What is differentiation in the context of Cambridge IGCSE Mathematics?

Differentiation is a fundamental concept in calculus, which is part of the Cambridge IGCSE Mathematics curriculum. It involves finding the derivative of a function, which represents the rate of change or the slope of the function at any given point. This is particularly useful in understanding how functions behave and in solving problems related to rates of change.

How do you differentiate a polynomial function in IGCSE Mathematics?

To differentiate a polynomial function, apply the power rule, which states that for any term ax^n, the derivative is n*ax^(n-1). For example, if you have the function f(x) = 3x^3 + 2x^2 - 5x + 7, the derivative f'(x) would be 9x^2 + 4x - 5. This method is a key part of the differentiation techniques taught in the Cambridge IGCSE Mathematics syllabus.

What are some common applications of differentiation in real-world problems?

Differentiation is used in various real-world applications, such as calculating velocity and acceleration in physics, optimizing business processes by finding maximum or minimum values, and analyzing trends in data. In the Cambridge IGCSE Mathematics curriculum, students learn to apply differentiation to solve practical problems involving rates of change and optimization.

What is the significance of the derivative's sign in a function's graph?

The sign of the derivative provides important information about the behavior of a function's graph. If the derivative is positive, the function is increasing at that point; if negative, the function is decreasing. A zero derivative indicates a potential local maximum or minimum. Understanding these concepts is crucial for Cambridge IGCSE Mathematics students when analyzing and sketching graphs.

How can I practice differentiation problems effectively for the IGCSE exam?

To practice differentiation effectively for the Cambridge IGCSE exam, start by mastering the basic rules, such as the power rule, product rule, and quotient rule. Work through past exam papers and practice questions to familiarize yourself with the types of problems you might encounter. Additionally, use online resources and textbooks to reinforce your understanding and seek help from teachers if needed.

Why is "Forgetting to decrement the power" a common mistake in Differentiation?

Why it happens: Students multiply by n but leave the power unchanged: d/dx(x^3) = 3x^3 instead of 3x^2. How to avoid it: Power rule has TWO parts: bring the index down AND subtract one from it.

Why is "Differentiating a constant to itself" a common mistake in Differentiation?

Why it happens: d/dx(7) = 7 instead of 0. How to avoid it: Constants always differentiate to 0.

Why is "Stating "max" or "min" without justification" a common mistake in Differentiation?

Why it happens: Students see the shape from a sketch and assert. How to avoid it: Always justify with the second-derivative test or a sign-change table. Source: 0580/42 — recurring

Why is "Using the symbolic gradient $\dfrac{dy}{dx}$ instead of evaluating it at the point" a common mistake in Differentiation?

Why it happens: Students plug dy/dx as the slope without substituting the x-value. How to avoid it: Step 1 = differentiate. Step 2 = substitute the point's x to get a NUMBER. That number is the slope.

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Detailed Study Notes

Detailed notes on Algebra for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Differentiation — Cambridge IGCSE 0580 Maths Extended (2026)

The gradient of a curve at any point. The single rule $\dfrac{d}{dx}(x^{n}) = nx^{n-1}$ unlocks gradients, tangents, normals, and turning points — a recurring 5-7 mark item on Paper 4.

At a glance

$\dfrac{dy}{dx}$ is the gradient function.
Power rule: $\dfrac{d}{dx}(x^{n}) = nx^{n-1}$ .
Constant: $\dfrac{d}{dx}(c) = 0$ .
Linear: $\dfrac{d}{dx}(kx) = k$ .
Sum rule: differentiate term by term.
Gradient at a point: substitute the $x$ -value.
Stationary points: solve $\dfrac{dy}{dx} = 0$ .
Maximum vs minimum: check the sign of the SECOND derivative or use a sign-change argument.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E2.14 — Use differentiation to find gradients of curves and the equations of tangents and normals; identify and classify stationary points.

The power rule

$\dfrac{d}{dx}(x^{n}) = nx^{n - 1}$ . Bring the power down, decrease it by one.

The single most important rule: $\dfrac{d}{dx}(x^{n}) = n x^{n - 1}.$

Bring the power down to multiply, then decrease the power by 1.

Worked.

$\dfrac{d}{dx}(x^{3}) = 3x^{2}$ .
$\dfrac{d}{dx}(x^{5}) = 5x^{4}$ .
$\dfrac{d}{dx}(x) = 1$ (since $x = x^{1}$ , derivative is $1 \cdot x^{0} = 1$ ).
$\dfrac{d}{dx}(7) = 0$ (a constant has zero gradient).

Coefficient rule. Constants pull through: $\dfrac{d}{dx}(k x^{n}) = k n x^{n - 1}.$

$\dfrac{d}{dx}(4x^{3}) = 12x^{2}$ .
$\dfrac{d}{dx}(5x) = 5$ .

Sum rule. Differentiate each term separately: $\dfrac{d}{dx}(f + g) = \dfrac{df}{dx} + \dfrac{dg}{dx}.$

Worked. Differentiate $y = 2x^{3} - 5x^{2} + 4x - 7$ .

$\dfrac{dy}{dx} = 6x^{2} - 10x + 4$ .

Negative and fractional powers. The same rule works.

$\dfrac{d}{dx}(x^{-2}) = -2x^{-3}$ .
$\dfrac{d}{dx}(\sqrt{x}) = \dfrac{d}{dx}(x^{1/2}) = \dfrac{1}{2}x^{-1/2}$ .

Bring power down, reduce by 1.
Constant: derivative is $0$ .
Differentiate term by term.
Negative and fractional powers: same rule, no change.

Gradient, tangent, normal

Substitute the $x$ -value into the derivative for the gradient. Use it to write tangent or normal equations.

Gradient at a point. To find the gradient of a curve $y = f(x)$ at $x = a$ :

Compute $\dfrac{dy}{dx}$ .
Substitute $x = a$ .

Worked. Find the gradient of $y = x^{2} + 3x$ at $x = 2$ .

$\dfrac{dy}{dx} = 2x + 3$ .
At $x = 2$ : $2(2) + 3 = 7$ .

Tangent line. Passes through the point and has the gradient you found. $y - y_{1} = m(x - x_{1}).$

Worked. Find the tangent to $y = x^{2} + 3x$ at $x = 2$ .

Point: $y = 4 + 6 = 10$ , so $(2, 10)$ .
Gradient: $7$ .
Equation: $y - 10 = 7(x - 2) \Rightarrow y = 7x - 4$ .

Normal line. Perpendicular to the tangent at the same point. Gradient of normal $= -\dfrac{1}{m_{\text{tangent}}}$ .

Worked. Find the normal at the same point.

Normal gradient: $-\dfrac{1}{7}$ .
Equation: $y - 10 = -\dfrac{1}{7}(x - 2)$ .

The tangent has the curve's gradient; the normal meets it at a right angle.

Gradient at $x = a$ : substitute $a$ into $\dfrac{dy}{dx}$ .
Tangent: $y - y_{1} = m(x - x_{1})$ .
Normal gradient: $-1/m_{\text{tangent}}$ .
Always find both the point AND the gradient before writing the equation.

Stationary points (turning points)

Where the gradient is zero. Use the second derivative or a sign change to classify max vs min.

Stationary points are where $\dfrac{dy}{dx} = 0$ — places where the curve momentarily has zero gradient.

Method to find them.

Differentiate to get $\dfrac{dy}{dx}$ .
Set $\dfrac{dy}{dx} = 0$ and solve.
Substitute each $x$ -solution back into $y$ to find the corresponding $y$ -coordinate.

Method to classify them.

Second derivative test. Compute $\dfrac{d^{2}y}{dx^{2}}$ $\frac{d ^{2} y}{d x ^{2}}$ and substitute the $x$ $x$ -coordinate.
- $\dfrac{d^{2}y}{dx^{2}} > 0$ : minimum.
- $\dfrac{d^{2}y}{dx^{2}} < 0$ : maximum.
Sign-change test. Check the gradient just before and just after the stationary point.
- $-, 0, +$ : minimum.
- $+, 0, -$ : maximum.
- $-, 0, -$ or $+, 0, +$ : point of inflexion.

At both turning points the gradient is zero; the second derivative tells you which is which.

Worked. Find and classify stationary points of $y = x^{3} - 6x^{2} + 9x + 1$ .

$\dfrac{dy}{dx} = 3x^{2} - 12x + 9$ .
Set to zero: $3x^{2} - 12x + 9 = 0 \Rightarrow x^{2} - 4x + 3 = 0 \Rightarrow (x - 1)(x - 3) = 0$ .
$x = 1$ or $x = 3$ .
$y$ -values: $y(1) = 1 - 6 + 9 + 1 = 5$ . $y(3) = 27 - 54 + 27 + 1 = 1$ .
$\dfrac{d^{2}y}{dx^{2}} = 6x - 12$ .
At $x = 1$ : $6 - 12 = -6 < 0 \Rightarrow$ maximum at $(1, 5)$ .
At $x = 3$ : $18 - 12 = 6 > 0 \Rightarrow$ minimum at $(3, 1)$ .

Stationary points: $\dfrac{dy}{dx} = 0$ .
Find $y$ -coordinates by substituting back.
Second derivative: positive → min, negative → max.
Sign-change test as backup.

Quick recap

Power rule: $\dfrac{d}{dx}(x^{n}) = nx^{n - 1}$ .
Constants vanish; coefficients pass through.
Differentiate term by term.
Gradient at $x = a$ : substitute $a$ .
Tangent: $y - y_{1} = m(x - x_{1})$ . Normal: gradient $= -1/m$ .
Stationary points: $\dfrac{dy}{dx} = 0$ . Classify with $\dfrac{d^{2}y}{dx^{2}}$ .

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Derivative — the gradient function $\dfrac{dy}{dx}$ .
Power rule — $\dfrac{d}{dx}(x^{n}) = nx^{n - 1}$ .
Stationary point — where $\dfrac{dy}{dx} = 0$ .
Maximum / minimum — local high / low point of a curve.
Tangent line — line touching a curve at one point with the same gradient.
Normal line — line perpendicular to the tangent at the same point.

How it’s examined

Differentiation appears every Paper 4 as a 5-7 mark question — usually a polynomial whose stationary points need finding and classifying, sometimes with tangent/normal equations. Examiner reports flag forgetting to find $y$ -coordinates of stationary points, and confusing maximum with minimum on the second-derivative test.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E2.14); 0580/42 Oct/Nov 2024 — Q16 (stationary points); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Differentiation

Step-by-step solutions to past-paper-style questions on differentiation, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Differentiation

Almost every differentiation exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single instruction — differentiate, find $\dfrac{dy}{dx}$ , find the gradient at $x = a$ — handled with the power rule alone.

How to approach it

Apply $\dfrac{d}{dx}(x^{n}) = nx^{n-1}$ to each term; rewrite fractions and roots as powers ( $\frac{4}{x} = 4x^{-1}$ ) first. To get a numerical gradient, substitute the $x$ -value into the derivative.

Common trap

Bringing the index down but not subtracting one, or differentiating a constant to itself instead of $0$ . Examiner reports flag both as routine slips.

Multi-step problem

Recognise it by

Several stages chained — find a tangent (point, then gradient, then line), or find and classify stationary points, or determine increasing/decreasing intervals.

How to approach it

Differentiate first, then carry out the remaining stages in order: substitute for the gradient, classify with $\dfrac{d^{2}y}{dx^{2}}$ or a sign table, and use $y - y_{1} = m(x - x_{1})$ for a tangent.

Common trap

Stating "maximum" or "minimum" with no justification, or using the symbolic $\dfrac{dy}{dx}$ as a tangent's slope instead of evaluating it at the point.

Word problem

Recognise it by

A real-world context — displacement, velocity, cost — with no derivative stated. You must recognise that a rate of change calls for differentiation.

How to approach it

Translate the context: the derivative of displacement is velocity, "at rest" means $v = 0$ . Differentiate, then substitute or solve as the question requires.

Common trap

Confusing displacement with velocity — setting $s = 0$ for "at rest" instead of $v = 0$ . Examiner reports flag this every series.

1Differentiate a polynomial
ExtendedDirect calculation• differentiation
Question
Differentiate $y = 4x^{3} - 5x^{2} + 7x - 2$ with respect to $x$ .
Step-by-step solution
1. Step 1
  Apply the power rule term by term: $\frac{d}{dx}(x^{n}) = nx^{n-1}$ .
  $\dfrac{dy}{dx} = 12x^{2} - 10x + 7$
2. Step 2
  The constant $-2$ differentiates to $0$ .
Answer
$\dfrac{dy}{dx} = 12x^{2} - 10x + 7$
2Find the gradient at a specific point
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2024 Q15• gradient
Question
Find the gradient of $y = x^{3} - 4x + 1$ at the point where $x = 2$ .
Step-by-step solution
1. Step 1
  Differentiate.
  $\dfrac{dy}{dx} = 3x^{2} - 4$
2. Step 2
  Substitute $x = 2$ .
  $\dfrac{dy}{dx}\bigg|_{x=2} = 3(4) - 4 = 8$
Answer
Gradient $= 8$
3Find the equation of a tangent
ExtendedMulti-step problem• Adapted from 0580/42 Oct/Nov 2023 Q14• tangent
Question
Find the equation of the tangent to $y = x^{2} - 3x + 2$ at the point $x = 4$ .
Step-by-step solution
1. Step 1
  Find $y$ at $x = 4$ .
  $y = 16 - 12 + 2 = 6.\ \text{Point: } (4, 6)$
2. Step 2
  Differentiate.
  $\dfrac{dy}{dx} = 2x - 3$
3. Step 3
  Gradient at $x = 4$ .
  $m = 2(4) - 3 = 5$
4. Step 4
  Use $y - y_{1} = m(x - x_{1})$ .
  $y - 6 = 5(x - 4) \implies y = 5x - 14$
Answer
$y = 5x - 14$
4Find and classify stationary points
ExtendedMulti-step problem• stationary points
Question
Find the stationary points of $y = x^{3} - 6x^{2} + 9x + 1$ and classify them.
Step-by-step solution
1. Step 1
  Differentiate and set to zero.
  $\dfrac{dy}{dx} = 3x^{2} - 12x + 9 = 0$
2. Step 2
  Divide by $3$ and factor.
  $x^{2} - 4x + 3 = 0 \implies (x - 1)(x - 3) = 0$
3. Step 3
  Compute $y$ for each.
  $x = 1: y = 5.\quad x = 3: y = 1.$
4. Step 4
  Classify with the second derivative: $\frac{d^{2}y}{dx^{2}} = 6x - 12$ .
  $x = 1: \frac{d^{2}y}{dx^{2}} = -6 < 0 \Rightarrow \text{maximum}.\quad x = 3: \frac{d^{2}y}{dx^{2}} = 6 > 0 \Rightarrow \text{minimum}.$
Answer
Maximum at $(1, 5)$ ; minimum at $(3, 1)$
Examiner tip
Always classify using either the second-derivative test OR a sign-change check on $\dfrac{dy}{dx}$ . Stating "max" or "min" without justification loses marks.
5Differentiate $y = 3x^{4} - 2x^{2} + 7$
CoreDirect calculation• differentiation, power rule
Question
Find $\dfrac{dy}{dx}$ for $y = 3x^{4} - 2x^{2} + 7$ .
Step-by-step solution
1. Step 1
  Apply the power rule to each term: bring the index down, subtract one.
  $\dfrac{d}{dx}(3x^{4}) = 12x^{3};\ \dfrac{d}{dx}(-2x^{2}) = -4x$
2. Step 2
  The constant $7$ differentiates to $0$ .
Answer
$\dfrac{dy}{dx} = 12x^{3} - 4x$
6Differentiate a term with a negative index
CoreDirect calculation• differentiation, negative index
Question
Find $\dfrac{dy}{dx}$ for $y = x^{2} + \dfrac{4}{x}$ .
Step-by-step solution
1. Step 1
  Rewrite $\dfrac{4}{x}$ as $4x^{-1}$ .
  $y = x^{2} + 4x^{-1}$
2. Step 2
  Apply the power rule to both terms.
  $\dfrac{dy}{dx} = 2x + 4(-1)x^{-2} = 2x - \dfrac{4}{x^{2}}$
Answer
$\dfrac{dy}{dx} = 2x - \dfrac{4}{x^{2}}$
Examiner tip
The examiner report flags candidates often drop the index manipulation and try to "differentiate the bottom". Rewrite as $x^{-1}$ first; the power rule then applies normally.
7Equation of a tangent at a specific point
ExtendedMulti-step problem• Adapted from 0580/42 Feb/Mar 2024 Q16• tangent
Question
Find the equation of the tangent to $y = 2x^{2} + x - 4$ at the point where $x = -1$ .
Step-by-step solution
1. Step 1
  Find $y$ at $x = -1$ .
  $y = 2(1) + (-1) - 4 = -3.\ \text{Point: } (-1, -3)$
2. Step 2
  Differentiate.
  $\dfrac{dy}{dx} = 4x + 1$
3. Step 3
  Gradient at $x = -1$ .
  $m = 4(-1) + 1 = -3$
4. Step 4
  Apply $y - y_{1} = m(x - x_{1})$ .
  $y + 3 = -3(x + 1) \implies y = -3x - 6$
Answer
$y = -3x - 6$
8Find the point with a given gradient
ExtendedDirect calculation• gradient
Question
On the curve $y = x^{2} - 8x + 5$ , find the coordinates of the point where the gradient is $-2$ .
Step-by-step solution
1. Step 1
  Differentiate.
  $\dfrac{dy}{dx} = 2x - 8$
2. Step 2
  Set the derivative equal to $-2$ and solve.
  $2x - 8 = -2 \implies x = 3$
3. Step 3
  Find $y$ .
  $y = 9 - 24 + 5 = -10$
Answer
$(3, -10)$
9Real-world rate of change (kinematics)
ChallengeWord problem• Adapted from 0580/42 May/Jun 2024 Q18• application, kinematics, rate of change
Question
The displacement of a particle (in metres) at time $t$ (in seconds) is $s = t^{3} - 6t^{2} + 9t$ . Find (a) the velocity at $t = 4$ , (b) the time(s) when the particle is at rest.
Step-by-step solution
1. Step 1
  Velocity = $\dfrac{ds}{dt}$ .
  $v = 3t^{2} - 12t + 9$
2. Step 2
  (a) Substitute $t = 4$ .
  $v(4) = 3(16) - 48 + 9 = 9\ \text{m/s}$
3. Step 3
  (b) At rest: $v = 0$ .
  $3t^{2} - 12t + 9 = 0 \implies t^{2} - 4t + 3 = 0$
4. Step 4
  Factor.
  $(t - 1)(t - 3) = 0 \implies t = 1\ \text{s or}\ t = 3\ \text{s}$
Answer
(a) $9$ m/s (b) $t = 1$ s and $t = 3$ s
Examiner tip
The examiner report flags candidates often confuse displacement and velocity. The first derivative of displacement gives velocity; "at rest" means $v = 0$ , not $s = 0$ .
10Curve behaviour from the sign of $f'(x)$
ChallengeMulti-step problem• curve sketching, increasing, decreasing
Question
For $f(x) = x^{3} - 12x$ , find the intervals on which $f$ is increasing and the intervals on which it is decreasing.
Step-by-step solution
1. Step 1
  Differentiate.
  $f'(x) = 3x^{2} - 12 = 3(x^{2} - 4) = 3(x - 2)(x + 2)$
2. Step 2
  Sign analysis of $f'(x)$ :
  $x < -2: f' > 0;\ -2 < x < 2: f' < 0;\ x > 2: f' > 0$
3. Step 3
  Increasing where $f' > 0$ , decreasing where $f' < 0$ .
Answer
Increasing on $x < -2$ and $x > 2$ ; decreasing on $-2 < x < 2$ .
Examiner tip
The mark scheme awards a method mark for factorising $f'(x)$ and another for the sign table. The examiner report flags candidates who skip the sign table and assert the answer.

Key Formulae — Differentiation

The formulae you need to memorise for differentiation on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Power rule
$\dfrac{d}{dx}\left(x^{n}\right) = nx^{n-1}$
$n$
any rational number
When to use
Differentiating any term of the form $ax^{n}$ .
Sum / constant rules
$\dfrac{d}{dx}\left(f + g\right) = f' + g',\quad \dfrac{d}{dx}(c) = 0$
$c$
any constant
When to use
Differentiate term by term; constants vanish.
Equation of a tangent at a point
$y - y_{1} = m(x - x_{1})$
$(x_{1}, y_{1})$
the point of contact on the curve
$m$
the gradient $\dfrac{dy}{dx}$ at $x_{1}$
When to use
After computing the gradient at the point of tangency.
Second-derivative test
$\dfrac{d^{2}y}{dx^{2}} > 0 \Rightarrow \min;\ \dfrac{d^{2}y}{dx^{2}} < 0 \Rightarrow \max$
When to use
Classifying a stationary point once you've found its $x$ -value.

Key Definitions and Keywords — Differentiation

Definitions to memorise and the exact keywords mark schemes credit for differentiation answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Derivative
Examiner keyword
The function $\dfrac{dy}{dx}$ giving the gradient of the curve $y$ at any point.
Gradient at a point
Examiner keyword
The numerical value of $\dfrac{dy}{dx}$ when a specific $x$ is substituted in.
Tangent
Examiner keyword
A straight line that touches a curve at exactly one point with the same gradient as the curve there.
Stationary point
Examiner keyword
A point on the curve where $\dfrac{dy}{dx} = 0$ — local maximum, local minimum, or point of inflection.
Second derivative
The derivative of the derivative, written $\dfrac{d^{2}y}{dx^{2}}$ .

Common Mistakes and Misconceptions — Differentiation

The traps other students keep falling into on differentiation questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Forgetting to decrement the power
Why it happens
Students multiply by $n$ but leave the power unchanged: $\dfrac{d}{dx}(x^{3}) = 3x^{3}$ instead of $3x^{2}$ .
How to avoid it
Power rule has TWO parts: bring the index down AND subtract one from it.
✕Differentiating a constant to itself
Why it happens
$\dfrac{d}{dx}(7) = 7$ instead of $0$ .
How to avoid it
Constants always differentiate to $0$ .
✕Stating "max" or "min" without justification
0580/42 — recurring
Why it happens
Students see the shape from a sketch and assert.
How to avoid it
Always justify with the second-derivative test or a sign-change table.
✕Using the symbolic gradient $\dfrac{dy}{dx}$ instead of evaluating it at the point
Why it happens
Students plug $\dfrac{dy}{dx}$ as the slope without substituting the $x$ -value.
How to avoid it
Step 1 = differentiate. Step 2 = substitute the point's $x$ to get a NUMBER. That number is the slope.

Differentiation — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

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Detailed notes on Algebra for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Differentiation — Cambridge IGCSE 0580 Maths Extended (2026)

The gradient of a curve at any point. The single rule $\dfrac{d}{dx}(x^{n}) = nx^{n-1}$ unlocks gradients, tangents, normals, and turning points — a recurring 5-7 mark item on Paper 4.

At a glance

$\dfrac{dy}{dx}$ is the gradient function.
Power rule: $\dfrac{d}{dx}(x^{n}) = nx^{n-1}$ .
Constant: $\dfrac{d}{dx}(c) = 0$ .
Linear: $\dfrac{d}{dx}(kx) = k$ .
Sum rule: differentiate term by term.
Gradient at a point: substitute the $x$ -value.
Stationary points: solve $\dfrac{dy}{dx} = 0$ .
Maximum vs minimum: check the sign of the SECOND derivative or use a sign-change argument.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E2.14 — Use differentiation to find gradients of curves and the equations of tangents and normals; identify and classify stationary points.

The power rule

$\dfrac{d}{dx}(x^{n}) = nx^{n - 1}$ . Bring the power down, decrease it by one.

The single most important rule: $\dfrac{d}{dx}(x^{n}) = n x^{n - 1}.$

Bring the power down to multiply, then decrease the power by 1.

Worked.

$\dfrac{d}{dx}(x^{3}) = 3x^{2}$ .
$\dfrac{d}{dx}(x^{5}) = 5x^{4}$ .
$\dfrac{d}{dx}(x) = 1$ (since $x = x^{1}$ , derivative is $1 \cdot x^{0} = 1$ ).
$\dfrac{d}{dx}(7) = 0$ (a constant has zero gradient).

Coefficient rule. Constants pull through: $\dfrac{d}{dx}(k x^{n}) = k n x^{n - 1}.$

$\dfrac{d}{dx}(4x^{3}) = 12x^{2}$ .
$\dfrac{d}{dx}(5x) = 5$ .

Sum rule. Differentiate each term separately: $\dfrac{d}{dx}(f + g) = \dfrac{df}{dx} + \dfrac{dg}{dx}.$

Worked. Differentiate $y = 2x^{3} - 5x^{2} + 4x - 7$ .

$\dfrac{dy}{dx} = 6x^{2} - 10x + 4$ .

Negative and fractional powers. The same rule works.

$\dfrac{d}{dx}(x^{-2}) = -2x^{-3}$ .
$\dfrac{d}{dx}(\sqrt{x}) = \dfrac{d}{dx}(x^{1/2}) = \dfrac{1}{2}x^{-1/2}$ .

Bring power down, reduce by 1.
Constant: derivative is $0$ .
Differentiate term by term.
Negative and fractional powers: same rule, no change.

Gradient, tangent, normal

Substitute the $x$ -value into the derivative for the gradient. Use it to write tangent or normal equations.

Gradient at a point. To find the gradient of a curve $y = f(x)$ at $x = a$ :

Compute $\dfrac{dy}{dx}$ .
Substitute $x = a$ .

Worked. Find the gradient of $y = x^{2} + 3x$ at $x = 2$ .

$\dfrac{dy}{dx} = 2x + 3$ .
At $x = 2$ : $2(2) + 3 = 7$ .

Tangent line. Passes through the point and has the gradient you found. $y - y_{1} = m(x - x_{1}).$

Worked. Find the tangent to $y = x^{2} + 3x$ at $x = 2$ .

Point: $y = 4 + 6 = 10$ , so $(2, 10)$ .
Gradient: $7$ .
Equation: $y - 10 = 7(x - 2) \Rightarrow y = 7x - 4$ .

Normal line. Perpendicular to the tangent at the same point. Gradient of normal $= -\dfrac{1}{m_{\text{tangent}}}$ .

Worked. Find the normal at the same point.

Normal gradient: $-\dfrac{1}{7}$ .
Equation: $y - 10 = -\dfrac{1}{7}(x - 2)$ .

The tangent has the curve's gradient; the normal meets it at a right angle.

Gradient at $x = a$ : substitute $a$ into $\dfrac{dy}{dx}$ .
Tangent: $y - y_{1} = m(x - x_{1})$ .
Normal gradient: $-1/m_{\text{tangent}}$ .
Always find both the point AND the gradient before writing the equation.

Stationary points (turning points)

Where the gradient is zero. Use the second derivative or a sign change to classify max vs min.

Stationary points are where $\dfrac{dy}{dx} = 0$ — places where the curve momentarily has zero gradient.

Method to find them.

Differentiate to get $\dfrac{dy}{dx}$ .
Set $\dfrac{dy}{dx} = 0$ and solve.
Substitute each $x$ -solution back into $y$ to find the corresponding $y$ -coordinate.

Method to classify them.

Second derivative test. Compute $\dfrac{d^{2}y}{dx^{2}}$ $\frac{d ^{2} y}{d x ^{2}}$ and substitute the $x$ $x$ -coordinate.
- $\dfrac{d^{2}y}{dx^{2}} > 0$ : minimum.
- $\dfrac{d^{2}y}{dx^{2}} < 0$ : maximum.
Sign-change test. Check the gradient just before and just after the stationary point.
- $-, 0, +$ : minimum.
- $+, 0, -$ : maximum.
- $-, 0, -$ or $+, 0, +$ : point of inflexion.

At both turning points the gradient is zero; the second derivative tells you which is which.

Worked. Find and classify stationary points of $y = x^{3} - 6x^{2} + 9x + 1$ .

$\dfrac{dy}{dx} = 3x^{2} - 12x + 9$ .
Set to zero: $3x^{2} - 12x + 9 = 0 \Rightarrow x^{2} - 4x + 3 = 0 \Rightarrow (x - 1)(x - 3) = 0$ .
$x = 1$ or $x = 3$ .
$y$ -values: $y(1) = 1 - 6 + 9 + 1 = 5$ . $y(3) = 27 - 54 + 27 + 1 = 1$ .
$\dfrac{d^{2}y}{dx^{2}} = 6x - 12$ .
At $x = 1$ : $6 - 12 = -6 < 0 \Rightarrow$ maximum at $(1, 5)$ .
At $x = 3$ : $18 - 12 = 6 > 0 \Rightarrow$ minimum at $(3, 1)$ .

Stationary points: $\dfrac{dy}{dx} = 0$ .
Find $y$ -coordinates by substituting back.
Second derivative: positive → min, negative → max.
Sign-change test as backup.

Quick recap

Power rule: $\dfrac{d}{dx}(x^{n}) = nx^{n - 1}$ .
Constants vanish; coefficients pass through.
Differentiate term by term.
Gradient at $x = a$ : substitute $a$ .
Tangent: $y - y_{1} = m(x - x_{1})$ . Normal: gradient $= -1/m$ .
Stationary points: $\dfrac{dy}{dx} = 0$ . Classify with $\dfrac{d^{2}y}{dx^{2}}$ .

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Derivative — the gradient function $\dfrac{dy}{dx}$ .
Power rule — $\dfrac{d}{dx}(x^{n}) = nx^{n - 1}$ .
Stationary point — where $\dfrac{dy}{dx} = 0$ .
Maximum / minimum — local high / low point of a curve.
Tangent line — line touching a curve at one point with the same gradient.
Normal line — line perpendicular to the tangent at the same point.

How it’s examined

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E2.14); 0580/42 Oct/Nov 2024 — Q16 (stationary points); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Differentiation

Step-by-step solutions to past-paper-style questions on differentiation, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Differentiation

Almost every differentiation exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single instruction — differentiate, find $\dfrac{dy}{dx}$ , find the gradient at $x = a$ — handled with the power rule alone.

How to approach it

Common trap

Bringing the index down but not subtracting one, or differentiating a constant to itself instead of $0$ . Examiner reports flag both as routine slips.

Multi-step problem

Recognise it by

Several stages chained — find a tangent (point, then gradient, then line), or find and classify stationary points, or determine increasing/decreasing intervals.

How to approach it

Common trap

Stating "maximum" or "minimum" with no justification, or using the symbolic $\dfrac{dy}{dx}$ as a tangent's slope instead of evaluating it at the point.

Word problem

Recognise it by

A real-world context — displacement, velocity, cost — with no derivative stated. You must recognise that a rate of change calls for differentiation.

How to approach it

Translate the context: the derivative of displacement is velocity, "at rest" means $v = 0$ . Differentiate, then substitute or solve as the question requires.

Common trap

Confusing displacement with velocity — setting $s = 0$ for "at rest" instead of $v = 0$ . Examiner reports flag this every series.

1Differentiate a polynomial
ExtendedDirect calculation• differentiation
Question
Differentiate $y = 4x^{3} - 5x^{2} + 7x - 2$ with respect to $x$ .
Step-by-step solution
1. Step 1
  Apply the power rule term by term: $\frac{d}{dx}(x^{n}) = nx^{n-1}$ .
  $\dfrac{dy}{dx} = 12x^{2} - 10x + 7$
2. Step 2
  The constant $-2$ differentiates to $0$ .
Answer
$\dfrac{dy}{dx} = 12x^{2} - 10x + 7$
2Find the gradient at a specific point
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2024 Q15• gradient
Question
Find the gradient of $y = x^{3} - 4x + 1$ at the point where $x = 2$ .
Step-by-step solution
1. Step 1
  Differentiate.
  $\dfrac{dy}{dx} = 3x^{2} - 4$
2. Step 2
  Substitute $x = 2$ .
  $\dfrac{dy}{dx}\bigg|_{x=2} = 3(4) - 4 = 8$
Answer
Gradient $= 8$
3Find the equation of a tangent
ExtendedMulti-step problem• Adapted from 0580/42 Oct/Nov 2023 Q14• tangent
Question
Find the equation of the tangent to $y = x^{2} - 3x + 2$ at the point $x = 4$ .
Step-by-step solution
1. Step 1
  Find $y$ at $x = 4$ .
  $y = 16 - 12 + 2 = 6.\ \text{Point: } (4, 6)$
2. Step 2
  Differentiate.
  $\dfrac{dy}{dx} = 2x - 3$
3. Step 3
  Gradient at $x = 4$ .
  $m = 2(4) - 3 = 5$
4. Step 4
  Use $y - y_{1} = m(x - x_{1})$ .
  $y - 6 = 5(x - 4) \implies y = 5x - 14$
Answer
$y = 5x - 14$
4Find and classify stationary points
ExtendedMulti-step problem• stationary points
Question
Find the stationary points of $y = x^{3} - 6x^{2} + 9x + 1$ and classify them.
Step-by-step solution
1. Step 1
  Differentiate and set to zero.
  $\dfrac{dy}{dx} = 3x^{2} - 12x + 9 = 0$
2. Step 2
  Divide by $3$ and factor.
  $x^{2} - 4x + 3 = 0 \implies (x - 1)(x - 3) = 0$
3. Step 3
  Compute $y$ for each.
  $x = 1: y = 5.\quad x = 3: y = 1.$
4. Step 4
  Classify with the second derivative: $\frac{d^{2}y}{dx^{2}} = 6x - 12$ .
  $x = 1: \frac{d^{2}y}{dx^{2}} = -6 < 0 \Rightarrow \text{maximum}.\quad x = 3: \frac{d^{2}y}{dx^{2}} = 6 > 0 \Rightarrow \text{minimum}.$
Answer
Maximum at $(1, 5)$ ; minimum at $(3, 1)$
Examiner tip
Always classify using either the second-derivative test OR a sign-change check on $\dfrac{dy}{dx}$ . Stating "max" or "min" without justification loses marks.
5Differentiate $y = 3x^{4} - 2x^{2} + 7$
CoreDirect calculation• differentiation, power rule
Question
Find $\dfrac{dy}{dx}$ for $y = 3x^{4} - 2x^{2} + 7$ .
Step-by-step solution
1. Step 1
  Apply the power rule to each term: bring the index down, subtract one.
  $\dfrac{d}{dx}(3x^{4}) = 12x^{3};\ \dfrac{d}{dx}(-2x^{2}) = -4x$
2. Step 2
  The constant $7$ differentiates to $0$ .
Answer
$\dfrac{dy}{dx} = 12x^{3} - 4x$
6Differentiate a term with a negative index
CoreDirect calculation• differentiation, negative index
Question
Find $\dfrac{dy}{dx}$ for $y = x^{2} + \dfrac{4}{x}$ .
Step-by-step solution
1. Step 1
  Rewrite $\dfrac{4}{x}$ as $4x^{-1}$ .
  $y = x^{2} + 4x^{-1}$
2. Step 2
  Apply the power rule to both terms.
  $\dfrac{dy}{dx} = 2x + 4(-1)x^{-2} = 2x - \dfrac{4}{x^{2}}$
Answer
$\dfrac{dy}{dx} = 2x - \dfrac{4}{x^{2}}$
Examiner tip
The examiner report flags candidates often drop the index manipulation and try to "differentiate the bottom". Rewrite as $x^{-1}$ first; the power rule then applies normally.
7Equation of a tangent at a specific point
ExtendedMulti-step problem• Adapted from 0580/42 Feb/Mar 2024 Q16• tangent
Question
Find the equation of the tangent to $y = 2x^{2} + x - 4$ at the point where $x = -1$ .
Step-by-step solution
1. Step 1
  Find $y$ at $x = -1$ .
  $y = 2(1) + (-1) - 4 = -3.\ \text{Point: } (-1, -3)$
2. Step 2
  Differentiate.
  $\dfrac{dy}{dx} = 4x + 1$
3. Step 3
  Gradient at $x = -1$ .
  $m = 4(-1) + 1 = -3$
4. Step 4
  Apply $y - y_{1} = m(x - x_{1})$ .
  $y + 3 = -3(x + 1) \implies y = -3x - 6$
Answer
$y = -3x - 6$
8Find the point with a given gradient
ExtendedDirect calculation• gradient
Question
On the curve $y = x^{2} - 8x + 5$ , find the coordinates of the point where the gradient is $-2$ .
Step-by-step solution
1. Step 1
  Differentiate.
  $\dfrac{dy}{dx} = 2x - 8$
2. Step 2
  Set the derivative equal to $-2$ and solve.
  $2x - 8 = -2 \implies x = 3$
3. Step 3
  Find $y$ .
  $y = 9 - 24 + 5 = -10$
Answer
$(3, -10)$
9Real-world rate of change (kinematics)
ChallengeWord problem• Adapted from 0580/42 May/Jun 2024 Q18• application, kinematics, rate of change
Question
The displacement of a particle (in metres) at time $t$ (in seconds) is $s = t^{3} - 6t^{2} + 9t$ . Find (a) the velocity at $t = 4$ , (b) the time(s) when the particle is at rest.
Step-by-step solution
1. Step 1
  Velocity = $\dfrac{ds}{dt}$ .
  $v = 3t^{2} - 12t + 9$
2. Step 2
  (a) Substitute $t = 4$ .
  $v(4) = 3(16) - 48 + 9 = 9\ \text{m/s}$
3. Step 3
  (b) At rest: $v = 0$ .
  $3t^{2} - 12t + 9 = 0 \implies t^{2} - 4t + 3 = 0$
4. Step 4
  Factor.
  $(t - 1)(t - 3) = 0 \implies t = 1\ \text{s or}\ t = 3\ \text{s}$
Answer
(a) $9$ m/s (b) $t = 1$ s and $t = 3$ s
Examiner tip
The examiner report flags candidates often confuse displacement and velocity. The first derivative of displacement gives velocity; "at rest" means $v = 0$ , not $s = 0$ .
10Curve behaviour from the sign of $f'(x)$
ChallengeMulti-step problem• curve sketching, increasing, decreasing
Question
For $f(x) = x^{3} - 12x$ , find the intervals on which $f$ is increasing and the intervals on which it is decreasing.
Step-by-step solution
1. Step 1
  Differentiate.
  $f'(x) = 3x^{2} - 12 = 3(x^{2} - 4) = 3(x - 2)(x + 2)$
2. Step 2
  Sign analysis of $f'(x)$ :
  $x < -2: f' > 0;\ -2 < x < 2: f' < 0;\ x > 2: f' > 0$
3. Step 3
  Increasing where $f' > 0$ , decreasing where $f' < 0$ .
Answer
Increasing on $x < -2$ and $x > 2$ ; decreasing on $-2 < x < 2$ .
Examiner tip
The mark scheme awards a method mark for factorising $f'(x)$ and another for the sign table. The examiner report flags candidates who skip the sign table and assert the answer.

Key Formulae — Differentiation

The formulae you need to memorise for differentiation on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Power rule
$\dfrac{d}{dx}\left(x^{n}\right) = nx^{n-1}$
$n$
any rational number
When to use
Differentiating any term of the form $ax^{n}$ .
Sum / constant rules
$\dfrac{d}{dx}\left(f + g\right) = f' + g',\quad \dfrac{d}{dx}(c) = 0$
$c$
any constant
When to use
Differentiate term by term; constants vanish.
Equation of a tangent at a point
$y - y_{1} = m(x - x_{1})$
$(x_{1}, y_{1})$
the point of contact on the curve
$m$
the gradient $\dfrac{dy}{dx}$ at $x_{1}$
When to use
After computing the gradient at the point of tangency.
Second-derivative test
$\dfrac{d^{2}y}{dx^{2}} > 0 \Rightarrow \min;\ \dfrac{d^{2}y}{dx^{2}} < 0 \Rightarrow \max$
When to use
Classifying a stationary point once you've found its $x$ -value.

Key Definitions and Keywords — Differentiation

Definitions to memorise and the exact keywords mark schemes credit for differentiation answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Derivative
Examiner keyword
The function $\dfrac{dy}{dx}$ giving the gradient of the curve $y$ at any point.
Gradient at a point
Examiner keyword
The numerical value of $\dfrac{dy}{dx}$ when a specific $x$ is substituted in.
Tangent
Examiner keyword
A straight line that touches a curve at exactly one point with the same gradient as the curve there.
Stationary point
Examiner keyword
A point on the curve where $\dfrac{dy}{dx} = 0$ — local maximum, local minimum, or point of inflection.
Second derivative
The derivative of the derivative, written $\dfrac{d^{2}y}{dx^{2}}$ .

Common Mistakes and Misconceptions — Differentiation

The traps other students keep falling into on differentiation questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Forgetting to decrement the power
Why it happens
Students multiply by $n$ but leave the power unchanged: $\dfrac{d}{dx}(x^{3}) = 3x^{3}$ instead of $3x^{2}$ .
How to avoid it
Power rule has TWO parts: bring the index down AND subtract one from it.
✕Differentiating a constant to itself
Why it happens
$\dfrac{d}{dx}(7) = 7$ instead of $0$ .
How to avoid it
Constants always differentiate to $0$ .
✕Stating "max" or "min" without justification
0580/42 — recurring
Why it happens
Students see the shape from a sketch and assert.
How to avoid it
Always justify with the second-derivative test or a sign-change table.
✕Using the symbolic gradient $\dfrac{dy}{dx}$ instead of evaluating it at the point
Why it happens
Students plug $\dfrac{dy}{dx}$ as the slope without substituting the $x$ -value.
How to avoid it
Step 1 = differentiate. Step 2 = substitute the point's $x$ to get a NUMBER. That number is the slope.

Differentiation — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Differentiation

Short Notes - Differentiation

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Differentiation

1Differentiate a polynomial

2Find the gradient at a specific point

3Find the equation of a tangent

4Find and classify stationary points

5Differentiate y=3x4−2x2+7y = 3x^{4} - 2x^{2} + 7y=3x4−2x2+7

6Differentiate a term with a negative index

7Equation of a tangent at a specific point

8Find the point with a given gradient

9Real-world rate of change (kinematics)

10Curve behaviour from the sign of f′(x)f'(x)f′(x)

Power rule

Sum / constant rules

Equation of a tangent at a point

Second-derivative test

Derivative

Gradient at a point

Tangent

Stationary point

Second derivative

✕Forgetting to decrement the power

✕Differentiating a constant to itself

✕Stating "max" or "min" without justification

✕Using the symbolic gradient dydx\dfrac{dy}{dx}dxdy​ instead of evaluating it at the point

Differentiation — frequently asked questions

Differentiation

Short Notes - Differentiation

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Differentiation

1Differentiate a polynomial

2Find the gradient at a specific point

3Find the equation of a tangent

4Find and classify stationary points

5Differentiate y=3x4−2x2+7y = 3x^{4} - 2x^{2} + 7y=3x4−2x2+7

6Differentiate a term with a negative index

7Equation of a tangent at a specific point

8Find the point with a given gradient

9Real-world rate of change (kinematics)

10Curve behaviour from the sign of f′(x)f'(x)f′(x)

Power rule

Sum / constant rules

Equation of a tangent at a point

Second-derivative test

Derivative

Gradient at a point

Tangent

Stationary point

Second derivative

✕Forgetting to decrement the power

✕Differentiating a constant to itself

✕Stating "max" or "min" without justification

✕Using the symbolic gradient dydx\dfrac{dy}{dx}dxdy​ instead of evaluating it at the point

Differentiation — frequently asked questions

5Differentiate $y = 3x^{4} - 2x^{2} + 7$

10Curve behaviour from the sign of $f'(x)$

✕Using the symbolic gradient $\dfrac{dy}{dx}$ instead of evaluating it at the point

5Differentiate $y = 3x^{4} - 2x^{2} + 7$

10Curve behaviour from the sign of $f'(x)$

✕Using the symbolic gradient $\dfrac{dy}{dx}$ instead of evaluating it at the point