复合函数的偏导

多变量函数的微分学

张瑞
中国科学技术大学数学科学学院

复合函数的偏导

复合函数偏导数的链式法则

类似单变量函数的链式法则。

设$D,D'$为$\mathbb{R}^2$中的区域，$f:D\to D'$，$g:D'\to\mathbb{R}$为两个映射。若$f(D)\subset D'$，则存在复合映射$g\circ f: D\to\mathbb{R}$，

\[D\xrightarrow {f}D'\xrightarrow {g}\mathbb{R} \]

记$f(x_1,x_2)=(y_1,y_2)$，$g(y_1,y_2)=z$，则

\[z=g\circ f(x_1,x_2)=g(y_1(x_1,x_2),y_2(x_1,x_2)) \]
% 链式法则 \begin{tikzpicture}[scale=1.5] \node (u) at (0,0) {$z$}; \node (xi) at (1,0.5) {$y_1$}; \node (eta) at (1,-0.5) {$y_2$}; \node (x) at (2,0.5) {$x_1$}; \node (y) at (2,-0.5) {$x_2$}; \draw[->] (u) -- (xi); \draw[->] (u) -- (eta); \draw[->] (eta) -- (x); \draw[->] (eta) -- (y); \draw[->] (xi) -- (x); \draw[->] (xi) -- (y); \end{tikzpicture}

定理 1.
设函数$f,g$如前所述，则

  1. 若$f,g$分别在$D, D'$中可微，则复合函数$g\circ f$在$D$中可微；
  2. 复合函数$z=g\circ f(x_1,x_2)$关于$x_1, x_2$的偏导数满足：
\[\begin{aligned} \dfrac{\partial z}{\partial x_1}=\frac{\partial z}{\partial y_1}\frac{\partial y_1}{\partial x_1}+\frac{\partial z}{\partial y_2}\frac{\partial y_2}{\partial x_1} \\ \dfrac{\partial z}{\partial x_2}=\frac{\partial z}{\partial y_1}\frac{\partial y_1}{\partial x_2}+\frac{\partial z}{\partial y_2}\frac{\partial y_2}{\partial x_2} \end{aligned} \]
% 链式法则 \begin{tikzpicture}[scale=1.5] \node (u) at (0,0) {$z$}; \node (xi) at (1,0.5) {$y_1$}; \node (eta) at (1,-0.5) {$y_2$}; \node (x) at (2,0.5) {$x_1$}; \node (y) at (2,-0.5) {$x_2$}; \draw[->] (u) -- (xi); \draw[->] (u) -- (eta); \draw[->] (eta) -- (x); \draw[->] (eta) -- (y); \draw[->] (xi) -- (x); \draw[->] (xi) -- (y); \end{tikzpicture}

证明：

例 1. (例6.3.3) 计算：

（1） 设$u=f(x,y,z),y=y(x), z=z(x)$，求$\frac{\partial u}{\partial x}$

\begin{tikzpicture}[scale=1.5] \node (u) at (0,0) {$u$}; \node (y) at (1,0.5) {$y$}; \node (z) at (1,-0.5) {$z$}; \node (x) at (2,0) {$x$}; %\node (r) at (2,0.5) {$r$}; % \draw[->] (u) -- (x); \draw[->] (u) -- (y); \draw[->] (u) -- (z); \draw[->] (y) -- (x); %\draw[->] (y) -- (r); %\draw[->] (z) -- (y); \draw[->] (z) -- (x); %\draw[->] (z) -- (r); \end{tikzpicture}

（2） 设$u=f(x,\phi(x,t),\psi(x,t))$，求$\frac{\partial u}{\partial x}$

\begin{tikzpicture}[scale=1.5] \node (u) at (0,0) {$u$}; \node (y) at (1,0.5) {$y=\phi$}; \node (z) at (1,-0.5) {$z=\psi$}; \node (x) at (2,0) {$x$}; \node (r) at (2,0.5) {$t$}; % \draw[->] (u) -- (x); \draw[->] (u) -- (y); \draw[->] (u) -- (z); \draw[->] (y) -- (x); \draw[->] (y) -- (r); %\draw[->] (z) -- (y); \draw[->] (z) -- (x); \draw[->] (z) -- (r); \end{tikzpicture}

例 2. 可微函数$u=f(x,y)$通过极坐标变换

\[\begin{cases} x=r\cos\theta \\ y=r\sin\theta \end{cases} \]

可以看作$r,\theta$的函数，证明：

\[\left(\frac{\partial u}{\partial x}\right)^2 +\left(\frac{\partial u}{\partial y}\right)^2 =\left(\frac{\partial u}{\partial r}\right)^2 +\frac1{r^2}\left(\frac{\partial u}{\partial \theta}\right)^2 \]

例2:

\[\begin{aligned} \frac{\partial u}{\partial r} =&\frac{\partial u}{\partial x}\frac{\partial x}{\partial r} +\frac{\partial u}{\partial y}\frac{\partial y}{\partial r} \\ =&\frac{\partial u}{\partial x}\cos\theta +\frac{\partial u}{\partial y}\sin\theta \\ \frac{\partial u}{\partial \theta} =&\frac{\partial u}{\partial x}\frac{\partial x}{\partial \theta} +\frac{\partial u}{\partial y}\frac{\partial y}{\partial \theta} \\ =&\frac{\partial u}{\partial x}(-r\sin\theta) +\frac{\partial u}{\partial y}(r\cos\theta) \\ \end{aligned} \]

则

\[\left(\frac{\partial u}{\partial r}\right)^2 +\left(\frac1r\frac{\partial u}{\partial \theta}\right)^2 =\left(\frac{\partial u}{\partial x}\right)^2 +\left(\frac{\partial u}{\partial y}\right)^2 \]

一阶微分的形式不变式

  • 对于一元函数$y=f(x)$，无论$x$是自变量，还是中间变量，都有
    \[dy=f'(x)dx \]
    这是一元微分的形式不变性。

  • 同样，对二元函数$z=f(x,y)$，无论$x,y$是自变量，还是中间变量，都有

    \[dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy \]

若有$x=\phi(r,s), y=\psi(r,s)$，则$z=f(\phi(r,s),\psi(r,s))$的微分

\[\begin{aligned} dz=&\frac{\partial z}{\partial r}dr+\frac{\partial z}{\partial s}ds \\ =&\left(\frac{\partial z}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial r}\right)dr +\left(\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s}\right)ds \\ =&\frac{\partial z}{\partial x}\left(\frac{\partial x}{\partial r}dr+\frac{\partial x}{\partial s}ds\right) +\frac{\partial z}{\partial y}\left(\frac{\partial y}{\partial r}dr+\frac{\partial y}{\partial s}ds\right) \\ =&\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy \end{aligned} \]

对于$z=f(x,y)$，有微分

\[dz = \begin{pmatrix} f'_x & f'_y \end{pmatrix} \begin{pmatrix} dx \\ dy \end{pmatrix} \]

若有$x=\phi(r,s), y=\psi(r,s)$，则

\[\begin{pmatrix} dx \\ dy \end{pmatrix} =\begin{pmatrix} x'_r & x'_s \\ y'_r & y'_s \end{pmatrix} \begin{pmatrix} dr \\ ds \end{pmatrix} \]

由一阶微分的形式不变性，可以得到

\[dz = \begin{pmatrix} f'_x & f'_y \end{pmatrix} \begin{pmatrix} x'_r & x'_s \\ y'_r & y'_s \end{pmatrix} \begin{pmatrix} dr \\ ds \end{pmatrix} \]

定理 2.
设$u=u(x,y), v=v(x,y)$为可微的二元函数，则

(1) $d(u+v)=du+dv$

(2) $d(uv)=udv+vdu$

(3) $d(\frac{u}v)=\frac{vdu-udv}{v^2} , v\neq 0$

证明与1维的情形类似。

例 3. (例6.3.6) 设$u=f(x,\xi,\eta)$,

$\xi=x^2+y^2$,$\eta=x^2+y^2+z^2$，求复合函数的全微分$du$

例 4. 设$u=f(\xi,\eta)=f(xy,\frac{x}{y})$，求$du$

\[du=f'_1d\xi+f'_2d\eta =f'_1\cdot(ydx+xdy)+f'_2\cdot\left(\frac1ydx-\frac{x}{y^2}dy\right) \]

例 5. 设$u=f(t,t^2,t^3)$，求$du$

\[du = f'_1dt+f'_2d(t^2)+f'_3d(t^3) \]

微分中值定理

定理 3. (微分中值定理)
设$f(x,y)$在区域$D$中可微。若连接$(x_0,y_0)$和$(x_0+h,y_0+k)$的线段包含在$D$中，则存在$\theta\in(0,1)$，使得

\[\begin{aligned} f(x_0+h,y_0+k)=f(x_0,y_0)+f'_x(x_0+\theta h,y_0+\theta k)h \\ +f'_y(x_0+\theta h,y_0+\theta k)k \end{aligned} \]

若记$\vec x=(x_1,x_2), \vec h=(h_1,h_2)$，则二元函数的中值定理可以写成

\[f(\vec x+\vec h)=f(\vec x)+Jf(\vec x+\theta \vec h)\cdot \vec h^T , \theta\in(0,1) \]

复合函数的高阶导数

反复运用复合函数求导的链式法则

例 6. (例6.3.4) 求复合函数$z=f(xy,\frac{y}x)$的三个二阶偏导数。

\[\frac{\partial z}{\partial x} =yf'_1-\frac{y}{x^2}f'_2, \quad \frac{\partial z}{\partial y} =xf'_1+\frac{1}{x}f'_2 \]
\[\begin{aligned} \frac{\partial^2 z}{\partial x^2} =&\frac{\partial}{\partial x}\left(yf'_1(xy,\frac{y}x)-\frac{y}{x^2}f'_2(xy,\frac{y}x)\right) \\ =&y(yf''_{11}-\frac{y}{x^2}f''_{12})-\frac{y}{x^2}(yf''_{21}-\frac{y}{x^2}f''_{22})+\frac{2y}{x^3}f'_2 \end{aligned} \]

例 7. (例6.3.5) $\phi(x),\psi(x)$具有二阶连续偏导数， 证明：函数$u=\phi(x-at)+\psi(x+at)$满足波动方程方程

\[\frac{\partial ^2u}{\partial t^2}=a^2\frac{\partial ^2u}{\partial x^2} \]
\[\begin{aligned} \frac{\partial u}{\partial x} =&\phi'(x-at)+\psi'(x+at) \\ \frac{\partial u}{\partial t} =&(-a)\phi'(x-at)+a\psi'(x+at) \\ \end{aligned} \]

Laplace方程是指

\[\frac{\partial ^2u}{\partial x^2}+\frac{\partial ^2u}{\partial y^2}+\frac{\partial ^2u}{\partial z^2}=0 \]

满足这个方程的解$u(x,y,z)$称为调和函数。

例 8. (例6.3.1) 证明：函数$u=\dfrac1r (r=\sqrt{x^2+y^2+z^2}\neq 0)$满足Laplace方程

例 9. $u=\ln\sqrt{(x-a)^2+(y-b)^2}$满足Laplace方程

一般的多维复合函数的偏导

若$z=f(y_1,y_2,\cdots,y_m)$为$D'\in\mathbb{R}^m$上的$m$元函数，

$(y_1,y_2,\cdots,y_m)=f(x_1,\cdots,x_n)$为$D\in\mathbb{R}^n$上的向量值函数，

则对$j=1,2,\cdots,n$，有

\[\begin{aligned} \frac{\partial z}{\partial x_j} =\frac{\partial z}{\partial y_1}\frac{\partial y_1}{\partial x_j} +\frac{\partial z}{\partial y_2}\frac{\partial y_2}{\partial x_j} +\cdots+\frac{\partial z}{\partial y_m}\frac{\partial y_m}{\partial x_j} \\ \end{aligned} \]

写成矩阵的形式，

\[\begin{pmatrix} \frac{\partial z}{\partial x_1} & \cdots & \frac{\partial z}{\partial x_n} \\ \end{pmatrix} =\begin{pmatrix} \frac{\partial z}{\partial y_1} & \cdots & \frac{\partial z}{\partial y_m} \\ \end{pmatrix} \begin{pmatrix} \frac{\partial y_1}{\partial x_1} & \cdots & \frac{\partial y_1}{\partial x_n} \\ \vdots & & \vdots \\ \frac{\partial y_m}{\partial x_1} & \cdots & \frac{\partial y_m}{\partial x_n} \end{pmatrix} \]

设$D,D'$分别为$\mathbb{R}^n,\mathbb{R}^m$中的区域，

\[f:D\to \mathbb{R}^m, \quad g:D'\to\mathbb{R}^l \]

为两个映射。若$f(D)\subset D'$，则存在复合映射

$g\circ f: D\to\mathbb{R}^l$，

\[(x_1,\cdots,x_n)\xrightarrow {f} (y_1,\cdots,y_m) \xrightarrow {g}(z_1,\cdots,z_l) \]

类似可得

\[\begin{aligned} \frac{\partial z_i}{\partial x_j}=\frac{\partial z_i}{\partial y_1}\frac{\partial y_1}{\partial x_j}+\frac{\partial z_i}{\partial y_2}\frac{\partial y_2}{\partial x_j}+\cdots+\frac{\partial z_i}{\partial y_m}\frac{\partial y_m}{\partial x_j} \\ i=1,2,\cdots,l; j=1,2,\cdots,n \end{aligned} \]
\[\begin{aligned} \frac{\partial z_i}{\partial x_j}=\frac{\partial z_i}{\partial y_1}\frac{\partial y_1}{\partial x_j}+\frac{\partial z_i}{\partial y_2}\frac{\partial y_2}{\partial x_j}+\cdots+\frac{\partial z_i}{\partial y_m}\frac{\partial y_m}{\partial x_j} \\ i=1,2,\cdots,l; j=1,2,\cdots,n \end{aligned} \]

写成矩阵形式，为

\[\begin{pmatrix} \frac{\partial z_1}{\partial x_1} & \cdots & \frac{\partial z_1}{\partial x_n} \\ \vdots & & \vdots \\ \frac{\partial z_l}{\partial x_1} & \cdots & \frac{\partial z_l}{\partial x_n} \end{pmatrix} =\begin{pmatrix} \frac{\partial z_1}{\partial y_1} & \cdots & \frac{\partial z_1}{\partial y_m} \\ \vdots & & \vdots \\ \frac{\partial z_l}{\partial y_1} & \cdots & \frac{\partial z_l}{\partial y_m} \end{pmatrix} \begin{pmatrix} \frac{\partial y_1}{\partial x_1} & \cdots & \frac{\partial y_1}{\partial x_n} \\ \vdots & & \vdots \\ \frac{\partial y_m}{\partial x_1} & \cdots & \frac{\partial y_m}{\partial x_n} \end{pmatrix} \]

即有

\[J(g\circ f)(x)=Jg(f(x))\cdot Jf(x) \]

注. 与单变量复合函数的链式法则是一致的

  • 多元函数或多元向量值函数的Jacobi矩阵具有一元函数的导数的功能。

    若$\vec z=g(\vec y)$, $\vec y=f(\vec x)$，则有

    \[dz=J_{\vec y}g \cdot dy = J_{\vec y}g \cdot J_{\vec x}f \cdot dx \]

    类比一维$z=g(y)$, $y=f(x)$，则

    \[dz= g'(y) dy = g'(y) f'(x) dx \]
  • 利用Jacobi矩阵，可以使多元向量值函数的结论表述得更为简洁。

目录

谢谢

例 10. (高阶微分) 设$u=f(x,y,z)$，求$du$, $d^2u$

例 11. 设$f(x,y)$在整个平面上有连续的二阶偏导数，且$f''_{xx}(x,y)=f''_{yy}(x,y)$。 已知$f(x,2x)=x^2$, $f'_x(x,2x)=x$，试求

\[f''_{xx}(x,2x), f''_{xy}(x,2x) \]