One-Sided Derivative¶

Definition¶

Let $X \subseteq \mathbb{R}\text{,}$ let $x_0 \in X\text{,}$ and let $f \colon X \to \mathbb{R}$ be a function.

Assume that there exists an $r > 0$ such that $f$ is defined for every $x \in [x_0, x_0 + r)\text{.}$ We say that $f$ is right-differentiable at $x = x_0$ with right derivative $L \in \mathbb{R}\text{,}$ and write $f'_+(x_0) = L\text{,}$ if the following limit holds true:

\lim_{x \to x_0^+} \frac{f(x) - f(x_0)}{x - x_0} = L.

Assume that there exists an $r > 0$ such that $f$ is defined for every $x \in (x_0 - r, x_0]\text{.}$ We say that $f$ is left-differentiable at $x = x_0$ with left derivative $L \in \mathbb{R}\text{,}$ and write $f'_-(x_0) = L\text{,}$ if the following limit holds true:

\lim_{x \to x_0^-} \frac{f(x) - f(x_0)}{x - x_0} = L.

Interpretation¶

One-sided derivatives measure the rate of change at a point when only one side of the point is used: the right derivative is the limit of the slopes of secants to points $x>x_0$ $x_{0}$ , the left derivative that of secants to points $x<x_0$ $x_{0}$ . They are particularly relevant at boundary points of an interval and at corners of a graph.

The graph of |x| with dashed tangent lines of slope -1 on the left of 0 and slope 1 on the right, so the derivative at the corner does not exist.

The graph has a corner at $0$ . The left derivative is $-1$ and the right derivative is $+1$ , as shown by the dashed tangent lines, so $f'(0)$ does not exist. Each one-sided derivative is the slope of a tangent from one side; the corner is the visible sign that the two slopes differ.

Approach $x_0=0$ from one side only: the one-sided quotients settle at $f'_+(0)$ and $f'_-(0)$ . For the presets $|x|$ they differ and for $\sqrt x$ the right quotients grow without bound. For custom formulas finite difference quotients do not determine a limit.

At an interior point, $f$ is differentiable if and only if both one-sided derivatives exist and are equal, because a two-sided limit exists exactly when both one-sided limits exist and agree. At an endpoint of a closed interval only one side is available, and the one-sided derivative takes the place of the derivative. A one-sided derivative can also fail to exist because the quotients grow without bound: for $\sqrt{x}$ at $0$ the right quotient is $\frac{1}{\sqrt{x}}$ , which tends to $+\infty$ , a vertical tangent.

A common slip is to compute $f'_+(x_0)$ as $\lim_{x\to x_0^+}f'(x)$ , the limit of the derivative function. The two can differ. For $f(x)=x^2\sin(1/x)$ with $f(0)=0$ $0$ , the right quotient $x\sin(1/x)$ tends to $0$ , so $f'_+(0)=0$ $0$ , but $f'(x)=2x\sin(1/x)-\cos(1/x)$ has no limit as $x\to0^+$ $0^{+}$ .

Key Results¶

Existence of both one-sided derivatives does not guarantee differentiability.
One-sided derivatives may exist even when the two-sided derivative does not.
Equality of left and right derivatives implies differentiability.
One-sided derivatives are not the same as the one-sided limits of the derivative function.

Examples¶

The function $f(x)=|x|$ has distinct one-sided derivatives at $x=0\text{.}$
Boundary points often admit only one-sided derivatives.

Prerequisites¶

One-Sided Continuity

Sources¶

Math I: Calculus · Lecture 6

Aliases: right derivative, left derivative, right-differentiable, left-differentiable Symbols: element of, infinity, real numbers, square root, subset

Last updated: 27 October 2026 11:52