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 geometric interpretation of one-sided derivatives is illustrated in Figure 1. Figure 1. The derivative of $$f$$ at $$x_{0}$$ from the right $$f’_{+}(x_{0})$$ is the limit of the secant line slopes connecting $$P(x_{0},f(x_{0}))$$ and $$Q(x,f(x))$$ as $$Q$$ approaches $$P$$ from the right. Similarly $$f’_{-}(x_{0})$$ is the limit of the secant line slopes as $$Q$$ approaches $$P$$ from the left.

• The Theorem of the Uniquness of a Limit that we learned in the previous chapter, states that $$\lim_{x\to a}f(x)=L$$ if and only if $$\lim_{x\to a^{+}}f(x)=\lim_{x\to a^{-}}f(x)=L$$. It follows from this 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
The graph of $$f$$ consists of two lines as shown in Fig. 2. To show that $$f$$ is continuous at $$x=2$$, we show that the three following conditions hold true
(i) $$f(2)=7-2=5$$.
(ii) $$\lim_{x\to2^{-}}f(x)=\lim_{x\to2^{-}}(3x-1)=6-1=5.$$
$$\lim_{x\to2^{+}}f(x)=\lim_{x\to2^{+}}(7-x)=7-2=5.$$
Therefore $$\lim_{x\to2}f(x)=5$$
(iii) $$\lim_{x\to2}f(x)=f(2)=5.$$
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. Figure 2. Graph of the piecewise-defined function $$y=f(x)=\begin{cases} 3x-1 & \text{if }x<2\\ 7-x & \text{if }2\leq x \end{cases}$$