设开集\(D \subset
\mathbb{R}^n\)，\(\boldsymbol{f}: D \to
\mathbb{R}^m\)。记\(\boldsymbol{f}\)的分量依次为\(f_1,f_2,\cdots,f_m\)，可以把\(\boldsymbol{f}(\boldsymbol{x})\)写成
\[
\boldsymbol{f(x)} = \left(
\begin{matrix}
f_1(\boldsymbol{x}) \\
f_2(\boldsymbol{x}) \\
\vdots \\
f_m(\boldsymbol{x})
\end{matrix}
\right) \quad (\boldsymbol{x} \in D)
\]
设点\(\boldsymbol{x}_0 \in D, \boldsymbol{h}
\in \mathbb{R}^n\)。由于\(\boldsymbol{x}_0\)是\(D\)的内点，所以总可以找到充分小的\(\Vert \boldsymbol{h} \Vert\)使得\(\boldsymbol{x}_0 + \boldsymbol{h} \in
D\)。
定义1
如果映射\(\boldsymbol{f}\)满足
\[ \boldsymbol{f}(\boldsymbol{x}_0 + \boldsymbol{h}) - \boldsymbol{f}(\boldsymbol{x}_0) = \boldsymbol{Ah} + \boldsymbol{r}(\boldsymbol{h}) \]
式中\(\boldsymbol{A}\)是一个\(m \times n\)矩阵，它的元素不依赖于\(\boldsymbol{h}\)，且
\[ \lim \limits_{\boldsymbol{h} \to \boldsymbol{0}} \frac{\Vert \boldsymbol{r}(\boldsymbol{h}) \Vert}{\Vert \boldsymbol{h} \Vert} = 0 \]
则称映射\(\boldsymbol{f}\)在点\(\boldsymbol{x}_0\)处可微，并称\(\boldsymbol{Ah}\)是\(\boldsymbol{f}\)在点\(\boldsymbol{x}_0\)处的微分，记作
\[ \mathrm{d}\boldsymbol{f}(\boldsymbol{x}_0) = \boldsymbol{Ah} \]
定理1
若映射\(\boldsymbol{f}\)在点\(\boldsymbol{x}_0\)处可微，则有
\[ \mathrm{d}\boldsymbol{f}(\boldsymbol{x}_0) = \boldsymbol{Jf}(\boldsymbol{x}_0)\boldsymbol{h} \]
其中
\[ \boldsymbol{Jf}(\boldsymbol{x}_0) = \left[ \begin{matrix} \frac{\partial f_1(\boldsymbol{x}_0)}{\partial x_1} & \cdots & \frac{\partial f_1(\boldsymbol{x}_0)}{\partial x_n} \\ \vdots & & \vdots \\ \frac{\partial f_m(\boldsymbol{x}_0)}{\partial x_1} & \cdots & \frac{\partial f_m(\boldsymbol{x}_0)}{\partial x_n} \end{matrix} \right] \]
称之为映射\(\boldsymbol{f}\)在点\(\boldsymbol{x}_0\)处的Jacobi矩阵，也称为导数。
证：有定义可知，
\[
\boldsymbol{f}(\boldsymbol{x}_0 + \boldsymbol{h}) -
\boldsymbol{f}(\boldsymbol{x}_0) = \boldsymbol{Ah} +
\boldsymbol{r}(\boldsymbol{h})
\]
设
\[
\boldsymbol{A} = \left[\begin{matrix}
a_{11} & \cdots & a_{1n} \\
\vdots & & \vdots \\
a_{m1} & \cdots & a_{mn}
\end{matrix}\right]
\]
将其代入上式中得
\[
f_i(\boldsymbol{x}_0 + \boldsymbol{h}) - f_i(\boldsymbol{x}_0) =
\sum_{j=1}^n a_{ij} h_j + r_i(\boldsymbol{h})
\]
其中\(r_i(\boldsymbol{h})\)表示\(\boldsymbol{r}(\boldsymbol{h})\)的第\(i\)个分量，由\(\boldsymbol{r}(\boldsymbol{h})\)的性质可知
\[
r_i(\boldsymbol{h}) = o(\Vert \boldsymbol{h} \Vert) \quad
(\boldsymbol{h} \to 0, i=1,2,\cdots,m)
\]
从而继续由函数导数八定理1可知
\[
a_{ij} = \frac{\partial f_i(\boldsymbol{x}_0)}{\partial x_j}
(i=1,2,\cdots,m;j=1,2,\cdots,n)
\]
从而得证。
Q.E.D.
定理2
若映射\(\boldsymbol{f}\)在点\(\boldsymbol{x}_0\)的某一邻域内存在Jacobi矩阵\(\boldsymbol{Jf}\)，且\(\boldsymbol{Jf}\)的各元素在\(\boldsymbol{x}_0\)处都连续，则映射\(\boldsymbol{f}\)在点\(\boldsymbol{x}_0\)处可微。
Q.E.D.
定义2
设开集\(D \subset \mathbb{R}^n\)，\(\boldsymbol{f}: D \to \mathbb{R}^m\)。如果\(\boldsymbol{f}\)在\(D\)上的每一点处都连续，则记\(\boldsymbol{f} \in C(D)\)；如果\(\boldsymbol{Jf}\)在\(\boldsymbol{D}\)上的每一点处都连续，则记\(\boldsymbol{f} \in C^1(D)\)。
定理3
设开集\(D \subset \mathbb{R}^n\)，\(\boldsymbol{g}: D \to \mathbb{R}^m\)，\(\boldsymbol{g}\)在点\(\boldsymbol{x}_0 \in D\)处可微，又设\(\boldsymbol{f}\)把包含\(\boldsymbol{g}(D)\)的一个开集映射至\(\mathbb{R}^l\)，并且\(\boldsymbol{f}\)在\(\boldsymbol{g}(\boldsymbol{x}_0)\)处可微，那么复合映射\(\boldsymbol{f} \circ \boldsymbol{g}\)在点\(\boldsymbol{x}_0\)处可微，并且
\[ \boldsymbol{J}(\boldsymbol{f}\circ \boldsymbol{g})(\boldsymbol{x}_0) = \boldsymbol{Jf}(\boldsymbol{g}(\boldsymbol{x}_0)) \boldsymbol{Jg}(\boldsymbol{x}_0) \]
证：令\(\boldsymbol{y}_0 =
\boldsymbol{g}(\boldsymbol{x}_0)\)，\(\boldsymbol{A} =
\boldsymbol{Jf}(\boldsymbol{y}_0)，\boldsymbol{B} =
\boldsymbol{Jg}(\boldsymbol{x}_0)\)，易知\(\boldsymbol{A}\)是\(l\times m\)矩阵，\(\boldsymbol{B}\)是\(m \times n\)矩阵，从而若能证明
\[
\lim \limits_{\Vert \boldsymbol{h} \Vert \to 0} \frac{\Vert
(\boldsymbol{f} \circ \boldsymbol{g})(\boldsymbol{x}_0 + \boldsymbol{h})
- (\boldsymbol{f} \circ \boldsymbol{g})(\boldsymbol{x}_0) -
\boldsymbol{ABh} \Vert}{\Vert \boldsymbol{h} \Vert} = 0 \tag{1}
\]
则由定义1可知，\(\boldsymbol{J}(\boldsymbol{f}\circ
\boldsymbol{g})(\boldsymbol{x}_0) = \boldsymbol{AB}\)。
由于\(\boldsymbol{g},\boldsymbol{f}\)分别在\(\boldsymbol{x}_0,\boldsymbol{y}_0\)处可微，从而有
\[
\begin{aligned}
\boldsymbol{g}(\boldsymbol{x}_0 + \boldsymbol{h}) -
\boldsymbol{g}(\boldsymbol{x}_0) = \boldsymbol{Bh} +
\boldsymbol{u}(\boldsymbol{h}) \\
\boldsymbol{f}(\boldsymbol{y}_0 + \boldsymbol{k}) -
\boldsymbol{f}(\boldsymbol{y}_0) = \boldsymbol{Ak} +
\boldsymbol{v}(\boldsymbol{k})
\end{aligned} \tag{2}
\]
其中\(\boldsymbol{h}，\boldsymbol{k}\)分别为\(n \times 1\)矩阵和\(m \times 1\)矩阵，且
\[
\begin{aligned}
\frac{\Vert \boldsymbol{u}(\boldsymbol{h}) \Vert}{\Vert
\boldsymbol{h} \Vert} \to 0 \quad (\Vert \boldsymbol{h} \Vert \to 0) \\
\frac{\Vert \boldsymbol{v}(\boldsymbol{k}) \Vert}{\Vert
\boldsymbol{k} \Vert} \to 0 \quad (\Vert \boldsymbol{k} \Vert \to 0)
\end{aligned}
\]
记
\[
\frac{\Vert \boldsymbol{u}(\boldsymbol{h}) \Vert}{\Vert
\boldsymbol{h} \Vert} = \varepsilon(\boldsymbol{h})， \quad \frac{\Vert
\boldsymbol{v}(\boldsymbol{h}) \Vert}{\Vert \boldsymbol{h} \Vert} =
\eta(\boldsymbol{h})
\]
则
\[
\Vert \boldsymbol{u}(\boldsymbol{h}) \Vert =
\varepsilon(\boldsymbol{h}) \Vert \boldsymbol{h} \Vert， \quad \Vert
\boldsymbol{v}(\boldsymbol{k}) \Vert = \eta(\boldsymbol{k}) \Vert
\boldsymbol{k} \Vert
\]
且
\[
\lim \limits_{\Vert \boldsymbol{h} \Vert \to 0}
\varepsilon(\boldsymbol{h}) = 0， \quad \lim \limits_{\Vert
\boldsymbol{k} \Vert \to 0} \eta(\boldsymbol{k}) = 0 \tag{3}
\]
对给定的\(\boldsymbol{h}\)，令\(\boldsymbol{k} = \boldsymbol{g}(\boldsymbol{x}_0 +
\boldsymbol{h}) -
\boldsymbol{g}(\boldsymbol{x}_0)\)，由式(2)可得
\[
\Vert \boldsymbol{k} \Vert \le \Vert \boldsymbol{Bh} \Vert + \Vert
\boldsymbol{u}(\boldsymbol{h})\Vert \le (\Vert \boldsymbol{B} \Vert +
\varepsilon(\boldsymbol{h})) \Vert \boldsymbol{h} \Vert
\]
从而有
\[
\begin{aligned}
& \Vert \boldsymbol{f} \circ \boldsymbol{g}
(\boldsymbol{x}_0 + \boldsymbol{h}) - \boldsymbol{f} \circ
\boldsymbol{g} (\boldsymbol{x}_0) - \boldsymbol{ABh} \Vert \\
& = \Vert \boldsymbol{f}(\boldsymbol{g} (\boldsymbol{x}_0 +
\boldsymbol{h})) - \boldsymbol{f}(\boldsymbol{g} (\boldsymbol{x}_0)) -
\boldsymbol{ABh} \Vert \\
& = \Vert \boldsymbol{f}(\boldsymbol{y}_0 + \boldsymbol{k})
- \boldsymbol{f}(\boldsymbol{y}_0) - \boldsymbol{ABh} \Vert \\
& = \Vert \boldsymbol{Ak} + \boldsymbol{v}(\boldsymbol{k}) -
\boldsymbol{ABh} \Vert \\
& = \Vert \boldsymbol{A}(\boldsymbol{k} - \boldsymbol{Bh}) +
\boldsymbol{v}{\boldsymbol{k}} \Vert \\
& \le \Vert \boldsymbol{A} \Vert u(\boldsymbol{h}) +
\eta(\boldsymbol{k})\Vert \boldsymbol{k} \Vert \\
& \le \Vert \boldsymbol{A} \Vert \varepsilon(\boldsymbol{h})
\Vert \boldsymbol{h} \Vert + \eta(\boldsymbol{k})(\Vert \boldsymbol{B}
\Vert + \varepsilon(\boldsymbol{h})) \Vert \boldsymbol{h} \Vert
\end{aligned}
\]
所以
\[
\frac{\Vert (\boldsymbol{f} \circ \boldsymbol{g})(\boldsymbol{x}_0 +
\boldsymbol{h}) - (\boldsymbol{f} \circ
\boldsymbol{g})(\boldsymbol{x}_0) - \boldsymbol{ABh} \Vert}{\Vert
\boldsymbol{h} \Vert} \le \Vert \boldsymbol{A} \Vert
\varepsilon(\boldsymbol{h}) + \eta(\boldsymbol{k})(\Vert \boldsymbol{B}
\Vert + \varepsilon(\boldsymbol{h}))
\]
再由式(3)可知式(1)成立。
Q.E.D.