Lesson 11: Introduction to Derivatives — The Tangent Problem

Here is the problem that created calculus. A curve has no single slope — it steepens and flattens from point to point — yet at any one point it clearly leans in some direction. The tangent problem asks: what is the slope of a curve at a single point? The answer, built in this lesson, is the derivative: a limit of ordinary slopes.

Average rate of change: slopes of secants

For a function f on [a, b], the average rate of change is (f(b) − f(a))/(b − a) — the slope of the secant line through the two points, and the direct descendant of every "average speed" you have ever computed. Numerical view: for f(x) = x^3 at x = 1, take the second point at x = 1 + h:

As h shrinks, the secant pivots towards the tangent and its slope closes in on 3. Nothing else is happening in this lesson — and in the derivative half of calculus — except making this table's limit exact.

The difference quotient and the derivative

Replace the two points a and a + h and the secant slope becomes the difference quotient (f(a + h) − f(a))/h. The derivative of f at a is its limit:

f'(a) = lim(h→0) (f(a + h) − f(a))/h

Worked example. f(x) = x^2 − 3x. The difference quotient (at a general point x) is ((x + h)^2 − 3(x + h) − (x^2 − 3x))/h = (2xh + h^2 − 3h)/h = 2x + h − 3, so f'(x) = 2x − 3. At x = 2: f'(2) = 1. The tangent line at (2, f(2)) = (2, −2) is y + 2 = 1·(x − 2), i.e. y = x − 4. Notice the workflow: expand, cancel h (legal because h ≠ 0 in a limit — the same licence as Lesson 9), then let h → 0. Every derivative you will ever compute from the definition follows these three steps.

Two more derivatives from the definition

f(x) = sqrt(x) at x = 1. Rationalise (Lesson 9's technique, now doing calculus): (sqrt(1 + h) − 1)/h = 1/(sqrt(1 + h) + 1) → 1/2. So f'(1) = 1/2 and the tangent at (1, 1) is y = (1/2)(x − 1) + 1. f(x) = 1/x. The difference quotient is (1/(x + h) − 1/x)/h = ((x − (x + h))/(x(x + h)))/h = −1/(x(x + h)) → −1/x^2. So the derivative of 1/x is −1/x^2 — negative, matching a graph that falls everywhere. These two results, with f'(x) = 2x − 3 above, are the first entries of the derivative table you will extend all through calculus.

Modelling: instantaneous velocity

Average velocity over an interval is a secant slope; instantaneous velocity is the derivative. A ball thrown upward has height s(t) = −16t^2 + 64t feet after t seconds. Its velocity is v(t) = s'(t) = −32t + 64 (from the definition: the difference quotient of −16t^2 + 64t is −32t − 16h + 64 → −32t + 64). So v(0) = 64 ft/s (launched upward), v(1) = 32 ft/s (slowing), v(2) = 0 — the top of the flight, where s(2) = −64 + 128 = 64 ft — and v(4) = −64 ft/s as it lands. The sign change of the derivative *is* the event "reaching the top": instantaneous rate zero, direction reversing. Speedometers, heart-rate monitors and marginal-cost calculations are all this same idea with different labels.

Common mistakes

• Substituting h = 0 into the difference quotient. That gives 0/0 — meaningless. The limit process (simplify first, then let h approach 0) is not a formality; it is the definition.
• Sign errors in f(x + h). For f(x) = x^2 − 3x, the term −3(x + h) = −3x − 3h, not −3x + 3h or −3h alone. Distribute over the whole of (x + h).
• Confusing average and instantaneous rates. The average velocity from t = 1 to t = 3 for s(t) = −16t^2 + 64t is (s(3) − s(1))/2 = (48 − 48)/2 = 0 ft/s — the ball returned to its t = 1 height. That says nothing about v(1) = 32 ft/s. Different questions, different mathematics.
• Reporting units wrongly. A derivative of position (feet) with respect to time (seconds) is in ft/s. If f is in dollars and x in items, f'(x) is dollars per item. The units are part of the interpretation and part of the answer.
• Thinking every function has a derivative everywhere. At a corner (like |x| at 0) the left and right difference quotients approach different limits (−1 and 1), so no single tangent exists. Differentiability is a property to check, not to assume.

Exam Experience

1. Using the limit definition, the derivative of f(x) = x^2 at x = 3 is

• (A) 9
• (B) 3
• (C) 6
• (D) 2

Answer: (C). lim(h→0) ((3 + h)^2 − 9)/h = lim(h→0) (6h + h^2)/h = lim (6 + h) = 6. (A) is f(3) — the function value, not the slope; the question tests exactly that confusion.

2. A car travels s(t) = t^2 metres in t seconds. Its average velocity on [1, 3] is

• (A) 2 m/s
• (B) 4 m/s
• (C) 6 m/s
• (D) 8 m/s

Answer: (B). (s(3) − s(1))/(3 − 1) = (9 − 1)/2 = 4 m/s. (A) is the *instantaneous* velocity at t = 1 (s'(1) = 2) — the average/instantaneous distinction in one question.

3. The slope of the tangent line to f(x) = 1/x at x = 2 is

• (A) 1/4
• (B) −1/4
• (C) −1/2
• (D) 4

Answer: (B). From the definition, f'(x) = −1/x^2, so f'(2) = −1/4. (A) is the function value f(2) = 1/2 halved by a slip, and also has the wrong sign — the graph of 1/x falls, so its tangents fall.

Constructed response. Let f(x) = x^2 − 3x. (a) Use the limit definition to find f'(x), showing the difference quotient and its simplification. (b) Write the equation of the tangent line to the graph at x = 2. (c) A ball has height s(t) = −16t^2 + 64t feet. Find v(1) using the derivative pattern of part (a), interpret it in words, and find the time and height at the top of the flight.

Scoring guide. (a) 3 points: difference quotient ((x + h)^2 − 3(x + h) − x^2 + 3x)/h — 1 point; simplifies to 2x + h − 3 after cancelling h — 1 point; limit as h → 0 is 2x − 3 — 1 point. (b) 1 point: f(2) = −2 and f'(2) = 1, giving y = x − 4. (c) 3 points: v(t) = −32t + 64 (applying the same definition to the quadratic) — 1 point; v(1) = 32 ft/s, the ball is rising at 32 ft/s after 1 second — 1 point; top when v = 0: t = 2 s, with s(2) = 64 ft — 1 point.

Going deeper: when derivatives fail, and tangent-line approximation

The derivative needs the two one-sided limits of the difference quotient to agree. For f(x) = |x| at 0 they do not: from the right the quotient is (h)/h = 1, from the left (−h)/h = −1. The graph has a corner, no single tangent, and no derivative at 0 — while being perfectly continuous. Continuity is necessary for differentiability but not sufficient; corners, cusps and vertical tangents are the three standard counterexamples, and recognising them from a graph is a core readiness skill.

Linear approximation is the tangent line put to work: near x = a, f(x) ≈ f(a) + f'(a)(x − a). For f(x) = sqrt(x) at a = 4: f(4) = 2 and f'(4) = 1/(2·sqrt(4)) = 1/4, so sqrt(4.2) ≈ 2 + (1/4)(0.2) = 2.05. The true value is 2.0494 — the tangent estimate errs by about 0.0006, because over a step of 0.2 the curve barely bends away from its tangent. This is not a parlour trick: it is how calculators approximated roots for decades, how engineers linearise sensors, and the seed of Newton's method and of all of differential calculus's local thinking: *curves are lines, if you stand close enough.*

Workshop: the sqrt table and the cubic's derivative

Secants closing in, once more. For f(x) = sqrt(x) at x = 4, secant slopes: h = 1 → (sqrt(5) − 2)/1 ≈ 0.2361; h = 0.1 → (sqrt(4.1) − 2)/0.1 ≈ 0.2485; h = 0.01 → (sqrt(4.01) − 2)/0.01 ≈ 0.2498. The target is visible: f'(4) = 1/4, consistent with f'(x) = 1/(2·sqrt(x)) (the pattern f'(1) = 1/2 from the main lesson). Tables like this are how derivatives were computed before the definition was routinised — and how you should sanity-check any derivative you compute by rule: one secant at small h, compared against your claimed slope, catches most sign and arithmetic errors.

A cubic from the definition. For f(x) = x^3, the difference quotient at a general x is (3x^2 h + 3xh^2 + h^3)/h = 3x^2 + 3xh + h^2 → 3x^2. At x = 1 the slope is 3 (matching the very first table of this lesson: 7, 3.31, 3.0301, 3.0030) and the tangent line at (1, 1) is y = 3x − 2. The expansion (x + h)^3 — binomial theorem again — is the only algebra the whole computation needed.

Final drills: a negative slope and an average-versus-instant check

Tangent at a negative point. For f(x) = x^2 at x = −1: f'(−1) = −2 (from f'(x) = 2x), the point is (−1, 1), and the tangent line is y − 1 = −2(x + 1), i.e. y = −2x − 1. Negative slopes on the left half of a parabola are the graph's symmetry made algebraic — the tangent at x = −1 mirrors the tangent y = 2x − 1 at x = 1. Average versus instantaneous, quantified: for f(x) = 1/x on [1, 2], the average rate is (1/2 − 1)/(2 − 1) = −1/2, while the instantaneous rate at x = 1 is f'(1) = −1. The average over the interval is *half* the instantaneous rate at its start — because the function flattens as x grows (f'(2) = −1/4), and the average blends the whole range of slopes. Derivatives vary; averages summarise; never substitute one for the other.