O Level Amath Differentiation Questions Singapore: A Complete Worksheet Practice Guide

If you’re doing O Level A-Math in Singapore, differentiation is one of those topics that can either pull your grade up… or quietly destroy your paper.

You already know the feeling: you kind of understand the formulas in class, but when it comes to full exam questions — especially those 6–8 mark ones with context — suddenly everything looks unfamiliar.

“Stuck on a question? See simple explanations that help you understand fast.”
👉 Give it a try and turn confusion into clarity in minutes.

This guide is written for you: a Secondary 3 or 4 student taking A-Math under the MOE syllabus, aiming to handle O Level Amath differentiation questions confidently.

I’ll walk you through:

• A clear step-by-step tutorial (from basic rules to common exam twists)
• A practical exam strategy guide (what to do in the first 10–20 seconds of each question)
• Worksheet-style practice, including hard variants similar to O Level / prelim standards
• A breakdown of common mistakes Singapore students make (and how to avoid them)

Along the way, I’ll show you how to use Tutorly.sg — a 24/7 AI tutor website built specifically for Singapore students — to get instant practice and explanations whenever you’re stuck.

Tutorly.sg has already been used by thousands of students in Singapore, and was even mentioned on Channel NewsAsia (CNA), so you’re in good company.

Useful links to keep open while you read:

• Learn more about the AI tutor: https://tutorly.sg/ai-tutor-singapore
• Go straight to practising questions: https://tutorly.sg/app

---

Step-by-step tutorial

Let’s build from the ground up. I’ll keep examples aligned to typical MOE / O Level style.

1. Core differentiation rules you must memorise

These are non-negotiable. You should be able to write them down from memory in under a minute.

1. Power rule

If $𝑦 = 𝑥^𝑛$, then
$\frac{dy}{dx} = nx^{𝑛-1}$

Examples:

• $𝑦 = 𝑥^3 \Rightarrow \frac{dy}{dx} = 3𝑥^2$
• $𝑦 = \frac{1}{𝑥^2} = 𝑥^{-2} \Rightarrow \frac{dy}{dx} = -2𝑥^{-3} = -\frac{2}{𝑥^3}$
• $𝑦 = \sqrt{𝑥} = 𝑥^{1/2} \Rightarrow \frac{dy}{dx} = \frac{1}{2}𝑥^{-1/2} = \frac{1}{2\sqrt{𝑥}}$

2. Constant rule

If 𝑦 = 𝑘 (where 𝑘 is a constant), then
$\frac{dy}{dx} = 0$

3. Constant multiple rule

If 𝑦 = 𝑘 𝑓(𝑥), then
$\frac{dy}{dx} = 𝑘 \frac{𝑑}{dx}[𝑓(𝑥)]$

Example: $𝑦 = 7𝑥^4 \Rightarrow \frac{dy}{dx} = 7 \cdot 4𝑥^3 = 28𝑥^3$

4. Sum and difference

If 𝑦 = 𝑓(𝑥) + 𝑔(𝑥), then
$\frac{dy}{dx} = 𝑓'(𝑥) + 𝑔'(𝑥)$

Same idea for subtraction.

5. Product rule

If 𝑦 = 𝑢(𝑥)𝑣(𝑥), then
$\frac{dy}{dx} = 𝑢'𝑣 + uv'$

6. Quotient rule

If $𝑦 = \dfrac{𝑢(𝑥)}{𝑣(𝑥)}$, then
$\frac{dy}{dx} = \frac{𝑣 𝑢' - 𝑢 𝑣'}{𝑣^2}$

7. Chain rule (composite function)

If 𝑦 = 𝑓(𝑔(𝑥)), then
$\frac{dy}{dx} = 𝑓'(𝑔(𝑥)) \cdot 𝑔'(𝑥)$

In O Level A-Math, you usually apply this to things like $(3𝑥 - 1)^5$, $\sqrt{2𝑥+3}$, $\dfrac{1}{𝑥^2+1}$ etc.

---

2. Differentiating standard algebraic expressions

Example 1: Expand then differentiate vs product rule

Question: Differentiate $𝑦 = (𝑥^2 + 3𝑥)(2𝑥 - 1)$ with respect to 𝑥.

Method A: Expand first (common in O Level)

1. Expand:
   $(𝑥^2 + 3𝑥)(2𝑥 - 1) = 2𝑥^3 - 𝑥^2 + 6𝑥^2 - 3𝑥 = 2𝑥^3 + 5𝑥^2 - 3𝑥$

2. Differentiate term by term:
   $\frac{dy}{dx} = 6𝑥^2 + 10𝑥 - 3$

Method B: Product rule

Let $𝑢 = 𝑥^2 + 3𝑥$, 𝑣 = 2𝑥 - 1
Then 𝑢' = 2𝑥 + 3, 𝑣' = 2

Using product rule:
$\frac{dy}{dx} = 𝑢'𝑣 + uv' = (2𝑥 + 3)(2𝑥 - 1) + (𝑥^2 + 3𝑥)(2)$

You can expand and simplify to get the same final answer.

Tip for exams:
If it’s easy to expand, expand first. Product rule is safer when expansion becomes messy or involves surds/fractions.

---

3. Chain rule in O Level style

You’ll see many expressions of the form $(ax + 𝑏)^𝑛$, $\sqrt{ax + 𝑏}$, $\dfrac{1}{ax^2 + 𝑏}$.

Example 2: $(3𝑥 - 2)^4$

Let $𝑦 = (3𝑥 - 2)^4$.

Think of 𝑢 = 3𝑥 - 2, so $𝑦 = 𝑢^4$.

• $\dfrac{dy}{du} = 4𝑢^3$
• $\dfrac{du}{dx} = 3$

By chain rule:
$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 4(3𝑥 - 2)^3 \cdot 3 = 12(3𝑥 - 2)^3$

In practice, you can do this mentally:
“Power comes down, reduce power by 1, then multiply by derivative of inside.”

---

4. Product and quotient rule with chain rule combined

This is where many O Level differentiation questions start to feel “tricky”.

Example 3: Quotient with inner function

Question: Differentiate $𝑦 = \dfrac{3𝑥 + 2}{(𝑥^2 + 1)^2}$.

Let 𝑢 = 3𝑥 + 2, $𝑣 = (𝑥^2 + 1)^2$.

• 𝑢' = 3
• For 𝑣: use chain rule. Let $𝑤 = 𝑥^2 + 1$, so $𝑣 = 𝑤^2$
  • $\dfrac{dv}{dw} = 2𝑤$
  • $\dfrac{dw}{dx} = 2𝑥$
  • So $\dfrac{dv}{dx} = 2(𝑥^2 + 1)\cdot 2𝑥 = 4𝑥(𝑥^2 + 1)$

Now use quotient rule:
$\frac{dy}{dx} = \frac{𝑣 𝑢' - 𝑢 𝑣'}{𝑣^2} = \frac{(𝑥^2 + 1)^2(3) - (3𝑥 + 2)\cdot 4𝑥(𝑥^2 + 1)}{(𝑥^2 + 1)^4}$

You can simplify by cancelling $(𝑥^2 + 1)$ factors, but in O Level marking schemes, partial simplification is usually accepted as long as it’s correct.

---

5. Tangents, normals, and gradient interpretation

This is heavily tested in O Levels, often combined with coordinate geometry.

Key ideas

• Gradient of curve at point: value of $\dfrac{dy}{dx}$ at that 𝑥-value
• Equation of tangent at point $(𝑥_1, 𝑦_1)$:
  $𝑦 - 𝑦_1 = 𝑚(𝑥 - 𝑥_1)$
  where $𝑚 = \left.\dfrac{dy}{dx}\right|_{𝑥=𝑥_1}$
• Normal line is perpendicular to tangent:
  $𝑚_{\text{normal}} = -\frac{1}{𝑚_{\text{tangent}}}$

Example 4: Equation of tangent

Curve: $𝑦 = 𝑥^3 - 2𝑥^2 + 1$
Find the equation of the tangent at 𝑥 = 1.

1. Find 𝑦-coordinate:
   $𝑦 = 1^3 - 2(1)^2 + 1 = 1 - 2 + 1 = 0$
   So point is (1, 0).

2. Differentiate:
   $\frac{dy}{dx} = 3𝑥^2 - 4𝑥$

3. Gradient at 𝑥 = 1:
   $𝑚 = 3(1)^2 - 4(1) = 3 - 4 = -1$

4. Equation of tangent:
   $𝑦 - 0 = -1(𝑥 - 1) \Rightarrow 𝑦 = -𝑥 + 1$

---

6. Stationary points and nature (max/min)

This is another major O Level A-Math differentiation topic.

Steps to find stationary points

1. Differentiate 𝑦 to get $\dfrac{dy}{dx}$.
2. Set $\dfrac{dy}{dx} = 0$ and solve for 𝑥 (gives stationary points).
3. Find corresponding 𝑦-values.
4. Use second derivative or gradient sign change to determine max/min.

Example 5: Stationary points

Curve: $𝑦 = 𝑥^3 - 6𝑥^2 + 9𝑥$

1. Differentiate:
   $\frac{dy}{dx} = 3𝑥^2 - 12𝑥 + 9$

2. Solve $\dfrac{dy}{dx} = 0$:
   $3𝑥^2 - 12𝑥 + 9 = 0 \Rightarrow 𝑥^2 - 4𝑥 + 3 = 0$
   $(𝑥 - 1)(𝑥 - 3) = 0 \Rightarrow 𝑥 = 1 \text{ or } 𝑥 = 3$

3. Find 𝑦-values:

   • At 𝑥 = 1: 𝑦 = 1 - 6 + 9 = 4 → (1, 4)
   • At 𝑥 = 3: 𝑦 = 27 - 54 + 27 = 0 → (3, 0)

4. Second derivative:
   $\frac{𝑑^2𝑦}{dx^2} = 6𝑥 - 12$

   • At 𝑥 = 1: 6(1) - 12 = -6 (negative) → maximum
   • At 𝑥 = 3: 6(3) - 12 = 6 (positive) → minimum

So (1, 4) is a maximum point, (3, 0) is a minimum point.

---

Exam strategy guide

You don’t just need to know differentiation; you need to perform under exam conditions — PSLE is over, this is O Level A-Math now.

“Access more than 1000+ past year papers to practice”
👉 Start a paper today and test yourself like it’s the real exam.

Here’s how to approach differentiation questions in your Paper 1 and Paper 2.

1. First 10 seconds: classify the question

Before you start differentiating, quickly ask:

• Is this basic differentiation (just find $\dfrac{dy}{dx}$)?
• Is it about tangent/normal?
• Is it about stationary points?
• Is it a rate of change word problem?
• Is there a quotient/product/chain rule involved?

This 10-second check helps you recall the correct formula and prevents you from blindly expanding or differentiating wrongly.

---

2. Highlight the “with respect to” variable

In many O Level questions, especially rate of change, you’ll see:

• “Find $\dfrac{dy}{dx}$”
• “Find $\dfrac{dA}{dt}$”
• “Given that 𝑥 and 𝑡 are related…”

Always underline/box the variable you’re differentiating with respect to.
This is crucial for questions involving $\dfrac{dy}{dt}$ and $\dfrac{dx}{dt}$.

Example: If you know $\dfrac{𝑦}{𝑥}$ relation and $\dfrac{dx}{dt}$, you might need chain rule:
$\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$

---

3. Choose your method wisely

• If the expression is simple to expand, expand then differentiate.
• If it has multiple brackets, surds, or powers, consider product/quotient + chain rule.
• If it’s in a fraction, check if you can simplify before applying quotient rule.

This decision can save you from heavy algebra mistakes.

---

4. Show clear working and structure

Markers in Singapore are used to our style of working. A good structure:

1. Write the original function clearly.
2. State the derivative rule if it’s a longer question (e.g. “Using quotient rule: …”).
3. Show intermediate steps neatly, especially when using product/quotient rule.
4. Box your final answer.

Even if your final answer is slightly off due to algebra, clear working can still earn method marks.

---

5. For tangent/normal questions: follow a fixed template

When you see “tangent” or “normal”, your brain should auto-run this:

1. Find the 𝑦-coordinate of the point (if not given).
2. Differentiate to get $\dfrac{dy}{dx}$.
3. Substitute 𝑥-value to get gradient 𝑚.
4. If normal, use $𝑚_{\text{normal}} = -\dfrac{1}{𝑚}$.
5. Use point-slope form: $𝑦 - 𝑦_1 = 𝑚(𝑥 - 𝑥_1)$.
6. Simplify to required form.

Write the steps in this order; it reduces careless mistakes.

---

6. Time management for long differentiation questions

Some A-Math questions combine:

• Differentiation
• Stationary points
• Nature of points
• Sketching or interpreting graph

For a 6–8 mark question:

• Spend 1–2 minutes getting $\dfrac{dy}{dx}$ and solving for stationary points.
• Spend 1 minute checking your algebra.
• Spend the remaining time on interpretation (max/min, coordinates, sketch).

If you’re stuck on the derivative for more than 3 minutes, move on first and come back later. Don’t let one messy quotient rule eat your whole Paper 2.

---

Worksheet practice

Here’s where you actually practise. Treat this section like a mini A-Math worksheet.

Try each question on your own first. After that, you can go to
https://tutorly.sg/app, choose your level and A-Math differentiation, and ask Tutorly to:

“Give me step-by-step solutions for similar O Level A-Math differentiation questions.”

Tutorly doesn’t read your working line-by-line, but it checks your final answer and then shows you how to get there step-by-step, so you can compare and learn.

---

A. Core skills practice (warm-up)

Q 1. Basic differentiation

Differentiate each of the following with respect to 𝑥:

1. $𝑦 = 5𝑥^3 - 4𝑥^2 + 7𝑥 - 3$
2. $𝑦 = \dfrac{2}{𝑥^3} - \sqrt{𝑥}$
3. $𝑦 = (3𝑥 - 1)^4$

Hints (not full solutions):

1. Use power rule term by term.
2. Rewrite as $2𝑥^{-3} - 𝑥^{1/2}$ first.
3. Use chain rule: outer $(\cdot)^4$, inner (3𝑥 - 1).

---

Q 2. Product and quotient rule

1. $𝑦 = (𝑥^2 + 1)(2𝑥 - 3)$
2. $𝑦 = \dfrac{4𝑥 - 5}{𝑥^2 + 2}$

Hints:

• For (1), either expand or use product rule.
• For (2), use quotient rule with 𝑢 = 4𝑥 - 5, $𝑣 = 𝑥^2 + 2$.

---

B. Tangents and normals (O Level style)

Q 3. Tangent to a curve

The curve $𝑦 = 𝑥^3 - 3𝑥^2 + 2$ passes through the point 𝑃(2, 2).

(a) Find $\dfrac{dy}{dx}$.
(b) Find the gradient of the tangent to the curve at 𝑃.
(c) Hence find the equation of the tangent at 𝑃.

Hint: Follow the fixed template from the exam strategy section.

---

Q 4. Normal to a curve

The curve $𝑦 = \dfrac{6}{𝑥} + 𝑥$ intersects the 𝑥-axis at the point 𝐴.

(a) Find the coordinates of 𝐴.
(b) Find $\dfrac{dy}{dx}$.
(c) Find the equation of the normal to the curve at 𝐴.

Hint: For (a), set 𝑦 = 0 and solve for 𝑥.

---

C. Stationary points and nature

Q 5. Stationary points

The curve $𝑦 = 2𝑥^3 - 9𝑥^2 + 12𝑥 + 1$.

“Doing Secondary Science? Pick a topic and practise like it’s a real exam — with clear answers right after.”
👉 Try Tutorly now and start a Science topic in seconds.


(a) Find $\dfrac{dy}{dx}$.
(b) Find the coordinates of the stationary points.
(c) Determine the nature of each stationary point.

Hint:

• For (b), solve $\dfrac{dy}{dx} = 0$.
• For (c), use the second derivative test.

---

D. Hard exam variants (prelim / O Level standard)

Now for the more challenging ones that often appear in school prelims in Singapore.

Q 6. Mixed quotient + chain rule + stationary point

The curve is given by
$𝑦 = \frac{2𝑥 - 1}{(𝑥 + 1)^2}.$

(a) Show that
$\frac{dy}{dx} = \frac{-2𝑥}{(𝑥 + 1)^3}.$

(b) Hence, or otherwise, find the coordinates of the stationary point of the curve.
(c) Determine whether this stationary point is a maximum or a minimum.

Suggested approach:

• For (a), use quotient rule carefully. Try to simplify your answer to match the given form.
• For (b), set $\dfrac{dy}{dx} = 0$ and solve for 𝑥.
• For (c), use second derivative or test sign change.

After you attempt, go to https://tutorly.sg/app and ask:

“Explain step-by-step how to differentiate $𝑦 = \dfrac{2𝑥 - 1}{(𝑥 + 1)^2}$ and find its stationary point, O Level A-Math style.”

You’ll get a full breakdown similar to what your teacher would write on the board.

---

Q 7. Tangent with a condition (harder algebra)

The curve $𝑦 = 𝑥^3 + ax^2 + bx + 4$ has a tangent with gradient $5$ at the point where 𝑥 = 1.

(a) Find $\dfrac{dy}{dx}$ in terms of 𝑥, 𝑎 and 𝑏.
(b) Use the information given to form an equation relating 𝑎 and 𝑏.
(c) Given that the curve passes through the point (2, 6), find the values of 𝑎 and 𝑏.

Why this is hard:
You’re not just differentiating — you’re using gradients and coordinates to form equations in unknown constants.

Hint outline:

• (a) Differentiate: $\dfrac{dy}{dx} = 3𝑥^2 + 2ax + 𝑏$.
• (b) At 𝑥 = 1, gradient is $5$:
  $3(1)^2 + 2𝑎(1) + 𝑏 = 5 \Rightarrow 3 + 2𝑎 + 𝑏 = 5$
• (c) Substitute (2, 6) into original curve to get a second equation in 𝑎 and 𝑏. Solve simultaneously.

---

Q 8. Rate of change (classic O Level context)

The length 𝑥 cm of a square is increasing at a constant rate of $0.5$ cm/s. The area of the square is 𝐴 cm$^2$.

(a) Express 𝐴 in terms of 𝑥.
(b) Find $\dfrac{dA}{dx}$.
(c) Hence find the rate of change of the area when 𝑥 = 10 cm.

Now a slightly harder twist (common in school exams):

A cube has side length 𝑥 cm, and its volume is 𝑉 cm$^3$. The side length is increasing at a rate of $0.2$ cm/s.

(d) Express 𝑉 in terms of 𝑥.
(e) Find $\dfrac{dV}{dx}$.
(f) Hence find $\dfrac{dV}{dt}$ when 𝑥 = 5 cm.

Key idea: Use chain rule for rates:
$\frac{dV}{dt} = \frac{dV}{dx} \cdot \frac{dx}{dt}.$

Again, after trying, you can ask Tutorly at https://tutorly.sg/app to walk you through the rate-of-change logic step-by-step.

---

Q 9. Combined skills: stationary point + tangent

The curve $𝑦 = 𝑥^3 - 4𝑥 + 1$ intersects the 𝑥-axis at point 𝐴 and has a stationary point at 𝐵.

(a) Find the coordinates of 𝐴.
(b) Show that the 𝑥-coordinate of 𝐵 is $𝑥 = \dfrac{2}{\sqrt{3}}$.
(c) Find the equation of the normal to the curve at 𝐵.

This type of question combines:

• Solving cubic equation (at least one root is usually simple, e.g. 𝑥 = 1 or 𝑥 = -1)
• Differentiation for stationary point
• Tangent/normal equation

Try to write your working as if you’re in an actual exam — neat, step-by-step, no skipping.

---

How to turn these into daily practice (without burning out)

You don’t need to do 50 questions a day. Instead, aim for:

• 10–15 minutes a day on differentiation, consistently
• Mix of basic and hard questions each session
• Always review your mistakes — don’t just mark wrong and move on

A simple routine using Tutorly.sg:

1. Go to https://tutorly.sg/app on your laptop or tablet.
2. Choose your level (Sec 3/4) and subject (A-Math).

---

“Practice PSLE Science questions and get clear, step-by-step answers instantly.”
👉 Try a question now and see how fast you can improve.

Ready to practise?

If you want a Singapore-focused AI tutor you can use immediately (website, no sign-up), try Tutorly here:

• https://tutorly.sg/ai-tutor-singapore
• https://tutorly.sg/app

---

Related Articles

• AI Tutor For Practice Questions In Singapore: How To Use It Properly (Not Lazily)
• How To Master O Level Math Algebra Questions (Singapore Secondary Level) With Smart Worksheet Practice
• How To Answer Geography Questions In Singapore O Levels (Step-By-Step Guide)