# 绝对值函数怎么求导？

## 一、题目

(A) $f(0) \neq 0$, $f^{\prime}(0)=0$

(C) $f(0)=0$, $f^{\prime}(0)=0$

(B) $f(0) \neq 0$, $f^{\prime}(0) \neq 0$

(D) $f(0)=0$, $f^{\prime}(0) \neq 0$

## 二、解析

$$\begin{cases} f(0) = 0 \\ f^{\prime} (0) = 1 \neq 0 \end{cases}$$

$$|f(x)| = \sqrt{f^{2}(x)}$$

\begin{aligned} (|f(x)|)^{\prime} = & (\sqrt{f^{2}(x)})^{\prime} \\ = & \frac{1}{2} \frac{1}{\sqrt{f^{2} (x)}} \cdot 2 f(x) \cdot f^{\prime}(x) \\ = & \textcolor{orangered}{\frac{f(x) f^{\prime}(x)}{\sqrt{f^{2} (x)}}, \ f(0) \neq 0} \end{aligned}

$$\textcolor{springgreen}{ f(0) = 0 }$$

\begin{aligned} g^{\prime}(0^{+}) = (|f(x)|)^{\prime} = & \lim_{x \rightarrow 0^{+}} \frac{g(x) – g(0)}{x – 0} \\ = & \lim_{x \rightarrow 0^{+}} \frac{|f(x)| – |f(0)|}{|x|} \\ = & \lim_{x \rightarrow 0^{+}} \frac{|f(x)| – 0}{|x|} \\ = & \lim_{x \rightarrow 0^{+}} \left| \frac{f(x)}{x} \right| \\ = & \lim_{x \rightarrow 0^{+}} \left| \frac{f(x) – f(0)}{x – 0} \right| = \left| f^{\prime}(0^{+}) \right| \end{aligned}

\begin{aligned} g^{\prime}(0^{-}) = (|f(x)|)^{\prime} = & \lim_{x \rightarrow 0^{-}} \frac{g(x) – g(0)}{x – 0} \\ = & \lim_{x \rightarrow 0^{+}} \frac{|f(x)| – |f(0)|}{-|x|} \\ = & \lim_{x \rightarrow 0^{-}} \frac{|f(x)| – 0}{-|x|} \\ = & – \lim_{x \rightarrow 0^{-}} \left| \frac{f(x)}{x} \right| \\ = & – \lim_{x \rightarrow 0^{-}} \left| \frac{f(x) – f(0)}{x – 0} \right| = – \left| f^{\prime}(0^{-}) \right| \end{aligned}

$$g^{\prime}(0^{+}) \neq g^{\prime}(0^{-})$$

$$\left| f^{\prime}(0^{+}) \right| \neq – \left| f^{\prime}(0^{-}) \right| \Rightarrow$$

$$\left| f^{\prime}(0) \right| \neq – \left| f^{\prime}(0) \right|$$

$$\textcolor{springgreen}{ f^{\prime}(0) \neq 0 }$$