Lesson 6: The Derivative: Definition

(a) Opening Question

A car's position along a straight road is given by s(t) = t² meters, where t is in seconds.

The average velocity from t = 2 to t = 2 + h is the change in position divided by the change in time:

[s(2+h) − s(2)] / h

Compute this average velocity for h = 1, then h = 0.1, then h = 0.01. (You found average rates of change like this back in Lesson 1.)

h = 1:    [s(3) − s(2)] / 1    = (9 − 4)/1     = 5
h = 0.1:  [s(2.1) − s(2)] / 0.1 = (4.41 − 4)/0.1 = 4.1
h = 0.01: [s(2.01) − s(2)]/0.01 = (4.0401 − 4)/0.01 = 4.01

The numbers are marching toward 4. As the time interval shrinks to zero, the average velocity approaches the instantaneous velocity at t = 2. That limiting value — 4 m/s — is the derivative of s at t = 2. This whole lesson is about that one idea: a derivative is a limit.

(b) Core Concepts

In Unit 1 you learned what a limit is. Today the limit does real work: it turns average rates of change into instantaneous ones. A derivative is nothing more than a limit of a difference quotient. If you understand limits, you already understand derivatives — we are just applying the machine to one specific expression.

The difference quotient

Pick a function f and a point x. Move a tiny step h away to x + h. The slope of the secant line through (x, f(x)) and (x+h, f(x+h)) is the difference quotient:

[f(x + h) − f(x)] / h

This is the average rate of change of f over the interval from x to x + h. It is also the average velocity, the average slope — same object, different words.

The derivative as a limit

The derivative of f is what the difference quotient approaches as the step h shrinks to zero:

f'(x) = lim_{h→0} [f(x + h) − f(x)] / h

This is the limit definition of the derivative. When this limit exists, we say f is differentiable at x. Geometrically, as h → 0 the secant line pivots until it becomes the tangent line, so:

f'(x) is the slope of the tangent line to the graph of f at the point (x, f(x)), and it equals the instantaneous rate of change of f at x.

The alternate (at-a-point) form

There is a second, equivalent way to write the derivative at a specific point x = a. Instead of stepping by h, let a second input x slide toward a:

f'(a) = lim_{x→a} [f(x) − f(a)] / (x − a)

Both forms give the same number. Use the first (h → 0) form when you want a formula for f'(x); use the second (x → a) form when a problem hands you a fixed point a and asks for the slope there — it often factors more cleanly.

Computing a derivative from the definition — every step

Let f(x) = 2x² − 3x + 1. Find f'(x) from the definition. Watch every algebra step, because this is exactly where AP students drop points.

Step 1 — Write f(x + h). Substitute x + h everywhere you see x:

f(x + h) = 2(x + h)² − 3(x + h) + 1
         = 2(x² + 2xh + h²) − 3x − 3h + 1
         = 2x² + 4xh + 2h² − 3x − 3h + 1

Step 2 — Form the numerator f(x + h) − f(x). Subtract the original f(x) = 2x² − 3x + 1. The non-h terms must cancel:

f(x+h) − f(x) = (2x² + 4xh + 2h² − 3x − 3h + 1) − (2x² − 3x + 1)
              = 4xh + 2h² − 3h

Notice 2x², −3x, and +1 all cancelled. If they don't cancel, you made an arithmetic error — go back.

Step 3 — Divide by h. Every surviving term has a factor of h, so we can cancel it. This is the whole point of the algebra: kill the h in the denominator before taking the limit.

[f(x+h) − f(x)] / h = (4xh + 2h² − 3h) / h = 4x + 2h − 3

Step 4 — Take the limit as h → 0. Now substitution is legal:

f'(x) = lim_{h→0} (4x + 2h − 3) = 4x − 3

So f'(x) = 4x − 3. Check one value: f'(2) = 4(2) − 3 = 5, matching the slope you'd see on the graph at x = 2. (You can confirm 4x − 3 with the power rule next lesson — but on the exam, "from the definition" means you must show Steps 1–4.)

The tangent line equation

The derivative gives you the slope of the tangent line. To get the line itself at x = a, you need two things: the slope m = f'(a) and a point (a, f(a)). Then use point-slope form:

y − f(a) = f'(a) · (x − a)

Mini-example. For f(x) = x², the derivative (from the definition) is f'(x) = 2x. Find the tangent line at x = 3.

• Point: f(3) = 9, so (3, 9).
• Slope: f'(3) = 2(3) = 6.
• Line: y − 9 = 6(x − 3), i.e. y = 6x − 9.

The most common error here is using the function value 9 where the slope 6 belongs. The slope is f'(a), never f(a).

• Parabola opening upward, vertex at (0, 0)
• Tangent line is straight, touching the parabola only at the point (3, 9)
• The tangent line has slope 6 and y-intercept (0, −9)
• Point of tangency (3, 9) marked with a dot]

Notation — four ways to say the same thing

The derivative of y = f(x) is written, interchangeably:

f'(x)      "f prime of x"           (Lagrange)
dy/dx      "dee y dee x"            (Leibniz)
y'         "y prime"                (Lagrange)
df/dx      "dee f dee x"            (Leibniz)

Leibniz notation dy/dx is not a fraction (yet — it behaves like one later), but it usefully reminds you it is a rate: units of y per unit of x. To show "the derivative evaluated at x = a," write f'(a) or dy/dx |_{x=a}.

Estimating a derivative from a table

If you only have data points, you can't take a true limit — but you can approximate f'(a) with a small difference quotient using the table values closest to a. The best estimate uses a value on each side of a (a symmetric difference):

f'(a) ≈ [f(a + h) − f(a − h)] / (2h)

This is still an approximation, not the exact derivative — always say "approximately" when you estimate from data.

Calculator: numerical derivative on the TI-84

On calculator-active parts you can get f'(a) numerically with nDeriv:

TI-84: MATH → 8:nDeriv( → nDeriv(2X²−3X+1, X, 2) → 5
Means f'(2) = 5 for f(x) = 2x² − 3x + 1

The newer template is nDeriv(expression, X, point). Important: nDeriv returns a numerical approximation, and it is only allowed on the calculator sections (Section I Part B, Section II Part A). On non-calculator parts you must compute by hand.

(c) Worked Examples

Example 1 — Derivative from the definition [NO CALC]

Problem. Let f(x) = x² − 4x. Use the limit definition to find f'(x), then find f'(1).

Strategy. Apply the four steps: write f(x+h), subtract, divide by h, take the limit.

Solution.

f(x+h) = (x+h)² − 4(x+h) = x² + 2xh + h² − 4x − 4h
f(x+h) − f(x) = (x² + 2xh + h² − 4x − 4h) − (x² − 4x) = 2xh + h² − 4h
[f(x+h) − f(x)] / h = 2x + h − 4
f'(x) = lim_{h→0} (2x + h − 4) = 2x − 4

Then f'(1) = 2(1) − 4 = −2.

Justification. Every term of the original f(x) cancelled in the numerator, and each remaining term carried a factor of h, so the difference quotient simplified to a polynomial in h. Direct substitution of h = 0 is then valid because the simplified expression is continuous.

Example 2 — Tangent line at a point [NO CALC]

Problem. For f(x) = x² − 4x, find the equation of the tangent line at x = 1.

Strategy. Need a point and a slope. Reuse f'(x) = 2x − 4 from Example 1.

Solution.

• Point: f(1) = 1² − 4(1) = −3, so (1, −3).
• Slope: f'(1) = −2.
• Tangent line: y − (−3) = −2(x − 1) → y = −2x − 1.

Justification. The slope of the tangent line at x = 1 is f'(1) = −2, the instantaneous rate of change of f at that point; point-slope form through (1, −3) gives the line.

Example 3 — Estimate f'(a) from a table [NO CALC]

Problem. A function g is given by the table below (x in seconds, g in meters). Estimate g'(4).

Strategy. Use a symmetric difference around x = 4, choosing the table values on each side: x = 2 and x = 6.

Solution.

g'(4) ≈ [g(6) − g(2)] / (6 − 2) = (17 − 5) / 4 = 12/4 = 3

So g'(4) ≈ 3 meters per second.

Justification. With only tabulated data, g'(4) cannot be computed exactly; the secant slope over the smallest symmetric interval around x = 4 available in the table, [2, 6], gives the best estimate, approximately 3.

Example 4 — Numerical derivative with nDeriv [CALC]

Problem. Let f(x) = x³ − 5x. Use a calculator to find f'(2), then write the tangent line at x = 2.

Strategy. Use nDeriv for the slope; compute f(2) directly for the point.

Solution.

TI-84: MATH → 8:nDeriv( → nDeriv(X³−5X, X, 2) → 7

So f'(2) ≈ 7. The point is f(2) = 8 − 10 = −2, i.e. (2, −2).

Tangent line: y − (−2) = 7(x − 2) → y = 7x − 16.

Justification. nDeriv reports the numerical value 7 for the instantaneous rate of change at x = 2; with the point (2, −2), point-slope form yields the tangent line. (By hand, the power rule gives f'(x) = 3x² − 5, so f'(2) = 12 − 5 = 7 — the calculator agrees, but this nDeriv value would only be acceptable on a calculator-active part.)

(d) Common Mistakes

Algebra slip in f(x + h). Students write f(x+h) = f(x) + h or forget to square the whole binomial: (x+h)² becomes x² + h² (missing the 2xh). Fix: expand (x+h)² fully as x² + 2xh + h² every time, and substitute x + h into every x, including inside −3x and constants.

Forgetting to take the limit. Some students simplify the difference quotient to 4x + 2h − 3 and stop — leaving an h in the answer. The difference quotient is the average rate; the derivative is its limit. Fix: the last line must always be lim_{h→0}, and the final answer can contain no h.

Cancelling h before simplifying. You cannot cancel the h in the denominator until every term in the numerator has a factor of h. Cancelling early (e.g., crossing out one h while another term has none) is illegal algebra. Fix: fully simplify the numerator first; the constant terms must cancel, leaving every surviving term with an h.

Using the point instead of the slope in the tangent line. Writing y − f(a) = f(a)(x − a) — using the function value as the slope. Fix: the slope is always f'(a) (the derivative), and the point is (a, f(a)). Two different objects: f(a) locates, f'(a) tilts.

Treating a table estimate as exact. Reporting g'(4) = 3 with an equals sign. From a table you can only estimate. Fix: write g'(4) ≈ 3 and, on an FRQ, name the interval you used.

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## (e) Practice Problems

Using the limit definition, f'(x) for f(x) = 3x² is:

- (A) 6x (B) 3x (C) 6x + h (D) 9x²

For f(x) = 5x − 2, f'(x) =

- (A) 5x (B) 5 (C) −2 (D) 0

The difference quotient [f(x+h) − f(x)] / h represents:

- (A) the slope of the tangent line at x

- (B) the instantaneous rate of change at x

- (C) the average rate of change of f from x to x + h

- (D) the value f'(x)

Using the alternate form, lim_{x→3} [f(x) − f(3)] / (x − 3) equals:

- (A) f(3) (B) f'(3) (C) 0 (D) the average rate of change on [0, 3]

The tangent line to f(x) = x² at x = −2 is:

- (A) y = −4x − 4 (B) y = 4x + 4 (C) y = −4x + 4 (D) y = 4x − 4

If f'(x) = 2x − 4 (from Example 1), at which x is the tangent line horizontal?

- (A) x = 0 (B) x = 2 (C) x = 4 (D) x = −2

For f(x) = x³ − 2x, nDeriv(X³−2X, X, 1) returns approximately:

- (A) −1 (B) 1 (C) 3 (D) 0

Use the table to estimate f'(3) with a symmetric difference:

| x | 1 | 2 | 3 | 4 | 5 |

|---|---|---|---|---|---|

| f(x) | 10 | 14 | 15 | 18 | 26 |

- (A) 2 (B) 4 (C) 1 (D) 3

Which notation does not represent the derivative of y = f(x)?

- (A) dy/dx (B) f'(x) (C) Δy/Δx (D) y'

(Short answer.) Use the limit definition to find f'(x) for f(x) = x² + 2x. Show all four steps.

(Short answer.) Find the equation of the tangent line to f(x) = x² + 2x at x = 1.

(Short answer.) For f(x) = sin(x), use nDeriv to estimate f'(0). State the value and what it represents.

(Justification.) A student claims that for f(x) = 7, the derivative is f'(x) = 7. Using the limit definition, explain whether this is correct.

(Justification.) The position of a particle is s(t) = t² − 6t (meters, t in seconds). Find s'(t) from the definition, and explain in a complete sentence what s'(2) tells you about the particle's motion, including units.

Given f(x) = x², the average rate of change on [2, 2+h] simplifies to 4 + h. As h → 0, this gives f'(2) =

- (A) 4 (B) 0 (C) 4 + h (D) 2

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## (f) AP Exam Focus — Free Response Question

This is the lesson's full AP-style FRQ. Total: 9 points. A calculator is permitted on parts that involve numerical evaluation; all justification must be shown in writing.

Problem. A storage tank is draining. The volume of water in the tank, V(t), is measured in gallons, where t is the time in minutes since draining began, 0 ≤ t ≤ 12. Selected values of V(t) are given in the table.

| t (minutes) | 0 | 2 | 5 | 9 | 12 |

|---|---|---|---|---|---|

| V(t) (gallons) | 240 | 220 | 198 | 164 | 150 |

(a) (2 points) Find the average rate of change of V over the interval 0 ≤ t ≤ 12. Indicate units of measure.

(b) (2 points) Use the data in the table to estimate V'(5). Show the computation that leads to your answer, and indicate units of measure.

(c) (3 points) Using your answer from part (b), write an equation for the line tangent to the graph of V at t = 5. Use this tangent line to approximate the volume of water in the tank at t = 6.

(d) (2 points) Using correct units, interpret the meaning of V'(5) in the context of this problem. Is the value of V'(5) consistent with a tank that is draining? Justify your answer.

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### Model Solution

(a) The average rate of change of V over [0, 12] is the difference quotient over the whole interval:

[V(12) − V(0)] / (12 − 0) = (150 − 240) / 12 = −90/12 = −7.5 gallons per minute

(b) Estimate V'(5) with a symmetric difference using the table values on each side of t = 5, namely t = 2 and t = 9:

V'(5) ≈ [V(9) − V(2)] / (9 − 2) = (164 − 220) / 7 = −56/7 = −8 gallons per minute

(c) The tangent line at t = 5 passes through the point (5, V(5)) = (5, 198) with slope V'(5) ≈ −8:

y − 198 = −8(t − 5)      i.e.   y = 198 − 8(t − 5)

Approximate V(6) by evaluating the tangent line at t = 6:

V(6) ≈ 198 − 8(6 − 5) = 198 − 8 = 190 gallons

(d) V'(5) ≈ −8 means that at t = 5 minutes, the volume of water in the tank is decreasing at a rate of approximately 8 gallons per minute (the instantaneous rate of change of volume with respect to time). This is consistent with a draining tank: a draining tank loses water, so the volume is decreasing, and a negative rate of change (V'(5) < 0) correctly indicates a decreasing volume.

Scoring Commentary — where students lose points

• Part (a) — 2 points: 1 point for the correct difference quotient setup (150 − 240)/12, 1 point for −7.5 with units (gallons per minute). Students lose the units point constantly. A bare −7.5 earns only 1 of 2.
• Part (b) — 2 points: 1 point for a valid difference-quotient setup using table values, 1 point for −8 with units. The strongest answer uses the symmetric interval [2, 9]. A one-sided estimate (e.g., [V(9)−V(5)]/(9−5)) is usually accepted, but students who average two table values, or who write V'(5) = ... with an equals sign instead of ≈, risk the communication point. Sign matters: dropping the negative loses the point.
• Part (c) — 3 points: 1 point for the correct point (5, 198), 1 point for using the slope from (b) in a correct tangent-line equation, 1 point for the approximation 190. The classic loss: using V(5) = 198 as the slope instead of V'(5) = −8. Also, students who never actually evaluate at t = 6 leave the third point on the table.
• Part (d) — 2 points: 1 point for interpreting V'(5) as a rate of decrease of volume with correct units ("the volume is decreasing at about 8 gallons per minute at t = 5"), 1 point for the justification tying the negative sign to draining. Students lose points for vague answers like "the tank is going down" (no units, no rate language) or for stating a value without the word "decreasing." Saying "the volume is 8" (a value, not a rate) earns zero — it confuses V with V'.

f(x+h) = (x+h)² + 2(x+h) = x² + 2xh + h² + 2x + 2h
f(x+h) − f(x) = 2xh + h² + 2h
[f(x+h) − f(x)]/h = 2x + h + 2
f'(x) = lim_{h→0} (2x + h + 2) = 2x + 2

s(t+h) = (t+h)² − 6(t+h) = t² + 2th + h² − 6t − 6h
s(t+h) − s(t) = 2th + h² − 6h
[s(t+h) − s(t)]/h = 2t + h − 6
s'(t) = lim_{h→0} (2t + h − 6) = 2t − 6