What Is a Derivative?
You have built up functions, trigonometry, and sequences. This topic opens the last big door in the Functions module: calculus. A derivative answers one question: how fast is something changing, right now? On a graph, that question becomes: how steep is the curve at this exact point? Everything in this topic — and the whole Applications topic after it — grows out of that one idea.
A word about the exam, so your effort goes where it pays. The standard dated papers barely test derivatives directly; the explicit questions live on the harder papers — Mock Exam Paper 20 asks for derivative values, tangent slopes, and a chain-rule derivative — and the official syllabus lists "basics of derivatives: definition, geometric meaning, and simple applications." So we will learn it properly, but efficiently. This page builds the idea itself: what a derivative is and what it means on a picture. No shortcut rules yet — those arrive on page 2 and turn everything here into fast exam technique.
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Average rate of change
Start somewhere familiar. A car drives 180 km in 2 hours. Its average speed is 180 ÷ 2 = 90 km/h. Notice what that number really is: the total change in position divided by the total change in time. It says nothing about any single moment — the speedometer almost certainly never sat exactly on 90 the whole trip. It is a summary of the journey, not a snapshot.
The same idea works for any function. Pick two inputs, a and b, and ask: on the way from a to b, how much did the output change per unit of input? Divide the change in output by the change in input. Graphically, this is the slope of the straight line drawn through the two points (a, f(a)) and (b, f(b)) — a line that cuts across the curve, called a secant line.
Find the average rate of change of f(x) = x² between x = 1 and x = 3.
Find the two outputs first: f(1) = 1² = 1 and f(3) = 3² = 9.
Change in output: 9 − 1 = 8. Change in input: 3 − 1 = 2.
Divide: 8 / 2 = 4.
So on the way from x = 1 to x = 3, the function climbs 4 units of output for each unit of input, on average. The secant line through (1, 1) and (3, 9) has slope 4.
From average to instantaneous
Now the interesting question. The car's speedometer does not show an average — it shows the speed right now, at this instant. That is an instantaneous rate of change. But there is a puzzle hiding here: an instant has no width. In zero time the car covers zero distance, and 0 / 0 is meaningless. So how can "speed at an instant" mean anything at all?
Here is the trick, and it is the founding idea of calculus: take averages over smaller and smaller intervals, and watch where the answers are heading. The average over one second is a decent estimate of the speed now. Over a tenth of a second, better. Over a thousandth, better still. The averages crowd toward one particular value — and that value is what we mean by the instantaneous rate.
Estimate the instantaneous rate of change of f(x) = x² at x = 1, by averaging over the small interval from 1 to 1 + h and letting h shrink.
Try h = 0.1, so the interval is [1, 1.1]: the average rate is (1.1² − 1²) / 0.1 = (1.21 − 1) / 0.1 = 0.21 / 0.1 = 2.1.
Try h = 0.01, interval [1, 1.01]: (1.0201 − 1) / 0.01 = 0.0201 / 0.01 = 2.01.
Each time h shrinks, the average lands closer to 2. The estimates are crowding toward 2 — that is our instantaneous rate at x = 1.
Laying out more values of h makes the pattern impossible to miss:
| Interval width h | Interval | Average rate of change of x² |
|---|---|---|
| 1 | [1, 2] | 3 |
| 0.5 | [1, 1.5] | 2.5 |
| 0.1 | [1, 1.1] | 2.1 |
| 0.01 | [1, 1.01] | 2.01 |
| 0.001 | [1, 1.001] | 2.001 |
The same story plays out on the graph. The average rate is the slope of a secant line through two points. As the second point slides toward the first, the secant tilts, settling toward one final position: the line that just grazes the curve at that point. Watch it happen:
The derivative: a definition
Time to give the idea its proper name and its proper formula. Fix a point x₀. Step a small distance h away from it, to x₀ + h. The average rate of change over that little step is [f(x₀ + h) − f(x₀)] / h — exactly the formula from the first section, with a = x₀ and b = x₀ + h. The instantaneous rate is what this average approaches as h shrinks to nothing. That target value is the derivative.
The formula looks busier than it is. In practice, computing a derivative "from the definition" (also called from first principles) is a three-step recipe — and the middle step always has the same goal: get rid of the lone h sitting in the denominator.
Use the definition to find f′(1) for f(x) = x².
Step 1 — the difference quotient at x₀ = 1: [(1 + h)² − 1²] / h.
Step 2 — expand the square: (1 + h)² = 1 + 2h + h², so the top is 1 + 2h + h² − 1 = 2h + h².
Factor out h on top: (2h + h²) / h = h(2 + h) / h = 2 + h. The troublesome h in the denominator is gone.
Step 3 — let h → 0: the expression 2 + h approaches 2. So f′(1) = 2.
Look back at the table in the last section: every row was 2 + h, and the rows crowded toward 2. The algebra has just explained the whole table in one line.
Even better: nothing in that computation cared that the point was 1. Run the same three steps at a general point x, and you get the derivative everywhere at once.
Use the definition to find f′(x) for f(x) = x², at a general point x.
Step 1 — the difference quotient: [(x + h)² − x²] / h.
Step 2 — expand: (x + h)² = x² + 2xh + h², so the top is x² + 2xh + h² − x² = 2xh + h².
Factor and cancel: (2xh + h²) / h = h(2x + h) / h = 2x + h.
Step 3 — let h → 0: the result is 2x. So f′(x) = 2x.
One computation, infinitely many answers: f′(1) = 2, f′(3) = 6, f′(−2) = −4 — the steepness at every point of the parabola, all from a single formula.
Two more functions are worth doing from the definition, because their answers become two of the most-used facts in the whole topic.
Use the definition to find f′(x) for the line f(x) = 3x + 1.
Step 1 — the difference quotient: [3(x + h) + 1 − (3x + 1)] / h.
Step 2 — simplify the top: 3x + 3h + 1 − 3x − 1 = 3h. So the quotient is 3h / h = 3.
Step 3 — let h → 0: there is no h left at all, so the limit is simply 3. f′(x) = 3.
That makes perfect sense: a straight line is equally steep everywhere, and its steepness is its slope. The derivative of a line is just its slope — no calculation needed once you see why.
Use the definition to find f′(x) for the constant function f(x) = 5.
Step 1 — the difference quotient: [5 − 5] / h, because the output is 5 no matter the input.
Step 2 — the top is 0, so the quotient is 0 / h = 0 for every h.
Step 3 — the limit of 0 is 0. So f′(x) = 0.
Again the picture agrees: the graph of f(x) = 5 is a horizontal line — perfectly flat, slope zero everywhere. A function that never changes has a rate of change of zero.
Notation: three ways you will see it written
You will meet the derivative in three slightly different costumes, and exam questions use all of them. f′(x) is the slope machine — a whole function. f′(x₀), like f′(3) or f′(−1), is that machine evaluated at one point — a single number. And when a function is written as y = …, its derivative is often written y′, which means exactly the same as f′(x). (Other notations exist in bigger calculus courses, but these three cover everything the CSCA asks.)
The geometric meaning: slope of the tangent
Go back to the sliding-secant picture for a moment. As Q slid into P, the secant lines settled into one final line — the line that just grazes the curve at P, matching its direction exactly at that point. That line is the tangent line at P. And since the secant slopes were the average rates, the line they settle into has slope equal to the value the average rates settle into: the derivative.
Here is the picture for our running example. We computed f′(1) = 2 for f(x) = x² — so the tangent at the point (1, 1) must climb 2 units for every 1 unit it runs. The triangle on the figure shows exactly that:
With the slope machine f′(x) = 2x in our pocket, tangent-slope questions about the parabola become instant — and this is precisely the shape of a real exam item. Mock Exam Paper 20 asks for the point on y = x³ where the tangent slope is 3; by page 4 you will dispatch that in seconds. For now, our parabola:
Find the slope of the tangent line to y = x² at x = 3.
The tangent slope at a point is the derivative there.
The slope machine for x² is f′(x) = 2x, so f′(3) = 2 × 3 = 6.
The tangent at (3, 9) has slope 6.
Sanity-check against the picture: x = 3 is well up the right branch of the parabola, where the curve is climbing steeply — a slope of 6 fits.
Find the slope of the tangent line to y = x² at x = −2.
Evaluate the slope machine at the point: f′(−2) = 2 × (−2) = −4.
The tangent at (−2, 4) has slope −4.
Read the sign and the size separately. The minus says the curve is falling there — and it is: x = −2 sits on the left branch, which descends as you read left to right. The 4 says it falls fairly steeply. A negative derivative is not an error; it is information.
Reading the sign of f′ from a graph
The last worked example hints at a powerful shortcut: very often you do not need the value of a derivative at all — only its sign. And the sign is visible directly on the graph, no formula required. Read the curve from left to right, like text. Wherever it is climbing, tangents point uphill: f′ > 0. Wherever it is falling, tangents point downhill: f′ < 0. And at the crest of a hill or the floor of a valley, the curve is — for one instant — neither rising nor falling: the tangent is horizontal, and f′ = 0.
Using the figure above, determine the sign of f′ at point A and at point D.
At A: reading left to right, the curve is climbing toward the hilltop. Rising curve means uphill tangent, so f′(A) > 0.
At D: the point sits at the very floor of the valley. For that one instant the curve has stopped falling and not yet started rising — the dashed tangent is horizontal.
A horizontal line has slope 0, so f′(D) = 0.
No formula, no algebra — the graph alone settles the sign. Exam questions love this skill because it tests understanding rather than computation.
Your turn! · Practice questions
Things to remember
| Concept | What to know |
|---|---|
| Average rate of change | [f(b) − f(a)] / (b − a): change in output per unit of input over a whole interval. Slope of the secant line through the two points. |
| Instantaneous rate | The value the average rates approach as the interval shrinks to nothing. The speedometer reading, not the trip summary. |
| The derivative | f′(x₀) = limit as h → 0 of [f(x₀ + h) − f(x₀)] / h. Read "f prime". |
| First-principles recipe | 1) Write the difference quotient. 2) Simplify until the h in the denominator cancels. 3) Let h → 0 — no h may survive in the answer. |
| Notation | f′(x) is a function (the slope machine); f′(x₀) is a single number; y′ means the same as f′(x). |
| Easy cases | A straight line's derivative is its slope; a constant's derivative is 0; for f(x) = x², f′(x) = 2x. |
| Geometric meaning | f′(x₀) is the slope of the tangent line to the graph at x₀. Height f(x₀) and steepness f′(x₀) are different things. |
| Signs from a graph | Reading left to right: rising → f′ > 0, falling → f′ < 0, hilltop or valley floor → f′ = 0. |
| Coming next | Page 2 replaces the limit recipe with fast shortcut rules for xⁿ, sin x, cos x, eˣ and ln x. Page 4 writes tangent-line equations. Exact uphill/downhill intervals live in the Applications of Differentiation topic. |