DWG NO. 10.5 — Lesson 5 of 5

Introduction to the Derivative as a Rate of Change

Unit 10: Limits and an Introduction to Calculus · ~30 min

Objective Compute an average rate of change over an interval and an instantaneous rate of change at a point using the limit definition of the derivative.

Every earlier unit that talked about "rate of change" meant an average — slope of a line, slope of a secant, change in y over change in x across some interval. But a car's speedometer reads a speed at one instant, not an average over a trip. Turning "average over an interval" into "exactly at one point" is precisely what a limit makes possible, and that's the derivative — the last new idea of this course, and the first big idea of calculus.

Average rate of change

The average rate of change of f over the interval [a, a+h] is the familiar slope formula, applied to two points on the curve:

Average rate of change [f(a+h) − f(a)\(\frac{]}{h}\)

Geometrically, this is the slope of the secant line — the line cutting through the curve at x=a and x=a+h. This exact ratio, written this way, is called the difference quotient.

From secant to tangent: the derivative

As h shrinks toward 0, the second point a+h slides back toward a, and the secant line rotates to hug the curve more and more closely at that single point — in the limit, it becomes the tangent line. The slope of that tangent line is the derivative of f at a, written f′(a):

The derivative (limit definition) f′(a) = li\(m_{h→0}\) [f(a+h) − f(a)\(\frac{]}{h}\)

f′(a) is the instantaneous rate of change of f at x = a — exactly what a speedometer or a marginal-cost calculation is really asking for. Applying the same limit at a general x instead of a fixed a gives f′(x), the derivative function, whose value at any point is the slope of f's tangent line there.

(a, f(a)) (a+h, f(a+h)) tangent as h→0 x y

As h→0, the secant line through (a, f(a)) and (a+h, f(a+h)) (dashed) rotates into the tangent line (solid gold) — its slope is f′(a)

Worked Example 1 · Average rate of change
ProblemFind the average rate of change of f(x) = x² on the interval [1, 3].
1f(1) = 1, f(3) = 9. Apply the slope formula: [f(3)−f(1)]/(3−1) = (9−1)/2.
Average rate of change = 4
Worked Example 2 · Derivative at a point (limit definition)
ProblemFind f′(2) for f(x) = x² using the limit definition.
1f′(2) = li\(m_{h→0}\) [f(2+h) − f(2)] / h = li\(m_{h→0}\) [(2+h)² − 4] / h.
2Expand: (2+h)² = 4 + 4h + h², so the numerator is 4+4h+h²−4 = 4h+h².
3[4h+h²]/h = 4+h for h≠0 (the 0/0 form resolves just like Lesson 10.2). As h→0, this approaches 4.
f′(2) = 4 — the tangent line to f(x)=x² at x=2 has slope 4
Worked Example 3 · The derivative function
ProblemFind f′(x) for f(x) = x² − 3x using the limit definition.
1f′(x) = li\(m_{h→0}\) [f(x+h)−f(x)]/h = li\(m_{h→0}\) [(x+h)²−3(x+h) − (x²−3x)]/h.
2Expand the numerator: x²+2xh+h²−3x−3h−x²+3x = 2xh+h²−3h.
3Divide by h: 2x+h−3 for h≠0. As h→0, this approaches 2x−3.
f′(x) = 2x − 3

Guided practice

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