5.5 One-sided Derivatives

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

Definition 1. If the function y=f(x) is defined for x=x0, then the derivative from the right of f at x0, denoted by f+′(x0), is defined by
f+′(x0)=limΔx→0+f(x0+Δx)−f(x0)Δx,
if the limit exists. Similarly, the derivative from the left of f at x0, denoted by f−′(x0), is defined by
f−′(x0)=limΔx→0−f(x0+Δx)−f(x0)Δx.

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

 

Geometrical Interpretation of one-sided derivatives
Figure 1. The derivative f at x0 from the right f+′(x0) is the limit of the secant line slopes connecting P(x0,f(x0)) and Q(x,f(x)) as Q approaches P from the right. Similarly f−′(x0) 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 limx→af(x)=L if and only if limx→a+f(x)=limx→a−f(x)=L. It follows from this theorem that f′(x0) exists if and only if f+′(x0)=f−′(x0).
Example
Let f be the function defined by
f(x)={3x−1if x<27−xif 2≤x.
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) limx→2−f(x)=limx→2−(3x−1)=6−1=5.
limx→2+f(x)=limx→2+(7−x)=7−2=5.
Therefore limx→2f(x)=5
(iii) limx→2f(x)=f(2)=5.
Because (i), (ii), and (iii) hold true, f is continuous at x=2.
f+′(2)=limΔx→0+f(2+Δx)−f(2)Δx=limΔx→0+[7−(2+Δx)]−5Δx=limΔx→0+−ΔxΔx=−1.
f−′(2)=limΔx→0−f(2+Δx)−f(2)Δx=limΔx→0−[3(2+Δx)−1]−5Δx=limΔx→0−3ΔxΔx=3.
Because
limΔx→0+f(2+Δx)−f(2)Δx≠limΔx→0−f(2+Δx)−f(2)Δx,
we conclude that
f′(2)=limΔx→0f(2+Δx)−f(2)Δx 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)={3x−1if x<27−xif 2≤x