Lesson 5: The Derivative as a LimitAP

Every derivative in this course is a limit in disguise. CED Unit 2 (topics 2.1–2.4) defines the derivative as the limit of average rates of change, and the exam returns to this definition constantly — in table problems, in graph problems, and in the famous questions that hand you a limit and ask you to recognise it as a derivative. This lesson builds the definition in both of its forms, computes derivatives directly from it, and settles exactly how differentiability and continuity interact.

Average rate of change becomes instantaneous

The average rate of change of f on [a, b] is (f(b) − f(a))/(b − a) — the slope of the secant line through the two graph points. Shrink the interval toward a single point a by writing b = a + h:

Definition (limit form). f'(a) = lim(h→0) (f(a + h) − f(a))/h, when this limit exists.

Definition (alternative form). f'(a) = lim(x→a) (f(x) − f(a))/(x − a). The two forms are identical (set x = a + h); the exam uses both. Geometrically f'(a) is the slope of the tangent line at a; physically it is the instantaneous rate of change. When the variable point matters more than the point a, we write the derivative function f'(x) = lim(h→0) (f(x + h) − f(x))/h.

Worked Example 1: derivative from the definition

Problem. Use the limit definition to find f'(3) for f(x) = x^2 − 4x.

f'(3) = lim(h→0) [f(3 + h) − f(3)]/h
f(3 + h) = (3 + h)^2 − 4(3 + h) = 9 + 6h + h^2 − 12 − 4h = h^2 + 2h − 3
f(3) = 9 − 12 = −3
Difference: (h^2 + 2h − 3) − (−3) = h^2 + 2h
Divide by h: h + 2
Limit as h → 0: 2

Answer: f'(3) = 2. The tangent line at (3, −3) is y + 3 = 2(x − 3). Notice the anatomy of every such computation: expand, subtract, watch the constant terms cancel (they must — at h = 0 the quotient is 0/0), factor out h, cancel, and only then substitute.

Worked Example 2: a square root, and a tangent line

Problem. Find f'(4) for f(x) = sqrt(x), and write the tangent line at x = 4.

Use the alternative form with the conjugate:
f'(4) = lim(x→4) (sqrt(x) − 2)/(x − 4)
Multiply by (sqrt(x) + 2)/(sqrt(x) + 2):  = lim(x→4) (x − 4)/[(x − 4)(sqrt(x) + 2)]
= lim(x→4) 1/(sqrt(x) + 2) = 1/(2 + 2) = 1/4

Answer: f'(4) = 1/4, so the tangent at (4, 2) is y − 2 = (1/4)(x − 4). The conjugate trick from Lesson 2 returns — limit techniques and derivative computations are the same subject.

Worked Example 3: estimating a derivative from a table (graphical/numerical)

Problem. A table gives values of a differentiable function f: f(1) = 2.0, f(1.5) = 2.9, f(2) = 4.1, f(2.5) = 5.6, f(3) = 7.4. Estimate f'(2).

Solution. The derivative is the limit of secant slopes around 2, so average the neighbouring secant slopes. Left secant on [1.5, 2]: (4.1 − 2.9)/0.5 = 2.4. Right secant on [2, 2.5]: (5.6 − 4.1)/0.5 = 3.0. A good estimate splits the difference: f'(2) ≈ 2.7. Using the symmetric interval [1.5, 2.5] directly: (5.6 − 2.9)/1 = 2.7 — the same answer, and usually the best table estimate available. The exam accepts a reasonable range around the secant averages; what it will not accept is f(2) itself, or a single far-away secant like [1, 3] when closer data exists.

Differentiability versus continuity

Theorem. If f is differentiable at a, then f is continuous at a. The converse is false: continuity does not guarantee differentiability. Differentiability fails at:

• Corners (|x| at 0): left derivative −1, right derivative 1 — the one-sided difference quotients disagree.
• Cusps (x^(2/3) at 0): the tangent steepens without bound from both sides in opposite directions.
• Vertical tangents (cube root of x at 0): the difference quotient grows without bound.
• Any discontinuity: a jump or hole automatically destroys the derivative — this is the contrapositive of the theorem, and a favourite true/false trap.

Real-world application: velocity from a position table

A drone's altitude s(t) metres is logged every half second: s(2.0) = 14.0, s(2.5) = 16.1, s(3.0) = 18.9, s(3.5) = 22.4. Ground control needs the climb rate at t = 3. The symmetric secant on [2.5, 3.5] gives (22.4 − 16.1)/1 = 6.3 m/s — the best estimate of the instantaneous velocity s'(3) from this data. Flight software does exactly this computation millions of times: every "instantaneous" sensor rate is a difference quotient over a small interval, i.e. a limit that engineering approximates but calculus defines.

Common mistakes

• Estimating f'(a) with the value f(a), or with the widest available secant. Use the closest symmetric data around a.
• Concluding "not differentiable" from a hole in the formula but a fine graph — or vice versa. Check the graph behaviour: corner, cusp, vertical tangent, discontinuity.
• Sign errors in f(a + h) for negative coefficients, as in f(3 + h) for x^2 − 4x: the term −4(3 + h) = −12 − 4h is where most algebraic slips happen. Expand term by term.
• "Proving" a derivative exists at a corner by computing one side. Both one-sided limits of the difference quotient must agree.

AP Exam Experience

1. (No calculator.) If f(x) = x^3, then f'(2) is the value of which limit?

• (A) lim(h→0) ((2 + h)^3 − 8)/h
• (B) lim(h→0) ((2 + h)^3 − 2)/h
• (C) lim(x→2) (x^3 − 2)/(x − 2)
• (D) lim(h→0) (8 + h)^3/h

Answer: (A). The definition needs f(2 + h) − f(2) = (2 + h)^3 − 8 over h. (B) subtracts 2 instead of f(2) = 8; (C) has the wrong numerator constant in alternative form (should be x^3 − 8).

2. The limit lim(h→0) (sqrt(9 + h) − 3)/h represents f'(a) for which f and a?

• (A) f(x) = sqrt(x), a = 3
• (B) f(x) = sqrt(x), a = 9
• (C) f(x) = sqrt(x + 9), a = 0 gives the same form
• (D) f(x) = x^2, a = 3

Answer: (B). Match the template (f(a + h) − f(a))/h: inside the root, 9 + h means a = 9, and f(9) = 3 is the subtracted constant. (Its value is 1/(2·3) = 1/6, as Lesson 6's rules will confirm.)

3. At x = 2, the function f(x) = |x − 2| − 2 has which property?

• (A) f is continuous and differentiable at 2.
• (B) f is continuous but not differentiable at 2.
• (C) f is not continuous at 2.
• (D) f has a removable discontinuity at 2.

Answer: (B). The graph is the V of |x − 2| shifted down 2, with its corner at (2, −2). Left and right values agree there (f is continuous), but the left-hand slope is −1 and the right-hand slope is 1, so the difference quotient has no single limit and f'(2) does not exist. The lesson: an absolute value's corner sits where its argument is zero — locate it, do not pattern-match.

Free-response question. A differentiable function f has the following values: f(0) = 1, f(1) = 3, f(2) = 6, f(3) = 8, f(4) = 9.

(a) Estimate f'(2). Show the calculation you use. (b) A student computes the average rate of change of f on [0, 4], obtaining (9 − 1)/4 = 2, and claims f'(2) must therefore equal 2 exactly. Explain the error in the student's reasoning. (c) Explain why there must be a point c in (1, 4) where the instantaneous rate of change of f equals 2. Name the theorem you use.

Scoring guideline.

• (a) 1 point. A symmetric secant around 2, or the average of the two adjacent secants: (f(3) − f(1))/(3 − 1) = (8 − 3)/2 = 2.5, or the average of (6 − 3)/1 = 3 and (8 − 6)/1 = 2, also 2.5. Award the point for f'(2) ≈ 2.5 with a shown difference quotient using neighbouring values.
• (b) 1 point. The student has computed an *average* rate over a four-unit interval; an average rate over [0, 4] need not equal the *instantaneous* rate at the single point x = 2, and a wide-interval secant is a poorer estimate than the neighbouring-data quotient of part (a). The number 2 is a coincidence of this data, not a deduction.
• (c) 2 points. 1 point for the computation (f(4) − f(1))/(4 − 1) = (9 − 3)/3 = 2. 1 point for invoking the Mean Value Theorem: f is differentiable on [1, 4] (given), hence continuous there, so there exists c in (1, 4) with f'(c) equal to that average rate, namely 2. (Lesson 11 develops the MVT fully; this preview rewards students who already see average rates as promises about instantaneous ones.)

Additional worked practice

A symmetric difference quotient is measurably better than a one-sided one. For f(x) = x^3 at x = 2 (true derivative f'(2) = 12), a table gives f(1.9) = 6.859 and f(2.1) = 9.261. The symmetric estimate is (9.261 − 6.859)/(2.1 − 1.9) = 2.402/0.2 = 12.01 — within a hundredth of the truth, because the errors of the left and right secants (which straddle the tangent slope for a convex curve) nearly cancel. The one-sided estimates, (8 − 6.859)/0.1 = 11.41 and (9.261 − 8)/0.1 = 12.61, are each off by about 0.6. Whenever a table offers data on *both* sides of the point, spend it symmetrically.

Extension: spotting derivatives in disguise, and a smoothness surprise

A family of exam questions presents a limit and asks what it *is*, rather than what it *equals*: lim(h→0) (e^(2 + h) − e^2)/h is, by template-matching against (f(a + h) − f(a))/h, the derivative of e^x at a = 2 — so its value is e^2, no computation required. Train the reflex: numerator = (something at a + h) − (same thing at a), denominator = h → derivative at a.

A calibration example for differentiability: let f(x) = x|x|, i.e. f(x) = x^2 for x ≥ 0 and f(x) = −x^2 for x < 0. Then f'(x) = 2x for x > 0 and f'(x) = −2x for x < 0; at 0, the difference quotient is (h|h|)/h = |h| → 0, so f'(0) = 0 and the two formulas stitch together as f'(x) = 2|x| everywhere. The graph has no corner — it flattens smoothly through the origin — yet f' itself has a corner at 0, so f is differentiable but not *twice* differentiable there. Differentiability comes in grades, and this function is the standard exhibit.