带有绝对值的一点处导数的定义怎么用？

一、题目

已知 $f(x), g(x)$ 在 $x=x_{0}$ 可导, $F(x)=g(x)|f(x)|$, 又 $f\left(x_{0}\right)=0$, 则 $F^{\prime}\left(x_{0}\right)$ 存在的充要条件是什么？

难度评级：

二、解析

由于 $F\left(x_{0}\right)=g\left(x_{0}\right)\left|f\left(x_{0}\right)\right|=0$, 于是：

$$
F^{\prime}\left(x_{0}\right)=\lim \limits_{x \rightarrow x_{0}^{\prime}} \frac{F(x)-F\left(x_{0}\right)}{x-x_{0}}=\lim \limits_{x \rightarrow x_{0}^{+}} \frac{g(x)|f(x)|}{x-x_{0}} =
$$

$$
\lim \limits_{x \rightarrow x_{0}^{+}} g(x) \cdot \frac{|f(x)|}{x-x_{0}}=\lim \limits_{x \rightarrow x_{0}^{+}} g(x) \cdot \frac{\left|f(x)-f\left(x_{0}\right)\right|}{x-x_{0}}
$$

则，当 $x \rightarrow x_{0}^{+}$ 时：

$$
F^{\prime}\left(x_{0}^{+}\right)=\lim \limits_{x \rightarrow x_{0}^{+}} g(x) \cdot \frac{\left|f(x)-f\left(x_{0}\right)\right|}{\textcolor{springgreen}{\left|x-x_{0}\right|}}=
$$

$$
\lim \limits_{x \rightarrow x_{0}^{+}} g(x) \cdot \left| \frac{f(x)-f\left(x_{0}\right)}{x-x_{0}} \right| =
$$

$$
\textcolor{springgreen}{
g\left(x_{0}\right)| f^{\prime}(x_{0})|
}
$$

当 $x \rightarrow x_{0}^{-}$ 时：

$$
F^{\prime}\left(x_{0}^{-}\right)=\lim \limits_{x \rightarrow x_{0}^{-}} g(x) \cdot \frac{\left|f(x)-f\left(x_{0}\right)\right|}{\textcolor{orangered}{-\left|x-x_{0}\right|}}=
$$

$$
-g\left(x_{0}\right)\left|f^{\prime}\left(x_{0}\right)\right|
$$

又：

$$
F^{\prime}\left(x_{0}^{+}\right)=F^{\prime}\left(x_{0}^{-}\right) \Rightarrow
$$

$$
g\left(x_{0}\right)\left|f^{\prime}\left(x_{0}\right)\right|=-g\left(x_{0}\right)\left|f^{\prime}\left(x_{0}\right)\right| \Rightarrow
$$

$$
g\left(x_{0}\right)\left|f^{\prime}\left(x_{0}\right)\right|=0
$$

于是：

$$
g\left(x_{0}\right) f^{\prime}\left(x_{0}\right)=0
$$

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