An open-topped box is made from a $20\text{ cm} \times 20\text{ cm}$ square piece of cardboard by cutting out identical squares of side length $x\text{ cm}$ from each corner and folding up the sides. Write an expression for the volume $V(x)$, then use your GDC to find the value of $x$ that maximises the volume and the maximum volume itself, to 3 s.f.

1. Determine the folded dimensions: base is $(20 - 2x)$ by $(20 - 2x)$, height is $x$. 2. Construct the volume function: $V(x) = x(20 - 2x)^2$. 3. Identify the practical domain: $x > 0$ and $20 - 2x > 0$, so $0 < x < 10$. 4. Graph $V(x)$ on the GDC (CG50: MENU 5, Y1 = x(20 - 2x)^2). 5. Use the maximum tool (G-Solv $\to$ MAX) on the valid domain. 6. State : $\mathbf{x = 3.33\text{ cm}}$ (3 s.f.) and maximum volume $\mathbf{593\text{ cm}^3}$ (3 s.f.).

Consider the curve $y = 2x^2 - 5x$ . Find the gradient of the tangent to the curve at $x = 1$ . Find the gradient of the normal to the curve at $x = 1$ . Find the equation of the normal line at $x = 1$ , giving your answer in the form $ax + by + d = 0$ , where $a, b,$ and $d$ are integers.

$y' = 4x - 5$ . Gradient at $x = 1$ : $y'(1) = 4(1) - 5 = -1$ . (M1)A1 Normal gradient $m_n = -\frac{1}{-1} = 1$ . A1 $y$ -coordinate at $x = 1$ : $y = 2(1)^2 - 5(1) = -3$ . Point is $(1, -3)$ . (M1) Equation: $y - (-3) = 1(x - 1) \implies y + 3 = x - 1$ . Rearranging to $ax + by + d = 0 \implies x - y - 4 = 0$ . A1

IB Maths AA SL · Unit 5: Calculus

Applications of Differentiation (IB Maths AA SL): notes, exam questions and worked examples

Q: How do I show a stationary point is a maximum?

Either show f″(x) < 0 there, or show f′(x) changes from positive to negative as x increases through the point. Write the conclusion in words.

Q: How many IB Maths AA SL applications of differentiation questions are there?

There are 42 exam-style applications of differentiation questions in the AA SL question bank (Paper 1: 22 · Paper 2: 20), graded 4 easy, 12 medium, 23 hard, 3 very hard. Every question has a full IB-style mark scheme (M, A and R marks).

Q: Is applications of differentiation on Paper 1 or Paper 2?

Both. In the bank, Paper 1: 22 · Paper 2: 20. Paper 1 is non-calculator, so practise the exact-value algebra as well as the GDC methods.

Updated October 2026 · By Pete Bromfield

The derivative gives the gradient, so it finds tangents and normals, shows where a function increases or decreases, and locates maximum and minimum points where $f^{'} (x) = 0$ . Optimisation turns a worded problem into a function of one variable, then finds and justifies its maximum or minimum.

42 questions
Paper 1: 22
Paper 2: 20
3 worked examples

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Notes
Formulas
Worked examples
Questions
FAQ

Applications of Differentiation notes

Tangents and normals

At $x = a$ the tangent has gradient $f^{'} (a)$ and equation $y - f (a) = f^{'} (a) (x - a)$ . The normal is perpendicular, with gradient $- \frac{1}{f^{'} (a)}$ . For $y = x^{2}$ at $x = 3$ : gradient 6, tangent $y = 6 x - 9$ , normal $y = - \frac{x}{6} + \frac{19}{2}$ .

Increasing and decreasing

$f$ is increasing where $f^{'} (x) > 0$ and decreasing where $f^{'} (x) < 0$ . Solve the inequality and give the answer as an interval of $x$ -values.

Stationary points

Solve $f^{'} (x) = 0$ , then classify. If $f^{″} (x) < 0$ it is a maximum, if $f^{″} (x) > 0$ a minimum; if $f^{″} (x) = 0$ , check the sign of $f^{'}$ either side. For $f (x) = x^{3} - 3 x$ : $f^{'} (x) = 3 x^{2} - 3 = 0$ at $x = \pm 1$ ; $f^{″} (1) = 6 > 0$ gives a minimum at $(1, - 2)$ and $f^{″} (- 1) = - 6$ gives a maximum at $(- 1, 2)$ .

Concavity and points of inflexion

Where $f^{″} (x) > 0$ the graph is concave up; where $f^{″} (x) < 0$ , concave down. A point of inflexion needs $f^{″} (x) = 0$ and a change of sign of $f^{″}$ .

Optimisation

Name the variables, use the constraint to write the quantity as a function of one variable, differentiate, set the derivative to zero, justify that it is a maximum or minimum, and answer in context. A rectangle with perimeter 20 has area $A = x (10 - x)$ ; $A^{'} (x) = 10 - 2 x = 0$ gives $x = 5$ and a maximum area of 25. Words such as "fastest" or "closest" are optimisation in disguise.

Key formulas

Tangent; normal gradient at $x = a$ Learn it: $y - f (a) = f^{'} (a) (x - a); m_{normal} = - \frac{1}{f^{'} (a)}$
Increasing / decreasing Learn it: $f^{'} (x) > 0; f^{'} (x) < 0$
Stationary points Learn it: $f^{'} (x) = 0 : f^{″} < 0 max, f^{″} > 0 min$
Concave up $f^{″} > 0$ ; point of inflexion: $f^{″} = 0$ and $f^{″}$ changes sign Learn it

The IB formula booklet is provided in every exam paper. In the booklet = printed there, so know where to find it; Learn it = not in the booklet, so learn it by heart. All AA SL formulas →

Applications of Differentiation worked examples

Worked example 1: Finding the equation of a tangent · easy

The equation of a curve is given by $f (x) = x^{2} - 4 x + 5$ . Find the exact gradient of the curve at $x = 3$ , and hence find the equation of the tangent to the curve at $x = 3$ . Give your answer in the form $y = m x + c$ .

Solution

6 steps · 1 to fill in

Examiner tip: A common error is substituting the

x

-coordinate into the original function

f (x)

when looking for the gradient. Always remember: substitute into

f^{'} (x)

for the gradient, and

f (x)

for the

y

-coordinate.

Worked example 2: Horizontal tangents on a cubic · medium

The curve $C$ is given by the equation $y = - x^{3} + 6 x^{2} - 5 x + 4$ . The tangent to the curve is parallel to the $x$ -axis at two specific points. Find the exact $x$ -coordinates of these two points.

Solution

6 steps

Examiner tip: "Parallel to the

x

-axis", "horizontal tangent", and "stationary point" all mean exactly the same thing mathematically: set the first derivative equal to zero.

Worked example 3: Optimisation of an open-topped box · hard

An open-topped box is made from a $20 cm \times 20 cm$ square piece of cardboard by cutting out identical squares of side length $x cm$ from each corner and folding up the sides. Write an expression for the volume $V (x)$ , then use your GDC to find the value of $x$ that maximises the volume and the maximum volume itself, to 3 s.f.

Solution

6 steps

Examiner tip: In Paper 2 optimisation questions do NOT differentiate manually unless the question says "Use calculus to...". Graph and use G-Solv — much faster with less risk of algebra error.

Applications of Differentiation exam-style questions

3 questions from our AA SL bank, written in the style of IB papers, easiest first. One has its full worked solution open here. Open the worked answer, or check your answer in the practice area.

Question 1 · easy · 4 marks · Paper 2

Let $y = 0.5 x^{3} - 2 x + 1$ .

Use the numerical derivative feature on your graphic display calculator to evaluate the gradient of the curve at $x = 2$ .
Hence, find the equation of the normal to the curve at the point where $x = 2$ , giving your answer in the form $y = m x + c$ .

Check your answer in practice →

Question 2 · medium · 5 marks · Paper 1

Consider the curve $y = 2 x^{2} - 5 x$ .

Find the gradient of the tangent to the curve at $x = 1$ .
Find the gradient of the normal to the curve at $x = 1$ .
Find the equation of the normal line at $x = 1$ , giving your answer in the form $a x + b y + d = 0$ , where $a, b,$ and $d$ are integers.

Full worked solution Free, no sign-in

$y^{'} = 4 x - 5$ . Gradient at $x = 1$ : $y^{'} (1) = 4 (1) - 5 = - 1$ . (M1)A1
Normal gradient $m_{n} = - \frac{1}{- 1} = 1$ . A1
$y$ -coordinate at $x = 1$ : $y = 2 (1)^{2} - 5 (1) = - 3$ . Point is $(1, - 3)$ . (M1)
Equation: $y - (- 3) = 1 (x - 1) ⟹ y + 3 = x - 1$ .
Rearranging to $a x + b y + d = 0 ⟹ x - y - 4 = 0$ . A1

Every question here has a full worked solution. See them all →

Question 3 · hard · 5 marks · Paper 1

The equation of a curve is $y = x^{3} - 4 x^{2} + k x$ , where $k$ is a constant. The equation of the tangent to the curve at the point where $x = 2$ is $y = - 3 x$ .

State the gradient of the tangent line.
Find $\frac{d y}{d x}$ in terms of $x$ and $k$ .
By substituting $x = 2$ into your derivative, find the exact value of $k$ .

Attempt it and see the mark scheme →

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All 42 applications of differentiation questions →

Where students lose marks on applications of differentiation

Examiner's noteModelling in context: not setting up the expression, and answers that ignore the context

Where marks are lost: Candidates did not recognise "closest" or "fastest" as an optimisation, solved h′(t) = 0 when the fastest rate needed h″(t) = 0, left units inside algebraic expressions, gave a time when a year was asked, forgot that a population was in millions, rounded a number of rows the wrong way, and kept two solutions when the context allowed one.

Examiners flagged this in 4 exam series (May 2024, November 2024, May 2025, November 2025). Summarised in our own words from IB subject reports, IB Programme Resource Centre. Read the full note →

Applications of Differentiation on your calculator

A GDC is allowed on Paper 2, not on Paper 1. Keystrokes for the Casio fx-CG50, TI-84 Plus CE and TI-Nspire, with a checked worked example:

Numerical derivative and integral on your calculator

Applications of Differentiation FAQ

Is the equation of a tangent in the IB formula booklet?

The booklet gives the forms of a straight line, y − y₁ = m(x − x₁), but not the tangent or normal rules. Learn that the tangent gradient is f′(a) and the normal gradient is −1/f′(a).

How do I show a stationary point is a maximum?

Either show f″(x) < 0 there, or show f′(x) changes from positive to negative as x increases through the point. Write the conclusion in words.

How many IB Maths AA SL applications of differentiation questions are there?

There are 42 exam-style applications of differentiation questions in the AA SL question bank (Paper 1: 22 · Paper 2: 20), graded 4 easy, 12 medium, 23 hard, 3 very hard. Every question has a full IB-style mark scheme (M, A and R marks).

Is applications of differentiation on Paper 1 or Paper 2?

Both. In the bank, Paper 1: 22 · Paper 2: 20. Paper 1 is non-calculator, so practise the exact-value algebra as well as the GDC methods.

Next steps

More AA SL Unit 5: Definite Integrals & Area · Graphical Behaviour and the Second Derivative · Kinematics · Limits and Rates of Change · Differentiation Rules

Teaching this topic? Free ready-to-teach lesson (slides and teacher notes): Optimisation in context.

Read more on the blog: How IB Maths is marked: M1, A1, R1 and AG explained

Keep revising AA SL: AA SL revision notes · IB Maths AA SL past papers · AA SL common exam mistakes · AA SL syllabus checklist · AA SL calculus knowledge organiser

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← All IB Maths AA SL topics

Applications of Differentiation (IB Maths AA SL): notes, exam questions and worked examples

Try it first

Applications of Differentiation notes

Tangents and normals

Increasing and decreasing

Stationary points

Concavity and points of inflexion

Optimisation

Key formulas

Applications of Differentiation worked examples

Worked example 1: Finding the equation of a tangent · easy

Solution

Worked example 2: Horizontal tangents on a cubic · medium

Solution

Worked example 3: Optimisation of an open-topped box · hard

Solution

Applications of Differentiation exam-style questions

Question 1 · easy · 4 marks · Paper 2

Question 2 · medium · 5 marks · Paper 1

Full worked solution Free, no sign-in

Question 3 · hard · 5 marks · Paper 1

Where students lose marks on applications of differentiation

Applications of Differentiation on your calculator

Applications of Differentiation FAQ

Next steps

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