# 数学分析笔记14：多元函数微分学

2020年06月05日 阅读数：1904

## 偏导数与全微分

### 偏导数与全微分的概念

$\lim_{\Delta x_i\to 0}{\frac{f(x_1^0,\cdots,x_{i-1}^0,x_i,x_{i+1}^0,\cdots,x_n^0)-f(x_1^0,\cdots,x_n^0)}{\Delta x_i}}$存在，称该极限为 $f(x_1,\cdots,x_n)$ $(x_1^0,\cdots,x_n^0)$处对 $x_i$的偏导数，\记为 $\frac{\partial f(x_1^0,\cdots,x_n^0)}{\partial x_i}$ $f_i(x_1^0,\cdots,x_n^0)$
web

$f(x)=f(x_0)+\sum_{k=1}^n{b_k\Delta x_k}+o(||\Delta x||)$算法

$f(x_1,\cdots,x_n)=f(x_1^0,\cdots,x_n^0)+\sum_{k=1}^n{A_k(x_k-x_k^0)}+ o(||x-x_0||)$其中 $A_1,\cdots,A_n$ $\Delta x = x-x_0$无关，则称 $f(x)$ $x_0$处可微，超平面 $\sum_{k=1}^n{A_kdx_k}$称为 $f(x)$ $x_0$处的全微分，记为 $df = \sum_{k=1}^n{A_kdx_k}$

$df = \sum_{k=1}^n{f_k(x_1^0,\cdots,x_n^0)dx_k}$
app

$f(x_1,\cdots,x_n)-f(x_1^0,\cdots,x_n^0) =f(x_1,\cdots,x_n)-\\f(x_1^0,\cdots,x_n^0)+\sum_{k=1}^{n-1}{( -f(x_1^0,\cdots,x_k^0,x_{k+1},\cdots,x_n) +f(x_1^0,\cdots,x_k^0,x_{k+1},\cdots,x_n))}\\ =\sum_{k=1}^n{[f(x_1^0,\cdots,x_{k-1}^0,x_k,x_{k+1},\cdots,x_n) -f(x_1^0,\cdots,x_k^0,x_{k+1},\cdots,x_n)]}$ 由拉格朗日中值定理，存在 $\xi_k$介于 $x_k$ $x_k^0$之间 $f(x_1,\cdots,x_n)=f(x_1^0,\cdots,x_n^0) +\sum_{k=1}^n{f_k(x_1^0,\cdots,x_{k-1}^0,\xi_k,x_{k+1},\cdots,x_n)\Delta x_k}\\ =\sum_{k=1}^n{f_k(x_1^0,\cdots,x_n^0)\Delta x_k} +\sum_{k=1}^n[f_k(x_1^0,\cdots,x_{k-1}^0,\xi_k,x_{k+1},\cdots,x_n)-f_k(x_1^0,\cdots,x_n^0)]\Delta x_k$考察余项： $\frac{f(x_1,\cdots,x_n)-f(x_1^0,\cdots,x_n^0)-\sum_{k=1}^n{f_k(x_1^0,\cdots,x_n^0)\Delta x_k}}{\sqrt{\sum_{k=1}^n{\Delta^2 x_k}}}\\ =\sum_{k=1}^n[f_k(x_1^0,\cdots,x_{k-1}^0,\xi_k,x_{k+1},\cdots,x_n)-f_k(x_1^0,\cdots,x_n^0)]\frac{\Delta x_k}{\sqrt{\sum_{i=1}^n\Delta^2 x_i}}$ $|\frac{\Delta x_k}{\sqrt{\sum_{i=1}^n\Delta^2 x_i}}|\le 1$所以： $|\frac{f(x_1,\cdots,x_n)-f(x_1^0,\cdots,x_n^0)-\sum_{k=1}^n{f_k(x_1^0,\cdots,x_n^0)\Delta x_k}}{\sqrt{\sum_{k=1}^n{\Delta^2 x_k}}}|\\ \le \sum_{k=1}^n|f_k(x_1^0,\cdots,x_{k-1}^0,\xi_k,x_{k+1},\cdots,x_n)-f_k(x_1^0,\cdots,x_n^0)|$再由偏导数的连续性，就有 $\lim_{(x_1,\cdots,x_n)\to(x_1^0,\cdots,x_n^0)}{|\frac{f(x_1,\cdots,x_n)-f(x_1^0,\cdots,x_n^0)-\sum_{k=1}^n{f_k(x_1^0,\cdots,x_n^0)\Delta x_k}}{\sqrt{\sum_{k=1}^n{\Delta^2 x_k}}}|}=0$所以， $f(x_1,\cdots,x_n)$ $(x_1^0,\cdots,x_n^0)$处可微orm

(1)连续可微必定可微
(2)可微必定可求偏导数
(3)可微必定连续
(4)连续不必定可求偏导
(5)可求偏导不必定可微

### 多元函数微分法则

$\left[\begin{matrix} \frac{\partial g_1(x_1^0,\cdots,x_n^0)}{\partial x_1}&\cdots&\frac{\partial g_1(x_1^0,\cdots,x_n^0)}{\partial x_n}\\ \cdots&\cdots&\cdots\\ \frac{\partial g_m(x_1^0,\cdots,x_n^0)}{\partial x_1}&\cdots&\frac{\partial g_m(x_1^0,\cdots,x_n^0)}{\partial x_n} \end{matrix}\right]$为了方便，咱们把Frechet导数记为 $g^\prime(x_0)$，对 $n$元函数来讲，Frechet导数就是 $n$维的行向量。实际上，Frechet导数就是一元导数的的一个推广，Frechet可导就等价于可微，在这层意义下，可微和可导是等价的。

（1） $g_1+g_2$ $(x_1^0,\cdots,x_n^0)$处可微，而且
$g_1^\prime(x_1^0,\cdots,x_n^0)+g_2^\prime(x_1^0,\cdots,x_n^0) =(g_1+g_2)^\prime(x_1^0,\cdots,x_n^0)$（2） $c$是任意实数， $cg_1$ $(x_1^0,\cdots,x_n^0)$处可微，而且
$cg_1^\prime(x_1^0,\cdots,x_n^0) = (cg_1)^\prime(x_1^0,\cdots,x_n^0)$

$F^\prime(x_1^0,\cdots,x_n^0)=g(x_1^0,\cdots,x_n^0)f^\prime(x_1^0,\cdots,x_n^0) +g^\prime(x_1^0,\cdots,x_n^0)f(x_1^0,\cdots,x_n^0)$

$(x_1^0,\cdots,x_n^0)$处的连续性，就有 $\lim_{||\Delta x|| \to 0}{\frac{[f(x_1,\cdots,x_n)-f(x_1^0,\cdots,x_n^0)]\Delta x}{||\Delta x||}}=0$所以 $\lim_{||\Delta x|| \to 0}{\frac{h(x_1,\cdots,x_n)}{||\Delta x||}}=0$ $F(x_1,\cdots,x_n)-F(x_1^0,\cdots,x_n^0)=[g(x_1^0,\cdots,x_n^0)f^\prime(x_1^0,\cdots,x_n^0) \\+g^\prime(x_1^0,\cdots,x_n^0)f(x_1^0,\cdots,x_n^0)]\Delta x + h(x_1,\cdots,x_n)$

$(g(f(x_1^0,\cdots,x_n^0)))^\prime = g^\prime(f(x_1^0,\cdots,x_n^0))f^\prime(x_1^0,\cdots,x_n^0)$

$g$ $(y_1^0,\cdots,y_m^0)$可微，则 $\tag{1} g(y_1,\cdots,y_m)-g(y_1^0,\cdots,y_m^0)=\\ g^\prime(y_1^0,\cdots,y_m^0)\Delta y + o_1(||\Delta y||)$再由 $f$ $(x_1^0,\cdots,x_n^0)$处可微，则 $\tag{2} f(x_1,\cdots,x_n)-(y_1^0,\cdots,y_m^0)= f^\prime(x_1^0,\cdots,x_n^0)\Delta x + o_2(||\Delta x||)$ $\frac{||f(x_1,\cdots,x_n)-(y_1,\cdots,y_m)||}{||\Delta x||} =||\frac{f^\prime(x_1^0,\cdots,x_n^0)\Delta x + o_2(||\Delta x||)}{||\Delta x||}||$由范数的性质，有 $||\frac{f^\prime(x_1^0,\cdots,x_n^0)\Delta x + o_2(||\Delta x||)}{||\Delta x||}|| \le ||\frac{f^\prime(x_1^0,\cdots,x_n^0)\Delta x}{||\Delta x||}||+||\frac{o_2(||\Delta x||)}{||\Delta x||}||$ $\lim_{||\Delta x||\to 0}{\frac{o_2(||\Delta x||)}{||\Delta x||}} = 0$同时设 $f_{ij}(x_1^0,\cdots,x_n^0)$ $f$的第 $i$个份量对第 $j$个变元的偏导数，则 $f^\prime(x_1^0,\cdots,x_n^0)\Delta x= \left[ \begin{matrix} \sum_{i=1}^n{f_{1i}(x_1^0,\cdots,x_n^0)\Delta x_i}\\ \cdots\\ \sum_{i=1}^n{f_{mi}(x_1^0,\cdots,x_n^0)\Delta x_i} \end{matrix} \right]$对任意的 $1\le i \le m$，都要 $|\sum_{j-1}^n f_{ij}(x_1^0,\cdots,x_n^0)\Delta x_j| \le \sqrt{\sum_{j=1}^n{f_{ij}^2(x_1^0,\cdots,x_n^0)}} ||\Delta x||$所以 $||f^\prime(x_1^0,\cdots,x_n^0)\Delta x|| \le \sqrt{\sum_{i=1}^{m}\sum_{j=1}^{n}{f_{ij}^2(x_1^0,\cdots,x_n^0)}}||\Delta x||$ $||\frac{f^\prime(x_1^0,\cdots,x_n^0)\Delta x}{||\Delta x||}|| \le \sqrt{\sum_{i=1}^{m}\sum_{j=1}^{n}{f_{ij}^2(x_1^0,\cdots,x_n^0)}}$所以， $||\frac{f^\prime(x_1^0,\cdots,x_n^0)\Delta x + o_2(||\Delta x||)}{||\Delta x||}||$局部有界，所以 $\lim_{||\Delta x||\to 0}{||\frac{f^\prime(x_1^0,\cdots,x_n^0)\Delta x + o_2(||\Delta x||)}{||\Delta x||}||}=0$再将(2)代入(1)就能够证得结论

### 高阶偏导数与高阶全微分

$\frac{\partial^2 f}{\partial x^2}$表示对 $x$求两次偏导， $\frac{\partial^2 f}{\partial x\partial y}$表示先对 $x$求偏导，再对 $y$求偏导，其余两个也能够相似写出。

$h^{(k)}(0)=\sum_{i=0}^{k}{C_k^i \frac{\partial^k f(x_0,y_0)}{\partial x^i \partial y^{(k-i)}}}$这就和二项式定理相似，其余高阶导数的求法，大多用到数学概括法，这里再也不赘述。

### 方向导数

$\frac{\partial f(x)}{\partial d}=\lim_{t\to 0^+}{\frac{f(x+td)-f(x)}{t}}$方向导数该如何计算呢？若是 $f(x)$ $f(x_0)$处可微，那么
$f(x_0+td)-f(x_0)=tf^\prime(x_0)d^T+o(t)$这样
$\frac{\partial f(x_0)}{\partial d} = f^\prime(x_0)d^T$实际上就是偏导按照方向进行加权。