Differentiation

Q: What is a derivative in calculus and how is it denoted?

The derivative is the formula for the gradient of the tangent to a curve at any point, written as \( \frac{\mathrm{d}y}{\mathrm{d}x} \) or \( f'(x) \).

Q: What is the Power Rule for differentiating terms of the form \( y = a x^n \)?

Multiply by the power, then reduce the power by 1: \( \frac{\mathrm{d}y}{\mathrm{d}x} = a \cdot n x^{n-1} \).

Q: What are the derivatives of a linear term \( y = ax \) and a constant term \( y = c \)?

The derivative of \( y = ax \) is \( a \), and the derivative of a constant \( y = c \) is \( 0 \).

Q: Why must you expand brackets before differentiating polynomial expressions?

Brackets cannot be differentiated term-by-term; they must first be expanded and simplified into separate polynomial terms.

Q: What is the geometric difference between a tangent and a normal to a curve at a point?

A tangent touches the curve with the same gradient, while a normal meets the curve at a right angle (\( 90^\circ \)) to the tangent.

Q: How are the gradients of the tangent and the normal found at a point \( x = x_1 \)?

The tangent gradient is \( m_{\text{tan}} = \left. \frac{\mathrm{d}y}{\mathrm{d}x} \right|_{x = x_1} \), and the normal gradient is \( m_{\text{norm}} = -\frac{1}{m_{\text{tan}}} \).

Q: What mathematical condition identifies a turning point (stationary point) on a curve?

A turning point occurs where the first derivative equals zero: \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \).

Q: How do you determine whether a turning point is a local maximum or a local minimum using the second derivative?

If \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0 \), it is a local minimum; if \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} < 0 \), it is a local maximum.

Q: What are the four key features required to sketch a quadratic or cubic curve?

The \( y \)-intercept, the \( x \)-intercepts (roots), the turning points with their nature, and the end behaviour.

Q: What are the main steps to solve a practical optimisation problem using differentiation?

Express the quantity as a single-variable formula, differentiate, set equal to zero to find the critical value, verify using \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \), and compute the final value.

Further Mathematics 2330 · CCEA GCSE · 11 min read · Updated: 31 Aug 2026

Welcome to Differentiation!

Have you ever wondered how a roller coaster designer calculates the exact steepness of a track at any second, or how a business works out the exact production level to maximise profit? The answer is Calculus, and more specifically, Differentiation!

In standard GCSE Maths, you learned how to find the gradient (steepness) of a straight line using \( \text{gradient} = \frac{\text{change in } y}{\text{change in } x} \). But curves are constantly changing direction, meaning their steepness is different at every single point. Differentiation is the mathematical tool that lets us find the instantaneous rate of change — the exact gradient of a curve at any point.

Don't worry if this seems a bit daunting at first. We will break it down into simple, manageable steps that will give you full confidence for your CCEA GCSE Further Mathematics (Unit 1: Pure Mathematics) exam.

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1. The Core Rules of Differentiation

What is the Derivative?

When you differentiate an equation \( y = f(x) \), you produce a new formula called the derivative or gradient function, written as:

\( \frac{\mathrm{d}y}{\mathrm{d}x} \) (pronounced "dee y by dee x") or \( f'(x) \) (pronounced "f-dashed of x").

This expression tells you the slope of the tangent to the curve at any value of \( x \).

The Power Rule

For CCEA GCSE Further Mathematics, you only need to differentiate polynomial terms. There is a simple, golden two-step rule to remember:

"Multiply by the power, then reduce the power by 1."

Mathematically, if \( y = a x^n \), then:

\( \frac{\mathrm{d}y}{\mathrm{d}x} = a \cdot n x^{n-1} \)

Special Cases to Remember

1. Linear terms (\( y = a x \)): The derivative is simply the coefficient \( a \).
Example: If \( y = 7x \), then \( \frac{\mathrm{d}y}{\mathrm{d}x} = 7 \).
2. Constant numbers (\( y = c \)): The derivative of any standalone number is always \( 0 \) because a horizontal line has zero gradient.
Example: If \( y = 9 \), then \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \).
3. Sums and differences: Differentiate each term one by one.
Example: If \( y = 4x^3 - 5x^2 + 6x - 2 \), then \( \frac{\mathrm{d}y}{\mathrm{d}x} = 12x^2 - 10x + 6 \).

Crucial Step: Expand Before Differentiating

You cannot differentiate brackets directly term-by-term. Always expand brackets and simplify fully into separate polynomial terms first!

Example: Find \( \frac{\mathrm{d}y}{\mathrm{d}x} \) when \( y = (x - 2)(x + 4) \).
• Step 1: Expand the brackets: \( y = x^2 + 4x - 2x - 8 = x^2 + 2x - 8 \)
• Step 2: Differentiate term-by-term: \( \frac{\mathrm{d}y}{\mathrm{d}x} = 2x + 2 \)

Key Takeaway for Section 1: Multiply by the existing power, decrease the power by \( 1 \), and remember that standalone numbers become \( 0 \).

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2. Tangents and Normals

What are Tangents and Normals?

• A Tangent is a straight line that touches a curve at a specific point without crossing it at that point. It has the exact same gradient as the curve at that point.
• A Normal is a straight line that meets the curve at a right angle (\( 90^\circ \)) to the tangent at that point of contact.

Step-by-Step Method to Find Equations of Tangents and Normals

To find the equation of a straight line, you always need two things: a point \( (x_1, y_1) \) and a gradient \( m \), used in the formula:

\( y - y_1 = m(x - x_1) \)

1. Gradient of the Tangent (\( m_{\text{tan}} \)):
Differentiate the curve to get \( \frac{\mathrm{d}y}{\mathrm{d}x} \), then substitute the given \( x \)-value (\( x_1 \)):
\( m_{\text{tan}} = \left. \frac{\mathrm{d}y}{\mathrm{d}x} \right|_{x = x_1} \)

2. Gradient of the Normal (\( m_{\text{norm}} \)):
Because the normal is perpendicular to the tangent, use the negative reciprocal rule:
\( m_{\text{norm}} = -\frac{1}{m_{\text{tan}}} \)

Worked Example

Find the equations of the tangent and normal to the curve \( y = x^2 - 3x + 5 \) at the point where \( x = 4 \).

Step 1: Find the \( y \)-coordinate:
Substitute \( x = 4 \) into the original curve equation:
\( y = (4)^2 - 3(4) + 5 = 16 - 12 + 5 = 9 \)
So the point is \( (4, 9) \).

Step 2: Find the gradient of the tangent:
\( \frac{\mathrm{d}y}{\mathrm{d}x} = 2x - 3 \)
When \( x = 4 \):
\( m_{\text{tan}} = 2(4) - 3 = 8 - 3 = 5 \)

Step 3: Equation of the Tangent:
Using \( y - y_1 = m_{\text{tan}}(x - x_1) \) with \( (4, 9) \) and \( m_{\text{tan}} = 5 \):
\( y - 9 = 5(x - 4) \)
\( y - 9 = 5x - 20 \)
\( y = 5x - 11 \)

Step 4: Equation of the Normal:
The normal gradient is \( m_{\text{norm}} = -\frac{1}{5} \).
Using \( y - 9 = -\frac{1}{5}(x - 4) \):
Multiply both sides by \( 5 \):
\( 5(y - 9) = -(x - 4) \)
\( 5y - 45 = -x + 4 \)
\( x + 5y - 49 = 0 \) (or \( y = -\frac{1}{5}x + \frac{49}{5} \))

Key Takeaway for Section 2: The tangent uses \( m = \frac{\mathrm{d}y}{\mathrm{d}x} \). The normal is perpendicular, so its gradient is \( -\frac{1}{m} \). Always remember to change the sign and flip the fraction!

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3. Turning Points and the Second Derivative

What is a Turning Point?

A turning point (also called a stationary point) is a point on the curve where the gradient momentarily stops changing and becomes flat. At any stationary point:

\( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \)

The Second Derivative (\( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \))

The second derivative is simply what you get when you differentiate \( \frac{\mathrm{d}y}{\mathrm{d}x} \) a second time. It measures how the gradient is changing and reveals the nature (type) of the turning point.

Determining the Nature: Maximum or Minimum?

• Local Minimum (Valley bottom): \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0 \) (Positive value)
• Local Maximum (Hill top): \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} < 0 \) (Negative value)

Memory Trick: It is the opposite of what you might naturally guess!
• Positive second derivative \( \implies \) Minimum turning point (Think: a positive person smiles \( \cup \), which forms a bottom/minimum).
• Negative second derivative \( \implies \) Maximum turning point (Think: a negative person frowns \( \cap \), which forms a peak/maximum).

Step-by-Step Method for Turning Point Questions

1. Find \( \frac{\mathrm{d}y}{\mathrm{d}x} \).
2. Set \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \) and solve for \( x \).
3. Substitute each \( x \)-value back into the original equation \( y = f(x) \) to find the corresponding \( y \)-coordinates.
4. Find the second derivative \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \).
5. Substitute your \( x \)-values into \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \) to test whether each point is a maximum or a minimum.

Worked Example

Find the coordinates of the turning points on the curve \( y = 2x^3 - 6x \) and determine their nature.

1. Differentiate:
\( \frac{\mathrm{d}y}{\mathrm{d}x} = 6x^2 - 6 \)

2. Set to zero and solve:
\( 6x^2 - 6 = 0 \)
\( 6(x^2 - 1) = 0 \implies x^2 = 1 \implies x = 1 \text{ or } x = -1 \)

3. Find the \( y \)-coordinates (using \( y = 2x^3 - 6x \)):
• When \( x = 1 \): \( y = 2(1)^3 - 6(1) = 2 - 6 = -4 \implies (1, -4) \)
• When \( x = -1 \): \( y = 2(-1)^3 - 6(-1) = -2 + 6 = 4 \implies (-1, 4) \)

4. Second derivative:
\( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12x \)

5. Test nature:
• At \( x = 1 \): \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12(1) = 12 > 0 \implies \) Local Minimum at \( (1, -4) \)
• At \( x = -1 \): \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12(-1) = -12 < 0 \implies \) Local Maximum at \( (-1, 4) \)

Key Takeaway for Section 3: Solve \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \) to locate turning points. Always use the original equation to get \( y \), and use \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \) to check the nature.

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4. Elementary Curve Sketching (Quadratics & Cubics)

When an exam question asks you to sketch a curve, you do not need to plot hundreds of points. Instead, you need to identify and clearly label the key features.

The 4-Step Curve Sketching Checklist

1. \( y \)-intercept: Set \( x = 0 \) in the original equation to find where the curve crosses the \( y \)-axis at \( (0, f(0)) \).
2. \( x \)-intercepts (roots): Set \( y = 0 \) and solve \( f(x) = 0 \) (by factorising) to find where the curve crosses the \( x \)-axis.
3. Turning Points: Use \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \) and \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \) to find the coordinates and nature of all peaks and valleys.
4. End Behaviour (General Shape):
• Positive cubic (\( y = +a x^3 \dots \)): Starts in the bottom-left and ends in the top-right (\( \nearrow \)).
• Negative cubic (\( y = -a x^3 \dots \)): Starts in the top-left and ends in the bottom-right (\( \searrow \)).

Sketching Tips

• Draw smooth, continuous curves — avoid pointy or jagged turning points.
• Clearly write the coordinates next to every intercept and turning point on your axes.

Key Takeaway for Section 4: Find the intercepts, find the stationary points, determine the cubic's general shape, and sketch smoothly with all coordinates clearly labelled.

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5. Practical Optimisation Problems

Optimisation means finding the maximum or minimum value of a real-world quantity (such as maximising the volume of a box or minimising the cost of materials).

How to Solve Optimisation Questions

1. Create the formula: Express the quantity to be optimised in terms of a single variable (e.g., \( V \) in terms of \( x \)).
2. Differentiate: Find the first derivative (e.g., \( \frac{\mathrm{d}V}{\mathrm{d}x} \)).
3. Set to zero: Solve \( \frac{\mathrm{d}V}{\mathrm{d}x} = 0 \) to find the critical value of \( x \).
4. Confirm with the second derivative: Check \( \frac{\mathrm{d}^2V}{\mathrm{d}x^2} \) to prove it is indeed a maximum (\( < 0 \)) or minimum (\( > 0 \)).
5. Answer the question: Substitute \( x \) back into the formula to find the requested maximum area, volume, or dimensions.

Worked Example

A farmer uses \( 40\text{ m} \) of fencing to make a rectangular pen against an existing stone wall (so only three sides need fencing). The area of the pen is given by \( A = 40x - 2x^2 \), where \( x \) is the width. Find the value of \( x \) that gives the maximum area, and calculate this maximum area.

Step 1: Differentiate the area formula:
\( \frac{\mathrm{d}A}{\mathrm{d}x} = 40 - 4x \)

Step 2: Set \( \frac{\mathrm{d}A}{\mathrm{d}x} = 0 \) and solve:
\( 40 - 4x = 0 \implies 4x = 40 \implies x = 10\text{ m} \)

Step 3: Confirm it is a maximum:
\( \frac{\mathrm{d}^2A}{\mathrm{d}x^2} = -4 \)
Since \( -4 < 0 \), this confirms that \( x = 10 \) produces a maximum area.

Step 4: Calculate the maximum area:
Substitute \( x = 10 \) into the area equation:
\( A = 40(10) - 2(10)^2 = 400 - 2(100) = 400 - 200 = 200\text{ m}^2 \)

Key Takeaway for Section 5: Optimisation is simply finding a turning point in a real-world scenario. Differentiate, set equal to zero, and verify using the second derivative.

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Common Examiner Pitfalls to Avoid!

The CCEA examiner reports highlight several recurring mistakes each year. Keep these in mind to secure top marks:

• Mixing up Tangent and Normal Gradients: If the tangent gradient is \( 3 \), the normal gradient is \( -\frac{1}{3} \). Do not forget the minus sign, and do not forget to invert the fraction!
• The Second Derivative Sign Trap: Remember that \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0 \) means a minimum, and \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} < 0 \) means a maximum.
• Finding \( y \)-coordinates: After finding \( x \) at a turning point, always substitute back into the original curve equation \( y = f(x) \). If you substitute \( x \) into \( \frac{\mathrm{d}y}{\mathrm{d}x} \), you will just get \( 0 \)!
• Differentiating Constants: Standalone numbers differentiate to \( 0 \). Differentiating \( 5 \) gives \( 0 \), not \( 5 \).
• Linear Terms: The derivative of \( 3x \) is \( 3 \), not \( 3x \).

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Quick Revision Summary Box

• Power Rule: If \( y = a x^n \implies \frac{\mathrm{d}y}{\mathrm{d}x} = a \cdot n x^{n-1} \)
• Tangent Gradient: \( m_{\text{tan}} = \frac{\mathrm{d}y}{\mathrm{d}x} \)
• Normal Gradient: \( m_{\text{norm}} = -\frac{1}{m_{\text{tan}}} \)
• Turning Points: Set \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \)
• Minimum: \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0 \)
• Maximum: \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} < 0 \)

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is a derivative in calculus and how is it denoted?

The derivative is the formula for the gradient of the tangent to a curve at any point, written as \( \frac{\mathrm{d}y}{\mathrm{d}x} \) or \( f'(x) \).

What is the Power Rule for differentiating terms of the form \( y = a x^n \)?

Multiply by the power, then reduce the power by 1: \( \frac{\mathrm{d}y}{\mathrm{d}x} = a \cdot n x^{n-1} \).

What are the derivatives of a linear term \( y = ax \) and a constant term \( y = c \)?

The derivative of \( y = ax \) is \( a \), and the derivative of a constant \( y = c \) is \( 0 \).

Why must you expand brackets before differentiating polynomial expressions?

Brackets cannot be differentiated term-by-term; they must first be expanded and simplified into separate polynomial terms.

What is the geometric difference between a tangent and a normal to a curve at a point?

A tangent touches the curve with the same gradient, while a normal meets the curve at a right angle (\( 90^\circ \)) to the tangent.

How are the gradients of the tangent and the normal found at a point \( x = x_1 \)?

The tangent gradient is \( m_{\text{tan}} = \left. \frac{\mathrm{d}y}{\mathrm{d}x} \right|_{x = x_1} \), and the normal gradient is \( m_{\text{norm}} = -\frac{1}{m_{\text{tan}}} \).

What mathematical condition identifies a turning point (stationary point) on a curve?

A turning point occurs where the first derivative equals zero: \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \).

How do you determine whether a turning point is a local maximum or a local minimum using the second derivative?

If \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0 \), it is a local minimum; if \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} < 0 \), it is a local maximum.

What are the four key features required to sketch a quadratic or cubic curve?

The \( y \)-intercept, the \( x \)-intercepts (roots), the turning points with their nature, and the end behaviour.

What are the main steps to solve a practical optimisation problem using differentiation?

Express the quantity as a single-variable formula, differentiate, set equal to zero to find the critical value, verify using \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \), and compute the final value.

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