Differentiating kx^n Revision Notes | GCSE AQA Higher Further Maths
kxⁿ is differentiated by multiplying the coefficient k by the power n, then reducing the power by one, giving knxⁿ⁻¹. Free notes and 11 questions answered.
Updated 14 Mar 2026 · Revise Right Team
Differentiating kx^n is part of Differentiation of kx^n and the sum of such functions in GCSE AQA Higher Further Maths.
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What this lesson covers
kxⁿ is differentiated by multiplying the coefficient k by the power n, then reducing the power by one, giving knxⁿ⁻¹. Free notes and 11 questions answered.
Apply the rule for differentiating $kx^n$ where $n$ is an integer to find derivatives of individual terms
Differentiate sums of functions of the form $kx^n$ by applying the rule term by term
Find the derivative of a polynomial expression and evaluate it at a given value of $x$
Identify and avoid common errors when differentiating, including handling constants and negative or fractional indices
Key ideas to keep in view
Use this lesson to stay inside Differentiation of kx^n and the sum of such functions while connecting the detail back to the wider Calculus topic.
Calculus
Differentiation of kx^n and the sum of such functions
GCSE AQA Higher Further Maths
Revision notes
Read the core explanations from Differentiating kx^n before testing yourself from memory.
1. Why Differentiation Matters
Imagine you are tracking a car's position over time. The position changes — but how fast is it changing right now? That question is exactly what differentiation answers. It turns a position function into a speed function, or a curve into a gradient at any chosen point.
At GCSE level, you will meet differentiation applied to expressions built from terms of the form $kx^n$, where $k$ is any constant and $n$ is an integer. The good news: there is one clean rule that handles every such term, and it works the same way whether the power is 2, 10, or $-3$.
2. The Power Rule for $kx^n$
Here is the core idea. When you differentiate $kx^n$, you multiply the coefficient $k$ by the power $n$, and then reduce the power by one. That single operation is the entire rule.
Two special cases are worth pinning down before you practise. A constant term like $7$ can be written as $7x^0$. Applying the rule: $7 \times 0 \times x^{-1} = 0$. So the derivative of any constant is zero — it contributes nothing to the gradient. A linear term like $5x$ is $5x^1$, so its derivative is $5 \times 1 \times x^0 = 5$. The gradient of a straight line is always constant, which matches what you already know.
3. Differentiating a Single Term — Worked Examples
The rule is quick to state, but seeing it applied step by step is what makes it stick. Work through each example below before revealing the next step.
Worked example: Differentiating $4x^3$
Find $\frac{dy}{dx}$ when $y = 4x^3$.
$k = 4$, $n = 3$ — Write down the coefficient and the power separately before applying the rule. [M1 — correct identification]
$k \times n = 4 \times 3 = 12$ — This becomes the new coefficient in the derivative.
$n - 1 = 3 - 1 = 2$ — The new power is 2.
$\frac{dy}{dx} = 12x^2$ — [A1 — correct final answer]
Answer: $\frac{dy}{dx} = 12x^2$
Worked example: Differentiating $x^5$
Find $f'(x)$ when $f(x) = x^5$.
$k = 1$, $n = 5$ — There is no written coefficient, so $k = 1$.
$1 \times 5 = 5$ — New coefficient is 5.
$5 - 1 = 4$ — New power is 4.
$f'(x) = 5x^4$ — [A1 — correct final answer]
Answer: $f'(x) = 5x^4$
Notice that when $k = 1$ the coefficient is invisible in the original expression, but you still multiply by it — it just gives you the power itself as the new coefficient. This is a common source of confusion, so always write $k = 1$ explicitly when you start.
4. Differentiating Sums of Functions
Most exam questions involve polynomials — expressions with several terms added together, such as $3x^4 - 2x^2 + 7$. The key principle here is that you differentiate each term independently and then collect the results.
This works because differentiation is a linear operation: the derivative of a sum is the sum of the derivatives. In practice, go through the expression term by term, apply the power rule to each, and write the results side by side.
Worked example: Differentiating a polynomial
Find $\frac{dy}{dx}$ when $y = 3x^4 - 2x^2 + 7$.
$\frac{d}{dx}(3x^4) = 3 \times 4 \times x^{4-1} = 12x^3$ — Multiply coefficient 3 by power 4 to get 12; reduce power from 4 to 3. [M1]
$\frac{d}{dx}(-2x^2) = -2 \times 2 \times x^{2-1} = -4x$ — Multiply $-2$ by 2 to get $-4$; reduce power from 2 to 1. Keep the negative sign.
$\frac{d}{dx}(7) = 0$ — Any constant differentiates to zero.
$\frac{dy}{dx} = 12x^3 - 4x + 0 = 12x^3 - 4x$ — [A1 — correct final answer with constant term removed]
Answer: $\frac{dy}{dx} = 12x^3 - 4x$
The graph above shows $y = 3x^4 - 2x^2 + 7$ (the original curve) alongside $\frac{dy}{dx} = 12x^3 - 4x$ (the derivative). Where the original curve is at a turning point — momentarily flat — the derivative curve crosses zero. That connection between the two graphs is exactly what differentiation captures.
5. Evaluating the Derivative at a Point
Finding $\frac{dy}{dx}$ gives you a general expression for the gradient at any $x$. To find the gradient at a specific point, substitute the $x$-value into your derivative expression.
Worked example: Finding the gradient at a specific point
Given $y = 2x^3 - 5x + 1$, find the gradient of the curve at $x = 2$.
$\frac{dy}{dx}\bigg|_{x=2} = 6(2)^2 - 5$ — Replace every $x$ with 2.
$2 \times 2 = 4$, so $6 \times 4 = 24$ — Always expand the square longhand to avoid errors.
$24 - 5 = 19$ — [A1 — gradient at $x = 2$ is 19]
Answer: Gradient at $x = 2$ is $19$.
6. Negative and Zero Powers
The power rule works for any integer $n$, including negative integers and zero. You apply exactly the same process: multiply by the power, then reduce the power by one.
Worked example: Differentiating with a negative power
Find $\frac{dy}{dx}$ when $y = 3x^{-2}$.
$k = 3$, $n = -2$ — The power is negative — the rule still applies.
$3 \times (-2) = -6$ — Multiplying a positive coefficient by a negative power gives a negative result.
$-2 - 1 = -3$ — Subtracting 1 from $-2$ gives $-3$, not $-1$.
$\frac{dy}{dx} = -6x^{-3}$ — [A1 — correct final answer]
Answer: $\frac{dy}{dx} = -6x^{-3}$
A common slip with negative powers is writing $-2 - 1 = -1$ instead of $-3$. Think of it on a number line: starting at $-2$ and moving one step further left lands you at $-3$.
7. Deliberate Practice
Work through these problems independently before checking the answers. For each one, write out the $k$ and $n$ values first — that habit prevents most errors.
8. Exam Technique and Common Mistakes
Differentiation questions at GCSE are reliable mark-earners — the method is always the same. The marks you lose are almost always down to a small number of predictable slips, so it is worth knowing exactly what they are.
Every differentiation question at GCSE reduces to this one rule applied repeatedly. Whether the expression has two terms or six, the approach is identical: work through each term, apply the power rule, and combine. Mastering this single technique covers the full specification requirement for differentiating $kx^n$ and sums of such functions.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-03-14.
The derivative of a constant is zero, because a horizontal line has zero gradient everywhere: 7 = 7x⁰, and 7 × 0 = 0. A linear term like 5x differentiates to 5, since 5 × 1 × x⁰ = 5, matching the constant gradient of a straight line.
How do you differentiate a polynomial?
Differentiate each term separately using the power rule, then combine the results. For y = 3x⁴ − 2x² + 7, 3x⁴ gives 12x³, −2x² gives −4x and the constant 7 gives 0, so dy/dx = 12x³ − 4x.
What is the derivative of 5x?
The derivative of 5x is 5. Writing 5x as 5x¹ and applying the rule gives 5 × 1 × x⁰ = 5, which matches the fact that a straight line has a constant gradient.
How do you differentiate kxⁿ?
kxⁿ is differentiated by multiplying the coefficient k by the power n, then reducing the power by one, giving knxⁿ⁻¹. For example, 4x³ differentiates to 4 × 3 × x² = 12x², and x⁵ (where k = 1) differentiates to 5x⁴.
How do you use differentiation to find the gradient at a point?
Differentiate to get dy/dx, then substitute the x-value. For y = 2x³ − 5x + 1, dy/dx = 6x² − 5, so at x = 2 the gradient is 6(2)² − 5 = 24 − 5 = 19.
How do you differentiate negative powers?
The power rule works the same for negative powers: multiply by the power, then reduce it by one. For y = 3x⁻², 3 × (−2) = −6 and −2 − 1 = −3, so dy/dx = −6x⁻³. A common slip is writing −2 − 1 = −1 instead of −3.
What is the derivative of 3x⁻²?
The derivative of 3x⁻² is −6x⁻³. Multiplying the coefficient by the power gives 3 × (−2) = −6, and reducing the power by one gives −2 − 1 = −3.
What is the most common mistake with negative powers in differentiation?
The most common mistake is writing −2 − 1 = −1 instead of −3 when reducing a negative power. On a number line, starting at −2 and moving one step further left lands at −3.
What is the gradient of y = 2x³ − 5x + 1 at x = 2?
The gradient of y = 2x³ − 5x + 1 at x = 2 is 19. Differentiating gives dy/dx = 6x² − 5, and substituting x = 2 gives 6 × 4 − 5 = 24 − 5 = 19.
Can you differentiate a polynomial term by term?
A polynomial is differentiated term by term, because differentiation is a linear operation: the derivative of a sum is the sum of the derivatives. For y = 3x⁴ − 2x² + 7, the terms give 12x³, −4x and 0, so dy/dx = 12x³ − 4x.
What is the derivative of x⁵?
The derivative of x⁵ is 5x⁴. There is no written coefficient, so k = 1; multiplying by the power gives 5, and reducing the power by one gives 4.