Defining Average and Instantaneous Rates of Change at a Point: Study Guide (Calculus AB 2.1)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation

This page has no equation renderer, so limits are written in a compact form:

lim (h → 0) g(h) means “the limit as h approaches 0 of g(h)”.

f′(a) is read “f prime of a”. It is the name for the instantaneous rate of change of f at x = a. The mark ′ is a prime, not an apostrophe. On paper, keep “lim” on every line until you substitute, exactly as in Topic 1.6.

From Topic 1.1 to here

In Topic 1.1 you saw the big idea. An average rate of change needs two different input values. A rate at an instant is the value that average rates approach as the interval shrinks. Topic 2.1 turns that idea into a precise definition with a name, f′(a), and shows you how to calculate it exactly.

Three things are new:

1. Two standard ways to write the average rate, called difference quotients.
2. The derivative at a point, defined as the limit of a difference quotient.
3. Reading a limit backwards: spotting which function and which point it describes.

Average rate of change: two difference quotients

The average rate of change of f over an interval is the change in output divided by the change in input. There are two common ways to label the interval.

Form 1 (the h-form). The interval runs from a to a + h, where h ≠ 0:

(f(a + h) − f(a)) / h

Here h is the change in input. If h > 0, the interval is [a, a + h]. If h < 0, it is [a + h, a].

Form 2 (the x-form). The interval runs from a to some other input x, with x ≠ a:

(f(x) − f(a)) / (x − a)

Here x − a is the change in input.

These are the same quantity with different labels. Put x = a + h. Then x − a = h and f(x) = f(a + h), so Form 2 turns into Form 1. Whichever form you use, the answer is the slope of the secant line through (a, f(a)) and the second point on the graph.

Quick example. Let p(x) = 8/x and a = 2. Over [2, 4]:

• h-form with h = 2: (p(4) − p(2))/2 = (2 − 4)/2 = −1
• x-form with x = 4: (p(4) − p(2))/(4 − 2) = −1

Same number, as expected. The negative sign means p decreases over this interval, by 1 unit of output per unit of input on average.

The instantaneous rate of change: the derivative at a point

Now let the interval shrink. In Form 1, h → 0. In Form 2, x → a. The instantaneous rate of change of f at x = a is

lim (h → 0) (f(a + h) − f(a)) / h or, equivalently, lim (x → a) (f(x) − f(a)) / (x − a)

provided the limit exists. This number is called the derivative of f at a and is written f′(a).

Four points to hold on to:

• Both limits are 0/0 forms. At h = 0 (or x = a), the top is f(a) − f(a) = 0 and the bottom is 0. So substitution never works directly. You rewrite first, using the algebra from Topic 1.6, and then substitute.
• The two forms always agree. They are the same limit with x = a + h. Use whichever makes the algebra easier.
• The answer is a number. f′(a) is the slope of the tangent line at (a, f(a)), and the rate at that single input. Its units are output units per input unit.
• “Provided the limit exists” matters. If the difference quotient does not approach a single finite value, f has no derivative at a.

When the limit does not exist. Take f(x) = |x − 3| at a = 3. Then f(3 + h) − f(3) = |h|, so the h-form quotient is |h|/h. For h > 0 this equals 1; for h < 0 it equals −1. The right-hand limit is 1 and the left-hand limit is −1. They differ, so the limit does not exist and f′(3) does not exist. On the graph, there is a sharp corner at (3, 0): no single tangent line fits. Topic 2.4 studies this in full.

Worked example 1: the h-form for a polynomial

Question. Let f(x) = x³ − 4x. (a) Find the average rate of change of f over [2, 2.1]. (b) Use the definition of the derivative to find f′(2).

1. Find f(2). f(2) = 8 − 8 = 0.
2. Expand f(2 + h). (2 + h)³ = 8 + 12h + 6h² + h³, and 4(2 + h) = 8 + 4h. So f(2 + h) = 8 + 12h + 6h² + h³ − 8 − 4h = 8h + 6h² + h³
3. Form the difference quotient. For h ≠ 0: (f(2 + h) − f(2))/h = (8h + 6h² + h³)/h = 8 + 6h + h²
4. (a) Average rate over [2, 2.1]. This interval has h = 0.1. So the average rate is 8 + 0.6 + 0.01 = 8.61.
5. (b) Take the limit. f′(2) = lim (h → 0) (8 + 6h + h²) = 8 + 0 + 0 = 8.

Answer. (a) 8.61. (b) f′(2) = 8.

Check with the other form. f(x) − f(2) = x³ − 4x = x(x − 2)(x + 2). Dividing by (x − 2) gives x(x + 2) for x ≠ 2. As x → 2 this approaches 2 × 4 = 8. Both forms agree.

Interpretation. Near x = 2 the outputs of f increase about 8 times as fast as the inputs. The average over [2, 2.1] is a little larger (8.61) because the graph gets steeper to the right of 2. Over [1.9, 2], with h = −0.1, the same formula gives 8 − 0.6 + 0.01 = 7.41, a little smaller. The true rate, 8, lies between the two.

Why step 3 is allowed. Cancelling h needs h ≠ 0. A limit only looks at h close to 0, never at h = 0, so this is fine (the key fact from Topic 1.6).

Worked example 2: the x-form with a square root

Question. In a fictional experiment, the radius of a circular ripple on a pond is r(t) = √(2t + 1) metres, t seconds after a stone lands. (a) Find the average rate of change of the radius over [4, 12]. (b) Find the rate at which the radius is increasing at t = 4, using the x-form of the definition.

1. (a) Average rate. r(4) = √9 = 3 and r(12) = √25 = 5. So the average rate is (5 − 3)/(12 − 4) = 2/8 = 0.25 m/s.
2. (b) Write the x-form. With the input called t: r′(4) = lim (t → 4) (√(2t + 1) − 3)/(t − 4) Substitution gives 0/0, so rewrite.
3. Multiply top and bottom by the conjugate √(2t + 1) + 3. The top becomes (2t + 1) − 9 = 2t − 8 = 2(t − 4). Keep the bottom factored: = lim (t → 4) 2(t − 4) / ((t − 4)(√(2t + 1) + 3))
4. Cancel (t − 4), valid for t ≠ 4: = lim (t → 4) 2/(√(2t + 1) + 3)
5. Substitute. 2/(3 + 3) = 2/6 = 1/3.

Answer. (a) 0.25 m/s. (b) r′(4) = 1/3 m/s, about 0.333 m/s.

Check (with a calculator, not part of the method). √9.02 ≈ 3.003331, so (r(4.01) − 3)/0.01 ≈ 0.3331. That is close to 1/3.

Interpretation. At t = 4 seconds, the ripple’s radius is growing at 1/3 metre per second. Over the longer interval [4, 12] it grew more slowly on average (0.25 m/s), because the growth slows down as time passes.

Worked example 3: reading a limit as a derivative

Exam questions often give you a limit and ask what it represents, or ask you to evaluate it by recognising it. Match it against the template lim (h → 0) (f(a + h) − f(a))/h or lim (x → a) (f(x) − f(a))/(x − a).

(a) lim (h → 0) ((5 + h)² − 25)/h.

1. The pattern f(a + h) is (5 + h)², so f(x) = x² and a = 5.
2. Check f(a): 5² = 25, which matches the number being subtracted.
3. So the limit is f′(5) for f(x) = x². To evaluate it: ((5 + h)² − 25)/h = (10h + h²)/h = 10 + h for h ≠ 0, which approaches 10.

(b) lim (x → 1) (x⁴ − 1)/(x − 1).

1. The bottom is x − a with a = 1. The top must be f(x) − f(1). Try f(x) = x⁴: then f(1) = 1. It matches.
2. So the limit is f′(1) for f(x) = x⁴.
3. To evaluate it: x⁴ − 1 = (x − 1)(x + 1)(x² + 1). Cancel (x − 1), valid for x ≠ 1, to get (x + 1)(x² + 1). As x → 1 this approaches 2 × 2 = 4.

The match is not always unique. In (a), you could also say f(x) = (5 + x)² with a = 0, since f(0 + h) = (5 + h)² and f(0) = 25. Both descriptions are correct; the value of the limit is the same. When a question gives you answer options, pick the one that fits the template exactly: check the function, the point and the value of f at that point.

Representations

The same instantaneous rate shows up in all four representations the course uses:

Representation	Average rate of change	Instantaneous rate (derivative at a)
Analytical	(f(a + h) − f(a))/h or (f(x) − f(a))/(x − a)	f′(a) = the limit of either quotient
Graphical	Slope of the secant line through two points	Slope of the tangent line at (a, f(a))
Numerical	One quotient from two table values	The value the quotients approach as the interval shrinks (estimating from tables is Topic 2.3)
Verbal	“On average, the radius grew 0.25 m each second between t = 4 and t = 12”	“At t = 4 the radius is growing at 1/3 m per second”

Common misconceptions

• “f′(a) is the same as f(a).” f(a) is the value of the function; f′(a) is its rate of change. In Worked example 1, f(2) = 0 but f′(2) = 8.
• Subtracting the wrong thing. The top is f(a + h) − f(a), not f(a + h) − a, and not f(a + h) − f(h). In (f(x) − f(a))/(x − a), the bottom is x − a, not f(x) − f(a).
• Expanding f(a) + h instead of f(a + h). For f(x) = x², f(3 + h) = (3 + h)² = 9 + 6h + h². It is not 9 + h.
• Putting h = 0 straight in. That gives 0/0 every time. Simplify and cancel first.
• Dropping the limit. (f(a + h) − f(a))/h is an average rate. Only its limit is the instantaneous rate. In Worked example 1 the quotient is 8 + 6h + h²; the derivative is 8.
• “The average over a long interval is a good enough rate at a point.” In Worked example 2, the average over [4, 12] is 0.25 m/s, but the rate at t = 4 is 1/3 m/s.
• “Every function has a derivative everywhere.” The definition says “provided the limit exists”. At the corner of |x − 3|, the one-sided limits are −1 and 1, so there is no derivative at 3.
• Forgetting units. A derivative has units of output per input, such as metres per second.

Where this leads

Topic 2.1 gives the derivative at one point. In the next topic, Defining the Derivative of a Function and Using Derivative Notation, you let the point vary, so the derivative becomes a function f′(x). You also meet the notations dy/dx and y′ and use f′(a) to write the equation of a tangent line. Topic 2.3 estimates derivatives from tables and graphs, and Topic 2.4 explains exactly when the limit fails to exist. For the end of Unit 1, see Working with the Intermediate Value Theorem. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.