An average rate is the slope of a secant; shrink the gap and it tips into the tangent — the speed of the moment.
Ask a car how fast it went and there are two honest answers. Over the whole trip it averaged, say, 50 miles per hour — total distance over total time, one number for the whole stretch. But the speedometer reads something else at every instant: 0 at the light, 65 on the highway, 20 in the parking lot. The average is the slope of the straight line joining the start and the end of a piece of the journey — a secant. The speedometer reading is what that slope settles onto when you shrink the piece of road down to a single moment. In this lesson we watch a secant tip over into a tangent as the gap Δx closes to nothing — and that limiting slope is the first object of calculus: the derivative f′(x₀).
You have measured rates all your life: miles per hour, dollars per year, degrees per minute. Each is a change in some quantity divided by the change in time (or in whatever the input is). If a quantity y = f(x) rises from f(a) to f(b) as x goes from a to b, the average rate of change over that stretch is
average rate over [a, b] = change in ychange in x = f(b) − f(a)b − a = Δy / Δx.
The units come along for the ride: if y is miles and x is hours, the ratio is miles per hour. The number tells you the typical pace across the whole interval — not what was happening at any one moment inside it.
A ball's height is f(x) = x² (meters, after x seconds). Over the stretch from x = 1 to x = 3 the average rate of change is f(3) − f(1)3 − 1 = 9 − 12 = 4 meters per second. On average the ball gained 4 m of height for every second over that window.
Plot the two points (a, f(a)) and (b, f(b)) and draw the straight line through them. That line is the secant (Latin secare, "to cut" — it cuts the curve at two places). Its slope is exactly rise over run, Δy / Δx — the very same fraction we just called the average rate. So the average rate of change is the slope of a secant line; the algebra and the picture are one idea.
The little dashed right triangle under the secant shows it: the run Δx along the bottom, the rise Δy up the side, and the secant's steepness is rise ÷ run. (Recall the idea of a limit we will lean on in a moment, and the derivative it builds toward.)
Move the two points a and b on y = x². The amber secant and its rise/run triangle redraw, and the readout quotes the average rate (f(b) − f(a)) / (b − a) — computed, never measured.
Average rate of change over [a, b] = slope of the secant through (a, f(a)) and (b, f(b)) = Δy / Δx. One fraction, two faces: an arithmetic and a line.
Now hold the left point still at a and slide the right point in toward it. As the gap Δx = b − a closes, the two cut-points merge into one, and the secant pivots — it stops cutting the curve and starts kissing it. The line it settles onto is the tangent at x = a, and its slope is what the average rates were heading for all along.
Watch the arithmetic. For f(x) = x² at a = 1, the secant slope over [1, 1 + Δx] is
f(1 + Δx) − f(1)Δx = (1 + Δx)² − 1Δx = 2Δx + Δx²Δx = 2 + Δx → 2 as Δx → 0.
The secant slopes 4, 3, 2.5, 2.1, 2.01, … march steadily toward 2. That single number 2 is the slope of the tangent at x = 1.
Hold a = 1 on y = x² and step the gap Δx down: 2 → 1 → 0.5 → 0.1 → 0.01. Each step redraws the amber secant and shows its slope from GX.diffQuotient marching toward the green tangent slope f′(1) = 2 (exact).
A secant slope is not the tangent slope for any finite Δx — it only becomes it in the limit. And you cannot just set Δx = 0 in f(a+Δx) − f(a)Δx; that is 0 / 0, undefined. The fix is to simplify first (the Δx cancels) and then let Δx → 0 — shrink, don't substitute.
That limiting number deserves a name. The instantaneous rate of change of f at x₀ is the limit of the average rates as the window collapses:
f′(x₀) = limΔx→0 f(x₀ + Δx) − f(x₀)Δx.
We read f′(x₀) as "f-prime of x₀." Geometrically it is the slope of the tangent at the point (x₀, f(x₀)); physically it answers "how fast, right now?" For our ball f(x) = x², the same algebra at a general point x₀ gives secant slope 2x₀ + Δx → 2x₀, so f′(x₀) = 2x₀: at x₀ = 1 the instant rate is 2, at x₀ = 3 it is 6. Every input gets its own slope — a hint of the next lesson, where collecting all of them turns f into a brand-new function.
Slide the point of contact x₀ along y = x². The green tangent redraws and the readout quotes f′(x₀) = 2x₀ — the exact slope from GX.polyDeriv, plus the tangent's equation y = m x + b.
The derivative f′(x₀) is the limit of secant slopes, the slope of the tangent line, and the instantaneous rate of change — three names for one number.
Put the idea on a clock. A dropped object falls s(t) = 5t² meters after t seconds. The average speed over a time window [t, t + Δx] is the secant slope (s(t + Δx) − s(t)) / Δx — total distance over total time. The instantaneous speed — what a speedometer would read at the single instant t — is the slope of the tangent, s′(t).
Since s(t) = 5t², the same cancellation gives average speed 10t + 5Δx → 10t, so s′(t) = 10t meters per second. At t = 2 s the speedometer reads 20 m/s, even though the average over the first two seconds was only 10 m/s. The object is speeding up, so the instant at the end beats the average.
On the position curve s(t) = 5t², fix the instant t = 2 s and step the window Δx down. The amber secant is the average speed over [2, 2 + Δx]; it closes onto the speedometer reading s′(2) = 20 m/s (exact).
Average speed of s(t) = 5t² over [2, 2.1]: s(2.1) − s(2)0.1 = 22.05 − 200.1 = 20.5 m/s. Over [2, 2.01] it is 20.05 m/s. The numbers are sliding toward the speedometer value s′(2) = 10·2 = 20 m/s — exactly what the limit predicts.
| Idea | Formula | Picture |
|---|---|---|
| Average rate of change over [a, b] | (f(b) − f(a)) / (b − a) | slope of the secant |
| Let the gap shrink | Δx → 0 | secant tips into the tangent |
| Instantaneous rate at x₀ (the derivative) | f′(x₀) = limΔx→0 (f(x₀+Δx) − f(x₀)) / Δx | slope of the tangent |
| For f(x) = x² | f′(x₀) = 2x₀ | steeper as |x₀| grows |
| Instant speed on a position curve | s′(t) (here 10t for s = 5t²) | speedometer = tangent slope |
(f(5) − f(2)) / (5 − 2) = (25 − 4) / 3 = 21 / 3 = 7. That is the slope of the secant joining (2, 4) and (5, 25).
((2 + Δx)² − 4) / Δx = (4 + 4Δx + Δx² − 4) / Δx = (4Δx + Δx²) / Δx = 4 + Δx. As Δx → 0 this → 4, so f′(2) = 4 — matching the rule f′(x) = 2x.
Average over [0, 4]: (s(4) − s(0)) / 4 = (80 − 0) / 4 = 20 m/s. Instant speed s′(t) = 10t, so s′(4) = 40 m/s. The car is accelerating, so the speed at the end is double the average over the whole stretch.
Plugging in Δx = 0 gives 0 / 0, which is undefined — division by zero. You must simplify the quotient first so the Δx in the denominator cancels, and only then let Δx → 0. Shrink, don't substitute.
Slope: f′(3) = 2·3 = 6. Point of contact: (3, f(3)) = (3, 9). Point-slope: y − 9 = 6(x − 3) ⟹ y = 6x − 9.
Secant slope = 2 + Δx, so the values are 2.5, 2.1, 2.01. They march toward 2, which is f′(1) — the slope of the tangent at x = 1.
Six questions to lock it in. Tap the answer you think is right.
This lesson builds the bridge from algebra's average rate of change to calculus's instantaneous rate. It develops CCSS HSF-IF.B.6 (calculate and interpret the average rate of change of a function over an interval, and estimate it from a graph) and the modeling ideas of HSF-LE (rates of change), then extends them past the standards into the limit that defines the derivative — the foundation of AP Calculus AB/BC. The recurring habit to reinforce is the difference between an average (a secant slope over a window) and an instant (the tangent slope, the limit as the window shrinks): the average is a single line through two points; the instant is what those slopes settle onto, never found by "plugging in zero."