IB Maths AA SL · Unit 5: Calculus
IB Maths AA SL Rules of Differentiation Questions
Exam-style IB Maths AA SL rules of differentiation 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.
- 26 questions
- Paper 1: 19
- Paper 2: 7
- 4 easy
- 11 medium
- 5 hard
- 3 very hard
- 3 starter
- 3 worked examples
Practise Rules of Differentiation questions →
AA SL formula booklet
What's examined in AA SL rules of differentiation
The question bank covers these rules of differentiation question types (number of questions in brackets):
- Chain Rule (9)
- Product and Quotient Rules (9)
- Power and Basic Derivatives (8)
Key formulas
- Chain rule
- \(\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}\)
- Product rule
- \((uv)' = u'v + uv'\)
- Quotient rule
- \(\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}\)
- Derivative of x^n
- \(\dfrac{d}{dx}\bigl(x^n\bigr) = n x^{\,n-1}\)
- Derivatives of standard functions
- \(\dfrac{d}{dx}(\sin x) = \cos x,\ \dfrac{d}{dx}(\cos x) = -\sin x,\ \dfrac{d}{dx}(\tan x) = \sec^2 x\)
- Exponential & log derivatives
- \(\dfrac{d}{dx}(e^x) = e^x,\ \dfrac{d}{dx}(\ln x) = \dfrac{1}{x}\)
In the same notation as the IB formula booklet. All AA SL formulas →
Rules of Differentiation worked examples
Worked example 1: Polynomial power rule and stationary points · easy
The equation of a curve is $y = \frac{3}{2}x^2 - 15x + 2$. Find an expression for $\frac{dy}{dx}$, and hence find the exact coordinates of the point on the curve where the gradient of the tangent is zero.
1. Apply the power rule: $\frac{dy}{dx} = 2 \cdot \frac{3}{2}x^{2-1} - 15x^0$.
2. Simplify: $\frac{dy}{dx} = 3x - 15$.
3. Set the gradient to zero: $3x - 15 = 0$.
4. Solve: $x = 5$.
5. Substitute back into the original curve: $y = \frac{3}{2}(5)^2 - 15(5) + 2$.
6. Evaluate: $y = 37.5 - 75 + 2 = -35.5$. Coordinates: $\mathbf{(5,\ -35.5)}$.
Examiner tip: The derivative of any constant is zero — the $+2$ has no effect on the gradient because a constant doesn't change the rate of change of the curve.
Worked example 2: Quotient rule differentiation · medium
A curve has equation $y = \dfrac{-7x}{x^3 - 1}$, for $x \neq 1$. Use the quotient rule to find a fully simplified expression for $\frac{dy}{dx}$.
1. Identify $u = -7x$ and $v = x^3 - 1$.
2. Differentiate both: $u' = -7$, $v' = 3x^2$.
3. State the quotient rule: $\frac{dy}{dx} = \frac{u'v - uv'}{v^2}$.
4. Substitute: $\frac{dy}{dx} = \frac{-7(x^3 - 1) - (-7x)(3x^2)}{(x^3 - 1)^2}$.
5. Expand carefully with signs: $\frac{-7x^3 + 7 + 21x^3}{(x^3 - 1)^2}$.
6. Simplify: $\mathbf{\dfrac{dy}{dx} = \dfrac{14x^3 + 7}{(x^3 - 1)^2}}$.
Examiner tip: The formula is $u'v - uv'$, NOT $uv' - u'v$. Memorise: "derivative of the top FIRST".
Worked example 3: Product rule with an embedded chain rule · hard
Let $f(x) = (12x^2 - 7)e^{-2x}$. Find $f'(x)$ and show it can be written in the fully factorised form $f'(x) = -2e^{-2x}(ax^2 + bx + c)$, where $a, b, c$ are integers.
1. Set up the product rule: $u = 12x^2 - 7$, $v = e^{-2x}$.
2. Differentiate (chain rule for $v$): $u' = 24x$, $v' = -2e^{-2x}$.
3. Apply the product rule: $f'(x) = (24x)e^{-2x} + (12x^2 - 7)(-2e^{-2x})$.
4. Identify the common factor from the target form: $-2e^{-2x}$.
5. Factorise: pulling $-2e^{-2x}$ out of $24xe^{-2x}$ leaves $-12x$; from the second term it leaves $(12x^2 - 7)$.
6. Combine: $f'(x) = -2e^{-2x}(-12x + 12x^2 - 7) = \mathbf{-2e^{-2x}(12x^2 - 12x - 7)}$.
Examiner tip: Questions that specify a factorised target form are giving you a huge hint. Do not expand — instead pull the required common factor out of every term.
Try these IB Maths AA SL rules of differentiation 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 2
Consider the function with rational exponents \(f(x) = x^{\frac{2}{3}} + \frac{5}{x}\) for \(x > 0\).
Rewrite \(f(x)\) so that all terms are in the form \(ax^n\). Hence, find \(f'(x)\).
Use your graphic display calculator to evaluate the exact gradient of the curve at \(x = 8\).
Question 2 · medium · 5 marks · Paper 1
Consider the function \(f(x) = e^x \sin x\), for \(0 \le x \le \pi\).
Use the product rule to find \(f'(x)\).
The curve has a local maximum point in the given domain. By setting \(f'(x) = 0\), find the exact \(x\)-coordinate of this maximum.
Question 3 · hard · 5 marks · Paper 1
A curve has the equation \(y = e^{-3x} + \ln x\), for \(x > 0\).
Find \(\frac{dy}{dx}\).
Find the exact gradient of the normal line to the curve at the point where \(x = 1\).
All 26 rules of differentiation questions with mark schemes →
FAQ
How many IB Maths AA SL rules of differentiation questions are there?
There are 26 exam-style rules of differentiation questions in the AA SL question bank (Paper 1: 19 · Paper 2: 7), graded 4 easy, 11 medium, 5 hard, 3 very hard, 3 starter. Every question has a full IB-style mark scheme (M, A and R marks).
Is rules of differentiation on Paper 1 or Paper 2?
Both. In the bank, Paper 1: 19 · Paper 2: 7. Paper 1 is non-calculator, so practise the exact-value algebra as well as the GDC methods.
Where can I get the mark schemes?
Open the AA SL 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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