多变量函数的微分学

3. 复合函数的偏导

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。

$D,D'$$\mathbb{R}^2$中的区域,$f:D\to D'$$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}$

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

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

\[z=g\circ f(x_1,x_2)=g(y_1(x_1,x_2),y_2(x_1,x_2)) \]

ex-6-3-1

定理 1.
设函数$f,g$如前所述,则

  1. $f,g$分别在$D, D'$中可微,则复合函数$g\circ f$$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:
    \[\begin{aligned} \dfrac{\partial z}{\partial x_1}=\frac{\partial z}{\partial y_1}\frac{\partial y_1}{\partial x_1}+\frac{\partial z}{\partial y_2}\frac{\partial y_2}{\partial x_1} \\ \dfrac{\partial z}{\partial x_2}=\frac{\partial z}{\partial y_1}\frac{\partial y_1}{\partial x_2}+\frac{\partial z}{\partial y_2}\frac{\partial y_2}{\partial x_2} \end{aligned} \]

证明:

一般的多维复合函数的偏导

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

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

则有

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

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

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

. 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。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。

例 1. 证明:函数$u=\dfrac1r (r=\sqrt{x^2+y^2+z^2}\neq 0)$满足方程

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

这个方程,称为Laplace方程,满足这个方程的解称为调和函数

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

例 3. 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:

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

例 4. 计算:

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

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

4:

微分中值定理

定理 2. (微分中值定理)
$f(x,y)$在区域$D$中可微。若连接$(x_0,y_0)$$(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 5. 求复合函数$z=f(xy,\frac{y}x)$的三个二阶偏导数。

例 6. 证明:函数$u=\phi(x-at)+\psi(x+at)$满足方程

\[\frac{\partial ^2u}{\partial t^2}=a^2\frac{\partial ^2u}{\partial x^2} \]

这个方程称为波动方程。对任意具有二阶导数的函数$\phi,\psi$$u$都是方程的解。

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。


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

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

若有$x=\phi(r,s), y=\psi(r,s)$,则

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

定理 3.
$u=u(x,y), v=v(x,y)$为可微的二元函数,则

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

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

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

例 7. $u=f(x,\xi,\eta)$, $\xi=x^2+y^2$,

$\eta=x^2+y^2+z^2$,求复合函数的全微分$du$

目录

谢谢

例 8. 本节读完

8.