Lesson 3: The Definition of the Derivative

Here is the question that creates the derivative: a speedometer shows instantaneous speed, but every measurement we can actually make is an average speed over some time interval. Calculus resolves the paradox by taking a limit — and that limit, applied to any function, is the single most important definition in this course. Everything later (rules, applications, even the integral's partner theorem) is bookkeeping built on top of it.

From average rates to the instantaneous rate

A car's position is s(t) = t^2 + 2t metres at time t seconds. Its average velocity on [1, 1 + h] is (s(1 + h) − s(1))/h. Shrink h:

   h        1       0.5      0.1      0.01
  avg vel   5       4.5      4.1      4.01     (m/s)

The averages converge to 4 m/s. That limiting value is the instantaneous velocity at t = 1 — a number no single measurement gives, but which the sequence of measurements pins down completely. Now replace "position" by "any function" and "velocity" by "rate of change," and you have the derivative.

The definition

The derivative of f at a is

f'(a) = lim(h→0) ( f(a + h) − f(a) ) / h

provided the limit exists. Geometrically it is the slope of the tangent line to the graph at a (the secants through (a, f(a)) and (a + h, f(a + h)) rotate into it as h→0). The equivalent form with x in place of a + h is f'(a) = lim(x→a) (f(x) − f(a))/(x − a). The derivative function f'(x) is what you get by letting a be a variable.

Worked Example 1 (the definition, used honestly). Find f'(2) for f(x) = x^3 directly from the definition.

( f(2 + h) − f(2) ) / h = ( (2 + h)^3 − 8 ) / h
(2 + h)^3 = 8 + 12h + 6h^2 + h^3, so the quotient is
(12h + 6h^2 + h^3)/h = 12 + 6h + h^2      (h ≠ 0)
As h→0:  12 + 0 + 0 = 12.

f'(2) = 12. Interpretation: at x = 2 the cube function is climbing at 12 units of output per unit of input, and the tangent line there has slope 12. Notice the method: expand, cancel the h (legal since h ≠ 0), then let h→0. Every derivative rule in the next lesson is this computation done once, in general, and then memorised.

Worked Example 2. Use the definition to find the derivative of f(x) = 1/x at x = 3.

( 1/(3 + h) − 1/3 ) / h = ( (3 − (3 + h)) / (3(3 + h)) ) / h
= ( −h / (3(3 + h)) ) / h = −1/(3(3 + h)) → −1/9 as h→0.

f'(3) = −1/9. The negative sign is information: 1/x is decreasing, and at x = 3 it falls gently. This matches the pattern f'(x) = −1/x^2 that the power rule will soon hand you for free.

Worked Example 3 (recognising a disguised derivative). Evaluate lim(x→2) (x^3 − 8)/(x − 2).

This is exactly the x-form of the definition with f(x) = x^3 and a = 2. By Worked Example 1, the value is f'(2) = 12 — no factoring needed once you see the shape. Exam questions love this disguise; the cue is "function minus its value at a point, over x minus a."

When the derivative does not exist

The limit can fail, and each failure has a picture: a corner (f(x) = |x| at 0: left secants have slope −1, right secants +1, so the two-sided limit fails), a cusp (slopes blow up with opposite signs), a vertical tangent (slopes grow without bound, same sign), and any discontinuity (if f is not continuous at a, it cannot be differentiable there — although the converse is false, as |x| shows). Differentiability is stronger than continuity: smooth implies unbroken, not the reverse.

Application: marginal analysis in economics

A firm's profit from selling x units is P(x) = −0.5x^2 + 40x − 100 dollars. What does one more unit earn at a production level of 10? The exact gain from unit 11 is P(11) − P(10) = (−60.5 + 440 − 100) − (−50 + 400 − 100) = 279.5 − 250 = 29.5 dollars. The derivative gives P'(x) = −x + 40, so P'(10) = 30 dollars per unit — the marginal profit. The derivative is an approximation to the one-step difference, improving as the unit size shrinks relative to the scale of the business. Economists run entire pricing decisions on this limit; it is the derivative doing its original job, measuring a rate, in a context with no graph in sight.

Common mistakes

• Dividing by h before taking the limit, then "setting h = 0" in the original quotient. Wrong: (f(a + 0) − f(a))/0 is 0/0, meaningless. Right: simplify until the h cancels, then let h→0.
• Treating a secant slope as the derivative. Wrong: with h = 0.5, the average 4.5 m/s in the opening table is not the instantaneous velocity. Right: 4.5 is an approximation; the derivative is the limit, 4.
• Believing continuity guarantees a derivative. |x| at 0 is the standard counterexample: perfectly continuous, sharply cornered, no derivative.
• Forgetting what the answer means. f'(2) = 12 is a rate with units (output per input) and a geometric meaning (tangent slope). Answers that report it as "the value of the function" signal a memorised procedure, not understanding.

Exam Experience

1. lim(h→0) ( (3 + h)^2 − 9 )/h equals

• (A) 0
• (B) 3
• (C) 6
• (D) The limit does not exist.

Answer: (C). This is the definition of the derivative of x^2 at 3: 2·3 = 6. Directly: ((9 + 6h + h^2) − 9)/h = 6 + h → 6.

2. Using the definition, the derivative of f(x) = sqrt(x) at x = 4 is

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

Answer: (B). Rationalising gives 1/( sqrt(4 + h) + 2 ) → 1/(2 + 2) = 1/4, matching the rule f'(x) = 1/(2 sqrt(x)) at x = 4.

3. At which point does f(x) = |x − 2| fail to be differentiable?

• (A) Everywhere
• (B) Only at x = 2
• (C) Only at x = 0
• (D) Nowhere — absolute value functions are smooth

Answer: (B). At x = 2 the left difference quotients equal −1 and the right ones equal +1; the corner kills the two-sided limit. Everywhere else the function is a straight line with a well-defined slope.

Free-response question. Let f(x) = x^2 − 3x. (a) Use the limit definition to find f'(x). (b) Write the equation of the tangent line to the graph at x = 2. (c) The tangent line in (b) crosses the graph at one other point besides the point of tangency. Find it, and explain in one sentence what this shows about tangent lines.

Scoring guideline.

• (a) 2 points. 1 point for correct setup and expansion: ((x + h)^2 − 3(x + h) − (x^2 − 3x))/h = (2xh + h^2 − 3h)/h = 2x + h − 3. 1 point for the limit f'(x) = 2x − 3. Reporting an unsimplified quotient with h still present earns at most 1.
• (b) 1 point. f(2) = −2 and f'(2) = 1, giving y + 2 = 1(x − 2), i.e. y = x − 4.
• (c) 1 point. Solving x^2 − 3x = x − 4 gives (x − 2)^2 = 0, so the "other point" is x = 2 itself — the tangent touches with multiplicity two and does not cross elsewhere; the point earned requires noting that a tangent line need not meet the curve only once in general (a cubic's tangent can cross again), with this quadratic as the degenerate case. Full credit also for the correct observation with a valid justification.

Going deeper: two more definition workouts

The square root at a general point. Find the derivative of f(x) = sqrt(x) from the definition, at a general x:

( sqrt(x + h) − sqrt(x) )/h, rationalise with the conjugate sqrt(x + h) + sqrt(x):
= h / ( h( sqrt(x + h) + sqrt(x) ) ) = 1/( sqrt(x + h) + sqrt(x) ) → 1/(2 sqrt(x))

So f'(x) = 1/(2 sqrt(x)) — the power rule's answer, earned honestly. Note where it collapses: at x = 0 the difference quotients grow without bound, the graph has a vertical tangent, and no derivative exists — which matches 1/(2 sqrt(0)) being undefined.

Where does the tangent meet the curve again? The tangent to y = x^3 at x = 2 is y = 12x − 16 (slope 12 from Worked Example 1, point (2, 8)). Intersections satisfy x^3 = 12x − 16, i.e. x^3 − 12x + 16 = 0, which factorises as (x − 2)^2 (x + 4) = 0. The double root x = 2 is the tangency itself; the simple root x = −4 is a genuine second meeting, at (−4, −64) (check the line: 12(−4) − 16 = −64 ✓). Tangent lines to cubics usually do cross the curve elsewhere — the touching is local, and the factorisation shows the tangency and the crossing in one glance: a double root and a single root.

One more existence case: the cusp

The power f(x) = x^(2/3) (the square of the cube root) is defined and continuous everywhere, including 0. Its difference quotient at 0 is h^(2/3)/h: from the right (h > 0) this is a positive number over a positive h, growing like h^(−1/3) → +∞; from the left (h < 0) the numerator h^(2/3) is still positive (it is a square) while h is negative, so the quotient → −∞. The two one-sided slope limits are +∞ and −∞ — the graph pinches into a cusp at the origin with a vertical tangent direction. Continuous, not differentiable, and for a third distinct reason in this lesson's catalogue: corners (finite one-sided slopes), vertical tangents (both sides +∞), cusps (opposite infinities). When a problem asks "is it differentiable?", the case work is the content.