AP®MathsCollege BoardCalculus BCStudy GuidesUnit 2: Differentiation Definition & Fundamental PropertiesDefinition of DifferentiationInstantaneous Rate of Change

Instantaneous Rate of Change (College Board AP® Calculus BC): Study Guide

Written by: Jamie Wood

Reviewed by: Dan Finlay

Updated on 4 May 2026

Instantaneous rate of change

What is the instantaneous rate of change?

The instantaneous rate of change is the slope of a graph at a specific point, rather than between two points

Graph of y = f(x) with curve, tangent at x = a showing instantaneous rate of change, and secant from a to x showing average rate of change. — **Visual representation of an instantaneous and average rate of change of a function**

How do I find the instantaneous rate of change?

Consider the graph of a function $f$ with two points on the graph, $A$ and $B$

Two graphs showing a positive gradient curve with two points A and B. The left graph shows distance x-a; the right shows distance h between a and a+h.

Using the labeling on the left,
- the average rate of change from $A$ to $B$ can be written as $\frac{f (x) - f (a)}{x - a}$
Using the labeling on the right,
- the average rate of change from $A$ to $B$ can be written as $\frac{f (a + h) - f (a)}{h}$
To find the instantaneous rate of change, consider finding the slope of the secant line between $A$ and $B$ as point $B$ moves closer to point $A$
- As the distance between the two points becomes smaller, the slope will become a more accurate estimate for the instantaneous rate of change at $x = a$
- Therefore the instantaneous rate of change will be the limit as this distance tends to zero
The instantaneous rate of change at $x = a$ can be written as
- $\lim_{x \to a} \frac{f (x) - f (a)}{x - a}$ or
- $\lim_{h \to 0} \frac{f (a + h) - f (a)}{h}$
These only give a valid answer if the relevant limit exists
They are both equivalent forms of the definition of the derivative of the function at $x = a$ , denoted by $f^{'} (a)$

Worked Example

A function $f (x)$ is defined by $f (x) = x^{2}$ .

Find, using limits, the instantaneous rate of change of $f (x)$ at the point where $x = 2$ .

Answer:

Method 1

Consider the formula

$\lim_{x \to a} \frac{f (x) - f (a)}{x - a}$

Substitute $a = 2$ into the formula

$\lim_{x \to 2} \frac{f (x) - f (2)}{x - 2}$

Evaluate the function, $f (x) = x^{2}$ , at the two points

$\lim_{x \to 2} \frac{x^{2} - 2^{2}}{x - 2}$

$\lim_{x \to 2} \frac{x^{2} - 4}{x - 2}$

Factor the numerator

$\lim_{x \to 2} \frac{(x - 2) (x + 2)}{(x - 2)}$

Simplify the fraction by cancelling terms

$\lim_{x \to 2} (x + 2)$

Evaluate the limit

$\lim_{x \to 2} (x + 2) = 2 + 2 = 4$

Instantaneous rate of change at $x = 2$ is $4$

Method 2

Consider the formula

$\lim_{h \to 0} \frac{f (a + h) - f (a)}{h}$

Substitute $a = 2$ into the formula

$\lim_{h \to 0} \frac{f (2 + h) - f (2)}{h}$

Evaluate the function, $f (x) = x^{2}$ , at $x = 2 + h$ and $x = 2$

$\lim_{h \to 0} \frac{{(2 + h)}^{2} - {(2)}^{2}}{h}$

Expand and simplify the numerator

$\lim_{h \to 0} \frac{4 + 4 h + h^{2} - 4}{h} \lim_{h \to 0} \frac{4 h + h^{2}}{h}$

Simplify the fraction by cancelling terms

$\lim_{h \to 0} (4 + h)$

Evaluate the limit

$\lim_{h \to 0} (4 + h) = 4 + 0 = 4$

Instantaneous rate of change at $x = 2$ is $4$

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