设$X$是$\mathbf{R}$的子集,$x_0$是$X$的极限点,设$f:X\to\mathbf{R},g:X\to\mathbf{R}$是函数.
如果$f$和$g$在$x_0$处可微，则$fg$在$x_0$处也可微，且
\begin{equation}
\label{eq:15.12.14}
(fg)'(x_0)=f'(x_0)g(x_0)+f(x_0)g'(x_0)
\end{equation}
证明:
\begin{align*}
(fg)'(x_0)&=\lim_{\Delta x\to 0;\Delta x\neq 0}\frac{f(x_0+\Delta x)g(x_0+\Delta x)-f(x_0)g(x_0)}{\Delta x}\\&=\lim_{\Delta x\to 0;\Delta x\neq 0}\frac{f(x_0+\Delta x)g(x_0+\Delta x)-f(x_0)g(x_0+\Delta x)+f(x_0)g(x_0+\Delta x)-f(x_0)g(x_0)}{\Delta x}\\&=\lim_{\Delta x\to 0;\Delta x\neq 0}g(x_0+\Delta x)f'(x_0)+f(x_0)g'(x_0)\\&=f'(x_0)g(x_0)+f(x_0)g'(x_0)
\end{align*}
如果$g$在$x_0$处可微，且$g$在$X$上不取零值，则$\frac{1}{g}$在$x_0$处也可微，且
\begin{equation}
\label{eq:15.12.29}
(\frac{1}{g})'(x_0)=-\frac{g'(x_0)}{g^2(x_0)}
\end{equation}
证明:
$$
(\frac{1}{g})'(x_0)=\lim_{\Delta x\to 0;\Delta x\neq
0}\frac{\frac{1}{g(x_0+\Delta x)}-\frac{1}{g(x_0)}}{\Delta x}=\lim_{\Delta x\to 0;\Delta x\neq 0}\frac{g(x_0)-g(x_0+\Delta x)}{g(x_0+\Delta x)g(x_0)\Delta x}=-\frac{g'(x_0)}{g^2(x_0)}
$$