$，则动点的运动方程为 $$ \vec r(t)=x(t)\vec i+y(t)\vec j+z(t)\vec k $$ 当$t$变动时，$\vec r$的模与方向一般都随之变动，即对每一个$t\in[\alpha,\beta]$，按上式都有唯一的矢量$\vec r$与之对应，这个对应规律称为矢值函数，记为$\vec r=\vec r(t)$，其动点$P(x,y,z)$画出的曲线，叫做矢值函数$\vec r=\vec r(t)$的矢端曲线，一般情况下$t$并不一定代表时间，而是在某一变化范围内的取值.

矢值函数的导数¶

设矢值函数$\vec r=\vec r(t)=x(t)\vec i+y(t)\vec j+z(t)\vec k,\Delta\vec r=\vec r(t+\Delta t)-\vec r(t)=\Delta x\vec i+\Delta y\vec j+\Delta z\vec k$.

定义：若极限$\lim_{\substack{\\\Delta t\to0}}\frac{\Delta x}{\Delta t},\lim_{\substack{\\\Delta t\to0}}\frac{\Delta y}{\Delta t},\lim_{\substack{\\\Delta t\to0}}\frac{\Delta z}{\Delta t}$都存在，则极限 $$ \lim_{\Delta t\to0}\frac{\Delta\vec r}{\Delta t}=x'(t)\vec i+y'(t)\vec j+z'(t)\vec k $$ 称为矢值函数$\vec r=\vec r(t)$在$t$处的导数（称矢量导数），记作$\vec r'(t)$或$\large\frac{d\vec r}{dt}$.

几何意义：$\large\frac{d\vec r}{dt}$为切线的方向向量$\iff$曲线在$P$点的切矢量.

7.2 空间曲线的切线与法平面¶

参数式方程的空间曲线的切线与法平面¶

设空间曲线的参数方程为$\begin{cases}x=x(t)\\y=y(t)\\z=z(t)\end{cases}$ 且$x'(t),y'(t),z'(t)$连续不同时为零，曲线上一点$P_0(x_0,y_0,z_0)\ \ (t=t_0)$处切矢量${\large\frac{d\vec r}{dt}}\bigg|_{t=t_0}=(x'(t_0),y'(t_0),z'(t_0))$，则曲线上点$P_0(x_0,y_0,z_0)$处的切线方程为 $$ \frac{x-x_0}{x'(t_0)}=\frac{y-y_0}{y'(t_0)}=\frac{z-z_0}{z'(t_0)} $$ 过点$P_0$且与该直线垂直的平面称为曲线在$P_0$点的法平面，则法平面方程为 $$ x'(t_0)(x-x_0)+y'(t_0)(y-y_0)+z'(t_0)(z-z_0)=0 $$

一般式方程的空间曲线的切线与法平面¶

曲线$\Gamma$：$\begin{cases}F(x,y,z)=0\\G(x,y,z)=0\end{cases}$ 在曲线上点$P_0(x_0,y_0,z_0)$处的切线方程.

曲线$\Gamma$在曲面$S_1:F(x,y,z)=0$上且经过曲面$S_1$上的点$P_0$

$\Longrightarrow$曲线$\Gamma$在$P_0$的切线必在曲面$S_1$在点$P_0$的切平面上

同理，曲线$\Gamma$在$P_0$的切线必在曲面$S_2:G(x,y,z)=0$在点$P_0$的切平面上.

故曲线$\Gamma$在点$P_0$处的切线方程为 $$ \begin{cases} F_x'(P_0)(x-x_0)+F_y'(P_0)(y-y_0)+F_z'(P_0)(z-z_0)=0\ G_x'(P_0)(x-x_0)+G_y'(P_0)(y-y_0)+G_z'(P_0)(z-z_0)=0 \end{cases} $$

7.3 空间曲面的切平面与法线¶

隐式方程的空间曲面的切平面与法向量¶

设已知曲面$S:F(x,y,z)=0$，$M_0(x_0,y_0,z_0)$是曲面$S$上一点，过点$M_0$在曲面$S$上可以作无数条曲线，这些曲线在该点分别都有切线，则这无数条切线都在同一个平面上（切平面）.

设过点$M_0$在曲面$S$上任作一曲线$\Gamma:\begin{cases}x=x(t)\\y=y(t)\\z=z(t)\end{cases}$，且$\begin{cases}x_0=x(t_0)\\y_0=y(t_0)\\z_0=z(t_0)\end{cases}$.

曲线$\Gamma$在曲面$S$上$\Longrightarrow F\big(x(t),y(t),z(t)\big)\equiv0$，对$t$求导： $$ F_x'\big(x(t),y(t),z(t)\big)x'(t)+F_y'\big(x(t),y(t),z(t)\big)y'(t)+F_z'\big(x(t),y(t),z(t))z'(t\big) $$ 把$t=t_0$代入上式得 $$ F_x'(x_0,y_0,z_0)x'(t_0)+F_y'(x_0,y_0,z_0)y'(t_0)+F_z'(x_0,y_0,z_0)z'(t_0)=0\ \iff\big(F_x'(x_0,y_0,z_0),F_y'(x_0,y_0,z_0),F_z'(x_0,y_0,z_0)\big)\big(x'(t_0),y'(t_0),z'(t_0)\big)=0 $$ $\Gamma$在点$M_0$处的切矢量$\vec v=\big(x'(t_0),y'(t_0),z'(t_0)\big)$，矢量$\vec n\stackrel{\small\triangle}{=}\big(F_x'(x_0,y_0,z_0),F_y'(x_0,y_0,z_0),F_z'(x_0,y_0,z_0)\big)$

$\Longrightarrow\vec v\cdot\vec n=0$.

$\Longrightarrow$曲面上过点$M_0$的任意一条曲线的切线都与矢量$\vec n$垂直.

$\Longrightarrow$矢量$\vec n$为曲面在点$M_0$的法线矢量（法矢量），切平面方程为 $$ F_x'(x_0,y_0,z_0)(x-x_0)+F_y'(x_0,y_0,z_0)(y-y_0)+F_z'(x_0,y_0,z_0)(z-z_0)=0 $$ 曲面在点$M_0$处存在切平面的充分条件：$F_x',F_y',F_z'$在$M_0$点具有连续的偏导数且$F_x'(x_0,y_0,z_0),F_y'(x_0,y_0,z_0),F_z'(x_0,y_0,z_0)$不同时为零.

显式方程的空间曲面的切平面与法向量¶

若曲面方程$z=f(x,y)$，则可将显式方程化为隐式方程求解.

参数式方程的空间曲面的切平面与法向量¶

设曲面参数方程$\begin{cases}x=x(u,v)\\y=y(u,v)\\z=z(u,v)\end{cases}$ 在$(u_0,v_0)$对应曲面上点$P_0(x_0,y_0,z_0)$，则由$\begin{cases}x=x(u,v)\\y=y(u,v)\end{cases}$ 可确定反函数组$\begin{cases}u=u(x,y)\\v=v(x,y)\end{cases}$ 代入$z=z(u,v)\Longrightarrow z$是$x,y$的函数.

由复合函数求导法则得 $$ {\large \begin{cases} \frac{\partial z}{\partial u} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial u} \ \frac{\partial z}{\partial v} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial v} \end{cases}} \Longrightarrow \frac{\partial z}{\partial x} = {\large\frac{\frac{\partial (z,y)}{\partial (u,v)}}{\frac{\partial (x,y)}{\partial (u,v)}}}, \frac{\partial z}{\partial y} = {\large\frac{\frac{\partial (x,z)}{\partial (u,v)}}{\frac{\partial (x,y)}{\partial (u,v)}}} $$ 令$F(x,y,z)=z[u(x,y),v(x,y)]-z$，则可转化为隐式方程求解.