IB DP Mathematics: Analysis and Approaches – Limits, derivatives, tangents, rules of differentiation and graph behaviour Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

This practice set covers the differentiation unit of IB Diploma Programme Mathematics: Analysis and Approaches. It is aligned to the IB Mathematics: analysis and approaches guide, first assessment 2021, syllabus sections 5.1–5.4 and 5.6–5.8, which are SL content required at both SL and HL. It follows the IB guide for first assessment 2021, which remains the examined syllabus until the new course is first assessed in May 2029, so it applies to the May and November 2026, 2027 and 2028 sessions.

Paper 1 allows no technology; Paper 2 requires a GDC. Questions are labelled “(calculator-free)” or “(calculator allowed)” to match. Give calculator-allowed answers exactly or to 3 significant figures unless the question says otherwise. The questions are new, not those in the calculus strand overview.

Learn the methods first in the differentiation study guide and the revision notes. The IB DP Maths AA course hub and the printable syllabus checklist show where this unit sits.

Questions

1. (calculator allowed) Let g(x) = (3ˣ − 1)/x, x ≠ 0.

(a) Complete the table, giving each value to 4 significant figures. [2]

x	0.1	0.01	0.001
g(x)

(b) Hence estimate lim (x→0) g(x), to 3 significant figures. [1]

2. (calculator-free) The temperature T °C of a cup of tea t minutes after it is poured is modelled by T = 90 − 12t + 0.5t², for 0 ≤ t ≤ 10.

(a) Find dT/dt. [1] (b) Find the value of dT/dt when t = 4, and interpret it in context. [2]

3. (calculator-free) Find dy/dx for each of the following.

(a) y = 2x⁴ − 3/x + 5x [2] (b) y = √(4x² + 9) [2] (c) y = x³ ln x [2]

4. (calculator-free) Let f(x) = (2x + 1)/(x² + 2).

(a) Show that f′(x) = −2(x + 2)(x − 1)/(x² + 2)². [3] (b) Hence find the interval on which f is increasing. [2]

5. (calculator-free) The curve y = ln(2x − 3) passes through the point P(2, 0).

(a) Find the equation of the tangent to the curve at P. [2] (b) Find the equation of the normal to the curve at P. [2] (c) The tangent meets the y-axis at A and the normal meets the y-axis at B. Find the area of triangle PAB. [2]

6. (calculator allowed) Let f(x) = e^(0.4x) − x².

(a) Find f′(1.5). [2] (b) Find the equation of the tangent to the graph of f at x = 1.5, in the form y = mx + c. [2] (c) Find the x-intercept of this tangent. [1]

7. (calculator-free) Let f(x) = 2x³ − 3x² − 12x + 7.

(a) Find the coordinates of the stationary points of f. [4] (b) Use the second derivative to determine the nature of each stationary point. [2] (c) Find the coordinates of the point of inflexion of the graph of f, justifying your answer. [3]

8. (calculator-free) A function f has derivative f′(x) = x² + x − 2.

(a) Find the x-coordinates of the stationary points of f, and determine the nature of each. [3] (b) Find the x-coordinate of the point of inflexion of the graph of f, justifying your answer. [2] (c) Write down the interval on which the graph of f is concave-up. [1]

9. (calculator allowed) An open-topped tank is a cuboid with a square base of side x m and height h m. Its volume is 4 m³. Material for the base costs 30 dollars per m² and material for the sides costs 20 dollars per m².

(a) Show that the total cost of material, C dollars, is C = 30x² + 320/x. [3] (b) Find dC/dx. [2] (c) Find the value of x that minimises the cost. [2] (d) Use the second derivative to justify that this value gives a minimum. [2] (e) Find the minimum cost. [1]

10. (calculator-free) Let f(x) = eˣ sin x, for 0 ≤ x ≤ π.

(a) Show that f′(x) = eˣ(sin x + cos x). [2] (b) Find the x-coordinate of the stationary point of f for 0 < x < π. [3] (c) Show that f″(x) = 2eˣ cos x. [2] (d) Hence determine the nature of the stationary point. [2] (e) Find the coordinates of the point of inflexion of the graph of f, justifying your answer. [3]

11. (calculator-free) Let f(x) = x − 4√x, for x > 0. Find the coordinates of the stationary point of f and show that it is a minimum. [5]

Answers

1. (a) Substituting each x into g [1]; 1.161, 1.105, 1.099 [1] (b) Values approach 1.10 [1] [3] Examiner insight: If part (a) is wrong but your estimate in (b) follows the trend in your own values, follow-through can still earn the (b) mark.

2. (a) dT/dt = t − 12 [1] (b) At t = 4, dT/dt = 4 − 12 = −8 [1]; the temperature is decreasing at 8 °C per minute at that moment [1] [3] Examiner insight: The interpretation mark needs both the direction (decreasing) and the units (°C per minute); “the gradient is −8” alone does not earn it.

3. (a) Rewrite −3/x = −3x⁻¹ [1]; dy/dx = 8x³ + 3x⁻² + 5 [1] (b) Chain rule: (1/2)(4x² + 9)^(−1/2) × 8x [1]; dy/dx = 4x/√(4x² + 9) [1] (c) Product rule: x³(1/x) + 3x² ln x [1]; dy/dx = 3x² ln x + x² [1] [6] Examiner insight: Write the unsimplified chain or product rule line first; it earns the method mark even if the simplification slips.

4. (a) Quotient rule with u = 2x + 1, u′ = 2, v = x² + 2, v′ = 2x: f′(x) = (2(x² + 2) − (2x + 1)(2x))/(x² + 2)² [1] Numerator = 2x² + 4 − 4x² − 2x = −2x² − 2x + 4 [1] = −2(x² + x − 2) = −2(x + 2)(x − 1), so f′(x) = −2(x + 2)(x − 1)/(x² + 2)² [1] (b) (x² + 2)² > 0, so f′(x) > 0 when −2(x + 2)(x − 1) > 0, that is (x + 2)(x − 1) < 0 [1]; −2 < x < 1 [1] [5] Examiner insight: In a “show that” every line must be visible, including the expanded numerator; jumping straight to the given factorised form loses the middle marks.

5. (a) dy/dx = 2/(2x − 3), so the gradient at x = 2 is 2 [1]; tangent y − 0 = 2(x − 2), y = 2x − 4 [1] (b) Normal gradient = −1/2 [1]; y = −x/2 + 1 [1] (c) A(0, −4) and B(0, 1) [1]; AB = 5, perpendicular distance from P to the y-axis is 2, so area = (1/2)(5)(2) = 5 [1] [6] Examiner insight: The chain rule factor 2 in d/dx ln(2x − 3) is where most wrong answers start; follow-through marks are then available for (b) and (c) if your working is consistent.

6. (a) f′(x) = 0.4e^(0.4x) − 2x [1]; f′(1.5) = −2.27 (−2.27115…) [1] (b) f(1.5) = −0.427881… [1]; y = −2.27115(x − 1.5) − 0.427881, so y = −2.27x + 2.98 [1] (c) 0 = −2.27115x + 2.97885, x = 1.31 [1] [5] Examiner insight: Carry the unrounded gradient and intercept from the GDC into part (c); using rounded values can shift the third significant figure and cost the accuracy mark.

7. (a) f′(x) = 6x² − 6x − 12 [1]; 6(x − 2)(x + 1) = 0 [1]; x = −1 or x = 2 [1]; (−1, 14) and (2, −13) [1] (b) f″(x) = 12x − 6; f″(−1) = −18 < 0, so (−1, 14) is a local maximum [1]; f″(2) = 18 > 0, so (2, −13) is a local minimum [1] (c) f″(x) = 0 when x = 1/2 [1]; f″ < 0 for x < 1/2 and f″ > 0 for x > 1/2, so f″ changes sign [1]; point of inflexion (1/2, 1/2) [1] [9] Examiner insight: The justification mark in (c) is for showing that f″ changes sign; stating only f″(1/2) = 0 is not enough, as the guide’s y = x⁴ example shows.

8. (a) (x + 2)(x − 1) = 0, so x = −2 or x = 1 [1]; at x = −2, f′ changes from + to −, so a maximum [1]; at x = 1, f′ changes from − to +, so a minimum [1] (b) f″(x) = 2x + 1 = 0 at x = −1/2 [1]; f″ changes from negative to positive there, so it is a point of inflexion [1] (c) x > −1/2 [1] [6] Examiner insight: You are given f′, not f, so the nature comes from the sign of f′ either side (or from f″); finding f first earns nothing extra.

9. (a) x²h = 4, so h = 4/x² [1]; C = 30x² + 20(4xh) [1]; = 30x² + 80x(4/x²) = 30x² + 320/x [1] (b) dC/dx = 60x [1] − 320/x² [1] (c) 60x − 320/x² = 0 gives x³ = 16/3 [1]; x = 1.75 m (1.7471…) [1] (d) d²C/dx² = 60 + 640/x³ [1]; at x = 1.7471…, d²C/dx² = 180 > 0, so a minimum [1] (e) C = 275 dollars (274.73…) [1] [10] Examiner insight: In (a) the cost of the four sides must be written as 4 × xh before substituting; a solution that starts from the printed answer earns no marks.

10. (a) Product rule: eˣ sin x + eˣ cos x [1]; = eˣ(sin x + cos x) [1] (b) eˣ > 0, so sin x + cos x = 0 [1]; tan x = −1 [1]; x = 3π/4 [1] (c) f″(x) = eˣ(sin x + cos x) + eˣ(cos x − sin x) [1]; = 2eˣ cos x [1] (d) f″(3π/4) = 2e^(3π/4)(−√2/2) = −√2 e^(3π/4) < 0 [1]; so a local maximum [1] (e) 2eˣ cos x = 0 gives cos x = 0, x = π/2 [1]; cos x > 0 for x < π/2 and cos x < 0 for x > π/2, so f″ changes sign [1]; point of inflexion (π/2, e^(π/2)) [1] [12] Examiner insight: State that eˣ is never zero before dividing by it; a decimal such as 2.36 instead of the exact 3π/4 loses the accuracy mark.

11. f′(x) = 1 − 2x^(−1/2) [1]; 1 − 2/√x = 0 gives √x = 2, x = 4 [1]; f(4) = 4 − 8 = −4, so (4, −4) [1]; f″(x) = x^(−3/2) [1]; f″(4) = 1/8 > 0, so it is a minimum [1] [5] Examiner insight: Rewrite √x as x^(1/2) before differentiating; the minimum must be justified with f″ or a sign change, not just stated.

Where marks are usually lost

• Leaving terms like 3/x or √x unrewritten and differentiating them wrongly.
• Missing the inner derivative in the chain rule, especially for ln(ax + b) and e^(kx).
• In “show that” questions, jumping to the printed result without the intermediate algebra.
• Using the tangent gradient for the normal, or forgetting the negative sign in −1/m.
• Finding x for a stationary point but not the y-coordinate when coordinates are asked for.
• Claiming a point of inflexion from f″(x) = 0 without a sign-change argument.
• Giving decimal answers on calculator-free questions where exact values such as 3π/4 are expected.
• Rounding GDC values too early, so later answers are wrong at 3 significant figures.
• Interpreting a rate of change without direction or units.

Next steps

• Differentiation revision notes
• Differentiation study guide
• IB DP Maths AA course hub
• Printable syllabus checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: analysis and approaches guide, first assessment 2021 (published February 2019, updated November 2020). This set covers syllabus sections 5.1, 5.2, 5.3, 5.4, 5.6, 5.7 and 5.8.