IB Maths AI HL · Unit 5: Calculus
IB Maths AI HL Advanced Differentiation and Related Rates Questions
Exam-style IB Maths AI HL advanced differentiation and related rates questions with worked solutions. Start with the 3 fully worked examples below — each is solved step by step the way an IB examiner expects — then try the practice questions and check your working against the mark scheme.
- 19 questions
- Paper 1: 19
- 5 easy
- 5 medium
- 5 hard
- 4 starter
- 3 worked examples
Practise Advanced Differentiation and Related Rates questions →
AI HL formula booklet
What's examined in AI HL advanced differentiation and related rates
The question bank covers these advanced differentiation and related rates question types (number of questions in brackets):
- Chain, Product, Quotient Rules (13)
- Related Rates Problems (5)
- Implicit Differentiation & Optimization (1)
Key formulas
- Separable differential equation
- \(\dfrac{dy}{dx} = f(x) g(y) \implies \int \dfrac{1}{g(y)}\, dy = \int f(x)\, dx\)
- Euler's method
- \(y_{n+1} = y_n + h\, f(x_n, y_n),\ x_{n+1} = x_n + h\)
- Trapezoidal rule (refresher)
- \(A \approx \tfrac{h}{2}\bigl(y_0 + y_n + 2\sum_{i=1}^{n-1} y_i\bigr)\)
In the same notation as the IB formula booklet. All AI HL formulas →
Advanced Differentiation and Related Rates worked examples
Worked example 1: Applying the Product Rule · easy
Consider the function $f(x) = x^2 \ln x$ for $x > 0$. Find the exact expression for the derivative $f'(x)$, giving your answer in a fully factorised form.
1. Identify the two functions to apply the product rule: let $u = x^2$ and $v = \ln x$.
2. Differentiate each component with respect to $x$: $u' = 2x$ and $v' = \frac{1}{x}$.
3. Apply the product rule formula: $f'(x) = uv' + vu'$.
4. Substitute the expressions: $f'(x) = (x^2)\left(\frac{1}{x}\right) + (\ln x)(2x)$.
5. Simplify and factorise: $f'(x) = x + 2x \ln x =$ $x(1 + 2\ln x)$.
Examiner tip: When using the product or quotient rule involving natural logarithms, always simplify expressions like $x^2 \times \frac{1}{x}$ immediately to avoid algebraic tangles in subsequent steps.
Worked example 2: Implicit Differentiation · medium
A curve is defined implicitly by the equation $x^2 y + y^3 = 10$. Find the exact value of the gradient of the curve at the point $(1, 2)$.
1. Differentiate the term $x^2 y$ using the product rule: $2xy + x^2 \frac{dy}{dx}$.
2. Differentiate the term $y^3$ using the chain rule: $3y^2 \frac{dy}{dx}$.
3. Set the derivative of the constant $10$ to $0$, forming the equation: $2xy + x^2 \frac{dy}{dx} + 3y^2 \frac{dy}{dx} = 0$.
4. Substitute the coordinates $x = 1$ and $y = 2$ directly into the equation: $2(1)(2) + (1)^2 \frac{dy}{dx} + 3(2)^2 \frac{dy}{dx} = 0$.
5. Simplify the terms: $4 + \frac{dy}{dx} + 12 \frac{dy}{dx} = 0 \implies 13 \frac{dy}{dx} = -4$.
6. Solve for the gradient exactly: $\frac{dy}{dx} = -\frac{4}{13}$.
Examiner tip: Remember to multiply by $\frac{dy}{dx}$ every single time you differentiate a term containing $y$ with respect to $x$. This is a mandatory application of the chain rule.
Worked example 3: Solving a Related Rates Problem · hard
Water pours into an inverted right circular conical tank at a constant rate of $2\text{ cm}^3\text{s}^{-1}$. The tank has a maximum base radius of $5\text{ cm}$ and a total height of $10\text{ cm}$. Let $h$ be the depth of the water. Find the exact rate at which the water level ($h$) is rising at the instant when the depth of the water is exactly $4\text{ cm}$. (The volume of a cone is $V = \frac{1}{3}\pi r^2 h$)
1. Relate the radius $r$ to the depth $h$ using similar triangles: $\frac{r}{h} = \frac{5}{10} \implies r = 0.5h$.
2. Substitute $r$ into the volume formula to express $V$ solely in terms of $h$: $V = \frac{1}{3}\pi (0.5h)^2 h = \frac{1}{12}\pi h^3$.
3. Differentiate both sides with respect to time $t$ using the chain rule: $\frac{dV}{dt} = \frac{3}{12}\pi h^2 \frac{dh}{dt} = \frac{1}{4}\pi h^2 \frac{dh}{dt}$.
4. Substitute the known rate $\frac{dV}{dt} = 2$ and the specific depth $h = 4$: $2 = \frac{1}{4}\pi (4^2) \frac{dh}{dt}$.
5. Simplify the equation: $2 = 4\pi \frac{dh}{dt}$.
6. Solve exactly for the rate of change of depth: $\frac{dh}{dt} = \frac{1}{2\pi}\text{ cm s}^{-1}$.
Examiner tip: In 3D related rate problems (like cones), you must substitute the geometrical ratio (e.g., $r = 0.5h$) to reduce the volume formula to a single variable before you attempt to differentiate.
Try these IB Maths AI HL advanced differentiation and related rates questions
Three questions from the bank, easiest first. Mark schemes and AI marking of your written working are in the practice area.
Question 1 · easy · 4 marks · Paper 1
Differentiate \(y = (3x^2 - 5x)^4\) with respect to \(x\). Express your answer in a fully factorised form.
Question 2 · medium · 5 marks · Paper 1
Consider the function \(y = \ln(\cos x)\).
(a) Find \(\frac{dy}{dx}\). [3 marks]
(b) State the continuous domain of values for \(x\) containing \(x=0\) for which this derivative is valid. [2 marks]
Question 3 · hard · 7 marks · Paper 1
Consider the curve given by the equation \(y = \frac{e^x - e^{-x}}{e^x + e^{-x}}\).
Use the quotient rule to differentiate this expression and prove algebraically that:
\[\frac{dy}{dx} = 1 - y^2\]
All 19 advanced differentiation and related rates questions with mark schemes →
FAQ
How many IB Maths AI HL advanced differentiation and related rates questions are there?
There are 19 exam-style advanced differentiation and related rates questions in the AI HL question bank (Paper 1: 19), graded 5 easy, 5 medium, 5 hard, 4 starter. Every question has a full IB-style mark scheme (M, A and R marks).
Is advanced differentiation and related rates on Paper 1 or Paper 2?
In the question bank these questions are set as Paper 1 questions.
Where can I get the mark schemes?
Open the AI HL Unit 5 practice page: every question has a step-by-step IB-style mark scheme, and you can photograph your working for instant AI marking. The worked examples on this page are free.
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