Differentiation guide: how to differentiate any function

Q: How do you differentiate?

First identify which differentiation rule matches the shape of the function: a simple polynomial — the power rule (multiply by the power and reduce it by 1). A product of two functions — the product rule. A quotient of two functions — the quotient rule. A function inside a function (for example a bracket raised to a power) — the chain rule. Most exercises combine several rules, so it matters to identify the structure of the expression before you start differentiating.

Q: How do you differentiate a polynomial?

Use the power rule on each term separately: the derivative of xⁿ is n·x^(n−1), a constant multiplier stays where it is, and a lone constant disappears (its derivative is 0). For example, the derivative of f(x) = 3x³ − 5x² + 2x − 9 is f'(x) = 9x² − 10x + 2.

Q: How do you differentiate a bracket?

When there is a function in a bracket raised to a power (like (2x + 1)⁵) or inside a root, use the chain rule: differentiate the outside (as if the bracket were a single variable) and multiply by the derivative of what is inside the bracket. For example, the derivative of (2x + 1)⁵ is 5·(2x + 1)⁴ times the inner derivative 2, that is 10(2x + 1)⁴. Forgetting to multiply by the inner derivative is the most common mistake here.

Q: What is the power rule for differentiation?

The power rule says that the derivative of xⁿ is n·x^(n−1) — multiply by the power and reduce it by 1. For example the derivative of x³ is 3x², and the derivative of x⁵ is 5x⁴. The rule works for negative and fractional powers too: the derivative of 1/x = x⁻¹ is −1/x², and the derivative of √x = x^(1/2) is 1/(2√x).

Q: When do you use the product rule and when the quotient rule?

The product rule when differentiating a product f(x)·g(x): the derivative is f'·g + f·g'. The quotient rule when differentiating a quotient f(x)/g(x): the derivative is (f'·g − f·g') over [g]². The difference: in the product the sign between the two terms is plus, in the quotient it is minus, and the quotient also has a square in the denominator.

Q: How do you differentiate a composite function like (2x + 1)⁵?

Use the chain rule: differentiate the outer function as if the inner one were a variable, then multiply by the inner derivative. For (2x + 1)⁵: first 5(2x + 1)⁴ (the power rule on the bracket), then multiply by the inner derivative which is 2, giving 10(2x + 1)⁴ altogether. Forgetting to multiply by the inner derivative is the most common mistake.

Q: How do you find stationary points using the derivative?

Solve the equation f'(x) = 0 to find the candidate x-values, then check the sign of the derivative around each point: a change from positive to negative = a maximum, from negative to positive = a minimum. Finally substitute the x-values into the original function f(x) to get the y-coordinates of the points.

Q: What is the most common mistake in differentiation questions?

Forgetting to multiply by the inner derivative in the chain rule, a wrong sign or a missing square in the denominator of the quotient rule, and forgetting to substitute back into the original function to find the y-values of the stationary points. These are all technical mistakes that lose marks even for those who understand the material — and the way to prevent them is repeated practice and self-checking.

The derivative is the central tool of calculus, and it is the basis of every curve-sketching and optimisation question at A level. In this guide we build the topic step by step: what a derivative means, how to apply the power rule, the product and quotient rules and the chain rule, how to find the equation of a tangent, and how to get from the derivative to stationary points and to where a function is increasing or decreasing — including the mistakes that lose the most marks and a practice plan that will take you to full command.

ℹ️ A note on level

Differentiation is A level content — there is no GCSE code for it. The power rule, tangents and stationary points are in the first year (AS); the product, quotient and chain rules arrive in the second year (A2). A-level practice on MathsUK is planned for a later phase; this guide covers the rules and worked examples in full so you can use it alongside your textbook.

To differentiate a function, you identify which differentiation rule matches the shape of the function and apply it: the power rule for a polynomial (bring the power down and multiply by it, then reduce the power by 1), the product or quotient rule when two functions are multiplied or divided, and the chain rule for a function 'inside brackets' — differentiate the outside and multiply by the derivative of the inside. There is no single universal formula for 'how to differentiate' — there are four fixed rules, and the moment you recognise which rule fits, the calculation is entirely mechanical.

The derivative is the heart of calculus, and it is the tool with which most function questions on the A-level Pure papers are solved. Every curve investigation — finding stationary points, intervals where the function is increasing or decreasing, and the gradient of a tangent — starts with differentiating. The good news: differentiating is a technical skill. Once you know a handful of rules by heart and practise them until they become a reflex, you can differentiate almost any function that appears in the exam. In this guide we go through all the rules in the order they are taught, see examples of each, and connect it all to a full curve investigation.

What is a derivative? — the gradient of the tangent

The derivative of a function f(x) at a point is the gradient of the tangent to the curve at that point — that is, the instantaneous rate of change of the function. If the derivative is positive, the function is increasing; if the derivative is negative, the function is decreasing; and if the derivative equals zero, there is a horizontal tangent there — a candidate for a stationary point. The notation for the derivative is f'(x) (read 'f dashed of x') or dy/dx. At A level you are expected to know the definition from first principles (the limit), but in the vast majority of questions you use the ready-made differentiation rules — and that is all you need to solve them.

💡 The idea in one line

The derivative measures how fast the function is changing. The sign of the derivative tells you whether the curve is going up or down, and where the derivative is zero — that is where the stationary points are. Every curve investigation is really a reading of the derivative.

The power rule — the most basic rule

The central rule to know by heart is the power rule: the derivative of xⁿ is n·x^(n−1). In words: multiply by the power, and reduce the power by 1. Direct examples: the derivative of x³ is 3x²; the derivative of x⁵ is 5x⁴; the derivative of x is 1 (because x is x¹, and the derivative is 1·x⁰ = 1); and the derivative of any constant (say 7) is 0, because a constant does not change.

Two simple rules join the power rule: the derivative of a sum is the sum of the derivatives (differentiate term by term), and a constant multiplier stays where it is (the derivative of 4x² is 4·2x = 8x). With these three rules you can differentiate any polynomial. A combined example: the derivative of f(x) = 3x³ − 5x² + 2x − 9 is f'(x) = 9x² − 10x + 2 (the constant term −9 disappears, because the derivative of a constant is zero).

The power rule also works for negative and fractional powers, which matters at A level. For example, 1/x is written as x^(−1), and its derivative is −1·x^(−2) = −1/x². And a root: √x is written as x^(1/2), and its derivative is (1/2)·x^(−1/2) = 1/(2√x). The trick is always the same: rewrite as a power, then apply the power rule.

Function f(x)	Derivative f'(x)	Short explanation
k (constant)	0	A constant does not change
x	1	Power rule with n = 1
x²	2x	Multiply by 2, reduce the power
xⁿ	n·x^(n−1)	The general power rule
1/x = x⁻¹	−1/x²	Negative power
√x = x^(1/2)	1/(2√x)	Fractional power
3x³ − 5x² + 2x − 9	9x² − 10x + 2	Term by term, constant disappears

The product rule and the quotient rule

When differentiating a product of two functions, it is not enough to differentiate each one separately. The product rule says: the derivative of f(x)·g(x) is f'(x)·g(x) + f(x)·g'(x). In words: differentiate the first times the second as it is, plus the first as it is times the derivative of the second. Example: the derivative of (x²)(x + 3) — you could expand the brackets, but with the product rule too: 2x·(x + 3) + x²·1 = 2x² + 6x + x² = 3x² + 6x.

The quotient rule deals with a quotient of two functions. The derivative of f(x)/g(x) is [f'(x)·g(x) − f(x)·g'(x)] / [g(x)]². Note the order: in the numerator it is 'the derivative of the numerator times the denominator, minus the numerator times the derivative of the denominator', and the denominator is the square of the original denominator. The sign and the order in the numerator are the most common mistake — always the top (the derivative of the numerator) first. Example: the derivative of (x²)/(x + 1) is [2x·(x + 1) − x²·1] / (x + 1)² = [2x² + 2x − x²]/(x + 1)² = (x² + 2x)/(x + 1)².

⚠️ Do not forget the square in the denominator

In the quotient rule the denominator of the result is always [g(x)]² — the square of the original denominator. Students forget the square and lose a mark. And another thing: the sign in the numerator is minus, not plus — unlike the product rule.

Differentiating a composite function — the chain rule

A composite function is a 'function inside a function', for example (2x + 1)⁵ or √(x² + 4). To differentiate it you use the chain rule: differentiate the 'outer' function as though the inner one were a single variable, then multiply by the derivative of the 'inner' function. Example: the derivative of (2x + 1)⁵ is 5·(2x + 1)⁴ · 2 = 10(2x + 1)⁴ — first the power rule on the whole bracket, then multiply by the derivative of what is inside the bracket (which is 2).

Another example with a root: the derivative of √(x² + 4) = (x² + 4)^(1/2) is (1/2)·(x² + 4)^(−1/2) · 2x = 2x / (2√(x² + 4)) = x/√(x² + 4). Again the same pattern: differentiate the outer (the power rule on the root), multiply by the derivative of the inner (2x). Remember: without multiplying by the inner derivative the answer is wrong.

From the derivative to a curve investigation

Once you can differentiate, you can carry out a full curve investigation — and that is exactly the big exam question. The process: (1) differentiate and get f'(x). (2) Solve the equation f'(x) = 0 to find the candidate stationary points. (3) Check the sign of the derivative to the left and right of each such point: if the derivative changes from positive to negative — it is a maximum; if from negative to positive — a minimum. (4) Substitute the x-values into the original function to get the y-coordinates of the stationary points.

A full example: investigate f(x) = x³ − 3x². Step 1 — differentiate: f'(x) = 3x² − 6x. Step 2 — set to zero: 3x² − 6x = 0, take out 3x: 3x(x − 2) = 0, giving x = 0 and x = 2. Step 3 — signs of the derivative: before 0 the derivative is positive (the function is increasing), between 0 and 2 it is negative (decreasing), after 2 it is positive (increasing). So x = 0 is a local maximum and x = 2 is a local minimum. Step 4 — y-values: f(0) = 0 (a maximum at (0, 0)), f(2) = 8 − 12 = −4 (a minimum at (2, −4)).

📐 The equation of the tangent

The gradient of the tangent to the curve at x₀ is f'(x₀). The equation of the tangent is y − f(x₀) = f'(x₀)·(x − x₀). Example: the tangent to f(x) = x² at x = 3: the gradient is f'(3) = 6, and the point is (3, 9), so the tangent is y − 9 = 6(x − 3), that is y = 6x − 9. The normal is perpendicular to the tangent, so its gradient is −1/6.

Common mistakes that lose marks

Mistake 1 — forgetting to multiply by the inner derivative in the chain rule. When differentiating (2x + 1)⁵ you must multiply by 2. Without it the answer is wrong. Mistake 2 — a wrong sign or order in the numerator of the quotient rule. Remember: the derivative of the numerator times the denominator, minus the numerator times the derivative of the denominator, over the denominator squared. Mistake 3 — zeroing a constant multiplier (the derivative of 5x is 5, not 0; only the derivative of a lone constant is 0).

Mistake 4 — confusing a maximum with a minimum. Always check the sign of the derivative around the point, or the second derivative: f'' > 0 → minimum, f'' < 0 → maximum. Mistake 5 — forgetting to substitute back into the original function. f'(x) = 0 gives only the x-values of the stationary points; to get the y-values you must substitute the x into f(x), not into f'(x).

A focused differentiation practice plan

Differentiating is a technical skill, so repeated practice is the key. The following plan assumes about an hour a day and is built to take you from the basics to a full curve investigation in about a fortnight.

Days 1-2: the power rule. Forty polynomial differentiation exercises, including negative and fractional powers, until differentiating becomes a reflex.
Days 3-4: the product rule and the quotient rule. Twenty-five exercises, with the emphasis on the sign and the square in the denominator of the quotient rule.
Days 5-6: the chain rule for composite functions. Twenty exercises — always multiply by the inner derivative.
Days 7-8: finding tangents, normals and stationary points. Fifteen exercises: f'(x) = 0, sign check, and substituting back.
Days 9-10: full curve investigations and past-paper questions, under timed conditions.

Summary

Differentiation is a skill that repays investment: it is technical, predictable, and full of safe marks for anyone who practises. The four rules you need to command are the power rule, the product rule, the quotient rule and the chain rule — and above them the curve-investigation process: differentiate, solve f'(x) = 0, check the sign, and substitute back. If you remember to multiply by the inner derivative, keep the sign and the square in the quotient rule, and substitute back into the original function — the curve-investigation question in the exam will become one of your strongest. The way there is through repeated practice.

Frequently asked questions

How do you differentiate?

First identify which differentiation rule matches the shape of the function: a simple polynomial — the power rule (multiply by the power and reduce it by 1). A product of two functions — the product rule. A quotient of two functions — the quotient rule. A function inside a function (for example a bracket raised to a power) — the chain rule. Most exercises combine several rules, so it matters to identify the structure of the expression before you start differentiating.

How do you differentiate a polynomial?

Use the power rule on each term separately: the derivative of xⁿ is n·x^(n−1), a constant multiplier stays where it is, and a lone constant disappears (its derivative is 0). For example, the derivative of f(x) = 3x³ − 5x² + 2x − 9 is f'(x) = 9x² − 10x + 2.

How do you differentiate a bracket?

When there is a function in a bracket raised to a power (like (2x + 1)⁵) or inside a root, use the chain rule: differentiate the outside (as if the bracket were a single variable) and multiply by the derivative of what is inside the bracket. For example, the derivative of (2x + 1)⁵ is 5·(2x + 1)⁴ times the inner derivative 2, that is 10(2x + 1)⁴. Forgetting to multiply by the inner derivative is the most common mistake here.

What is the power rule for differentiation?

The power rule says that the derivative of xⁿ is n·x^(n−1) — multiply by the power and reduce it by 1. For example the derivative of x³ is 3x², and the derivative of x⁵ is 5x⁴. The rule works for negative and fractional powers too: the derivative of 1/x = x⁻¹ is −1/x², and the derivative of √x = x^(1/2) is 1/(2√x).

When do you use the product rule and when the quotient rule?

The product rule when differentiating a product f(x)·g(x): the derivative is f'·g + f·g'. The quotient rule when differentiating a quotient f(x)/g(x): the derivative is (f'·g − f·g') over [g]². The difference: in the product the sign between the two terms is plus, in the quotient it is minus, and the quotient also has a square in the denominator.

How do you differentiate a composite function like (2x + 1)⁵?

Use the chain rule: differentiate the outer function as if the inner one were a variable, then multiply by the inner derivative. For (2x + 1)⁵: first 5(2x + 1)⁴ (the power rule on the bracket), then multiply by the inner derivative which is 2, giving 10(2x + 1)⁴ altogether. Forgetting to multiply by the inner derivative is the most common mistake.

How do you find stationary points using the derivative?

Solve the equation f'(x) = 0 to find the candidate x-values, then check the sign of the derivative around each point: a change from positive to negative = a maximum, from negative to positive = a minimum. Finally substitute the x-values into the original function f(x) to get the y-coordinates of the points.

What is the most common mistake in differentiation questions?

Forgetting to multiply by the inner derivative in the chain rule, a wrong sign or a missing square in the denominator of the quotient rule, and forgetting to substitute back into the original function to find the y-values of the stationary points. These are all technical mistakes that lose marks even for those who understand the material — and the way to prevent them is repeated practice and self-checking.

A graded set of exercises — from the power rule to a full curve investigation in exam style

Practise differentiation and curve sketching ←

Links that might help

→ Algebra and functions — lessons and practice for every level → Integration and finding areas for A level — the complete guide → Sequences and series for A level → GCSE Higher algebra worksheets — the foundation for A level

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