Differentiation (Pure 3) — A-Level Mathematics Pure 3 Study Guide

For: A-Level Mathematics candidates sitting Paper 3 (Pure Mathematics 3).

Covers: Product, quotient and chain rules, standard derivatives of exponential, logarithmic and trigonometric functions, implicit differentiation, parametric differentiation, and stationary points for implicit and parametric curves.

You should already know: A-Level Mathematics Pure 1 (functions, calculus, trig).

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 (Pure 3)?

Differentiation in Pure 3 extends the basic calculus toolkit you learned in P1 to handle non-explicit function forms, including composites of exponential, logarithmic and trigonometric functions, implicit curves where y is not isolated as a function of x, and parametric curves defined with a third independent variable. This topic accounts for 15-20% of total marks on Paper 3, and is a foundational skill for integration, differential equations, and applied calculus topics in mechanics and statistics papers.

2. Product, Quotient and Chain Rules

You first learned these three core differentiation rules in P1, but in P3 you will apply them to nested, mixed function types, so it is critical to memorise and apply them accurately.

• Product Rule: For a function written as the product of two functions of x, $y=u(x)v(x)$: $$\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$$
• Quotient Rule: For a function written as a fraction of two functions of x, $y=\frac{u(x)}{v(x)}$: $$\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$$
• Chain Rule: For a composite function $y=f(g(x))$, substitute $u=g(x)$ so $y=f(u)$: $$\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}$$

Worked Example

Differentiate $y=e^{2x}\sin (3x)$ with respect to x.

1. Label $u=e^{2x}$ and $v=\sin (3x)$.
2. Apply chain rule to find individual derivatives: $\frac{du}{dx}=2e^{2x}$, $\frac{dv}{dx}=3\cos (3x)$.
3. Substitute into product rule: $$\frac{dy}{dx}=e^{2x}(3\cos (3x))+\sin (3x)(2e^{2x})=e^{2x}\left(3\cos (3x)+2\sin (3x)\right)$$

Exam tip: Always explicitly label u and v before applying product or quotient rules to avoid mixing up terms, which is one of the most common mark-losing mistakes on P3 calculus questions.

3. Derivatives of $e^x,\ln x,\sin x,\cos x,\tan x$

The table below lists standard derivatives for core function types, including their composite forms, which you must memorise for the exam. All trigonometric derivatives apply only when the argument is in radians, which is the required unit for all A-Level Mathematics calculus questions.

Worked Example

Differentiate $y=\ln (\cos (4x))$ for $0<x<\frac{\pi }{8}$.

1. Substitute $u=\cos (4x)$ so $y=\ln (u)$.
2. Find derivatives: $\frac{dy}{du}=\frac{1}{u}$, $\frac{du}{dx}=-4\sin (4x)$.
3. Apply chain rule: $$\frac{dy}{dx}=\frac{1}{\cos (4x)}\times (-4\sin (4x))=-4\tan (4x)$$

4. Implicit Differentiation — finding $\frac{dy}{dx}$ from $F(x,y)=0$

Implicit functions are curves where y cannot easily be rearranged to be a standalone function of x, e.g., $x^2+3xy+y^3=5$. To differentiate these functions:

1. Differentiate every term on both sides of the equation with respect to x.
2. When differentiating a term containing y, apply the chain rule and multiply by $\frac{dy}{dx}$ (since y is a function of x).
3. Collect all terms containing $\frac{dy}{dx}$ on one side of the equation, factor out $\frac{dy}{dx}$, and rearrange to solve for the derivative.

Worked Example

Find $\frac{dy}{dx}$ for the curve $x^2+3xy+y^3=5$.

1. Differentiate each term:

• $\frac{d}{dx}(x^2)=2x$
• $\frac{d}{dx}(3xy)=3x\frac{dy}{dx}+3y$ (product rule applied)
• $\frac{d}{dx}(y^3)=3y^2\frac{dy}{dx}$
• $\frac{d}{dx}(5)=0$

2. Combine terms: $2x+3x\frac{dy}{dx}+3y+3y^2\frac{dy}{dx}=0$
3. Collect and factor $\frac{dy}{dx}$: $$\frac{dy}{dx}(3x+3y^2)=-2x-3y$$ $$\frac{dy}{dx}=\frac{-2x-3y}{3(x+y^2)}$$

5. Parametric Differentiation — $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$

Parametric curves define x and y as separate functions of a third parameter, usually t or θ, e.g., $x=3\cos \theta$, $y=2\sin \theta$. To find the gradient of a parametric curve, rearrange the chain rule to eliminate the parameter: $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ Note that this is only valid when $\frac{dx}{dt}\ne 0$ (to avoid division by zero). The resulting derivative will usually be expressed in terms of the parameter, not x and y, unless the question explicitly asks you to eliminate the parameter.

Worked Example

A curve is defined parametrically by $x=3\cos \theta$, $y=2\sin \theta$ for $0\le \theta <2\pi$. Find the gradient of the curve at $\theta =\frac{\pi }{4}$.

1. Differentiate x and y with respect to θ: $$\frac{dx}{d\theta }=-3\sin \theta ,\frac{dy}{d\theta }=2\cos \theta$$
2. Apply parametric differentiation rule: $$\frac{dy}{dx}=\frac{2\cos \theta }{-3\sin \theta }=-\frac{2}{3}\cot \theta$$
3. Substitute $\theta =\frac{\pi }{4}$: $\cot \left(\frac{\pi }{4}\right)=1$, so $\frac{dy}{dx}=-\frac{2}{3}$.

6. Stationary Points and Curvature in Implicit / Parametric Form

Stationary points on any curve occur where $\frac{dy}{dx}=0$ (for horizontal tangents, maxima or minima) or $\frac{dy}{dx}$ is undefined (for vertical tangents):

• For implicit curves: Set the numerator of your $\frac{dy}{dx}$ expression equal to 0, solve for (x,y) pairs, and verify the pair satisfies the original implicit equation.
• For parametric curves: Set $\frac{dy}{dt}=0$ (since $\frac{dy}{dx}=0$ only when the numerator is 0, as long as $\frac{dx}{dt}\ne 0$), solve for the parameter value, then find the corresponding x and y coordinates.
• To find the nature of a stationary point, use the second derivative test: if $\frac{d^{2}y}{d{x}^2}>0$, the point is a minimum; if $\frac{d^{2}y}{d{x}^2}<0$, it is a maximum. For parametric curves, calculate the second derivative using: $$\frac{d^{2}y}{d{x}^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$$

Worked Example

Find the stationary points of the parametric curve $x=3t+\frac{1}{t}$, $y=t^2-\ln t$ for $t>0$, and determine their nature.

1. Calculate first derivatives: $\frac{dx}{dt}=3-\frac{1}{t^2}$, $\frac{dy}{dt}=2t-\frac{1}{t}$.
2. Set $\frac{dy}{dt}=0$: $2t-\frac{1}{t}=0⟹2t^2=1⟹t=\frac{1}{\sqrt{2}}$ (since $t>0$).
3. Check $\frac{dx}{dt}\ne 0$ at $t=\frac{1}{\sqrt{2}}$: $\frac{dx}{dt}=3-2=1\ne 0$, so the point is valid.
4. Find coordinates: $x=3\left(\frac{1}{\sqrt{2}}\right)+\sqrt{2}=\frac{5}{\sqrt{2}}$, $y=\frac{1}{2}-\ln \left(\frac{1}{\sqrt{2}}\right)=\frac{1}{2}+\frac{1}{2}\ln 2$.
5. Calculate second derivative: $$\frac{dy}{dx}=\frac{2t-\frac{1}{t}}{3-\frac{1}{t^2}}=\frac{t(2t^2-1)}{3t^2-1}$$ $$\frac{d}{dt}\left(\frac{dy}{dx}\right){|}_{t=\frac{1}{\sqrt{2}}}=4$$ $$\frac{d^{2}y}{d{x}^2}=\frac{4}{1}=8>0$$
6. Conclusion: The stationary point at $\left(\frac{5}{\sqrt{2}},\frac{1+\ln 2}{2}\right)$ is a minimum.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Mixing up the order of terms in the quotient rule, writing $u\frac{dv}{dx}-v\frac{du}{dx}$ as the numerator. Why it happens: Misremembering the rule structure. Correct move: Explicitly label u (top function) and v (bottom function) before substituting, and remember the denominator term comes first in the numerator: "v du minus u dv".
• Wrong move: Flipping the fraction in parametric differentiation, writing $\frac{dx/dt}{dy/dt}$ for $\frac{dy}{dx}$. Why it happens: Confusion rearranging the chain rule. Correct move: Use the mnemonic "Y comes before X, so Y goes on top", or write $\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}$ to confirm the order.
• Wrong move: Forgetting to multiply by $\frac{dy}{dx}$ when differentiating y terms in implicit differentiation, e.g., writing $\frac{d}{dx}(y^2)=2y$ instead of $2y\frac{dy}{dx}$. Why it happens: Habit of differentiating with respect to y instead of x. Correct move: Add a $\frac{dy}{dx}$ factor immediately every time you differentiate a function of y.
• Wrong move: Using degrees instead of radians for trigonometric derivatives. Why it happens: Muscle memory from pre-A Level trigonometry. Correct move: Set your calculator to radian mode before starting the exam, and confirm all trigonometric arguments are in radians for calculus questions.
• Wrong move: Failing to verify stationary point candidates are valid. Why it happens: Rushing to solve $\frac{dy}{dx}=0$ without checking constraints. Correct move: For implicit curves, substitute (x,y) pairs back into the original equation to confirm they lie on the curve; for parametric curves, confirm $\frac{dx}{dt}\ne 0$ at the parameter value.

8. Practice Questions (Paper 3 Style)

Question 1

Differentiate the following functions with respect to x, simplifying your answers where possible: a) $y=e^{3x}\cos (2x)$ b) $y=\frac{\ln (x^2+1)}{\tan x}$ for $x>0$ and $0<x<\frac{\pi }{2}$

Solution 1

a) Use product rule: $u=e^{3x}$, $\frac{du}{dx}=3e^{3x}$; $v=\cos (2x)$, $\frac{dv}{dx}=-2\sin (2x)$ $$\frac{dy}{dx}=e^{3x}(-2\sin 2x)+\cos 2x(3e^{3x})=e^{3x}(3\cos 2x-2\sin 2x)$$

b) Use quotient rule: $u=\ln (x^2+1)$, $\frac{du}{dx}=\frac{2x}{x^2+1}$; $v=\tan x$, $\frac{dv}{dx}={\sec }^{2}x$ $$\frac{dy}{dx}=\frac{\tan x\times \frac{2x}{x^2+1}-\ln (x^2+1){\sec }^{2}x}{{\tan }^{2}x}$$

Question 2

A curve is defined implicitly by $x\ln y+y^2=2x$ for $y>0$. a) Find $\frac{dy}{dx}$ in terms of x and y. b) Find the gradient of the curve at the point (1, 1).

Solution 2

a) Differentiate term by term: $$\ln y+\frac{x}{y}\frac{dy}{dx}+2y\frac{dy}{dx}=2$$ Collect and rearrange: $$\frac{dy}{dx}\left(\frac{x}{y}+2y\right)=2-\ln y$$ $$\frac{dy}{dx}=\frac{y(2-\ln y)}{x+2y^2}$$

b) Substitute x=1, y=1 (note $\ln 1=0$): $$\frac{dy}{dx}=\frac{1(2-0)}{1+2(1{)}^2}=\frac{2}{3}$$

Question 3

A parametric curve is given by $x=t+e^{-t}$, $y=t-e^{-t}$ for $t\in \mathbb{R}$. Find the exact value of $\frac{d^{2}y}{d{x}^2}$ at $t=0$, and state the curvature of the curve at this point.

Solution 3

First derivatives: $\frac{dx}{dt}=1-e^{-t}$, $\frac{dy}{dt}=1+e^{-t}$ $$\frac{dy}{dx}=\frac{1+e^{-t}}{1-e^{-t}}=\frac{e^t+1}{e^t-1}$$ Differentiate $\frac{dy}{dx}$ with respect to t: $$\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{e^t(e^t-1)-e^t(e^t+1)}{(e^t-1{)}^2}=\frac{-2e^t}{(e^t-1{)}^2}$$ Second derivative: $$\frac{d^{2}y}{d{x}^2}=\frac{\frac{-2e^t}{(e^t-1{)}^2}}{1-e^{-t}}=\frac{-2e^{2t}}{(e^t-1{)}^3}$$ At t=0: $\frac{d^{2}y}{d{x}^2}=\frac{-2}{0}$ which is undefined, so the curve has a vertical tangent at t=0.

9. Quick Reference Cheatsheet

10. What's Next

Mastery of P3 differentiation is a non-negotiable prerequisite for two high-weight topics later in the A-Level Mathematics syllabus: integration (including integration by substitution, by parts, and solving first-order differential equations) and kinematics/optimisation problems in Mechanics Paper 4. Complex differentiation questions also appear regularly in Section B of Paper 3, which contains extended response questions worth 10-12 marks each, so practicing these rules until they are automatic will directly boost your exam score.

If you struggle with any of the rules, worked examples, or practice questions in this guide, you can ask Ollie for step-by-step clarification, extra practice problems, or custom quizzes tailored to your weak spots at any time. Head to the homepage to start practicing with AI-generated, exam-aligned questions for A-Level Mathematics Paper 3, and track your progress to target the marks you need for your university offer.

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