Calculus · Study Notes · Edexcel IGCSE 4MA1 Higher

Differentiation: find gradients of curves and stationary points. The only IGCSE that includes calculus — A* topic.

At a glance

Power rule: $\dfrac{d}{dx}(x^{n}) = nx^{n-1}$ .
Constants disappear: $\dfrac{d}{dx}(c) = 0$ .
Multiply by power, reduce power by 1.
Stationary points: where $\dfrac{dy}{dx} = 0$ .
Maximum vs minimum: second derivative test.
Gradient at point: substitute $x$ into derivative.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

3.7 — Differentiate polynomial functions.
3.7 — Find gradient at a specific point.
3.7 — Find stationary points and determine their nature.

The power rule

▼

Multiply by power, reduce by 1.

Power rule: $\dfrac{d}{dx}(x^{n}) = nx^{n-1}$ .

Examples:

$y$	$\dfrac{dy}{dx}$
$x^{2}$	$2x$
$x^{3}$	$3x^{2}$
$x^{4}$	$4x^{3}$
$x$	$1$ (since $x = x^{1}$ , $1 \cdot x^{0} = 1$ )
$1/x = x^{-1}$	$-x^{-2} = -1/x^{2}$
$\sqrt{x} = x^{1/2}$	$\frac{1}{2}x^{-1/2}$
$5$ (constant)	$0$

Constants multiply through:

$\dfrac{d}{dx}(kf(x)) = k \cdot \dfrac{d}{dx}(f(x))$ .

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

Sums and differences:

$\dfrac{d}{dx}(f + g) = \dfrac{df}{dx} + \dfrac{dg}{dx}$ .

So $\dfrac{d}{dx}(x^{3} - 2x + 5) = 3x^{2} - 2 + 0 = 3x^{2} - 2$ .

Worked qualitative. Why does the constant disappear?

Derivatives measure RATE OF CHANGE.
A constant doesn't change.
So its rate of change is $0$ .

Edexcel tip. Show every term's derivative explicitly. Mark schemes credit each term separately.

Power rule: $nx^{n-1}$ .
Multiply, then reduce.
Constants → $0$ .
Sum: differentiate term by term.

Gradient at a specific point

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Differentiate, then substitute the $x$ -value.

Method:

Differentiate $y$ to get $\dfrac{dy}{dx}$ .
Substitute the $x$ -value of the point.
The result is the gradient at that point.

Example. $y = x^{3} - 2x + 5$ . Find gradient at $x = 2$ .

$\dfrac{dy}{dx} = 3x^{2} - 2$ .
At $x = 2$ : $3(4) - 2 = 10$ .
Gradient = $10$ .

The value of dy/dx at x = 3 is the gradient of the tangent touching the curve there.

Tangent line at a point.

If you have the point $(a, b)$ and gradient $m$ at that point:

Equation of tangent: $y - b = m(x - a)$ .

Example. Tangent to $y = x^{2}$ at $(3, 9)$ .

Gradient: $\dfrac{dy}{dx} = 2x = 6$ at $x = 3$ .
Equation: $y - 9 = 6(x - 3) \to y = 6x - 9$ .

Worked qualitative. What's the gradient of $y = x^{2}$ at the origin?

$\dfrac{dy}{dx} = 2x$ .
At $x = 0$ : gradient = $0$ .
The curve has a horizontal tangent at the vertex.

Edexcel tip. Show the differentiation step AND the substitution step separately. Mark schemes credit both.

Differentiate, then substitute.
Gradient = $\dfrac{dy}{dx}$ at $x$ .
Tangent: $y - y_1 = m(x - x_1)$ .
Always show both steps.

Stationary points

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Set $\dfrac{dy}{dx} = 0$ , solve. Find $y$ . Determine nature.

Stationary point: where $\dfrac{dy}{dx} = 0$ . The curve has zero gradient (horizontal tangent).

Could be:

MINIMUM (low point, curve goes up either side).
MAXIMUM (high point, curve goes down either side).
POINT OF INFLECTION (rare at IGCSE).

Stationary points have a horizontal tangent; the second derivative tells you which is which.

Method:

Differentiate $y$ to get $\dfrac{dy}{dx}$ .
Set $\dfrac{dy}{dx} = 0$ . Solve for $x$ .
Substitute back into ORIGINAL $y$ equation to find $y$ .
Determine nature.

Determining nature.

Method A: Second derivative test.

If $\dfrac{d^{2}y}{dx^{2}} > 0$ at the point: MINIMUM.
If $\dfrac{d^{2}y}{dx^{2}} < 0$ : MAXIMUM.
If $= 0$ : inconclusive (rare at IGCSE).

Method B: Sign change test.

Test $\dfrac{dy}{dx}$ either side of the point.
$-$ then $+$ : MINIMUM.
$+$ then $-$ : MAXIMUM.
Same sign: point of inflection.

Worked example. $y = x^{3} - 3x^{2} - 9x + 5$ .

$\dfrac{dy}{dx} = 3x^{2} - 6x - 9$ .
Set $= 0$ : $x^{2} - 2x - 3 = 0$ . Factor: $(x-3)(x+1) = 0$ . $x = 3$ or $x = -1$ .
Find $y$ $y$ :
- $x = 3$ : $27 - 27 - 27 + 5 = -22$ . Point: $(3, -22)$ .
- $x = -1$ : $-1 - 3 + 9 + 5 = 10$ . Point: $(-1, 10)$ .
Second derivative: $\dfrac{d^{2}y}{dx^{2}} = 6x - 6$ $\frac{d ^{2} y}{d x ^{2}} = 6 x - 6$ .
- At $x = 3$ : $12 > 0$ → MIN.
- At $x = -1$ : $-12 < 0$ → MAX.

So: minimum at $(3, -22)$ , maximum at $(-1, 10)$ .

Worked qualitative. Why does the second derivative tell us min vs max?

Second derivative = rate of change of GRADIENT.
$> 0$ : gradient is INCREASING (going from $-$ to $+$ ) → curve curves UP → MIN.
$< 0$ : gradient is DECREASING ( $+$ to $-$ ) → curve curves DOWN → MAX.

Edexcel tip. Always determine NATURE — Edexcel asks for it. Use second derivative or sign change test.

Set $\dfrac{dy}{dx} = 0$ .
Solve for $x$ .
Find $y$ from original.
Determine nature: $\dfrac{d^{2}y}{dx^{2}}$ .
$> 0$ min, $< 0$ max.

Quick recap

Power rule: $nx^{n-1}$ .
Constants: $0$ .
Term by term.
Stationary: $\dfrac{dy}{dx} = 0$ .
Nature: second derivative.
Tangent: point-slope form.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

$\dfrac{d}{dx}(x^{n}) = nx^{n-1}$ .
$\dfrac{d}{dx}(c) = 0$ .
Stationary: $\dfrac{dy}{dx} = 0$ .
Min: $\dfrac{d^{2}y}{dx^{2}} > 0$ .
Max: $\dfrac{d^{2}y}{dx^{2}} < 0$ .

How it’s examined

Calculus appears Higher Tier (4-7 marks). Differentiation, gradient at a point, stationary points, sometimes tangents. Examiner reports flag (1) not multiplying by the power, (2) skipping nature determination, (3) stating only $x$ for stationary points.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H Jan 2024, 4MA1/2H May/Jun 2024. Last reviewed 2026-05-07.

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

1Differentiate a power
Higher• differentiation
▼
Question
Differentiate $y = 3x^{4}$ .
Step-by-step solution
1. Step 1
  Power rule: $\dfrac{d}{dx}(x^{n}) = nx^{n-1}$ .
2. Step 2
  Apply: $\dfrac{d}{dx}(3x^{4}) = 3 \cdot 4x^{3} = 12x^{3}$ .
Answer
$\dfrac{dy}{dx} = 12x^{3}$
2Differentiate a polynomial
Higher• Adapted from 4MA1/2H May/Jun 2024 Q21• differentiation, polynomial
▼
Question
Differentiate $y = 4x^{3} - 5x^{2} + 7x - 2$ .
Step-by-step solution
1. Step 1
  Differentiate term by term.
2. Step 2
  $4x^{3} \to 12x^{2}$ . $-5x^{2} \to -10x$ . $7x \to 7$ . Constant $-2 \to 0$ .
3. Step 3
  Combine: $\dfrac{dy}{dx} = 12x^{2} - 10x + 7$ .
Answer
$12x^{2} - 10x + 7$
Examiner tip
Differentiate each term separately. Constants ( $-2$ here) differentiate to $0$ .
3Find gradient at a specific point
Higher• gradient, differentiation
▼
Question
$y = x^{3} - 2x + 5$ . Find the gradient at the point where $x = 2$ .
Step-by-step solution
1. Step 1
  Differentiate: $\dfrac{dy}{dx} = 3x^{2} - 2$ .
2. Step 2
  Substitute $x = 2$ : $3(4) - 2 = 10$ .
Answer
Gradient = $10$
4Find stationary points
Higher• Adapted from 4MA1/1H Jan 2024 Q18• stationary points
▼
Question
Find the stationary points of $y = x^{3} - 3x^{2} - 9x + 5$ and determine their nature.
Step-by-step solution
1. Step 1
  Differentiate: $\dfrac{dy}{dx} = 3x^{2} - 6x - 9$ .
2. Step 2
  Set $\dfrac{dy}{dx} = 0$ : $3x^{2} - 6x - 9 = 0 \to x^{2} - 2x - 3 = 0$ .
3. Step 3
  Factor: $(x - 3)(x + 1) = 0$ . So $x = 3$ or $x = -1$ .
4. Step 4
  Find $y$ : at $x = 3$ , $y = 27 - 27 - 27 + 5 = -22$ . At $x = -1$ , $y = -1 - 3 + 9 + 5 = 10$ .
5. Step 5
  Nature: test second derivative or values either side. $\dfrac{d^{2}y}{dx^{2}} = 6x - 6$ . At $x = 3$ : $12 > 0$ → minimum. At $x = -1$ : $-12 < 0$ → maximum.
Answer
$(3, -22)$ minimum; $(-1, 10)$ maximum
Examiner tip
Stationary point: where $\dfrac{dy}{dx} = 0$ . Use second derivative test for nature: $> 0$ → min, $< 0$ → max.
5Differentiating a reciprocal [1 mark]
Getting started• Paper 1H/2H style — 1 mark• differentiation, negative powers
▼
Question
$y = \dfrac{3}{x}$

Find $\dfrac{dy}{dx}$ [1]
Step-by-step solution
1. Step 1
  Write the fraction as a power of $x$ .
  $y = 3x^{-1}$
2. Step 2
  B1 Multiply by the power $-1$ and reduce the power from $-1$ to $-2$ .
  $\dfrac{dy}{dx} = -3x^{-2} = -\dfrac{3}{x^{2}}$
Answer
$\dfrac{dy}{dx} = -3x^{-2}$ or $-\dfrac{3}{x^{2}}$
Examiner tip
Examiner tip: Either form is accepted: $-3x^{-2}$ or $-\dfrac{3}{x^{2}}$ . Do not try to simplify further.

Common mistake: Writing $3x^{0} = 3$ by reducing the power from $-1$ to $0$ instead of to $-2$ , or writing $\dfrac{3}{1} = 3$ by differentiating the denominator. Both score B0.
6Gradient at a negative value of x [2 marks]
Getting started• Paper 1H/2H style — 2 marks• differentiation, gradient
▼
Question
A curve has equation $y = x^{4} - 2x^{3}$

Find the gradient of the curve at the point where $x = -1$ [2]
Step-by-step solution
1. Step 1
  M1 Differentiate term by term.
  $\dfrac{dy}{dx} = 4x^{3} - 6x^{2}$
2. Step 2
  A1 Substitute $x = -1$ , with brackets.
  $4(-1)^{3} - 6(-1)^{2} = -4 - 6 = -10$
Answer
$-10$
Examiner tip
Examiner tip: An odd power of a negative number is negative and an even power is positive: $(-1)^{3} = -1$ , $(-1)^{2} = 1$ .

Common mistake: Typing $-1^{2}$ into a calculator without brackets. The calculator squares $1$ first and gives $-1$ , which leads to $-4 + 6 = 2$ .
7Time for a ball to reach its greatest height [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• rate of change, maximum
▼
Question
A ball is thrown upwards. Its height, $h$ metres, above the ground $t$ seconds after it is thrown is given by

$h = 1.5 + 14t - 5t^{2}$

(a) Find $\dfrac{dh}{dt}$ [1]

(b) Use your answer to part (a) to find the time at which the ball reaches its greatest height. [2]
Step-by-step solution
1. Step 1
  B1 (a) Differentiate with respect to $t$ . The constant $1.5$ goes to $0$ .
  $\dfrac{dh}{dt} = 14 - 10t$
2. Step 2
  M1 (b) At the greatest height the ball stops rising, so the rate of change of height is $0$ .
  $14 - 10t = 0$
3. Step 3
  A1 Solve for $t$ .
  $t = 1.4$
Answer
(a) $\dfrac{dh}{dt} = 14 - 10t$ (b) $1.4$ seconds
Examiner tip
Examiner tip: $\dfrac{dh}{dt}$ is the vertical velocity of the ball. It is positive while the ball rises and negative while it falls, and it is $0$ at the top.

Common mistake: Giving the greatest height ( $11.3$ m) when the question asks for the time. Read the last line of the question again before you write your answer.
8Two turning points from x² = 9 [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• stationary points, differentiation
▼
Question
The curve $C$ has equation $y = x^{3} - 27x + 5$

Use calculus to find the coordinates of the turning points of $C$ . [4]
Step-by-step solution
1. Step 1
  M1 Differentiate.
  $\dfrac{dy}{dx} = 3x^{2} - 27$
2. Step 2
  M1 Set the gradient equal to $0$ and solve.
  $3x^{2} - 27 = 0 \quad \Rightarrow \quad x^{2} = 9$
3. Step 3
  A1 A square has two square roots.
  $x = 3 \quad \text{or} \quad x = -3$
4. Step 4
  A1 Substitute each value into the equation of $C$ .
  $x = 3: \ y = 27 - 81 + 5 = -49 \qquad x = -3: \ y = -27 + 81 + 5 = 59$
Answer
$(3, -49)$ and $(-3, 59)$
Examiner tip
Examiner tip: A cubic has up to two turning points. If you find only one, check whether a square root has given you a second value.

Common mistake: Taking only the positive square root. The June 2024 examiner report notes candidates who ignored the negative value of $x$ when two points were required. Here that loses the $(-3, 59)$ point and both A marks.
9Finding two unknown coefficients [5 marks]
Stretch• Paper 1H/2H style — 5 marks• differentiation, simultaneous equations, gradient
▼
Question
The curve $C$ has equation $y = ax^{3} + bx$ , where $a$ and $b$ are constants.

The point $(1, -2)$ lies on $C$ . The gradient of $C$ at this point is $6$

(a) Find the value of $a$ and the value of $b$ . [3]

(b) Find the $x$ -coordinates of the turning points of $C$ . Give your answers correct to $3$ significant figures. [2]
Step-by-step solution
1. Step 1
  M1 (a) The point is on the curve, so substitute $x = 1$ , $y = -2$ .
  $a + b = -2$
2. Step 2
  M1 Differentiate and use gradient $6$ at $x = 1$ .
  $\dfrac{dy}{dx} = 3ax^{2} + b \quad \Rightarrow \quad 3a + b = 6$
3. Step 3
  A1 Subtract the first equation from the second.
  $2a = 8 \quad \Rightarrow \quad a = 4, \quad b = -6$
4. Step 4
  M1 (b) Set $\dfrac{dy}{dx} = 12x^{2} - 6$ equal to $0$ .
  $12x^{2} = 6 \quad \Rightarrow \quad x^{2} = 0.5$
5. Step 5
  A1 Take both square roots.
  $x = \pm\sqrt{0.5} = \pm 0.70710\ldots$
Answer
(a) $a = 4$ , $b = -6$ (b) $x = 0.707$ and $x = -0.707$
Examiner tip
Examiner tip: Two unknowns need two equations. One comes from the point lying on the curve, the other from the gradient.

Common mistake: Using the gradient condition in $y$ , writing $a + b = 6$ . The gradient is a property of $\dfrac{dy}{dx}$ , not of $y$ . In (b), giving only $0.707$ loses the A1.
10Largest cuboid from a fixed length of wire [6 marks]
Stretch• Paper 1H/2H style — 6 marks• optimisation, stationary points, volume
▼
Question
A frame in the shape of a cuboid is made from $60$ cm of wire. The cuboid has a square base of side $x$ cm and a height of $h$ cm. All $12$ edges of the cuboid are made from the wire.

The volume of the cuboid is $V$ cm³.

(a) Show that $V = 15x^{2} - 2x^{3}$ [2]

(b) Use calculus to find the maximum value of $V$ . [4]
Step-by-step solution
1. Step 1
  M1 (a) There are $8$ edges of length $x$ and $4$ edges of length $h$ .
  $8x + 4h = 60 \quad \Rightarrow \quad h = 15 - 2x$
2. Step 2
  A1 Volume is base area times height.
  $V = x^{2}(15 - 2x) = 15x^{2} - 2x^{3}$
3. Step 3
  M1 (b) Differentiate.
  $\dfrac{dV}{dx} = 30x - 6x^{2}$
4. Step 4
  M1 Set $\dfrac{dV}{dx} = 0$ and factorise.
  $6x(5 - x) = 0 \quad \Rightarrow \quad x = 0 \text{ or } x = 5$
5. Step 5
  A1 $x = 0$ gives no cuboid. The $x^{3}$ term is negative, so the curve falls, turns at $x = 0$ (a minimum), rises, and turns at $x = 5$ (a maximum).
  $x = 5$
6. Step 6
  A1 Substitute $x = 5$ .
  $V = 15 \times 25 - 2 \times 125 = 375 - 250 = 125$
Answer
$125$ cm³ (the cuboid is a cube of side $5$ cm)
Examiner tip
Examiner tip: Count the edges of a cuboid carefully: a square base and a square top give $8$ edges of length $x$ , and there are $4$ vertical edges.

Common mistake: Writing $4x + 4h = 60$ , which counts only one square face, and then being unable to reach the given result. In (b), stopping at $x = 5$ without finding $V$ loses the final A1.

Question 1
Paper 1H/2H style — 1 mark1 mark
$s = 4t^{3} - t$

Find $\dfrac{ds}{dt}$ [1]
Model answer
$\dfrac{ds}{dt} = 12t^{2} - 1$
Why this scores
Mark scheme:
- B1 for $12t^{2} - 1$ cao
Why this scores full marks: The candidate differentiates with respect to $t$ exactly as they would with respect to $x$ . The $-t$ term becomes $-1$ , not $0$ and not $-t$ .
Question 2
Paper 1H/2H style — 2 marks2 marks
$y = 5x^{2} - \dfrac{2}{x^{3}}$

Find $\dfrac{dy}{dx}$ [2]
Model answer
$y = 5x^{2} - 2x^{-3}$

$\dfrac{dy}{dx} = 10x + 6x^{-4}$
Why this scores
Mark scheme:
- M1 for $-2x^{-3}$ seen, or for one term differentiated correctly
- A1 for $10x + 6x^{-4}$ oe
Why this scores full marks: The fraction is rewritten as $-2x^{-3}$ before differentiating. Then $(-3) \times (-2) = +6$ and the power falls from $-3$ to $-4$ . Weaker scripts write $-6x^{-4}$ or $6x^{-2}$ .
Question 3
Paper 1H/2H style — 2 marks2 marks
A curve has equation $y = x^{3} - 3x^{2} + 3x$

Show that the gradient of the curve is $0$ at the point where $x = 1$ [2]
Model answer
$\dfrac{dy}{dx} = 3x^{2} - 6x + 3$

$x = 1: \quad 3(1)^{2} - 6(1) + 3 = 3 - 6 + 3 = 0$

So the gradient is $0$ when $x = 1$
Why this scores
Mark scheme:
- M1 for $3x^{2} - 6x + 3$
- A1 for substitution of $x = 1$ shown, reaching $0$ , with a conclusion
Why this scores full marks: In a show-that question the answer is given, so the marks are for the working. The candidate writes the derivative, shows every value in the substitution and ends with a sentence. Writing only " $= 0$ " does not show anything.
Question 4
Paper 1H/2H style — 3 marks3 marks
The curve with equation $y = (x + 3)(x - 7)$ has a turning point.

Use calculus to find the coordinates of the turning point. [3]
Model answer
$y = x^{2} - 4x - 21$

$\dfrac{dy}{dx} = 2x - 4$

$2x - 4 = 0 \quad \Rightarrow \quad x = 2$

$y = (2 + 3)(2 - 7) = 5 \times (-5) = -25$

Answer: $(2, -25)$
Why this scores
Mark scheme:
- M1 for $x^{2} - 4x - 21$ and an attempt to differentiate
- M1 for $2x - 4 = 0$
- A1 for $(2, -25)$
Why this scores full marks: The brackets are expanded before differentiating, and the $y$ -coordinate is found by substituting back. The question says "use calculus", so completing the square with no derivative would not score, even though it gives the same point.
Question 5
Paper 1H/2H style — 3 marks3 marks
A particle moves along a straight line. Its velocity, $v$ m/s, at time $t$ seconds is given by

$v = 2t^{2} - 8t + 7 \qquad 0 \leqslant t \leqslant 5$

(a) Find the value of $t$ when the acceleration of the particle is $0$ [2]

(b) Find the velocity of the particle at this time. [1]
Model answer
(a)

$a = \dfrac{dv}{dt} = 4t - 8$

$4t - 8 = 0 \quad \Rightarrow \quad t = 2$

(b)

$v = 2 \times 4 - 16 + 7 = -1$

Answer: $-1$ m/s
Why this scores
Mark scheme:
- M1 for $a = 4t - 8$
- A1 for $t = 2$
- B1 for $-1$ ft their $t$
Why this scores full marks: The candidate differentiates velocity once to get acceleration; there is no displacement here. In (b) the negative velocity is kept: it means the particle is moving in the negative direction at $1$ m/s.
Question 6
Paper 1H/2H style — 4 marks4 marks
A curve has equation $y = 3x - \dfrac{12}{x}$

Find the exact coordinates of the two points on the curve where the gradient is $7$ [4]
Model answer
$y = 3x - 12x^{-1} \quad \Rightarrow \quad \dfrac{dy}{dx} = 3 + 12x^{-2}$

$3 + \dfrac{12}{x^{2}} = 7 \quad \Rightarrow \quad \dfrac{12}{x^{2}} = 4 \quad \Rightarrow \quad x^{2} = 3 \quad \Rightarrow \quad x = \pm\sqrt{3}$

$x = \sqrt{3}: \quad y = 3\sqrt{3} - \dfrac{12}{\sqrt{3}} = 3\sqrt{3} - 4\sqrt{3} = -\sqrt{3}$

$x = -\sqrt{3}: \quad y = -3\sqrt{3} + 4\sqrt{3} = \sqrt{3}$

Answer: $(\sqrt{3}, -\sqrt{3})$ and $(-\sqrt{3}, \sqrt{3})$
Why this scores
Mark scheme:
- M1 for $3 + 12x^{-2}$
- M1 for $3 + 12x^{-2} = 7$ (their derivative equated to $7$ )
- A1 for $x = \sqrt{3}$ and $x = -\sqrt{3}$
- A1 for both points, exact
Why this scores full marks: The derivative is set equal to $7$ , the given gradient, and both square roots of $3$ are kept. The candidate rationalises $\dfrac{12}{\sqrt{3}} = 4\sqrt{3}$ to give exact coordinates, as the question asks. Decimals such as $1.73$ lose the final A1.
Question 7
Paper 1H/2H style — 4 marks4 marks
The curve with equation $y = x^{3} + kx^{2} + 7$ , where $k$ is a constant, has a turning point where $x = 4$

(a) Find the value of $k$ . [3]

(b) Find the coordinates of the other turning point of the curve. [1]
Model answer
(a)

$\dfrac{dy}{dx} = 3x^{2} + 2kx$

$x = 4: \quad 48 + 8k = 0 \quad \Rightarrow \quad k = -6$

(b)

$3x^{2} - 12x = 0 \quad \Rightarrow \quad 3x(x - 4) = 0 \quad \Rightarrow \quad x = 0 \text{ or } x = 4$

$x = 0: \quad y = 7$

Answer: $(0, 7)$
Why this scores
Mark scheme:
- M1 for $3x^{2} + 2kx$
- M1 for $3 \times 4^{2} + 2k \times 4 = 0$ oe
- A1 for $k = -6$
- B1 for $(0, 7)$
Why this scores full marks: The candidate treats $k$ as a number when differentiating, so $kx^{2}$ becomes $2kx$ . A turning point means gradient $0$ , which gives an equation in $k$ . Part (b) reuses the derivative with $k = -6$ and picks the root that is not $4$ .
Question 8
Paper 1H/2H style — 5 marks5 marks
The curve $C$ has equation $y = x + \dfrac{9}{x}$ , $x \neq 0$

$C$ has two turning points.

Find the coordinates of the turning points of $C$ and state, with a reason, which is a maximum and which is a minimum. Show clear algebraic working. [5]
Model answer
$y = x + 9x^{-1} \quad \Rightarrow \quad \dfrac{dy}{dx} = 1 - 9x^{-2}$

$1 - \dfrac{9}{x^{2}} = 0 \quad \Rightarrow \quad x^{2} = 9 \quad \Rightarrow \quad x = 3 \text{ or } x = -3$

$x = 3: \ y = 3 + 3 = 6 \qquad x = -3: \ y = -3 - 3 = -6$

Gradient either side of $x = -3$ : at $x = -4$ , $1 - \dfrac{9}{16} > 0$ ; at $x = -2$ , $1 - \dfrac{9}{4} < 0$ . Positive then negative, so a maximum.

Gradient either side of $x = 3$ : at $x = 2$ , $1 - \dfrac{9}{4} < 0$ ; at $x = 4$ , $1 - \dfrac{9}{16} > 0$ . Negative then positive, so a minimum.

Answer: $(-3, -6)$ is a maximum and $(3, 6)$ is a minimum
Why this scores
Mark scheme:
- M1 for at least one term differentiated correctly
- A1 for $1 - 9x^{-2}$ oe
- M1 for their derivative $= 0$ leading to $x^{2} = 9$ oe
- A1 for $(3, 6)$ and $(-3, -6)$
- B1 for $(-3, -6)$ maximum and $(3, 6)$ minimum, with a valid reason (sign of the gradient either side, or the shape of each branch of the curve)
Why this scores full marks: Both square roots of $9$ are kept, so both turning points are found. The nature of each is justified by the sign of the gradient either side, tested on the same branch of the curve. An equally valid reason is the shape: for $x > 0$ the curve is high near $x = 0$ and high for large $x$ , so its turning point there is a minimum; for $x < 0$ it is low at both ends, so that turning point is a maximum.

Why weaker scripts lose the B1: They call the point with the larger $y$ -coordinate the maximum. Here the maximum, $-6$ , is lower than the minimum, $6$ , because the two turning points lie on separate branches of the curve.
Question 9
Paper 1H/2H style — 5 marks5 marks
A company makes $x$ hundred games each week. The weekly profit, $P$ thousand pounds, is modelled by

$P = -2x^{3} + 21x^{2} - 36x - 10 \qquad x > 0$

Use calculus to find the value of $x$ that gives the maximum weekly profit, and work out this maximum profit. [5]
Model answer
$\dfrac{dP}{dx} = -6x^{2} + 42x - 36$

$-6x^{2} + 42x - 36 = 0 \quad \Rightarrow \quad x^{2} - 7x + 6 = 0 \quad \Rightarrow \quad (x - 1)(x - 6) = 0$

$x = 1$ or $x = 6$

The $x^{3}$ term is negative, so the first turning point ( $x = 1$ ) is a minimum and the second ( $x = 6$ ) is the maximum.

$P = -2(216) + 21(36) - 36(6) - 10 = -432 + 756 - 216 - 10 = 98$

Answer: $x = 6$ ( $600$ games); maximum profit £ $98\,000$
Why this scores
Mark scheme:
- M1 for $-6x^{2} + 42x - 36$
- M1 for their derivative $= 0$ and a correct method to solve
- A1 for $x = 1$ and $x = 6$
- B1 for choosing $x = 6$ with a reason
- A1 for £ $98\,000$ (or $98$ thousand pounds)
Why this scores full marks: The candidate finds both turning points and uses the shape of a negative cubic to choose the maximum, rather than guessing. The answer is given in the units of the question: $P = 98$ means $98$ thousand pounds. For comparison, $P = -27$ at $x = 1$ , a loss.
Question 10
Paper 1H/2H style — 6 marks6 marks
A closed box is a cuboid with length $2x$ cm, width $x$ cm and height $h$ cm.

The total surface area of the box is $108$ cm². The volume of the box is $V$ cm³.

(a) Show that $V = 36x - \dfrac{4}{3}x^{3}$ [3]

(b) Use calculus to find the maximum volume of the box. [3]
Model answer
(a)

$2(2x \times x) + 2(2x \times h) + 2(x \times h) = 108$

$4x^{2} + 6xh = 108 \quad \Rightarrow \quad h = \dfrac{108 - 4x^{2}}{6x}$

$V = 2x \times x \times h = 2x^{2} \times \dfrac{108 - 4x^{2}}{6x} = \dfrac{x(108 - 4x^{2})}{3} = 36x - \dfrac{4}{3}x^{3}$

(b)

$\dfrac{dV}{dx} = 36 - 4x^{2}$

$36 - 4x^{2} = 0 \quad \Rightarrow \quad x^{2} = 9 \quad \Rightarrow \quad x = 3 \quad (x > 0)$

$V = 36 \times 3 - \dfrac{4}{3} \times 27 = 108 - 36 = 72$

Answer: $72$ cm³
Why this scores
Mark scheme:
- M1 for $4x^{2} + 6xh = 108$ oe
- M1 for $h = \dfrac{108 - 4x^{2}}{6x}$ oe and substituting into $V = 2x^{2}h$
- A1 for reaching the given result with no errors seen
- M1 for $36 - 4x^{2}$
- M1 for their derivative $= 0$ leading to $x = 3$
- A1 for $72$ cao
Why this scores full marks: The six faces are listed in three pairs, so no face is missed, and $h$ is made the subject before substituting. In (b) the negative root is rejected because a length must be positive; the negative $x^{3}$ term means the turning point at $x = 3$ is the maximum. The box is $6$ cm by $3$ cm by $4$ cm.

4MA1 Calculus

At a glance

What you’ll learn

The power rule

Gradient at a specific point

Stationary points

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Differentiate a power

2Differentiate a polynomial

3Find gradient at a specific point

4Find stationary points

5Differentiating a reciprocal [1 mark]

6Gradient at a negative value of x [2 marks]

7Time for a ball to reach its greatest height [3 marks]

8Two turning points from x² = 9 [4 marks]

9Finding two unknown coefficients [5 marks]

10Largest cuboid from a fixed length of wire [6 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10