\begin{equation*}
\frac{\mathrm{d} y}{\mathrm{~d} x}=f^{\prime}(u) \cdot \varphi^{\prime}(x)
\end{equation*}
即
\begin{equation}
\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d} y}{\mathrm{~d} u} \cdot \frac{\mathrm{d} u}{\mathrm{~d} x} .\tag{2.2.2}
\end{equation}
证 设 \(x\) 有增量 \(\Delta x \neq 0\text{,}\) 相应地 \(u=\varphi(x)\) 有增量 \(\Delta u\text{,}\) 当 \(\Delta u \neq 0\) 时, 函数 \(y\) 有增量 \(\Delta y\text{.}\) 因为 \(y=f(u)\) 在点 \(u\) 处可导, 所以有
\begin{equation*}
\lim\limits_{\Delta u \rightarrow 0} \frac{\Delta y}{\Delta u}=f^{\prime}(u)
\end{equation*}
故
\begin{equation*}
\frac{\Delta y}{\Delta u}=f^{\prime}(u)+\alpha \text {, 其中 } \lim\limits_{\Delta u \rightarrow 0} \alpha=0 \text {, }
\end{equation*}
从而
\begin{equation}
\Delta y=f^{\prime}(u) \Delta u+\alpha \Delta u \text {. }\tag{2.2.3}
\end{equation}
当
\(\Delta u=0\) 时,
\(\Delta y=f(u+\Delta u)-f(u)=0\text{,}\) (2.2.3) 仍成立, 两边除以
\(\Delta x \neq 0\text{,}\) 有
\begin{equation*}
\frac{\Delta y}{\Delta x}=f^{\prime}(u) \frac{\Delta u}{\Delta x}+\alpha \frac{\Delta u}{\Delta x},
\end{equation*}
于是
\begin{equation}
\frac{\mathrm{d} y}{\mathrm{~d} x}=\lim\limits_{\Delta x \rightarrow 0}\left[f^{\prime}(u) \frac{\Delta u}{\Delta x}+\alpha \frac{\Delta u}{\Delta x}\right] .\tag{2.2.4}
\end{equation}
当 \(\Delta x \rightarrow 0\) 时, \(\Delta u \rightarrow 0\text{,}\)
\begin{equation*}
\lim\limits_{\Delta x \rightarrow 0}\left[f^{\prime}(u) \frac{\Delta u}{\Delta x}\right]=f^{\prime}(u) \frac{\mathrm{d} u}{\mathrm{~d} x},
\end{equation*}
\begin{equation*}
\frac{\mathrm{d} y}{\mathrm{~d} x}=f^{\prime}(u) \frac{\mathrm{d} u}{\mathrm{~d} x}=f^{\prime}(u) \varphi^{\prime}(x) .
\end{equation*}
