IB Maths AI SLDifferentiationPaper 1 & 2dy/dx = f′(x)~6 min read
Introduction to Derivatives
Calculus is the mathematics of change. The derivative of a function gives its gradient — its rate of change — at any point. This note builds the idea from scratch: what a limit is, how the gradient of a curve is defined by chords closing in on a tangent, and the notation used throughout differentiation.
📘 What you need to know
A limit is the value a function approaches as x approaches a particular value from both sides — useful even where the function is undefined.
Calculus studies rates of change. The derivative of a function is its gradient function — input an x-value, get the gradient there.
The gradient of a curve at a point P equals the gradient of the tangent to the curve at P.
The tangent’s gradient is the limit of the gradients of chords PQ as Q slides along the curve toward P.
The gradient of a chord between two points is (change in y) ÷ (change in x) = (y2 − y1)/(x2 − x1).
Notation: for y = f(x) the derivative is dy/dx = f′(x); other variables follow the same pattern.
Limits
A limit is the value a function heads toward as x gets close to some value — approaching from below and from above. Limits matter most when the function is undefined at that exact value.
Take f(x) = (x2 − 9)/(x − 3). At x = 3 this is 0 ÷ 0 — undefined. But for every other x it simplifies to x + 3, so as x → 3 the function approaches 6. That limiting value is 6, even though f(3) itself does not exist. You can estimate a limit from a table of values either side of the point, or from a GDC graph.
The derivative: a gradient function
A straight line has one fixed gradient. A curve does not — its steepness changes from point to point. The derivative is the function that records this: feed in a value of x and it returns the gradient of the curve there.
Because calculus is the study of rates of change, the derivative is a rate of change — of position it gives velocity, of velocity it gives acceleration.
Derivative notationdydx = f′(x)
the derivative (gradient function) of y = f(x) · other variables follow the pattern, e.g. V = f(s) gives dV/ds = f′(s)
The letters just track which quantity changes with respect to which — dy/dx and f′(x) mean exactly the same thing.
From chords to the tangent
How is the gradient of a curve at a single point even defined? Take a point P on the curve and a second point Q. The straight line PQ is a chord, and its gradient is easy: (y2 − y1)/(x2 − x1). Now slide Q along the curve toward P. The chord rotates, and its gradient closes in on a single value — the gradient of the tangent at P. That limiting value is the gradient of the curve at P.
As the point Q slides along the curve toward P, the chord PQ rotates toward the tangent at P — so the chord’s gradient closes in on the gradient of the curve at P.
Gradient of a chord
gradient = y2 − y1x2 − x1as the second point slides toward the first, the chord gradient approaches the gradient of the tangent
A chord gradient is only an estimate of the tangent gradient — but the closer Q sits to P, the better that estimate becomes.
🧠Recipe — estimating a gradient from chords
Mark the point P(x1, y1) on the curve where you want the gradient.
Pick points Q on the curve, each one closer to P than the last.
Find each chord gradient: (yQ − yP) ÷ (xQ − xP).
Look at the sequence of chord gradients — see what value they close in on.
State the estimate: that limiting value is the gradient of the curve (the derivative) at P.
Worked examples
WE 1
Gradient of a chord
The curve y = x2 passes through P(3, 9) and Q(5, 25). Find the gradient of the chord [PQ].
chord gradient = (change in y) ÷ (change in x)= (25 − 9) / (5 − 3) = 16 / 2gradient of [PQ] = 8a chord is just a straight line between two points on the curve — its gradient is the usual (y₂−y₁)/(x₂−x₁).
WE 2
Estimating a gradient from chords
The curve y = x2 passes through P(3, 9). By finding the gradients of chords from P to points ever closer to P, estimate the gradient of the curve at P.
chord to (4, 16): (16 − 9)/(4 − 3) = 7chord to (3.5, 12.25): (12.25 − 9)/(3.5 − 3) = 6.5chord to (3.1, 9.61): (9.61 − 9)/(3.1 − 3) = 6.1the chord gradients 7, 6.5, 6.1 close in on 6gradient of the curve at P ≈ 6as Q slides toward P the chord gradient approaches a limit — here clearly 6. That limit is the derivative at P.
WE 3
Estimating a limit from a table
A function is given by f(x) = (x2 − 9)/(x − 3). (a) Explain why f(x) is undefined at x = 3. (b) Use a table of values to estimate the limit of f(x) as x → 3.
(a) at x = 3 the denominator x − 3 = 0dividing by zero is undefined, so f(3) does not exist(b) table either side of 3:x = 2.9, 2.99 → f(x) = 5.9, 5.99x = 3.01, 3.1 → f(x) = 6.01, 6.1from both sides f(x) → 6, so the limit is 6for x ≠ 3, (x²−9)/(x−3) = x + 3, which confirms the limit — but f is still undefined exactly at x = 3.
WE 4
Derivative notation
A balloon is being inflated. Its volume V is a function of time t, written V = f(t). (a) Write the derivative of V with respect to t in two notations. (b) State in words what this derivative represents.
(a) two notations for the same derivative:dV/dt and f′(t)(b) what it representsthe rate of change of volume with respect to timehow fast the balloon’s volume is increasingthe letters change with the variables, but dV/dt and f′(t) mean the same thing — the gradient of the V–t graph.
WE 5
A derivative as a rate of change
The distance s metres travelled by a cyclist after t seconds is given by s = f(t). It is known that f′(12) = 7. (a) State what f′(12) = 7 tells you. (b) State the units of this derivative.
(a) f′(12) is the rate of change of distance at t = 12 sdistance is increasing at 7 metres each secondthe cyclist’s speed at t = 12 s is 7 m/s(b) units of the derivativemetres per second (m/s)a derivative is always a rate of change — here distance per unit time, which is speed.
WE 6
Full question: chords, estimate and meaning
A drone’s height h metres after t seconds follows a curve h = f(t). The curve passes through P(4, 32), and also through (6, 72), (5, 50) and (4.5, 40.5). (a) Find the gradient of each chord from P. (b) Estimate the gradient of the curve at P. (c) Interpret your answer in context.
(a) chord gradients from P(4, 32):to (6, 72): (72 − 32)/(6 − 4) = 20to (5, 50): (50 − 32)/(5 − 4) = 18to (4.5, 40.5): (40.5 − 32)/(4.5 − 4) = 17(b) the gradients 20, 18, 17 close in on 16gradient of the curve at P ≈ 16(a) 20, 18, 17 · (b) ≈ 16 · (c) the drone climbs at ≈ 16 m/s when t = 4 seach Q is closer to P than the last, so the chord gradients home in on the true gradient at P — the drone’s vertical speed.
💡 Top tips
The gradient of a curve changes along it — that’s why the derivative is a whole function, not a single number.
A chord gradient only estimates the tangent gradient — the closer the second point, the better the estimate.
The gradient of a curve at a point equals the gradient of the tangent there.
A limit can exist even where the function is undefined — it describes what a function approaches, not what it equals.
dy/dx and f′(x) mean the same thing — and the pattern holds with other letters: dV/dr, ds/dt, and so on.
âš Common mistakes
Thinking a curve has one fixed gradient — only straight lines do; a curve’s gradient varies with x.
Treating a chord gradient as the exact answer — it estimates the tangent gradient, it does not equal it.
Assuming a function must be defined at x = a for a limit to exist there — it need not be.
Mixing up the order in (y2 − y1)/(x2 − x1) — keep the points in the same order top and bottom.
Reading dy/dx as a fraction to “cancel” — it is one symbol meaning the derivative, not a division.
Next up: Differentiating Powers of x — the quick rule that turns a function straight into its derivative, no chords needed. For now, hold on to the big picture: the derivative is a gradient function, built from chords closing in on a tangent. Every technique that follows is just a shortcut to that same idea.
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