One-Sided Derivatives

The standard derivative requires the difference quotient to have the same limit from both sides. One-sided derivatives relax this to allow a limit from only one side, which is essential for functions defined on closed intervals or with corners at interior points.

Concept	Quick Reference
Derivative from the right	f'_+(x_0) = \lim_{\Delta x \to 0^+} \dfrac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}
Derivative from the left	f'_-(x_0) = \lim_{\Delta x \to 0^-} \dfrac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}
Two-sided derivative exists	if and only if f'_+(x_0) = f'_-(x_0)

Why One-Sided Derivatives?

When we first introduced the derivative of

f

x_{0}

, we considered

x_{0}

to be an interior point of the domain of

f

. This meant that whether

Δ x

was positive or negative,

x_{0} + Δ x

stayed in the domain of

f

, and

f (x_{0} + Δ x)

remained defined for all sufficiently small

| Δ x |

It is practical to broaden the definition of the derivative to encompass derivatives at endpoints of intervals.

When the domain of $f$ is a closed interval $[a, b]$ , to determine the derivative at $a$ , one must consider the limit of the difference quotient

\frac{f (a + Δ x) - f (a)}{Δ x}

as $Δ x$ approaches zero through positive values, $Δ x \to 0^{+}$ . If $Δ x$ were negative, $a + Δ x$ would fall outside the domain of $f$ , and $f (a + Δ x)$ would be undefined. Conversely, to find the derivative of $f$ at $b$ , $Δ x$ must be negative, requiring a one-sided limit as $Δ x \to 0^{-}$ .

One-sided derivatives are also relevant for interior points where only the one-sided limits of $\frac{Δ y}{Δ x}$ exist as $Δ x$ approaches zero from either side. If these one-sided limits both exist but are unequal, the graph at that point will have two distinct one-sided tangents, creating a sharp angle.

Definition

One-Sided Derivatives. If the function $y = f (x)$ is defined at $x = x_{0}$ , then:

The derivative from the right of $f$ at $x_{0}$ , denoted f'_+(x_0), is defined by f'_+(x_0) = \lim_{\Delta x \to 0^+} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x},
if the limit exists.

The derivative from the left of $f$ at $x_{0}$ , denoted f'_-(x_0), is defined by f'_-(x_0) = \lim_{\Delta x \to 0^-} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x},
if the limit exists.

Geometric Interpretation

The geometric interpretation of one-sided derivatives is illustrated below.

Geometric illustration of right and left derivatives as limits of secant slopes approaching P from the right and from the left

The derivative from the right f'_+(x_0) is the limit of the secant line slopes connecting $P (x_{0}, f (x_{0}))$ and a nearby point $Q$ on the curve, as $Q$ approaches $P$ from the right. Similarly f'_-(x_0) is the limit of these secant slopes as $Q$ approaches $P$ from the left.

Relationship to the Two-Sided Derivative

Recall that $lim_{x \to a} f (x) = L$ if and only if $lim_{x \to a^{+}} f (x) = lim_{x \to a^{-}} f (x) = L$ . This leads to the following key conclusion:

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 rays as shown below.

Graph of a piecewise linear function consisting of two line segments meeting at x=2 with a corner

To show that

f

is continuous at

x = 2

, we verify the three conditions:

$f (2) = 7 - 2 = 5$ .
One-sided limits: $lim_{x \to 2^{-}} f (x) = lim_{x \to 2^{-}} (3 x - 1) = 6 - 1 = 5,$ $lim_{x \to 2^{+}} f (x) = lim_{x \to 2^{+}} (7 - x) = 7 - 2 = 5,$ so $lim_{x \to 2} f (x) = 5$ .
$lim_{x \to 2} f (x) = f (2) = 5$ .

Because all three conditions hold,

f

is continuous at

x = 2

. Now we compute the one-sided derivatives: \begin{aligned} f'_+(2) &= \lim_{\Delta x \to 0^+} \frac{f(2 + \Delta x) - f(2)}{\Delta x} \\ &= \lim_{\Delta x \to 0^+} \frac{[7 - (2 + \Delta x)] - 5}{\Delta x} \\ &= \lim_{\Delta x \to 0^+} \frac{-\Delta x}{\Delta x} \\ &= -1. \end{aligned} \begin{aligned} f'_-(2) &= \lim_{\Delta x \to 0^-} \frac{f(2 + \Delta x) - f(2)}{\Delta x} \\ &= \lim_{\Delta x \to 0^-} \frac{[3(2 + \Delta x) - 1] - 5}{\Delta x} \\ &= \lim_{\Delta x \to 0^-} \frac{3\Delta x}{\Delta x} \\ &= 3. \end{aligned} Because

lim_{Δ x \to 0^{+}} \frac{f (2 + Δ x) - f (2)}{Δ x} \neq lim_{Δ x \to 0^{-}} \frac{f (2 + Δ x) - f (2)}{Δ x},

we conclude that f'(2) does not exist and the function is not differentiable at

x = 2

. However, the one-sided derivatives f'_+(2) = -1 and f'_-(2) = 3 both exist. The graph of f' is shown below.

Graph of the derivative of the piecewise function, equal to 3 for x less than 2 and equal to negative 1 for x greater than 2, undefined at x=2

Differentiability on a Closed Interval

A function $f$ is differentiable on a closed interval $[a, b]$ if $f$ is differentiable on the interior $(a, b)$ (that is, f'(x) exists and is finite for every $x$ in the open interval $(a, b)$ ) and f'_+(a) and f'_-(b) both exist.

One-Sided Derivatives vs. One-Sided Limits of the Derivative

It is important not to confuse f'_+(x_0) and f'_-(x_0) with f'(x_0+) and f'(x_0-).

f'_+(x_0) and f'_-(x_0) are the derivatives from the right and left, defined by one-sided limits of the difference quotient at $x_{0}$ .
f'(x_0+) and f'(x_0-) are the one-sided limits of the derivative function as $x$ approaches $x_{0}$ from the right and left, respectively.

Even when both f'(x_0+) and f'(x_0-) exist and are equal, it does not necessarily mean that f'(x_0) exists. Consider the function

f

defined by

f (x) = x - ⌊ x ⌋,

where

⌊ x ⌋

is the greatest integer less than or equal to

x

(the floor function). The graph of

f

comprises a series of straight segments at a

45^{\circ}

angle to the

x

-axis:

Graph of f(x) = x minus floor(x), a sawtooth function with jump discontinuities at each integer

Since

f

is discontinuous at each integer

x = n

, f'(n) does not exist. The function is continuous from the right at

x = n

with f'_+(n) = 1, but discontinuous from the left, making f'_-(n) undefined. However, the one-sided limits of the derivative are both f'(n+) = \lim_{x \to n^+} f'(x) = 1, \qquad f'(n-) = \lim_{x \to n^-} f'(x) = 1. The graph of f' illustrates this behavior:

Graph of the derivative of the sawtooth function, equal to 1 everywhere it is defined but undefined at each integer

Frequently Asked Questions

When should I use one-sided derivatives?

Use one-sided derivatives whenever the domain of

f

includes an endpoint (e.g.,

f

is defined on

[a, b]

), or when you suspect that the two-sided derivative fails to exist at an interior point because the function has a corner there.

If both one-sided derivatives exist but are unequal, what does the graph look like?

The graph has a corner or sharp angle at

x_{0}

. The left-hand slope and right-hand slope are both well-defined, but they point in different directions. The function is continuous at

x_{0}

but not differentiable there.

Can a function have a one-sided derivative at a point where it is not continuous?

No. Just as the ordinary derivative requires continuity, a one-sided derivative also requires the function to be defined and continuous from the relevant side. If

f

is discontinuous from the right at

x_{0}

, then f'_+(x_0) does not exist.