IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 ~7 min read

Introduction to Derivatives

A derivative is a function that gives the gradient of another function at any point. The idea is built on a limit: take a chord between two points on a curve, then slide one point closer to the other — the chord gradient approaches a fixed value, the gradient of the tangent. The derivative f′(x) outputs that gradient for every input x.

📘 What you need to know

Limit: the value f(x) approaches as x → some value (even if f is undefined there).
Chord gradient: f(x₂) − f(x₁)x₂ − x₁ — rise over run between two points on the curve.
Tangent gradient: the limit of the chord gradient as x₂ → x₁; the slope of the line that just touches the curve.
Derivative: the function f′(x) that outputs the tangent gradient at every x; also called the gradient function.
Two notations: f′(x) (Lagrange) and dydx (Leibniz) — same thing, used interchangeably.
Variable can be anything: for V = f(s), the derivative is dVds = f′(s).
GDC limitation: it can compute a derivative AT a specific value of x but won’t return the derivative function itself.
Real-world meaning: rate of change — speed = derivative of position with respect to time, etc.

Limits — what a function approaches

The limit of a function at a point is the value it approaches as x gets closer and closer — even if the function isn’t defined there. Limits matter because the chord-to-tangent argument relies on one.

Limit notation lim_{x → a} f(x) = L

Read this as: as x approaches a, the value of f(x) approaches L.

From chord to tangent — the derivative as a limit

Pick a fixed point P on the curve y = f(x) and a nearby point Q. The chord PQ has gradient (rise over run). As Q slides along the curve toward P, the chord gradient approaches the tangent gradient at P.

As Q slides toward P along the curve, the chord [PQ] flattens onto the tangent. The tangent gradient is the limit of the chord gradients.

Two-sided check: a smooth function has the same chord-gradient limit whether Q approaches P from the right (x₂ > x₁) or the left (x₂ < x₁). If the two sides disagree, the derivative doesn’t exist at that point.

Notation and what it means

Lagrange notation

f′(x)

“f-prime of x”; emphasises function form

Leibniz notation

dydx

“dee y by dee x”; emphasises rate of change

Both mean exactly the same thing. Leibniz form is handy when the variables aren’t x and y — e.g. for V = f(s) the derivative is dVds.

🧭 Recipe — estimate a derivative numerically

Pick the point P(x₁, f(x₁)) where you want the gradient.
Choose a nearby point Q at x₂ slightly away from x₁; compute the chord gradient.
Repeat with Q closer (e.g. x₁ ± 0.1, ±0.01, ±0.001) — chord gradients should converge.
Estimate the tangent gradient as the limit of the chord gradients.
Cross-check by computing f′ algebraically (next sub-section) or with the GDC’s d/dx tool.

Worked examples

WE 1

Estimate a limit from values approaching a point

The function g(x) = x³ − 8x − 2 is undefined at x = 2 (zero in the denominator). Use the values g(2.1), g(2.01), g(2.001), g(1.999), g(1.99), g(1.9) to estimate lim_{x → 2} g(x).

Compute g at points either side of x = 2 g(2.1) = 12.6100 g(2.01) = 12.0601 g(2.001) = 12.006001 g(1.999) = 11.994001 g(1.99) = 11.9401 g(1.9) = 11.4100 Both sides converge on the same value lim_{x→2} g(x) = 12 algebraic check: x³ − 8 = (x − 2)(x² + 2x + 4), so g simplifies to x² + 2x + 4 (with hole at x = 2); at x = 2 this is 4 + 4 + 4 = 12 ✓

WE 2

Chord gradients converging to a tangent — quadratic

Consider the curve y = f(x) where f(x) = x² + 1, at the point P(3, 10). Find the gradients of the chords [PA], [PB], [PC] where A is at x = 3.5, B at x = 3.1, C at x = 3.01. Hence estimate the tangent gradient at P.

Apply chord gradient formula (y₂ − y₁)/(x₂ − x₁) f(3.5) = 13.25 → [PA] = (13.25 − 10)/0.5 = 6.50 f(3.1) = 10.61 → [PB] = (10.61 − 10)/0.1 = 6.10 f(3.01) = 10.0601 → [PC] = (10.0601 − 10)/0.01 = 6.01 As x → 3, chord gradient approaches a clear limit [PA] = 6.50 → [PB] = 6.10 → [PC] = 6.01 → … Estimate of tangent gradient at P: 6 spoiler: f′(x) = 2x, so f′(3) = 6 exactly — your numerical estimate is bang-on

WE 3

Chord gradients converging — cubic

Consider f(x) = x³ + x at the point P(1, 2). Find the chord gradients to A(x = 1.2), B(x = 1.1) and C(x = 1.05). Estimate the tangent gradient at P.

Compute f at each x and the chord gradients f(1.2) = 1.728 + 1.2 = 2.928 → [PA] = (2.928 − 2)/0.2 = 4.64 f(1.1) = 1.331 + 1.1 = 2.431 → [PB] = (2.431 − 2)/0.1 = 4.31 f(1.05) ≈ 2.207625 → [PC] = (2.207625 − 2)/0.05 ≈ 4.15 Convergence pattern 4.64 → 4.31 → 4.15 → … Estimate of tangent gradient at P: 4 algebraic answer: f′(x) = 3x² + 1 → f′(1) = 4; gradients are converging from above

WE 4

Notation and substitution

The function h(t) = t² − 4t + 5 has derivative h′(t) = 2t − 4. (a) Write the derivative in Leibniz notation. (b) Find h′(3).

(a) Leibniz form uses the input variable dh/dt = 2t − 4 (b) Substitute t = 3 h′(3) = 2(3) − 4 = 6 − 4 = 2 (a) dh/dt = 2t − 4; (b) h′(3) = 2 notation choice is just preference — both forms communicate the same idea

WE 5

Compute derivative AT a point — using the GDC

The function f(x) = 4x − x² has derivative function f′(x) = 4 − 2x. Use this to find f′(0), f′(2), f′(4) and interpret.

Substitute each x into f′(x) = 4 − 2x f′(0) = 4 − 0 = 4 → positive → curve increasing here f′(2) = 4 − 4 = 0 → zero → stationary point at x = 2 (the peak) f′(4) = 4 − 8 = −4 → negative → curve decreasing here f′(0) = 4; f′(2) = 0; f′(4) = −4 on a GDC: use d/dx(4x − x²)|_{x=0}, etc — same answer, no algebra needed

WE 6

Real-world rate of change — average to instantaneous velocity

A particle moves so its displacement (in metres) at time t (in seconds) is s(t) = t² + 2t. (a) Find s(3) and s(4). (b) Find the average velocity over [3, 4]. (c) Find the average velocities over [3, 3.1] and [3, 3.01]. (d) Estimate the instantaneous velocity at t = 3.

(a) Substitute s(3) = 9 + 6 = 15 m s(4) = 16 + 8 = 24 m (b) Average velocity = change in displacement / time (24 − 15)/(4 − 3) = 9/1 = 9 m/s (c) Shorter intervals s(3.1) = 9.61 + 6.2 = 15.81 → (15.81 − 15)/0.1 = 8.10 m/s s(3.01) = 15.0801 → (15.0801 − 15)/0.01 = 8.01 m/s (d) Convergence pattern 9 → 8.10 → 8.01 → … → 8 Instantaneous velocity at t = 3 is 8 m/s ds/dt = 2t + 2 = 8 at t = 3 — your numerical estimate matches the calculus answer

💡 Top tips

A limit is what f(x) approaches, not necessarily what f(x) equals — the function may be undefined at that point.
Chord gradient = rise / run — keep x₂ in the same position in both numerator and denominator.
Take chords on BOTH sides of the point when estimating numerically — both should converge to the same value.
The derivative is a function, not a number — once you have f′(x), plug in any x for that point’s gradient.
GDC d/dx gives the value AT a chosen x; great for cross-checking algebraic work but won’t show the formula.

⚠ Common mistakes

Confusing f(a) with the limit — f might be undefined at a while the limit is perfectly fine.
Reversing rise and run in the chord gradient — it’s (Δy) / (Δx), not the other way.
Using f(x) when you mean f′(x) — these are different functions.
Plugging x into f instead of f′ when asked for a gradient — gradient comes from the DERIVATIVE.
Confusing average and instantaneous rates of change — average uses two points; instantaneous is the limit as the points coincide.

Next: Differentiating Powers of x. Estimating gradients numerically (chord-by-chord) is fine for understanding, but very slow. The power rule — if f(x) = xⁿ then f′(x) = nxⁿ⁻¹ — turns differentiation into a one-line algebraic operation that works for any rational power.

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