设$E$是$\mathbf{R}$的子集，$F$是$\mathbf{R}^m$的子集，设$f:E\to F$是函数，$g:F\to \mathbf{R}^p(p\in\mathbf{N}^+)$是另一个函数.设$x_0$是$E$的内点，假设$f$在$x_0$处可微，且$f(x_0)$是$F$的内点,还假设$g$在$f(x_0)$处可微，那么$g\circ f:E\to\mathbf{R}^p$在$x_0$处可微，且
\begin{equation}
\label{eq:16.13.20}
(g\circ f)'(x_0)=g'(f(x_0))f'(x_0)
\end{equation}
证明:不妨设点$f(x_0)$的坐标是$(a_1,\cdots,a_m)$,点$g(f(x_0))$的坐标是$(b_1,\cdots,b_p)$.设映射$g$的坐标形式为$(g_1,\cdots,g_p)$.则
\begin{equation}
\label{eq:18.21.42}
(g\circ f)'(x_0)=\begin{pmatrix}\frac{\partial g\circ f }{\partial b_1}(x_0)&\cdots&\frac{\partial g\circ f}{\partial b_p}(x_0)\\
\end{pmatrix}
\end{equation}
根据多元链法则(1)，可知
\begin{equation}
\label{eq:18.23.55}
\frac{\partial g\circ f}{\partial b_i}(x_0)=g_i'(f(x_0))f'(x_0)
\end{equation}
将\ref{eq:18.23.55}代入\ref{eq:18.21.42},可得
\begin{equation}
\label{eq:19.00.21}
(g\circ f)'(x_0)=\begin{pmatrix}g_1'(f(x_0))f'(x_0)&\cdots&g_p'(f(x_0))f'(x_0))\\\end{pmatrix}
\end{equation}
易得
\begin{equation}
\label{eq:19.00.32}
(g\circ f)'(x_0)=g'(f(x_0))f'(x_0)
\end{equation}(为什么?提示:利用\ref{eq:19.00.21}以及矩阵的乘法)因此\ref{eq:16.13.20}成立.
