设$E$是$\mathbf{R}^n$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:16.13.222}
(g\circ f)'(x_0)=g'(f(x_0))f'(x_0)
\end{equation}
证明:设点$x_0$的坐标是$(a_1,\cdots,a_n)$.则
\begin{equation}
\label{eq:21.16.1}
(g\circ f)'(x_0)=\begin{pmatrix}
\frac{\partial g\circ f}{\partial a_1}(x_0)&\cdots&\frac{\partial g\circ f}{\partial a_n}(x_0)\\
\end{pmatrix}
\end{equation}
根据多元链法则(2)，可得
\begin{equation}
\label{eq:21.16.3}
\frac{\partial g\circ f}{\partial a_i}(x_0)=g'(f(x_0))\frac{\partial f}{\partial a_i}(x_0)
\end{equation}
把\ref{eq:21.16.3}代入\ref{eq:21.16.1},可得
\begin{equation}
\label{eq:21.17.4}
(g\circ f)'(x_0)=\begin{pmatrix}
g'(f(x_0))\frac{\partial f}{\partial a_1}(x_{0})&\cdots&g'(f(x_0))\frac{\partial f}{\partial a_n}(x_0)
\end{pmatrix}
\end{equation}
我们知道，
\begin{equation}
\label{eq:21.17.5}
f'(x_0)=\begin{pmatrix}
\frac{\partial f}{\partial a_1}(x_0)&\cdots&\frac{\partial f}{\partial a_n}(x_{0})
\end{pmatrix}
\end{equation}
因此\ref{eq:16.13.222}和\ref{eq:21.17.4}是相等的(根据的是分块矩阵的乘法).
