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Defining Average and Instantaneous Rates of Change at a Point

Build a finite instantaneous rate from either form of the limiting secant quotient.

College Board Topic 2.1 · CED p. 55 (PDF p. 60) · CED effective Fall 2020; project course outline 2026–27.

Two points give a secant slope, the average rate [f(b)−f(a)]/(b−a). One point can have an instantaneous rate when nearby secant slopes approach a finite common value. Write that value as f′(a). Its units are output units per input unit.

The derivative at a can be written in two equivalent ways: f′(a)=lim(h→0)[f(a+h)−f(a)]/h or f′(a)=lim(x→a)[f(x)−f(a)]/(x−a). Setting x=a+h connects them. The limit must exist as a finite real number for this derivative to exist.

For s(t)=t² meters, the average speed from 1 to 3 seconds is (9−1)/2=4 m/s. At t=1, the difference quotient is 2+h, so the instantaneous speed is 2 m/s. The average over a long interval does not describe every instant inside it.

Let's work through it

  1. For f(x)=x² at a=3, form [(3+h)²−9]/h. Reason: the derivative is a limit of secant slopes.
  2. For h≠0, simplify to 6+h. Reason: expansion exposes the common factor h.
  3. Let h→0 to find f′(3)=6. Reason: both sides approach the same finite slope.

Give the quotient its interval before taking a limit

The expressions [f(a+h) − f(a)]/h for h ≠ 0 and [f(x) − f(a)]/(x − a) for x ≠ a describe the same secant slopes when x = a+h. They are average rates over nonzero input changes, with output-units per input-unit. At h = 0, the original quotient is undefined; a finite limiting value can still exist.

To represent an instantaneous rate, write f′(a) = lim(h → 0)[f(a+h) − f(a)]/h, or f′(a) = lim(x → a)[f(x) − f(a)]/(x − a). The same finite limit must be reached from both sides. Letting the base input vary gives the derivative function f′(x), defined wherever its limiting quotient exists. A finite secant estimate, a tangent drawing and an explicit limiting definition provide related information, but they are different mathematical operations.

CHA-2.A.1CHA-2.B.1

An interval rate is not its starting tangent rate

A position model is s(t) = 2t² + 1 meters, with t in seconds. Compare the average rate on [2,4] with the instantaneous rate at 2 using a difference quotient.

  1. Compute s(2) = 9 m and s(4) = 33 m. The average rate is (33 − 9)/(4 − 2) = 12 m/s.
  2. For h ≠ 0, [s(2+h) − s(2)]/h = [9 + 8h + 2h² − 9]/h = 8 + 2h m/s.
  3. In the alternate form, [s(t) − s(2)]/(t − 2) = 2(t² − 4)/(t − 2) = 2(t+2) m/s for t ≠ 2.
  4. Taking either h → 0 or t → 2 gives s′(2) = 8 m/s. The original zero-length quotient has not been evaluated; its nearby slopes have a finite limit.

Answer and check

Average rate on [2,4] is 12 m/s; instantaneous rate at 2 is 8 m/s. The substitutions t = 2+h and h = t−2 connect both limiting forms.

CHA-2.ACHA-2.A.1CHA-2.B.1

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Question 1 of 20 self-assessed

Question 1

What is the average rate of x² from 2 to 5?

See every recall question with its answer

These references preserve each prompt and its answer for direct evidence links. Try the interactive questions above before opening the worked reasoning.

Retrieval 1

What is the average rate of x² from 2 to 5?

Worked answer and reason

7.

The secant slope is (25−4)/(5−2).

Retrieval 2

Which substitution connects the h and x derivative definitions?

Worked answer and reason

x=a+h.

Then x−a=h and approaching x=a is equivalent to approaching h=0.

Catch the tempting mistake

The average and instantaneous rates at an interval endpoint are always equal.
They describe different quantities; equality happens for special models such as a constant slope.

Make the picture move

Explore → predict → explain

A derivative is a limit of slopes.

Try first. At the selected a, predict the tangent slope. Which side’s secant lies above it, and which lies below?

f(x)=x2f′(a)=limh→0f(a+h)−f(a)h
A derivative is a limit of slopes.A solid function graph, a dashed secant through the two labeled points, and a dotted exact tangent. Right secant slope 2.5 at a = 1. The derivative is 2. The numerical table and explanation follow the graph.-2.6-1.301.32.6-11.053.15.157.2aa+hxy

Scroll sideways to see the full graph on a small screen. The table below gives the numerical values.

  • Function: x²
  • Selected secant
  • Exact tangent
A solid function graph, a dashed secant through the two labeled points, and a dotted exact tangent. Right secant slope 2.5 at a = 1. The derivative is 2.
Input a1
Selected secant slope2.5
Exact derivative2

Right secant slope 2.5 at a = 1. The derivative is 2.

Compare the numbers
Computed values for this scenario (rounded for display)
Distance |h|Left secant slopeRight secant slope
0.51.52.5
0.251.752.25
0.11.92.1
0.051.952.05

What’s going on? For x², the difference quotient simplifies to 2a + h. As h approaches zero from either side, the secant slope approaches 2a.

Keep in mind. The controls keep h nonzero so that the quotient is defined. A small h is an approximation; the limiting calculation supplies the exact derivative.

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Represent the interval rate and its limiting tangent

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A position model is s(t) = t² − 4t + 7 meters, for t in seconds. Consider the average rate on [3,6], the instantaneous rate at 3, and the derivative function. Use limiting difference quotients rather than the power rule.

(a) Calculate the average rate on [3,6] from endpoint values. Write both [s(3+h) − s(3)]/h and [s(t) − s(3)]/(t − 3) as average-rate expressions and simplify them on their nonzero-increment domains. Explain why setting h = 0 in the original quotient does not calculate an average rate.

CHA-2.ACHA-2.A.1CHA-1.A.2

(b) Represent s′(3) in both limiting difference-quotient forms and evaluate the finite two-sided limits. Then represent the derivative function s′(t) using lim(h → 0)[s(t+h) − s(t)]/h and derive it by expansion. Explain the difference between the derivative function and its value at 3.

CHA-2.BCHA-2.B.1CHA-2.B.2

(c) On 2 ≤ t ≤ 6, sketch the position curve, the secant through its points at 3 and 6, and the tangent at 3. Give both line equations and label their slopes with units. Explain why the average rate 5 m/s does not make the instantaneous rate at 3 equal to 5 m/s.

CHA-2.A.1CHA-2.B.4

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For the teacher
  • Keep secant and tangent vocabulary attached to their formulas.
  • Ask for units and the interpretation before arithmetic.

Unit planning and formative guidance →

Glossary and deeper reading · 6 refreshers

These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.

  • Instantaneous rate of change

    An instantaneous rate of change is the limiting average rate around one input when that limit exists. For a differentiable function, it is its derivative at that input. For example, position s(t)=t² meters has instantaneous rate 4 meters per second at t=2 seconds. The rate’s units are output units per input unit.

  • Derivative

    A derivative gives the instantaneous rate of change of a function with respect to its input. At a point, it is the finite limit of the difference quotient when that limit exists. For example, the derivative of x² is 2x, so the tangent slope at x=3 is 6. A derivative function collects those rates across the inputs where the function is differentiable.

    Read more: 2.2 · Derivative →

  • Limit

    A limit describes the value approached by outputs as inputs approach a specified target or grow without bound. For an input approaching a finite target, its value at that exact input does not determine the limit. For example, (x²−1)/(x−1) approaches 2 as x approaches 1 even though the original expression is undefined at 1. A finite two-sided limit requires the left and right approaches to agree.

    Read more: 1.2 · Limit →

  • Difference quotient

    A difference quotient divides a change in a function’s output by the associated nonzero input change. The form [f(a+h)−f(a)]/h is an average rate over an interval of length h. For example, for f(x)=x² it simplifies to 2a+h when h≠0. Taking its limit as h approaches 0 defines the derivative at a if the finite limit exists.

    Read more: 2.2 · Difference quotient →

  • Function

    A mathematical function assigns exactly one output to each input in its domain. Inputs can share an output, but one input cannot have two different outputs. For example, f(x)=x² assigns the output 9 to both 3 and −3. Function notation names the rule and its input; f(3) means the output when the input is 3.

  • Power rule

    The derivative power rule gives d(xⁿ)/dx=n xⁿ⁻¹ wherever the real power function is differentiable. Constant multiples remain as constant factors. For example, the derivative of 3x⁴ is 12x³. A composite power such as (x²+1)³ also requires the chain rule.

    Read more: 2.5 · Power rule →

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