Differentiation (Pure 1) — A-Level Mathematics Pure 1 Study Guide

For: A-Level Mathematics candidates sitting Paper 1 (Pure Mathematics 1).

Covers: Derivative as a limit, power/constant multiple/sum rules, chain rule for composite functions, increasing/decreasing function intervals, stationary point classification, optimisation word problems, and connected rates of change.

You should already know: IGCSE / Add-Maths algebra, sketching basic curves, solving linear and simple quadratic equations.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Mathematics style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is Differentiation?

Differentiation is the mathematical process of calculating the gradient of a curve at any given point, and describing how a quantity changes relative to another. It is one of the highest-weighted topics in A-Level Mathematics Paper 1, making up 15-20% of total marks, and is often combined with coordinate geometry, functions, or mensuration in multi-part questions. Common notation for the derivative (gradient function) includes $f^{\prime }(x)$ (read as "f prime of x") and $\frac{dy}{dx}$ (read as "dy by dx") for a function $y=f(x)$.

2. Derivative as a limit — $f^{\prime }(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$

This formula is known as the first principles definition of the derivative, and it is the foundation of all differentiation rules. It works by calculating the gradient of a secant line connecting two points on the curve: $(x,f(x))$ and $(x+h,f(x+h))$, where $h$ is a very small increment in $x$. As $h$ approaches 0 (gets infinitely small without ever reaching 0), the secant line becomes the tangent to the curve at $x$, giving the exact gradient at that point.

Worked Example

Find the derivative of $f(x)=x^2$ from first principles.

1. Calculate $f(x+h)=(x+h{)}^2=x^2+2xh+h^2$
2. Find the difference: $f(x+h)-f(x)=2xh+h^2$
3. Divide by $h$: $\frac{f(x+h)-f(x)}{h}=2x+h$
4. Take the limit as $h\to 0$: $f^{\prime }(x)=2x$

Exam tip: Examiners regularly set 2-3 mark first principles questions for simple polynomials. You must show all four steps, including writing the limit statement, to score full marks.

3. Power rule, sum/difference, constant multiple

First principles is slow for complex functions, so we use these three standard rules for 90% of Paper 1 differentiation questions:

1. Power Rule: For any real constant $n$, $\frac{d}{dx}\left[x^n\right]=n{x}^{n-1}$. Constants differentiate to 0, since $\frac{d}{dx}[k]=\frac{d}{dx}[k{x}^0]=0$.
2. Constant Multiple Rule: For any constant $k$, $\frac{d}{dx}\left[kf(x)\right]=k{f}^{\prime }(x)$
3. Sum/Difference Rule: $\frac{d}{dx}\left[f(x)\pm g(x)\right]=f^{\prime }(x)\pm g^{\prime }(x)$

Worked Example

Differentiate $y=4x^3+2\sqrt{x}-\frac{3}{x^2}+7$

1. Rewrite all terms as powers of $x$: $y=4x^3+2x^{1/2}-3x^{-2}+7$
2. Differentiate term by term: $$\frac{dy}{dx}=4(3x^2)+2\left(\frac{1}{2}{x}^{-1/2}\right)-3(-2x^{-3})+0=12x^2+x^{-1/2}+6x^{-3}$$
3. Simplify if required: $\frac{dy}{dx}=12x^2+\frac{1}{\sqrt{x}}+\frac{6}{x^3}$

Exam tip: Always rewrite roots and reciprocals as powers of $x$ before applying the power rule to avoid arithmetic errors.

4. Composite functions — chain rule introduction

Composite functions are functions of functions, e.g. $y=(3x+2{)}^5$, where the inner function $u=3x+2$ is substituted into the outer function $y=u^5$. The chain rule lets you differentiate composite functions without expanding them: $$\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}$$ Or for written as $f(g(x))$, the derivative is $f^{\prime }(g(x))\times g^{\prime }(x)$.

Worked Example

Differentiate $y=\sqrt{2x^2-5}$

1. Let inner function $u=2x^2-5$, outer function $y=u^{1/2}$
2. Calculate separate derivatives: $\frac{dy}{du}=\frac{1}{2}{u}^{-1/2}$, $\frac{du}{dx}=4x$
3. Multiply and substitute back $u$: $$\frac{dy}{dx}=\frac{1}{2}(2x^2-5{)}^{-1/2}\times 4x=\frac{2x}{\sqrt{2x^2-5}}$$

Exam tip: For Paper 1, you only need to apply the chain rule for single composite functions. Product and quotient rules are only tested in Paper 3, so you do not need to learn them for this exam.

5. Increasing / decreasing intervals

A function is increasing on an interval if its gradient is non-negative for all points in that interval, and strictly increasing if the gradient is positive. Similarly, a function is decreasing if its gradient is non-positive, and strictly decreasing if the gradient is negative:

• Increasing: $f^{\prime }(x)\ge 0$
• Strictly increasing: $f^{\prime }(x)>0$
• Decreasing: $f^{\prime }(x)\le 0$
• Strictly decreasing: $f^{\prime }(x)<0$

To find intervals, first solve $f^{\prime }(x)=0$ to find critical points, then test values of $f^{\prime }(x)$ on either side of each critical point to determine its sign.

Worked Example

Find the intervals where $f(x)=x^3-6x^2+9x+1$ is strictly decreasing.

1. Differentiate: $f^{\prime }(x)=3x^2-12x+9=3(x-1)(x-3)$
2. Critical points at $x=1$ and $x=3$
3. Test values:

• $x=0$: $f^{\prime }(0)=9>0$ (positive, increasing)
• $x=2$: $f^{\prime }(2)=3(1)(-1)=-3<0$ (negative, decreasing)
• $x=4$: $f^{\prime }(4)=9>0$ (positive, increasing)

4. The function is strictly decreasing on the interval $1<x<3$

Exam tip: Examiners accept both inequality and interval notation, but be consistent. Omit critical points if the question asks for strictly increasing/decreasing.

6. Stationary points — finding and classifying (maximum, minimum, inflection)

Stationary points are points on a curve where the gradient is 0 ($f^{\prime }(x)=0$), so the tangent is horizontal. There are three types:

1. Local maximum: Gradient changes from positive to negative as $x$ increases
2. Local minimum: Gradient changes from negative to positive as $x$ increases
3. Stationary point of inflection: Gradient sign does not change, but the gradient itself reaches a maximum or minimum

You can classify stationary points using two methods:

• First derivative test: Test the sign of $f^{\prime }(x)$ just left and right of the stationary point
• Second derivative test: Differentiate $f^{\prime }(x)$ to get $f^{\prime \prime }(x)$. If $f^{\prime \prime }(a)<0$, $x=a$ is a maximum; if $f^{\prime \prime }(a)>0$, $x=a$ is a minimum; if $f^{\prime \prime }(a)=0$, use the first derivative test.

Worked Example

Find and classify the stationary points of $f(x)=2x^3+3x^2-12x+4$

1. $f^{\prime }(x)=6x^2+6x-12=6(x+2)(x-1)$, stationary points at $x=-2$ and $x=1$
2. Calculate coordinates: $f(-2)=24$, $f(1)=-3$, so points $(-2,24)$ and $(1,-3)$
3. Second derivative: $f^{\prime \prime }(x)=12x+6$

• $f^{\prime \prime }(-2)=-18<0$ → $(-2,24)$ is a local maximum
• $f^{\prime \prime }(1)=18>0$ → $(1,-3)$ is a local minimum

Exam tip: Never assume a point is an inflection point just because $f^{\prime \prime }(a)=0$. Always use the first derivative test to confirm the gradient sign does not change.

7. Optimisation — translating word problems into max/min

Optimisation uses stationary points to find the maximum or minimum value of a real-world quantity (e.g. maximum area, minimum cost) described in a word problem. Follow these steps:

1. Define variables for the quantities in the question
2. Write an expression for the quantity to optimise, then rearrange it to be in terms of a single variable using given constraints
3. Differentiate the expression, set the derivative equal to 0, and solve for the variable
4. Classify the stationary point (use the second derivative test or context of the question)
5. Substitute back to find the required value, and include units in your final answer

Worked Example

A rectangular garden is fenced on 3 sides, with the 4th side against a wall. 20m of fencing is used in total. Find the maximum possible area of the garden.

1. Let $x$ = width perpendicular to wall, $y$ = length parallel to wall. Constraint: $2x+y=20$ → $y=20-2x$
2. Area to maximise: $A=x\ast y=x(20-2x)=20x-2x^2$
3. $\frac{dA}{dx}=20-4x$, set to 0: $20-4x=0$ → $x=5m$
4. $\frac{d^{2}A}{d{x}^2}=-4<0$, so this is a maximum
5. $y=10m$, maximum area = $5\ast 10=50m^2$

Exam tip: You will lose 1 mark for missing units in your final answer, even if your calculation is fully correct.

8. Connected rates of change — $\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$

Rates of change are derivatives with respect to time $t$. The connected rates rule (a version of the chain rule) lets you calculate the rate of change of one quantity if you know the rate of change of a related quantity. Follow these steps:

1. Write down the given rate and the rate you need to find, including signs (negative for decreasing quantities, positive for increasing)
2. Write a formula linking the two quantities
3. Differentiate the formula with respect to $t$ using the chain rule
4. Substitute known values to find the unknown rate, and state if the quantity is increasing or decreasing

Worked Example

The radius of a sphere is increasing at 2 cm/s. Find the rate of increase of its volume when $r=5cm$. (Volume of sphere: $V=\frac{4}{3}\pi r^3$)

1. Given $\frac{dr}{dt}=+2cm/s$, need $\frac{dV}{dt}$
2. $V=\frac{4}{3}\pi r^3$
3. $\frac{dV}{dt}=\frac{dV}{dr}\times \frac{dr}{dt}=4\pi r^2\times \frac{dr}{dt}$
4. Substitute $r=5$, $\frac{dr}{dt}=2$: $\frac{dV}{dt}=4\pi (25)(2)=200\pi c{m}^3/s$

Exam tip: Always check for words like "leaking", "decreasing" or "cooling" — these mean the related rate is negative.

9. Common Pitfalls (and how to avoid them)

• Pitfall: Differentiating roots/reciprocals incorrectly, e.g. writing $\frac{d}{dx}[\sqrt{x}]=\frac{1}{\sqrt{x}}$ instead of $\frac{1}{2\sqrt{x}}$. Why? Students rush and skip rewriting terms as $x^n$ first. Correct move: Always rewrite all terms in the form $k{x}^n$ before applying the power rule.
• Pitfall: Forgetting to multiply by the inner derivative in the chain rule, e.g. writing $\frac{d}{dx}[(2x+1{)}^4]=4(2x+1{)}^3$ instead of $8(2x+1{)}^3$. Why? Students skip writing the inner $u$ function when practising. Correct move: Explicitly write inner and outer functions until you are fully confident with the rule.
• Pitfall: Assuming $f^{\prime \prime }(a)=0$ means a stationary point of inflection. Why? Students memorise the second derivative test but forget its limitations. Correct move: If $f^{\prime \prime }(a)=0$, test the sign of $f^{\prime }(x)$ on either side of the point to confirm the gradient sign does not change.
• Pitfall: Using positive rates for decreasing quantities in connected rates questions. Why? Students ignore context words like "leaking" or "shrinking". Correct move: Write the sign of the given rate as soon as you read the question, before starting calculations.
• Pitfall: Optimising the wrong quantity in word problems. Why? Students rush to solve equations without reading the question carefully. Correct move: Circle the quantity you are asked to maximise/minimise before you start working.

10. Practice Questions (Paper 1 Style)

Question 1

(a) Simplify $y=\frac{3x^4-2x^2+5x-1}{x}$ and find $\frac{dy}{dx}$. [2 marks] (b) Find the x-coordinates of the points on the curve $y=x^3-3x^2+2$ where the gradient is equal to 9. [3 marks]

Solution 1

(a) Simplify first: $y=3x^3-2x+5-x^{-1}$ Differentiate term by term: $\frac{dy}{dx}=9x^2-2+x^{-2}=9x^2-2+\frac{1}{x^2}$

(b) Gradient function: $\frac{dy}{dx}=3x^2-6x$ Set equal to 9: $3x^2-6x=9$ → $x^2-2x-3=0$ → $(x-3)(x+1)=0$ Solutions: $x=3$ and $x=-1$

Question 2

A closed cylindrical can has a fixed surface area of $54\pi c{m}^2$. Find the maximum possible volume of the can, giving your answer in terms of $\pi$. (Surface area of cylinder: $S=2\pi r^2+2\pi rh$, volume: $V=\pi r^{2}h$) [5 marks]

Solution 2

1. Use surface area constraint: $54\pi =2\pi r^2+2\pi rh$ → divide by $2\pi$: $27=r^2+rh$ → $h=\frac{27-r^2}{r}=\frac{27}{r}-r$
2. Rewrite volume in terms of $r$: $V=\pi r^{2}h=\pi r^2\left(\frac{27}{r}-r\right)=\pi (27r-r^3)$
3. Differentiate: $\frac{dV}{dr}=\pi (27-3r^2)$, set to 0: $27-3r^2=0$ → $r^2=9$ → $r=3cm$ (r>0)
4. Second derivative: $\frac{d^{2}V}{d{r}^2}=\pi (-6r)=-18\pi <0$ at $r=3$, so maximum
5. Substitute back: $V=\pi (27\ast 3-27)=54\pi c{m}^3$

Question 3

The side length of a cube is decreasing at a rate of 0.4 mm/s. Find the rate of change of the surface area of the cube when the side length is 12 mm, and state if the surface area is increasing or decreasing. [4 marks]

Solution 3

1. Let $x$ = side length, given $\frac{dx}{dt}=-0.4mm/s$ (negative because length is decreasing), need $\frac{dA}{dt}$ where $A$ is surface area
2. Surface area of cube: $A=6x^2$
3. Apply chain rule: $\frac{dA}{dt}=\frac{dA}{dx}\times \frac{dx}{dt}=12x\times \frac{dx}{dt}$
4. Substitute values: $\frac{dA}{dt}=12\ast 12\ast (-0.4)=-57.6m{m}^2/s$ The negative sign means the surface area is decreasing at $57.6m{m}^2/s$.

11. Quick Reference Cheatsheet

12. What's Next

The differentiation skills you have learned in this guide are foundational for almost all remaining topics in A-Level Mathematics. In Paper 1, you will use these skills to find equations of tangents and normals to curves, sketch polynomial functions with labelled stationary points, and solve mixed coordinate geometry problems. In Paper 3 Pure Mathematics, you will extend this knowledge to product rule, quotient rule, implicit differentiation, and differentiation of trigonometric, exponential, and logarithmic functions, all of which build directly on the core rules covered here.

If you are struggling with any part of this topic, whether it is setting up optimisation word problems, classifying stationary points, or applying the chain rule correctly, you can ask Ollie, our AI tutor, for personalised explanations, extra practice questions, or step-by-step walkthroughs of any problem type. You can also find more Paper 1 study guides and past paper practice on the homepage to test your knowledge and prepare for your exam.

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