In this section, we introduce the concept of a one-sided derivative.

Definition 1. If the function $$y=f(x)$$ is defined for $$x=x_{0}$$, then the derivative from the right of $$f$$ at $$x_{0}$$, denoted by $$f’_{+}(x_{0})$$, is defined by
$f’_{+}(x_{0})=\lim_{\Delta x\to0^{+}}\frac{f(x_{0}+\Delta x)-f(x_{0})}{\Delta x},$

if the limit exists. Similarly, the derivative from the left of $$f$$ at $$x_{0}$$, denoted by $$f’_{-}(x_{0})$$, is defined by
$f’_{-}(x_{0})=\lim_{\Delta x\to0^{-}}\frac{f(x_{0}+\Delta x)-f(x_{0})}{\Delta x},$

The geometrical interpretation of one-sided derivatives is illustrated in Figure 1.

• In the previous chapter, we learned that $$\lim_{x\to a}f(x)=L$$ if and only if $$\lim_{x\to a^{+}}f(x)=\lim_{x\to a^{-}}f(x)=L$$ (Theorem 8). It follows from that theorem that $$f'(x_{0})$$ exists if and only if $$f’_{+}(x_{0})=f’_{-}(x_{0})$$.
Example
Let $$f$$ be the function defined by
$f(x)=\begin{cases} 3x-1 & \text{if }x<2\\ 7-x & \text{if }2\leq x \end{cases}.$
Draw a sketch of the graph of $$f$$. Show that $$f$$ is continuous at $$x=2$$, and find the derivative from the right and from the left of $$f$$ at $$x=2$$.
Solution
Graph of $$f$$ consists of two lines as shown in Fig. 2. To show that $$f$$ is continuous at $$x=2$$, we show three conditions hold true
(i) $$f(2)=7-2=5$$.
(ii) $${\displaystyle \lim_{x\to2^{-}}f(x)=\lim_{x\to2^{-}}(3x-1)=6-1=5.}$$
$${\displaystyle \lim_{x\to2^{+}}f(x)=\lim_{x\to2^{+}}(7-x)=7-2=5.}$$
Therefore $${\displaystyle \lim_{x\to2}f(x)=5}$$
(iii) $${\displaystyle \lim_{x\to2}f(x)=f(2)=5.}$$
Because (i), (ii), and (iii) hold true, $$f$$ is continuous at $$x=2$$.
\begin{aligned} f_{+}'(2) & =\lim_{\Delta x\to0^{+}}\frac{f(2+\Delta x)-f(2)}{\Delta x}\\ & =\lim_{\Delta x\to0^{+}}\frac{[7-(2+\Delta x)]-5}{\Delta x}\\ & =\lim_{\Delta x\to0^{+}}\frac{-\Delta x}{\Delta x}\\ & =-1.\end{aligned}
\begin{aligned} f_{-}'(2) & =\lim_{\Delta x\to0^{-}}\frac{f(2+\Delta x)-f(2)}{\Delta x}\\ & =\lim_{\Delta x\to0^{-}}\frac{[3(2+\Delta x)-1]-5}{\Delta x}\\ & =\lim_{\Delta x\to0^{-}}\frac{3\Delta x}{\Delta x}\\ & =3.\end{aligned}
Because
$\lim_{\Delta x\to0^{+}}\frac{f(2+\Delta x)-f(2)}{\Delta x}\neq\lim_{\Delta x\to0^{-}}\frac{f(2+\Delta x)-f(2)}{\Delta x},$
we conclude that
$f'(2)=\lim_{\Delta x\to0}\frac{f(2+\Delta x)-f(2)}{\Delta x}$ does not exists and the function is not differentiable at $$x=2$$. However, the derivative from the right and from the left of $$f$$ at $$x=2$$, $$f’_{+}(2)$$ and $$f’_{-}(2)$$, exist.