【数学二】多元函数微积分学-多元函数的微分

考试要求

1、了解多元函数的概念，了解二元函数的几何意义.
2、了解二元函数的极限与连续的概念，了解有界闭区域上二元连续函数的性质.
3、了解多元函数偏导数与全微分的概念，会求多元复合函数一阶、二阶偏导数，会求全微分，了解隐函数存在定理，会求多元隐函数的偏导数.
4、了解多元函数极值和条件极值的概念，掌握多元函数极值存在的必要条件，了解二元函数极值存在的充分条件，会求二元函数的极值，会用拉格朗日乘数法求条件极值，会求简单多元函数的最大值和最小.值，并会解决一些简单的应用问题.
5、了解二重积分的概念与基本性质，掌握二重积分的计算方法(直角坐标、极坐标).

二元函数的偏导数与全微分

偏导数定义 设函数$z=f(x,y)$在点$(x_0,y_0)$的某一领域内有定义，如果$$\underset{△x\to 0}{\lim }\frac{f(x_0+△x,y_0)-f(x_0,y_0))}{△x}$$存在，则称此极限为函数$z=f(x,y)$在点$(x_0,y_0)$处对$x$的偏导数，记为$f_x^{{}^{\prime }}(x_0,y_0)或\frac{\partial f(x_0,y_0)}{\partial x}$

类似地可定义$$f_y^{{}^{\prime }}(x_0,y_0)=\underset{△y\to 0}{\lim }\frac{f(x_0,y_0+△y)-f(x_0,y_0)}{△y}$$也可记为$\frac{\partial f(x_0,y_0)}{\partial y}$


偏导数的几何意义 偏导数$f_x^{{}^{\prime }}(x_0,y_0)$在几何上表示曲面$z=f(x,y)$与平面$y=y_0$的交线在点$M_0(x_0,y_0,f(x_0,y_0))$处的切线$T_x$对$x$轴的斜率，如下图，$f_x^{{}^{\prime }}(x_0,y_0)=\tan \alpha$

偏导数$f_y^{{}^{\prime }}(x_0,y_0)$在几何上表示曲面$z=f(x,y)$与平面$x=x_0$的交线在点$M_0(x_0,y_0,f(x_0,y_0))$处的切线$T_y$对$y$轴的斜率，如下图，$f_y^{{}^{\prime }}(x_0,y_0)=\tan \beta$

全微分定义 如果函数$z=f(x,y)$在点$(x,y)$处的全增量$△z=f(x+△x,y+△y)-f(x,y)$可表示为$$△z=A△x+B△y+o(\rho )$$
其中$A,B$不依赖于$△x，△y$，仅于$x,y$有关，$\rho =\sqrt{(△x{)}^2+(△y{)}^2}$，则称函数$z=f(x,y)$在点$(x,y)$可微，$A△x+B△y$称为函数 $z=f(x,y)$在点$(x,y)$的全微分，记为$$dz=A△x+B△y$$

$△z=dz+o(\rho )$

全微分存在的必要定理 如果函数$z=f(x,y)$在点$(x,y)$处可微，则该函数在点$(x,y)$处的偏导数$\frac{\partial z}{\partial x}，\frac{\partial z}{\partial y}$ 必定存在，且$$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$$

简记：$可微\Rightarrow 偏导数\frac{\partial z}{\partial x}，\frac{\partial z}{\partial y}必定存在$

全微分存在的充分定理 如果函数$z=f(x,y)$的偏导数$\frac{\partial z}{\partial x}，\frac{\partial z}{\partial y}$ 在点$(x,y)$处连续，则函数$z=f(x,y)$在该点可微

简记：$偏导数存在+偏导连续\Rightarrow 可微$

练习1：设$z=(\sin y^3+x^2)(x+\sin y^5{)}^{\frac{y^3}{x^2}+e^{x^{3}y}}$，求$\frac{\partial z}{\partial x}{|}_{(1,0)}$

解：$$\begin{gathered} z(x,0)=(\sin y^3+x^2)(x+\sin y^5{)}^{\frac{y^3}{x^2}+e^{x^{3}y}}=x^3 \\ \frac{\partial z}{\partial x}=3x^2 \\ \frac{\partial z}{\partial x}{|}_{(1,0)}=3 \end{gathered}$$

练习2：二元函数$f(x,y)=\{\begin{array}{l}\frac{xy}{x^3+y^2},x^2+y^2\ne 0\\ \\ 0,x^2+y^2=0\end{array}$在$(0,0)$处连续、偏导数情况？

解：$$\begin{gathered} 当(x,y)沿y=kx趋向于(0,0)是，\underset{x\to 0,y\to 0}{\lim }f(x,y)=\underset{x\to 0}{\lim }\frac{k{x}^2}{x^3+k^2{x}^2}=\frac{1}{k} \\ f(x,y)在k取不同值时，极限不唯一 \\ 故在(0,0)极限不存在，所以不连续 \\ 由偏导定义可知：f_x^{{}^{\prime }}(0,0)=\underset{△x\to 0}{\lim }\frac{f(0+△x,0)-f(0,0)}{△x}=\underset{△x\to 0}{\lim }\frac{y△x}{△x((△x{)}^3+y^2)}=0 \\ 同理可得：f_y^{{}^{\prime }}(0,0)=0\Rightarrow f(x,y)在(0,0)偏导数存在 \\ 综上可知：f(x,y)在(0,0)处不连续，偏导数存在 \end{gathered}$$

练习3：极限${\lim }_{(x,y)\to (0,0)}\frac{f(x,y)-f(0,0)}{\sqrt{x^2+y^2}}=0$是二元函数$f(x,y)$处可微的?
$$\begin{gathered} A、充分非必要条件B、必要非充分条件 \\ C、充分必要条件D、即不充分也不必要 \end{gathered}$$

解：$$\begin{gathered} 全微分存在的充分定理：偏导存在且连续\Rightarrow 可微 \\ {f}_y^{{}^{\prime }}(0,0)=\underset{y\to 0}{\lim }\frac{f(0,y)-f(0,0)}{y}=\underset{y\to 0}{\lim }\frac{f(0,y)-f(0,0)}{\sqrt{y^2+0}}\frac{\sqrt{y^2+0}}{y}=0 \\ 同理可得：f_x^{{}^{\prime }}(0,0)=0\Rightarrow f(x,y)处可微 \\ 由可微偏导数必存在可知： \\ 取f(x,y)=x+y，函数f(x,y)在(0,0)处可微 \\ \underset{(x,y)\to (0,0)}{\lim }\frac{x+y}{\sqrt{x^2+y^2}} \\ (x,y)沿着y=kx趋向于(0,0)时： \\ \underset{(x,y)\to (0,0)}{\lim }\frac{x+y}{\sqrt{x^2+y^2}}=\frac{1+k}{\sqrt{1+k^2}} \\ 在k取不同值时没有唯一常数，故极限不存在 \\ 故：\underset{(x,y)\to (0,0)}{\lim }\frac{f(x,y)-f(0,0)}{\sqrt{x^2+y^2}}=0不是必要条件 \\ \underset{(x,y)\to (0,0)}{\lim }\frac{f(x,y)-f(0,0)}{\sqrt{x^2+y^2}}=0是f(x,y)在点(0,0)处充分非必要条件，选A \end{gathered}$$

练习4：函数$f(x,y)=\{\begin{array}{l}(x^2+y^2)\sin \frac{1}{x^2+y^2},x^2+y^2\ne 0\\ \\ 0,x^2+y^2=0\end{array}$在$(0,0)$点是否可微？

解：$$\begin{gathered} 由全微分存在的充分定理可知：（0，0）处偏导存在且偏导在（0，0）连续 \\ 由偏导定义可知：f_y^{{}^{\prime }}(0,0)=\underset{y\to 0}{\lim }\frac{f(0,y)-f(0,0)}{y}=\underset{y\to 0}{\lim }\frac{y^2\sin \frac{1}{y^2}}{y}=0(无穷小与有界函数积为无穷小) \\ 同理可得：f_x^{{}^{\prime }}(0,0)=0，综上可得：f(x,y)在(0,0)点可微 \end{gathered}$$

复合函数的偏导数与全微分

复合函数求导法则

多元函数与一元函数的复合 如果函数$u=\phi (t),v=\psi (t)$都在点$t$可导，函数$z=f(u,v)$在对应点$(u,v)$具有连续一阶偏导数，则复合函数$z=f(\phi (t),\psi (t))$在点$t$可导，且$$\frac{dz}{dt}=\frac{dz}{du}\frac{du}{dt}+\frac{dz}{dv}\frac{dv}{dt}$$

函数$z=f(\phi (t),\psi (t))$仅是$t$的一元函数，这里的$\frac{dz}{dt}$称为全导数

多元函数与多元函数的复合

如果函数$u=\phi (x,y),v=\psi (x,y)$都在点$(x,y)$可导，函数$z=f(u,v)$在对应点$(u,v)$具有连续一阶偏导数，则复合函数$z=f(\phi (x,y),\psi (x,y))$在点$(x,y)$对$x,y$的偏导数，且$$\begin{gathered} \frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial x} \\ \frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial y} \end{gathered}$$

练习1：设$z=f(xy,x^2-y^2)$，其中$f$可微，$\frac{\partial z}{\partial x}=$?

解: $$\begin{gathered} 由多元函数与多元函数的复合定义可知 \\ 令u(x,y)=xy,v(x,y)=x^2-y^2 \\ 由公式：\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial x} \\ \frac{\partial z}{\partial x}=y\frac{\partial z}{\partial u}+2x\frac{\partial z}{\partial v} \end{gathered}$$

全微分形式不变性

设$z=f(u,v)$和$u=\phi (x,y),v=\psi (x,y)$都具有连续一阶偏导数，则复合函数$z=f(\phi (x,y),\psi (x,y))$可微，且$$\begin{gathered} dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy \\ 由多元函数与多元函数复合求导规则可知 \\ dz=\frac{\partial z}{\partial u}du+\frac{\partial z}{\partial v}dv \end{gathered}$$
由此可见，无论把$z$看作自变量$x$和$y$的函数，它的微分$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy和dz=\frac{\partial z}{\partial u}du+\frac{\partial z}{\partial v}dv$具有同样的形式，这个性质叫全微分形式不变性

高阶偏导数

设函数$z=f(x,y)$在区域$D$内具有偏导数，$$\frac{\partial z}{\partial x}=f_x^{{}^{\prime }}(x,y)，\frac{\partial z}{\partial y}=f_y^{{}^{\prime }}(x,y)$$

如果$f_x^{{}^{\prime }}(x,y)$和$f_y^{{}^{\prime }}(x,y)$的偏导数也存在，则称它们是函数$z=f(x,y)$的二阶导数，二阶导数有以下四个：$$\begin{gathered} \frac{{\partial }^{2}z}{\partial x^2}=\frac{\partial }{\partial x}\frac{\partial z}{\partial x}=f_{xx}^{{}^{\prime \prime }}(x,y)，\frac{{\partial }^{2}z}{\partial x\partial y}=\frac{\partial }{\partial y}\frac{\partial z}{\partial x}=f_{xy}^{{}^{\prime \prime }}(x,y) \\ \frac{{\partial }^{2}z}{\partial y^2}=\frac{\partial }{\partial y}\frac{\partial z}{\partial y}=f_{yy}^{{}^{\prime \prime }}(x,y)，\frac{{\partial }^{2}z}{\partial y\partial x}=\frac{\partial }{\partial x}\frac{\partial z}{\partial y}=f_{yx}^{{}^{\prime \prime }}(x,y) \end{gathered}$$
其中$\frac{{\partial }^{2}z}{\partial x\partial y}$和$\frac{{\partial }^{2}z}{\partial y\partial x}$称为混合偏导数，二阶及二阶以上的偏导数统称为高阶偏导数。

混合偏导数相等判定定理 若函数$z=f(x,y)$的两个混合偏导数$\frac{{\partial }^{2}z}{\partial x\partial y}$和$\frac{{\partial }^{2}z}{\partial y\partial x}$在点$(x_0,y_0)$都连续，则在$(x_0,y_0)$点$\frac{{\partial }^{2}z}{\partial x\partial y}=\frac{{\partial }^{2}z}{\partial y\partial x}$

练习1： 设$z=\frac{y}{x}+xyf(\frac{y}{x})，f$二阶可微，则$\frac{{\partial }^{2}z}{\partial x\partial y}=$?

解:$$\begin{gathered} \frac{\partial z}{\partial x}=-\frac{y}{x^2}+yf(\frac{y}{x})+xy{f}^{{}^{\prime }}(\frac{y}{x})(-\frac{y}{x^2}) \\ \frac{{\partial }^{2}z}{\partial x\partial y}=\frac{\partial }{\partial y}\frac{\partial z}{\partial x}=-\frac{1}{x^2}+f(\frac{y}{x})+y{f}^{{}^{\prime }}(\frac{y}{x})\frac{1}{x}-[(\frac{2y}{x})f^{{}^{\prime }}(\frac{y}{x})+(\frac{y^2}{x^2})f^{{}^{\prime \prime }}(\frac{y}{x})] \\ =-\frac{1}{x^2}+f(\frac{y}{x})-f^{{}^{\prime }}(\frac{y}{x})\frac{y}{x}-(\frac{y^2}{x^2})f^{{}^{\prime \prime }}(\frac{y}{x}) \end{gathered}$$

练习2：设函数$F(x,y)={\int }_0^{xy}\frac{\sin t}{1+t^2}dt$,则$\frac{{\partial }^{2}F}{\partial x^2}{|}_{x=0,y=2}=$?
知识点：$d(\int f(x)dx)=f(x)dx$

解：$$\begin{gathered} \frac{\partial F}{\partial x}=y\frac{\sin xy}{1+(xy{)}^2} \\ \frac{{\partial }^{2}F}{\partial x^2}=\frac{y^2\cos xy(1+(xy{)}^2)-y\sin xy(2x{y}^2)}{(1+(xy{)}^2{)}^2} \\ \frac{{\partial }^{2}F}{\partial x^2}{|}_{x=0,y=2}=4 \end{gathered}$$

练习3：设$\mu =e^{-x}\sin \frac{x}{y}$，则$\frac{{\partial }^2\mu }{\partial x\partial y}{|}_{x=2,y=\frac{1}{\pi }}=$？

解： $$\begin{gathered} \frac{\partial \mu }{\partial x}=-e^{-x}\sin \frac{x}{y}+\frac{1}{y}{e}^{-x}\cos \frac{x}{y} \\ \frac{\partial \mu }{\partial x}{|}_{x=2,y}=e^{-2}(\frac{1}{y}\cos \frac{2}{y}-\sin \frac{2}{y}) \\ \frac{{\partial }^2\mu }{\partial x\partial y}{|}_{x=2,y=\frac{1}{\pi }}=e^{-2}(-\frac{1}{y^2}\cos \frac{2}{y}+\frac{2}{y^3}\sin \frac{2}{y}+\frac{2}{y^2}\cos \frac{2}{y}) \\ ={\pi }^2{e}^{-2} \end{gathered}$$

隐函数的偏导数与全微分

1、由一个方程确定的隐函数 (一元函数) 求导法
设$F(x,y)$有连续一阶偏导数，且$F_y^{{}^{\prime }}\ne 0$，则由方程$F(x,y)=0$ 确定的函数$y=y(x)$可导，且$$\frac{dy}{dx}=-\frac{F_x^{{}^{\prime }}}{F_y^{{}^{\prime }}}$$

2、 由一个方程式确定的隐函数（二元函数）求导法
设$F(x,y,z)$有连续一阶偏导数，且$F_z^{{}^{\prime }}\ne 0,z=z(x,y)$由方程$F(x,y,z)=0$所确定，则$$\frac{\partial z}{\partial x}=-\frac{F_x^{{}^{\prime }}}{F_z^{{}^{\prime }}},\frac{\partial z}{\partial y}=-\frac{F_y^{{}^{\prime }}}{F_z^{{}^{\prime }}}$$

3、由方程组所确定的隐函数 (一元函数) 求导法
设$u=u(x)，v=v(x)$由方程组$\{\begin{array}{l}F(x,u,v)=0\\ \\ G(x,u,v)=0\end{array}$所确定，要求$\frac{du}{dx}$和$\frac{dv}{dx}$，可通过原方程组两端对$x$求导得到，即$$\{\begin{array}{l}{F}_x^{{}^{\prime }}+F_u^{{}^{\prime }}\frac{du}{dx}+F_v^{{}^{\prime }}\frac{dv}{dx}=0，\\ \\ G_x^{{}^{\prime }}+G_u^{{}^{\prime }}\frac{du}{dx}+G_v^{{}^{\prime }}\frac{dv}{dx}=0\end{array}$$然后从上方程组中解出$\frac{du}{dx}$和$\frac{dv}{dx}$


4、由方程组所确定的隐函数 (二元函数) 求导法
设$u=u(x,y)，v=v(x,y)$由方程组$\{\begin{array}{l}F(x,y,u,v)=0\\ \\ G(x,y,u,v)=0\end{array}$所确定，要求$\frac{\partial u}{\partial x}$和$\frac{\partial v}{\partial x}$，可通过原方程组两端对$x$求偏导得到，即$$\{\begin{array}{l}{F}_x^{{}^{\prime }}+F_u^{{}^{\prime }}\frac{\partial u}{\partial x}+F_v^{{}^{\prime }}\frac{\partial v}{\partial x}=0，\\ \\ G_x^{{}^{\prime }}+G_u^{{}^{\prime }}\frac{\partial u}{\partial x}+G_v^{{}^{\prime }}\frac{\partial v}{\partial x}=0\end{array}$$然后从上方程组中解出$\frac{\partial u}{\partial x}$和$\frac{\partial v}{\partial x}$,同理可求得$\frac{\partial u}{\partial y}$和$\frac{\partial v}{\partial y}$

练习1 ：设函数$z=z(x,y)$由方程$xyz+\sqrt{x^2+y^2+z^2}=\sqrt{2}$确定，求$z(x,y)$在$(1,0,-1)$点处的全微分

解：$$\begin{gathered} 令F(x,y,z)=xyz+\sqrt{x^2+y^2+z^2}-\sqrt{2} \\ \{\begin{array}{l}\frac{\partial z}{\partial x}=-\frac{F_x^{{}^{\prime }}}{F_z^{{}^{\prime }}}=-\frac{yz+\frac{x}{\sqrt{x^2+y^2+z^2}}}{yx+\frac{z}{\sqrt{x^2+y^2+z^2}}}=1,\\ \\ \frac{\partial z}{\partial y}=-\frac{F_y^{{}^{\prime }}}{F_z^{{}^{\prime }}}=-\frac{xz+\frac{y}{\sqrt{x^2+y^2+z^2}}}{yx+\frac{z}{\sqrt{x^2+y^2+z^2}}}=-\sqrt{2}\end{array} \\ dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy \\ z(x,y)在(1,0,-1)点处的全微分dz=dx-\sqrt{2}dy \end{gathered}$$

练习2：若函数$z=z(x,y)$由方程$e^{x+2y+3z}+xyz=1$确定，则$dz{|}_{(}0,0)=$?

解：$$\begin{gathered} 令F(x,y,z)=e^{x+2y+3z}+xyz-1由x=y=0解得z=0 \\ \{\begin{array}{l}\frac{\partial z}{\partial x}=-\frac{F_x^{{}^{\prime }}}{F_z^{{}^{\prime }}}=-\frac{e^{x+2y+3z}+yz}{3e^{x+2y+3z}+yx},\\ \\ \frac{\partial z}{\partial y}=-\frac{F_y^{{}^{\prime }}}{F_z^{{}^{\prime }}}=-\frac{2e^{x+2y+3z}+xz}{3e^{x+2y+3z}+yx}\end{array} \\ dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy \\ z(x,y)在(0,0,0)点处的全微分dz=-\frac{1}{3}dx-\frac{2}{3}dy \end{gathered}$$

练习3： 设函数$z=z(x,y)$由参数方程 $\{\begin{array}{l}x=u\cos v\\ \\ y=u\sin v\\ \\ z=uv\end{array}$给出，求 $\frac{\partial z}{\partial x}$

解：$$\begin{gathered} 由参数方程可得出z=uv可知\frac{\partial z}{\partial x}=u\frac{\partial v}{\partial x}+v\frac{\partial u}{\partial x} \\ \{\begin{array}{l}x-u\cos v=0\Rightarrow 1-\frac{\partial u}{\partial x}\cos v+u\sin v\frac{\partial v}{\partial x}=0\\ \\ y-u\sin v=0\Rightarrow 0-\frac{\partial u}{\partial x}\sin v-u\cos v\frac{\partial v}{\partial x}=0\end{array} \\ 解得：\frac{\partial v}{\partial x}=-\frac{\sin v}{u}，\frac{\partial u}{\partial x}=\cos v \\ 故：\frac{\partial z}{\partial x}=-\sin v+v\cos v \end{gathered}$$

练习4：设$z=f(x+y,x-y),f(x,y)$的二阶偏导连续，则$\frac{{\partial }^{2}z}{\partial x\partial y}=$?

解： $$\begin{gathered} 令u(x,y)=x+y，v(x,y)=x-y，则z=f(u,v) \\ \frac{\partial z}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}=f_u{}^{\prime }+f_v{}^{\prime } \\ \frac{{\partial }^{2}z}{\partial x\partial y}=\frac{\partial }{\partial y}\frac{\partial z}{\partial x}=\frac{\partial }{\partial y}(f_u{}^{\prime }(u,v)+f_v{}^{\prime }(u,v)) \\ =\frac{\partial f_u{}^{\prime }(u,v)}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f_u{}^{\prime }(u,v)}{\partial v}\frac{\partial v}{\partial y}+\frac{\partial f_v{}^{\prime }(u,v)}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f_v{}^{\prime }(u,v)}{\partial v}\frac{\partial v}{\partial y} \\ =f_{uu}{}^{\prime \prime }(u,v)-f_{uv}{}^{\prime \prime }(u,v)+f_{vu}{}^{\prime \prime }(u,v)-f_{vv}{}^{\prime \prime }(u,v) \\ 由二阶偏导连续，混合偏导数相等：f_{vu}{}^{\prime \prime }(u,v)=f_{uv}{}^{\prime \prime }(u,v) \\ 故：\frac{{\partial }^{2}z}{\partial x\partial y}=f_{uu}{}^{\prime \prime }(u,v)-f_{vv}{}^{\prime \prime }(u,v) \\ 由全微分形式不变性可调整为： \\ \frac{{\partial }^{2}z}{\partial x\partial y}=f_{xx}{}^{\prime \prime }(x,y)-f_{yy}{}^{\prime \prime }(x,y) \end{gathered}$$