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)}

$$