【高等数学基础进阶】多元函数微分学-多元函数微分法

一、复合函数微分法

定理：设$u=u(x,y)$及$v=v(x,y)$在点$(x,y)$具有对$x$及对$y$的偏导数，函数$z=f(u,v)$在对应点$(u,v)$具有连续偏导数，那么复合函数$z=f[u(x,y),v(x,y)]$在点$(x,y)$的两个偏导数都存在，且有

$$

\begin{aligned}\frac{dz}{dx}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial x},\frac{dz}{dy}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial y}\end{aligned}

$$

graph TD

A(z) -- 这里 --- B(u)

A --- C(v)

B -- 这里 --- D((x))

B --- E(y)

C --- F((x)) & G(y)

           

例如该式$z=f[u(x,y),v(x,y)]$，画出树形图，做对$x$的偏导，树形图底部有几个$x$，就有几项（最底下两个圆的$x$）；连接到某个$x$有几条线，该式就有几个导数（标有这里的两条线表示该项有两个导数组成，即$\begin{aligned} \frac{dz}{dy} \frac{du}{dx}\end{aligned}$）

全微分形式的不变性

设函数$z=f(u,v),u=u(x,y),v=v(x,y)$都有连续的一阶偏导数，则复合函数$z=f[u(x,y),v(x,y)]$的全微分

$$

\begin{aligned}

dz&=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}y\

&=\left(\frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial x}\right)dx+\left(\frac{\partial z}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial y}\right)dy\

&=\frac{\partial z}{\partial u}\left(\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy\right)+\frac{\partial z}{\partial v}\left(\frac{\partial v}{\partial x}dx+\frac{\partial v}{\partial y}dy\right)\

&=\frac{\partial z}{\partial u}du+\frac{\partial z}{\partial v}dv

\end{aligned}

$$

二、隐函数微分法

由方程$F(x,y)=0$确定的隐函数$y=y(x)$

$$

y'=-\frac{F'{x}}{F'{y}}

$$

由方程$F(x,y,z)=0$确定的隐函数$z=z(x,y)$

若$F(x,y,z)$在点$P(x_{0},y_{0},z_{0})$的某一邻域内有连续偏导数，且$F(x_{0},y_{0},z_{0})=0,F'{z}(x{0},y_{0},z_{0})\ne 0$，则方程$F(x,y,z)=0$在点$(x_{0},y_{0},z_{0})$的某邻域可唯一确定一个有连续偏导数的函数$z=z(x,y)$，并有

$$

\frac{\partial z}{\partial x}=-\frac{F'{x}}{F'{z}},\frac{\partial z}{\partial y}=-\frac{F'{y}}{F'{z}}

$$

如果$F'{z}(x{0},y_{0},z_{0})\ne 0$，那么$z$就是$x,y$的函数，如果$F'{y}(x{0},y_{0},z_{0})\ne 0$。那么$y$就是$x,z$的函数，其余同理

这个很重要，这里先不解释，要看的话直接看最后

常考题型与典型例题

复合函数的偏导数与全微分

例1：设函数$\begin{aligned} F(x,y)=\int_{0}^{xy}\frac{\sin t}{1+t^{2}}dt\end{aligned}$，则$\begin{aligned} \frac{\partial F}{\partial x}=()\end{aligned}$

$$

\begin{aligned}

\frac{\partial F}{\partial x}&=\frac{y \sin xy}{1+x^{2}y^{2}}\

\frac{\partial ^{2}F}{\partial x^{2}}&=\frac{y^{2}\cos (xy)(1+x^{2}y^{2})-2xy^{3}\sin xy}{(1+x^{2}y^{2})^{2}}\

\frac{\partial ^{2}F}{\partial x^{2}}\Big|_{\substack{x=0\y=2}}^{}&=4

\end{aligned}

$$

由于是对$x$偏导，因此可以将$y$先带进去

$$

\begin{aligned}

F(x,2)&=\int_{0}^{2x}\frac{\sin t}{1+t^{2}}dt\

\frac{\partial F}{\partial x}&=\frac{2\sin 2x}{1+4x^{2}}\

\frac{\partial^{2} F}{\partial x^{2}}\Big|{\substack{x=0\ y=2}}^{}&=F{xx}(0,2)=\lim\limits_{x\to0}\frac{2\sin 2x}{x(1+4x^{2})}=4

\end{aligned}

$$

这里设$\begin{aligned} \frac{\partial F}{\partial x}=\phi (x)\end{aligned}$，由于$\phi (x)=0$，因此

$$\begin{aligned} \frac{\partial^{2} F}{\partial x^{2}}\Big|{\substack{x=0\ y=2}}^{}&=\phi (x)'\Big|{x=0}^{}=\phi '(0)\&=\lim\limits_{x\to0}\frac{\phi (x)-\phi (0)}{x}=\lim\limits_{x\to0}\frac{\phi (x)}{x}\end{aligned}$$

例2：设$\begin{aligned} z=\left(1+ \frac{x}{y}\right)^{\frac{x}{y}}\end{aligned}$，则$dz \Big|_{(1,3)}^{}=()$

$$

\begin{aligned}

dz \Big|{(1,1)}^{}&=z{x}(1,1)dx+z_{y}(1,1)dy\

z_{x}(x,1)&=\frac{d(1+ x)^{x}}{dx}=e^{x \ln (1+x)}\left(\ln (1+x)+ \frac{x}{1+x}\right)\

z_{x}(1,1)&=1+2\ln 2\

z_{y}(1,y)&=\frac{d(1+ \frac{1}{y})^{\frac{1}{y}}}{dy}\overset{令u=\frac{1}{y}}{=}\frac{d(1+u)^{u}}{du} \frac{du}{dy}\

z_{y}(1,1)&=(1+2 \ln 2)\cdot \frac{du}{dy}=(1+2 \ln 2)\cdot (- \frac{1}{y^{2}})=-1-2\ln 2\

dz \Big|_{(1,1)}^{}&=(1+2\ln 2)dx-(1+2\ln 2)dy

\end{aligned}

$$

例3：设函数$f(u,v)$具有$2$阶连续导数$y=f(e^{x},\cos x)$，求$\begin{aligned} \frac{dy}{dx}\Big|{x=0}^{},\frac{d^{2}y}{dx^{2}}\Big|{x=0}^{}\end{aligned}$

graph TD

A(y) --- B(u)

A --- C(v)

B --- D(x)

C --- F(x)

           

$$

\begin{aligned}

\frac{dy}{dx}&=f_{1}e^{x}+f_{2}(-\sin x)\

\frac{dy}{dx}\Big|{x=0}^{}&=f{1}(1,1)\

\frac{d^{2}y}{dx^{2}}&=e^{x}f_{1}+e^{x}f_{11}e^{x}+e^{x}f_{12}(-\sin x)-\cos x f_{2}-\sin x f_{21}(-\sin x)-\sin x f_{21}e^{x}\

\frac{d^{2}y}{dx^{2}}\Big|{x=0}^{}&=f{1}(1,1)+f_{11}(1,1)-f_{2}(1,1)

\end{aligned}

$$

一定注意$f_{1}$仍然是$f_{1}(e^{2},\cos x)$，因此对$x$求导，也是两项

例4：设函数$z=f(xy,yg(x))$，其中函数$f$具有二阶连续偏导数，函数$g(x)$可导且在$x=1$处取得极值$g(1)=1$，求$\begin{aligned} \frac{\partial^{2} z}{\partial x \partial y}\Big|_{\substack{x=1\ y=1}}^{}=()\end{aligned}$

graph TD

A(z) --- xy

xy --- 1(x)

xy --- 2(y)

A --- 3("yg(x)")

3 --- y

3 --- 4("g(x)")

4 --- x

           

$$

\begin{aligned}

\frac{\partial z}{\partial x}&=y f_{1}+yg'(x)f_{2}\

\frac{\partial^{2} z}{\partial x \partial y}&=f_{1}+(x f_{11}+g(x)f_{12})+g'(x)f_{2}+yg'(x)(xf_{21}+g(x)f_{22})\

\frac{\partial^{2} z}{\partial x \partial y}\Big|{\substack{x=1\ y=1}}^{}&=f{1}(1,1)+f_{11}(1,1)+f_{12}(1,1)

\end{aligned}

$$

也可以用先带后求

$$

\begin{aligned}

\frac{\partial z}{\partial x}&=y f_{1}+yg'(x)f_{2}\

\frac{\partial z}{\partial x}\Big|{\substack {x=1\g(1)=1\g'(1)=0}}^{}&=y f{1}(y,y)\

\frac{\partial^{2} z}{\partial x \partial y}&=\frac{\partial y f_{1}(y,y)}{\partial y}=f_{1}+y(f_{11}+f_{12})\

\frac{\partial^{2} z}{\partial x \partial y}\Big|{\substack{x=1\ y=1}}^{}&=f{1}(1,1)+f_{11}(1,1)+f_{12}(1,1)

\end{aligned}

$$

隐函数的偏导数与全微分

例5：若函数$z=z(x,y)$由方程$e^{x+2y+3z}+xyz=1$确定，则$dz \Big|_{(0,0)}^{}=()$

由$x=0,y=0$，知$z=0$。方程$e^{x+2y+3z}+xyz=1$两端微分，得

$$

\begin{aligned}

d(e^{x+2y+3z}+xyz)&=d1\

de^{x+2y+3z}+dxyz&=0\

e^{x+2y+3z}d(x+2y+3z)+dxyz&=0\

e^{x+2y+3z}(dx+2dy+3dz)+yzdx+xzdy+xydz&=0

\end{aligned}

$$

将$x=0,y=0,z=0$代入上式得

$$

dx+2dy+3dz=0

$$

则

$$

dz \Big|_{(0,0)}^{}=- \frac{1}{3}(dx+2dy)

$$

用该方法显然需要微分好求，如果第二项是$\begin{aligned} \frac{xyz}{\sqrt{1+x^{2}+y^{2}+z^{2} }}\end{aligned}$，显然该方法并不好

补充微分四则运算

$$\begin{aligned} d(f(x)+g(x))&=df(x)+dg(x)\d(f(x)-g(x))&=df(x)-dg(x)\d(f(x) \times g(x))&=g(x) \times df(x)+f(x) \times d g(x)\d \frac{f(x)}{g(x)}&=\frac{g(x) \times df(x)-f(x) \times dg(x)}{g(x)^{2}}\end{aligned}$$

有点类似于导数的四则运算

也可以分别求出$z_{x}(0,0),z_{y}(0,0)$

这里只是对$x$求一次偏导，显然$y$可以先代入

由$x=0,y=0$知$z=0$

$$

dz \Big|{(0,0)}^{}=z{x}(0,0)dx+z_{y}(0,0)dy

$$

在$e^{x+2y+3z}+xyz=1$中令$y=0$得，$e^{x+3z}=1$，两边对$x$求导得

$$

\begin{aligned}

e^{x+3z}(1+3z_{x})&=0\

z_{x}(0,0)&=- \frac{1}{3}

\end{aligned}

$$

同理可得$z_{y}(0,0)= - \frac{2}{3}$

则

$$

dz \Big|_{(0,0)}^{}=- \frac{1}{3}(dx+2dy)

$$

例6：已知$u+e^{u}=xy$，求$\begin{aligned} \frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial^{2} u}{\partial x \partial y}\end{aligned}$

求一阶偏导，一般有三种方法

两端同时对$x$求偏导

$$

\begin{aligned}

(1+e^{u})\frac{\partial u}{\partial x}&=y\

\frac{\partial u}{\partial x}&=\frac{y}{1+e^{u}}\

同理\quad \frac{\partial u}{\partial y}&=\frac{x}{1+e^{u}}

\end{aligned}

$$

注意此处由于是对$x$求偏导，显然$u$是$x,y$的函数，因此$u$不能看做常数

也可用定义法

$$

\frac{\partial u}{\partial x}=-\frac{F_{x}}{F_{u}}=\frac{y}{1+e^{u}}

$$

这里根据定义法要求，需要将所有式子移到一边，即

$$F(x,y,u)=u+e^{u}-xy$$

对任何一个变量求导，其他变量看做常数，例如

$$F_{x}=-y,F_{y}=-x,F_{u}=1+e^{u}$$

也可两边同时求微分，主要应用微分形式不变性

$$

(1+e^{u})du=ydx+xdy \Rightarrow du=\frac{y}{1+e^{u}}dx+ \frac{x}{1+e^{u}}dy

$$

个人理解，这里微分形式不变性，是指等式两边同时取微分，计算时看等式一边出现了哪些变量，则就有$d$哪些变量，然后前面的式子就是对该侧等式对该变量的微分。例如本题，两边同时取微分

$$d(u+e^{u})=dxy$$

看等式左边，出现了$u$，因此，就有$du$，然后左边的式子对$u$求偏导，即

$$1+e^{u}du=dxy$$

看等式右边，出现了$x,y$，因此，就有$dx,dy$，然后右边的式子对$x$求偏导，写在$dx$前，对$y$求偏导，写在$dy$前

$$(1+e^{u})du=ydx+xdy$$

整理可得

$$du=\frac{y}{1+e^{u}}dx+ \frac{x}{1+e^{u}}dy$$

$$

\frac{\partial^{2} u}{\partial x \partial y}=\frac{(1+e^{u})-e^{u}\frac{\partial u}{\partial y}y}{(1+e^{u})^{2}}=\frac{1}{1+e^{u}}- \frac{xye^{u}}{(1+e^{u})^{3}}

$$

例7：设函数$z=z(x,y)$由方程$\begin{aligned} F\left(\frac{y}{x},\frac{z}{x}\right)=0\end{aligned}$确定，其中$F$为可微函数，且$F_{2}'\ne 0$，则$\begin{aligned} x \frac{\partial z}{\partial x}+y \frac{\partial z }{\partial y}=()\end{aligned}$

该类题一般用定义法求偏导比较简单

例如对$x$，$\begin{aligned} F\left(\frac{y}{x},\frac{z}{x}\right)\end{aligned}$有三个位置都需要求导，因此无论是两边同时求导还是求微分，都需要对着三项求导，会比较麻烦

$$

\begin{aligned}

\frac{\partial z}{\partial x}&=-\frac{- \frac{y}{x^{2}}F_{1}- \frac{z}{x^{2}}F_{2}}{\frac{1}{x}F_{2}}\

\frac{\partial z}{\partial y}&=-\frac{\frac{1}{x}F_{1}}{\frac{1}{x}F_{2}}\

x \frac{\partial z}{\partial x}+y \frac{\partial z }{\partial y}&=-\frac{- \frac{y}{x^{2}}F_{1}- \frac{z}{x^{2}}F_{2}}{\frac{1}{x}F_{2}}-\frac{\frac{1}{x}F_{1}}{\frac{1}{x}F_{2}}=z

\end{aligned}

$$

例8：设$u=f(x,y,z)$有连续的一阶偏导数，又函数$y=y(x)$及$z=z(x)$分别由下列两式确定：$e^{xy}-xy=2$和$\begin{aligned} e^{x}=\int_{0}^{x-z}\frac{\sin t}{t}dt\end{aligned}$，求$\begin{aligned} \frac{du}{dx}\end{aligned}$

画树形图两边对$x$求导可以，这里不赘述

这里考虑两边同时取微分，利用微分形式不变性。微分形式不变性不需要考虑各变量之间的关系，要什么留什么

$$

du=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz

$$

如果用$f$对$x$，偏导，由于$y,z$是$x$的函数，写出来会比较麻烦，根据微分形式不变性，只需要同时取微分，不需要研究各变量之间的关系

例如本题，因为$dy$前是$\begin{aligned} \frac{\partial f}{\partial y}\end{aligned}$不涉及$x$，如果后面需要$x$，按照题目给出的，想办法将$dy$换成$dx$即可，不需要在该式上继续计算

$e^{xy}-xy=2$两端取微分

$$

e^{xy}(ydx+xdy)-(ydx+xdy)=0\Rightarrow dy=- \frac{y}{x}dx

$$

$\begin{aligned} e^{x}=\int_{0}^{x-z}\frac{\sin t}{t}dt\end{aligned}$两边取微分

$$

e^{x}dx=\frac{\sin (x-z)}{x-z}(dx-dz)\Rightarrow dz=\left(1-\frac{e^{x}(x-z)}{\sin (x-z)}\right)dx

$$

因此

$$

du=\left[\frac{\partial f}{\partial x}- \frac{y}{x}\frac{\partial f}{\partial y}+\left[1-\frac{e^{x}(x-z)}{\sin (x-z)}\right]\frac{\partial f}{\partial z}\right]dx

$$

例9：设$z=z(x,y)$是由方程$x^{2}+y^{2}-z=\phi (x+y+z)$所确定的函数，其中$\phi$具有二阶导数，且$\phi '\ne 1$

• 求$dz=()$
• $\begin{aligned} u(x,y)=\frac{1}{x-y}\left(\frac{\partial z}{\partial x} - \frac{\partial z}{\partial y}\right)\end{aligned}$，求$\begin{aligned} \frac{\partial u}{\partial x}\end{aligned}$

虽然$\phi (x+y+z)$其中的变量是$x+y+z$，但求导的时候仍然正常求，对$x$求导只需要看成符合了两次就行

设$F(x,y,z)=x^{2}+y^{2}-z-\phi (x+y+z)=0$

$$

\begin{aligned}

\frac{\partial z}{\partial x}&=-\frac{F_{x}}{F_{z}}=\frac{2x-\phi '}{1+\phi '}\

\frac{\partial z}{\partial y}&=-\frac{F_{y}}{F_{z}}=\frac{2y-\phi '}{1+\phi '}\

dz&=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy=\frac{1}{1+\phi '}[(2x-\phi ')dx+(2y-\phi ')dy]

\end{aligned}

$$

还可以用两边同时取微分，思路和上面基本相同

$$

\begin{aligned}

2xdx+2ydy-dz&=\phi '\cdot (dx+dy+dz)\

dz&=\frac{2x-\phi '}{1+\phi '}dx+\frac{2y-\phi '}{1+\phi '}dy

\end{aligned}

$$

由于 $\begin{aligned} u(x,y)=\frac{1}{x-y}\left(\frac{\partial z}{\partial x} - \frac{\partial z}{\partial y}\right)\end{aligned}$，所以

$$

\frac{\partial u}{\partial x}=\frac{-2}{(1+\phi ')^{2}}\left(1+\frac{\partial z}{\partial x}\right)\phi ''=-\frac{2(2x+1)\phi ''}{(1+\phi ')^{3}}

$$

说一下

$$\frac{\partial z}{\partial x}=-\frac{F_{x}}{F_{z}}$$

怎么来的

已知$z=(x,y)$由$F(x,y,z)=0$，或只给了$F(x,y,z)=0$

$$\begin{aligned} F(x,y,z)&=0\&两边同时对x求导\F_{x}+F_{z}\frac{\partial z}{\partial x}&=0\\frac{\partial z}{\partial x}&=-\frac{F_{x}}{F_{z}}\end{aligned}$$

这里的$F_{x}$可以理解为

$$\begin{aligned} \frac{\partial F}{\partial x}(x,y,z(x,y))\end{aligned}$$

而$\begin{aligned} F_{x}+F_{z}\frac{\partial z}{\partial x}\end{aligned}$可以理解为

$$\frac{\partial }{\partial x}(F(x,y,z(x,y)))$$

因此$F_{x}$是要把$y,z$看做常量

思路来源：

作者：盗铃人

链接：函数u=f(x,y,z),u对x的偏导与f对x的偏导有什么区别？ - 知乎 (zhihu.com)

如果题目中给出已知$z=z(x,y)$由$F(x,y,z)=0$确定，求偏导$\begin{aligned} \frac{\partial z}{\partial x}\end{aligned}$一定能用公式法，求$\begin{aligned} \frac{\partial F}{\partial x}\end{aligned}$就要把$y,z$看做常数

已知$z=z(x,y)$由$F(x,y,z)=0$确定，$z=z(x,y)$可以看做移项得来的（移项可能得不到），因此地位也相同，所以求$\begin{aligned} \frac{\partial F}{\partial x}\end{aligned}$也把$y,z$看做常数

如果只有$F(x,y,z)=0$，需要自己判断关系，是否有$x,y,z$其中一个符合定理，如果符合定理，求偏导$\begin{aligned} \frac{\partial z}{\partial x}\end{aligned}$就能用公式法，$\begin{aligned} \frac{\partial F}{\partial x}\end{aligned}$无论有没有隐函数关系，都要把$y,z$看做常数

上面提到的定理为

由方程$F(x,y,z)=0$确定的隐函数$z=z(x,y)$

若$F(x,y,z)$在点$P(x_{0},y_{0},z_{0})$的某一邻域内有连续偏导数，且$F(x_{0},y_{0},z_{0})=0,F'{z}(x{0},y_{0},z_{0})\ne 0$，则方程$F(x,y,z)=0$在点$(x_{0},y_{0},z_{0})$的某邻域可唯一确定一个有连续偏导数的函数$z=z(x,y)$，并有

$$\frac{\partial z}{\partial x}=-\frac{F'{x}}{F'{z}},\frac{\partial z}{\partial y}=-\frac{F'{y}}{F'{z}}$$

如果给出$z=z(x,y)$和$F(x,y,z)$，那么此时不能用公式法，此时$z$是$x,y$的因变量，求偏导$\begin{aligned} \frac{\partial F}{\partial x}\end{aligned}$时，$y$依旧可以看做常数，但$z$不是，需要求$\begin{aligned} \frac{\partial F}{\partial z}\frac{\partial z}{\partial x}\end{aligned}$，且$\begin{aligned} \frac{\partial z}{\partial x}\end{aligned}$不能用定义法