设$E$是$\mathbf{R}$的子集，$F$是$\mathbf{R}^m$的子集，设$f:E\to F$是函数，$g:F\to \mathbf{R}$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:16.13.22}
(g\circ f)'(x_0)=g'(f(x_0))f'(x_0)
\end{equation}
证明:设点$f(x_0)$的坐标是$(a_1,\cdots,a_m)$. 则
\begin{equation}
\label{eq:16.22.29}
g'(f(x_0))(a_1,\cdots,a_m)^T=a_{1}\frac{\partial g}{\partial x_1}(f(x_0))+\cdots +a_m \frac{\partial g}{\partial x_m}(f(x_0))
\end{equation}
因此
\begin{equation}
\label{eq:16.23.32}
g'(f(x_0))=\begin{pmatrix}
\frac{\partial g}{\partial x_1}(f(x_0))&\cdots&\frac{\partial g}{\partial x_m}(f(x_0))
\end{pmatrix}
\end{equation}
因此
\begin{equation}
\label{eq:16.00.22}
g'(f(x_0))f'(x_0)= \begin{pmatrix}\frac{\partial g}{\partial x_1}(f(x_0))&\cdots&\frac{\partial g}{\partial x_m}(f(x_0))\end{pmatrix}f'(x_0)
\end{equation}
而
\begin{equation}
\label{eq:16.22.6}
(g\circ f)'(x_0)=\lim_{\Delta x\to 0;\Delta x\neq 0}\frac{(g\circ
f)(x_0+\Delta x)-(g\circ f)(x_0)}{\Delta x}
\end{equation}
令$f(x_0+\Delta x)=f(x_0)+\varepsilon(\Delta x)$,其中当$\Delta x\to 0$时，$\varepsilon(\Delta x)\to 0$.则\ref{eq:16.22.6}可以变为
\begin{equation}
\label{eq:16.23.7}
\lim_{\Delta x\to 0;\Delta x\neq 0} \frac{g(f(x_0)+\varepsilon(\Delta x))-g(f(x_0))}{\Delta x}
\end{equation}
设$\varepsilon(\Delta x)=(\Delta x_1,\cdots,\Delta x_m)$.我们变\ref{eq:16.23.7}式为
\begin{equation}
\label{eq:17.11.57}
\lim_{\Delta x\to 0;\Delta x\neq 0}\frac{g((a_1+\Delta
x_1,\cdots,a_m+\Delta x_m))-g((a_1,\cdots,a_m))}{\Delta x}
\end{equation}
根据中间人技巧,易得当$\Delta x_1,\cdots,\Delta x_n\neq 0$时，可以将\ref{eq:17.11.57}变成\ref{eq:16.00.22}(为什么?)当存在$i$,使得$\Delta x_i=0$时，处理方法和单变元链法则一样.完毕.
