Differentiation

Q: How should roots and algebraic fractions be prepared before applying the power rule?

Rewrite them into individual index terms of the form \(a x^n\) using fractional and negative indices.

Q: What is the second derivative and how is it denoted?

The derivative of the first derivative with respect to \(x\), denoted by \(\frac{d^2y}{dx^2}\) or \(f''(x)\).

Q: How are the gradients of the tangent and normal to a curve at \(x = x_1\) related?

The tangent gradient is \(m_T = \left.\frac{dy}{dx}\right|_{x=x_1}\), and the normal gradient is \(m_N = -\frac{1}{m_T}\).

Q: What derivative conditions define where a function is increasing or decreasing?

A function is increasing when \(\frac{dy}{dx} \ge 0\) and decreasing when \(\frac{dy}{dx} \le 0\).

Q: How do you find the coordinates of a stationary point on a curve \(y = f(x)\)?

Set \(\frac{dy}{dx} = 0\), solve for \(x\), and substitute those \(x\)-values into \(y = f(x)\) to get the corresponding \(y\)-values.

Q: How is the nature of a stationary point determined using the second derivative test?

\(\frac{d^2y}{dx^2} > 0\) indicates a local minimum, \(\frac{d^2y}{dx^2}

Q: What is the process for solving real-world optimisation problems with differentiation?

Formulate the target function in one variable, solve \(\frac{d(\text{quantity})}{dx} = 0\), and verify the maximum/minimum using \(\frac{d^2y}{dx^2}\).

Mathematics 2210 · CCEA AS Level · 9 min read · Updated: 31 Aug 2026

Welcome to Differentiation (AS 1: Pure Mathematics)

Welcome to one of the most powerful and exciting areas of mathematics! In basic algebra and coordinate geometry, you learned how to calculate the steepness (gradient) of straight lines. But what happens if you are looking at a curve, like a roller coaster or the trajectory of a rocket? Its steepness changes at every single point.

Differentiation is the mathematical tool that lets us find the exact gradient and instantaneous rate of change at any point on a curve. In this chapter for CCEA AS 1 Pure Mathematics, you will learn the foundations of calculus, from first principles to real-world optimisation.

Did you know? The speedometer in a car is a real-life differentiator: instead of telling you your average speed over the whole journey, it measures your instantaneous rate of change of distance with respect to time right at that very second!

1. The Derivative and Differentiation from First Principles

What is a Derivative?

The derivative, written as \(\frac{dy}{dx}\) or \(f'(x)\), represents the gradient function of a curve \(y = f(x)\). When you plug an \(x\)-value into \(\frac{dy}{dx}\), it outputs the exact gradient of the tangent to the curve at that point.

Understanding First Principles

To find the gradient between two points on a curve, \(A(x, f(x))\) and \(B(x+h, f(x+h))\), we use the standard gradient formula:

\(\text{Gradient of chord } AB = \frac{f(x+h) - f(x)}{(x+h) - x} = \frac{f(x+h) - f(x)}{h}\)

As point \(B\) slides closer and closer to point \(A\), the horizontal gap \(h\) shrinks towards zero (\(h \to 0\)). The chord becomes the tangent line at \(A\).

The First Principles Formula

\(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\)

Step-by-Step Example: Differentiate \(f(x) = x^2\) from first principles.

Step 1: Write down the definition formula:
\(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\)

Step 2: Substitute \(f(x) = x^2\) and \(f(x+h) = (x+h)^2\):
\(f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h}\)

Step 3: Expand the bracket carefully:
\(f'(x) = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h}\)
\(f'(x) = \lim_{h \to 0} \frac{2xh + h^2}{h}\)

Step 4: Factorise \(h\) from the numerator and cancel with the denominator:
\(f'(x) = \lim_{h \to 0} \frac{h(2x + h)}{h} = \lim_{h \to 0} (2x + h)\)

Step 5: Evaluate the limit as \(h \to 0\):
\(f'(x) = 2x + 0 = 2x\)

Examiner Warning & Pitfall: You must write \(\lim_{h \to 0}\) on every single line of working until the step where you actually let \(h = 0\). Dropping the limit notation early will cost you easy accuracy marks!

Key Takeaway: Differentiation from first principles finds the gradient of a tangent by shrinking the distance \(h\) between two points on a curve to zero using clear limit notation.

2. The Power Rule & Algebraic Preparation

The Standard Power Rule

For any term where \(n\) is a rational number (\(n \in \mathbb{Q}\)):

If \(y = a x^n\), then \(\frac{dy}{dx} = a n x^{n-1}\)

Memory Trick: "Multiply by the power, then reduce the power by 1."

Important Linearity Properties

• Differentiating term-by-term: \(\frac{d}{dx}[a f(x) + b g(x)] = a f'(x) + b g'(x)\)
• Constants differentiate to zero: \(\frac{d}{dx}(c) = 0\) (because a horizontal line has zero gradient!)

Algebraic Preparation (Before Differentiating)

You cannot directly apply the power rule to roots, fractions with \(x\) in the denominator, or bracketed expressions. You must rewrite them in the form \(a x^n\) first:

1. Fractional Powers / Roots:
• \(\sqrt{x} = x^{1/2}\)
• \(\frac{1}{\sqrt{x}} = x^{-1/2}\)
• \(\sqrt[n]{x^m} = x^{m/n}\)

2. Negative Powers:
• \(\frac{1}{x^n} = x^{-n}\)
• \(\frac{5}{x^3} = 5x^{-3}\)
• Common Trap: \(\frac{1}{2x^3} = \frac{1}{2}x^{-3}\) (The \(2\) stays in the denominator! It is not \(2x^{-3}\)).

3. Splitting Multi-Term Numerators:
If you have a fraction with a single monomial denominator, divide every term in the numerator by that denominator:
\(\frac{4x^3 - 5x}{x^2} = \frac{4x^3}{x^2} - \frac{5x}{x^2} = 4x - 5x^{-1}\)

Worked Example:
Find \(\frac{dy}{dx}\) for \(y = 3\sqrt{x} + \frac{2}{x^4} - 7\).
Step 1: Prepare into index form:
\(y = 3x^{1/2} + 2x^{-4} - 7\)
Step 2: Differentiate each term:
\(\frac{dy}{dx} = 3\left(\frac{1}{2}\right)x^{-1/2} + 2(-4)x^{-5} - 0\)
\(\frac{dy}{dx} = \frac{3}{2}x^{-1/2} - 8x^{-5} = \frac{3}{2\sqrt{x}} - \frac{8}{x^5}\)

Examiner Warning: Be extra careful when subtracting \(1\) from negative numbers: \(-4 - 1 = -5\), not \(-3\)!

Key Takeaway: Always rewrite expressions as separate terms in the form \(a x^n\) before differentiating.

3. The Second Derivative

The second derivative is found by differentiating the first derivative \(\frac{dy}{dx}\) once more with respect to \(x\).

• Leibniz notation: \(\frac{d^2y}{dx^2}\)
• Function notation: \(f''(x)\)

What does it mean? While the first derivative tells you the slope of the curve, the second derivative tells you how rapidly that slope is changing (the curvature or concavity of the graph).

Quick Example:
If \(y = 2x^4 - 5x^2 + 3x\):
\(\frac{dy}{dx} = 8x^3 - 10x + 3\)
\(\frac{d^2y}{dx^2} = 24x^2 - 10\)

Key Takeaway: The second derivative is simply the derivative of the derivative.

4. Tangents and Normals

Definitions

• Tangent: A straight line that touches a curve at a given point without crossing it there, sharing the exact same gradient as the curve at that point.
• Normal: A straight line that is perpendicular (\(90^\circ\)) to the tangent at the point of contact.

Method for Finding Tangents and Normals

Step 1: Find the gradient of the tangent, \(m_T\), by evaluating the derivative at the given \(x_1\):
\(m_T = \left. \frac{dy}{dx} \right|_{x = x_1}\)

Step 2: Find the \(y_1\) coordinate (if not already provided) by substituting \(x_1\) into the original curve equation \(y = f(x)\).

Step 3: Write the equation of the Tangent using the straight-line formula:
\(y - y_1 = m_T(x - x_1)\)

Step 4: Find the gradient of the Normal, \(m_N\), using the perpendicular gradient rule:
\(m_N = -\frac{1}{m_T}\)

Step 5: Write the equation of the Normal:
\(y - y_1 = m_N(x - x_1) \implies y - y_1 = -\frac{1}{m_T}(x - x_1)\)

Worked Example:
Find the equation of the normal to the curve \(y = x^2 - 3x + 4\) at the point where \(x = 2\).

1. Find \(y_1\): \(y = (2)^2 - 3(2) + 4 = 4 - 6 + 4 = 2\). Point is \((2, 2)\).
2. Differentiate: \(\frac{dy}{dx} = 2x - 3\).
3. Tangent gradient at \(x=2\): \(m_T = 2(2) - 3 = 1\).
4. Normal gradient: \(m_N = -\frac{1}{m_T} = -\frac{1}{1} = -1\).
5. Equation of normal:
\(y - 2 = -1(x - 2)\)
\(y - 2 = -x + 2\)
\(y = -x + 4\) or \(x + y - 4 = 0\)

Key Takeaway: The tangent has gradient \(m_T = \frac{dy}{dx}\); the normal has perpendicular gradient \(m_N = -\frac{1}{m_T}\).

5. Increasing Functions, Decreasing Functions & Stationary Points

Increasing and Decreasing Functions

• A function is increasing on an interval if \(\frac{dy}{dx} \ge 0\) (strictly increasing if \(\frac{dy}{dx} > 0\)). As \(x\) goes right, the graph goes uphill.
• A function is decreasing on an interval if \(\frac{dy}{dx} \le 0\) (strictly decreasing if \(\frac{dy}{dx} < 0\)). As \(x\) goes right, the graph goes downhill.

Stationary Points (Turning Points)

A stationary point occurs wherever the gradient is zero:

\(\frac{dy}{dx} = 0\)

Classifying Stationary Points (The Nature of Turning Points)

To determine whether a stationary point is a local maximum or local minimum, use the Second Derivative Test:

• Local Minimum: \(\frac{d^2y}{dx^2} > 0\)
Visual Aid: Positive second derivative means a positive, happy "smiley face" curve \(\cup\). The bottom is a minimum!

• Local Maximum: \(\frac{d^2y}{dx^2} < 0\)
Visual Aid: Negative second derivative means a negative, sad "frowning face" curve \(\cap\). The top is a maximum!

• Inconclusive Test (\(\frac{d^2y}{dx^2} = 0\)):
If \(\frac{d^2y}{dx^2} = 0\), the test fails to give an answer. You must test the sign of the first derivative \(\frac{dy}{dx}\) slightly to the left (\(x - \epsilon\)) and slightly to the right (\(x + \epsilon\)) of the stationary point:

• Gradient changes from positive (\(+\)) to zero (\(0\)) to negative (\(-\)): Local Maximum
• Gradient changes from negative (\(-\)) to zero (\(0\)) to positive (\(+\)): Local Minimum
• Gradient does not change sign (e.g. \(+ \to 0 \to +\) or \(- \to 0 \to -\)): Stationary Point of Inflection

Examiner Pitfall: When a question asks for the coordinates of a stationary point, students frequently find the \(x\)-values where \(\frac{dy}{dx} = 0\) but forget to calculate the \(y\)-values. Always substitute \(x\) back into the original curve equation \(y = f(x)\).

Key Takeaway: Stationary points occur when \(\frac{dy}{dx} = 0\). Test their nature using \(\frac{d^2y}{dx^2}\) (positive = minimum, negative = maximum).

6. Real-World Contexts & Optimisation

Optimisation means finding the maximum or minimum possible value of a quantity (e.g., maximising the volume of a box, minimising manufacturing material, or minimising cost).

The 5-Step Optimisation Strategy

Step 1: Set up the equations. Identify the quantity to optimise (e.g., Area \(A\)) and any given constraint (e.g., fixed perimeter or volume).
Step 2: Eliminate extra variables. Use the constraint equation to express the formula in terms of a single variable (e.g., \(A\) purely in terms of \(x\)).
Step 3: Differentiate and set to zero. Find \(\frac{dA}{dx}\) and solve \(\frac{dA}{dx} = 0\) to find the stationary value.
Step 4: Prove the nature. Calculate the second derivative \(\frac{d^2A}{dx^2}\) to prove it is genuinely a maximum (\(< 0\)) or minimum (\(> 0\)).
Step 5: Answer the specific question. Calculate the optimal dimensions, maximum area, or minimum cost as requested.

Worked Example:
A farmer has \(40\text{ m}\) of fencing to create a rectangular pen against an existing stone wall (so only 3 sides need fencing).
Let \(x\) be the width perpendicular to the wall, and \(y\) be the length parallel to the wall.

1. Constraint equation: \(2x + y = 40 \implies y = 40 - 2x\)
2. Quantity to maximise: Area \(A = x \times y = x(40 - 2x) = 40x - 2x^2\)
3. Differentiate: \(\frac{dA}{dx} = 40 - 4x\)
4. Set to zero: \(40 - 4x = 0 \implies 4x = 40 \implies x = 10\text{ m}\)
5. Prove it is a maximum: \(\frac{d^2A}{dx^2} = -4\). Since \(-4 < 0\), the area is confirmed to be a local maximum.
6. Maximum Area: \(A = 40(10) - 2(10)^2 = 400 - 200 = 200\text{ m}^2\).

Examiner Warning: In exam questions, marks are specifically allocated for showing that your answer is a maximum or minimum. If you do not evaluate the second derivative (or sign change), you will lose method and accuracy marks!

Key Takeaway: Express the quantity in one variable, solve \(\frac{d(\text{quantity})}{dx} = 0\), and always verify the nature with the second derivative.

Quick Review: AS 1 Differentiation Checklist

Before sitting your AS 1 exam, make sure you can confidently:
• Differentiate polynomials from first principles using \(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\) without dropping limit notation.
• Convert roots and fractions to index form \(a x^n\) before applying \(\frac{dy}{dx} = a n x^{n-1}\).
• Find equations of tangents (\(m_T = \frac{dy}{dx}\)) and normals (\(m_N = -\frac{1}{m_T}\)) using \(y - y_1 = m(x - x_1)\).
• Identify increasing intervals (\(f'(x) \ge 0\)) and decreasing intervals (\(f'(x) \le 0\)).
• Find and fully classify stationary points using \(\frac{dy}{dx} = 0\) and \(\frac{d^2y}{dx^2}\).
• Complete contextual optimisation problems including the mandatory second derivative verification test.

Quick check

Can you answer these now?

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

What is the definition formula for differentiating a function \(f(x)\) from first principles?

\(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\)

What is the standard power rule for differentiating \(y = a x^n\)?

\(\frac{dy}{dx} = a n x^{n-1}\)

How should roots and algebraic fractions be prepared before applying the power rule?

Rewrite them into individual index terms of the form \(a x^n\) using fractional and negative indices.

What is the second derivative and how is it denoted?

The derivative of the first derivative with respect to \(x\), denoted by \(\frac{d^2y}{dx^2}\) or \(f''(x)\).

How are the gradients of the tangent and normal to a curve at \(x = x_1\) related?

The tangent gradient is \(m_T = \left.\frac{dy}{dx}\right|_{x=x_1}\), and the normal gradient is \(m_N = -\frac{1}{m_T}\).

What derivative conditions define where a function is increasing or decreasing?

A function is increasing when \(\frac{dy}{dx} \ge 0\) and decreasing when \(\frac{dy}{dx} \le 0\).

How do you find the coordinates of a stationary point on a curve \(y = f(x)\)?

Set \(\frac{dy}{dx} = 0\), solve for \(x\), and substitute those \(x\)-values into \(y = f(x)\) to get the corresponding \(y\)-values.

How is the nature of a stationary point determined using the second derivative test?

\(\frac{d^2y}{dx^2} > 0\) indicates a local minimum, \(\frac{d^2y}{dx^2} < 0\) indicates a local maximum, and \(\frac{d^2y}{dx^2} = 0\) is inconclusive.

What is the process for solving real-world optimisation problems with differentiation?

Formulate the target function in one variable, solve \(\frac{d(\text{quantity})}{dx} = 0\), and verify the maximum/minimum using \(\frac{d^2y}{dx^2}\).

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