# Differentiation

> CIE A-Level Mathematics · 9709 Pure Mathematics 2 & 3
> Source: https://www.owlsprep.com/study/cie-9709-u2-differentiation/

This subtopic covers core advanced differentiation techniques for CIE A-Level Pure 2 & 3, including product, quotient, chain rules, implicit and parametric differentiation, and rates of change, a high-weight exam topic. The derivative of $\tan^{-1}x$ is the only outcome here required for Pure 3 but not Pure 2.

**Prerequisites:** [Basic differentiation of polynomials and standard functions (P1)](https://www.owlsprep.com/study/cie-9709-u1-basic-differentiation/); Knowledge of exponential, logarithmic and trigonometric functions

## Learning objectives

- Differentiate standard advanced functions including exponentials, logarithms, and trigonometric functions
- Apply product, quotient and chain rules to combinations of functions
- Carry out implicit and parametric differentiation for non-explicit functions
- Solve rates of change problems using the chain rule
- Differentiate $\tan^{-1}x$ and composite forms $\tan^{-1}(ax+b)$ (Pure 3 only)

## How often is this tested?

Based on OwlsPrep's analysis of official CIE 9709 P2 past papers (2016–2025): **Differentiation** appears **257 times** in the last 10 years — **28.4%** of all P2 questions (257 of 455).

Most-tested forms: Product, quotient, and chain rule (extended) (130), Parametric differentiation; tangent and normal (58), Implicit differentiation (50), Derivatives of e^x, ln x, sin x, cos x, tan x (20), Related rates of change (1).

## Core Differentiation Rules

For combinations of functions, we use three core rules to find derivatives, building on basic differentiation of standard functions.

**Chain Rule** — Rule for differentiating composite functions $y = f(g(x))$: differentiate the outer function, multiply by derivative of the inner function.

*Notation:* \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

- Product Rule: For $y = uv$, $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$
- Quotient Rule: For $y = \frac{u}{v}$, $\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

**Worked example:** Differentiate $y = e^{2x} \sin(3x)$ with respect to $x$

1. Identify that $y$ is a product of two functions: $u = e^{2x}$, $v = \sin(3x)$
2. Differentiate $u$ and $v$ using the chain rule:
3. $$\frac{du}{dx} = 2e^{2x}, \quad \frac{dv}{dx} = 3\cos(3x)$$
4. Apply the product rule:
5. $$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} = e^{2x}(3\cos 3x) + \sin(3x)(2e^{2x})$$
6. Factor out the common term $e^{2x}$ to simplify:
7. $$\frac{dy}{dx} = e^{2x}(3\cos 3x + 2\sin 3x)$$

**Derivative of Inverse Tangent (Pure 3 only)** — This standard result is required for Pure 3 only. Composite form: $\frac{d}{dx}\left(\tan^{-1}(ax + b)\right) = \frac{a}{1 + (ax + b)^2}$. It is the only differentiation outcome in this topic examined in Pure 3 but not Pure 2.

*Notation:* \frac{d}{dx}\left(\tan^{-1} x\right) = \frac{1}{1 + x^2}

**Worked example:** Differentiate $y = \tan^{-1}(3x)$ with respect to $x$ (Pure 3 only)

1. Apply the composite inverse-tangent rule with $a = 3$, $b = 0$:
2. $$\frac{dy}{dx} = \frac{a}{1 + (ax + b)^2} = \frac{3}{1 + (3x)^2}$$
3. Simplify the denominator:
4. $$\frac{dy}{dx} = \frac{3}{1 + 9x^2}$$

> **info**
>
> The derivatives of $\sin^{-1} x$ and $\cos^{-1} x$ are not required by the 9709 syllabus; among the inverse trigonometric functions, only $\tan^{-1} x$ is examined (Pure 3 only).

> **Exam tip:** Always simplify and factor your final derivative; exam markers award full marks only for simplified expressions.

## Implicit Differentiation

Many relations between $x$ and $y$ cannot be rearranged to the explicit form $y = f(x)$. We use implicit differentiation to find $\frac{dy}{dx}$ for these cases.

**Implicit Differentiation** — Differentiate every term on both sides of the equation with respect to $x$, applying the chain rule to all terms containing $y$.

**Worked example:** Find $\frac{dy}{dx}$ at the point $(3, 4)$ for the circle $x^2 + y^2 = 25$

1. Differentiate each term with respect to $x$:
2. $$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)$$
3. Apply the chain rule to $y^2$: $\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}$
4. $$2x + 2y \frac{dy}{dx} = 0$$
5. Rearrange to isolate $\frac{dy}{dx}$:
6. $$\frac{dy}{dx} = -\frac{x}{y}$$
7. Substitute $x=3$, $y=4$:
8. $$\frac{dy}{dx} = -\frac{3}{4}$$

> **warning**
>
> You must add $\frac{dy}{dx}$ when differentiating any term with $y$; forgetting this is the most common mistake in implicit differentiation.

## Parametric Differentiation

When a curve is defined by parametric equations $x = x(t)$, $y = y(t)$, where $t$ is the parameter, we rearrange the chain rule to find $\frac{dy}{dx}$.

**Parametric Differentiation** — Find derivatives of $x$ and $y$ with respect to $t$, then divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ to get $\frac{dy}{dx}$.

*Notation:* \frac{dy}{dx} = \frac{dy/dt}{dx/dt}

**Worked example:** Find $\frac{dy}{dx}$ for the parametric curve $x = 2\cos t$, $y = 3\sin t$, $0 < t < \frac{\pi}{2}$

1. Differentiate $x$ and $y$ with respect to $t$:
2. $$\frac{dx}{dt} = -2\sin t, \quad \frac{dy}{dt} = 3\cos t$$
3. Apply the parametric differentiation formula:
4. $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3\cos t}{-2\sin t} = -\frac{3}{2}\cot t$$

> **Exam tip:** For the second derivative $\frac{d^2y}{dx^2}$, remember that $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}$, do not just differentiate $\frac{dy}{dx}$ directly with respect to $t$.

## Applications: Rates of Change

We use the chain rule to relate the rates of change of two quantities that both change with time. This is a common exam application of differentiation.

**Worked example:** The radius of a spherical balloon increases at $0.5 \text{ cm s}^{-1}$. Find the rate of increase of volume when radius is $10 \text{ cm}$.

1. Volume of a sphere is $V = \frac{4}{3}\pi r^3$. We know $\frac{dr}{dt} = 0.5$, need $\frac{dV}{dt}$.
2. Apply the chain rule:
3. $$\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}$$
4. Differentiate $V$ with respect to $r$:
5. $$\frac{dV}{dr} = 4\pi r^2$$
6. Substitute $r = 10$ and $\frac{dr}{dt} = 0.5$:
7. $$\frac{dV}{dt} = 4\pi (10)^2 (0.5) = 200\pi \text{ cm}^3\text{s}^{-1}$$

## Common pitfalls

- **Wrong:** Forgetting to multiply by the derivative of the inner function when using the chain rule
  - Why it fails: You only differentiate the outer function and miss the inner derivative factor
  - Correct: Always explicitly identify inner and outer functions, do not skip writing the inner derivative step
- **Wrong:** Mixing up the order of terms in the quotient rule
  - Why it fails: Most people incorrectly reverse the numerator terms
  - Correct: Remember the mnemonic: 'top derivative times bottom minus bottom derivative times top, all over bottom squared'
- **Wrong:** Forgetting the $\frac{dy}{dx}$ factor when differentiating $y$-terms in implicit differentiation
  - Why it fails: You accidentally treat $y$ as a constant instead of a function of $x$
  - Correct: Every time you differentiate a term containing $y$, add a $\frac{dy}{dx}$ factor from the chain rule
- **Wrong:** Calculating the second derivative for parametric curves by just differentiating $\frac{dy}{dx}$ with respect to $t$
  - Why it fails: You forget $\frac{dy}{dx}$ is still a function of $t$, not $x$
  - Correct: After differentiating $\frac{dy}{dx}$ with respect to $t$, divide the result by $\frac{dx}{dt}$ to get $\frac{d^2y}{dx^2}$
- **Wrong:** Using mismatched units in rates of change problems
  - Why it fails: You forget to convert all quantities to consistent units before starting calculations
  - Correct: Convert all lengths, times and volumes to matching units at the start of the problem

## Cheatsheet

| Rule | Formula |
| --- | --- |
| $e^x$ | $\frac{d}{dx}(e^x) = e^x$;  $\frac{d}{dx}(e^{ax+b}) = a e^{ax+b}$ |
| $\ln x$ | $\frac{d}{dx}(\ln x) = \frac{1}{x}$;  $\frac{d}{dx}(\ln(ax+b)) = \frac{a}{ax+b}$ |
| $\sin x$ | $\frac{d}{dx}(\sin x) = \cos x$ |
| $\cos x$ | $\frac{d}{dx}(\cos x) = -\sin x$ |
| $\tan x$ | $\frac{d}{dx}(\tan x) = \sec^2 x$ |
| $\tan^{-1} x$ (Pure 3 only) | $\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$;  $\frac{d}{dx}(\tan^{-1}(ax+b)) = \frac{a}{1+(ax+b)^2}$ |
| Chain Rule | $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ |
| Product Rule | $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$ |
| Quotient Rule | $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$ |
| Implicit Differentiation | Differentiate all terms, add $\frac{dy}{dx}$ for $y$-terms |
| Parametric First Derivative | $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ |
| Parametric Second Derivative | $\frac{d^2y}{dx^2} = \frac{d/dt(dy/dx)}{dx/dt}$ |
| Rates of Change | $\frac{dA}{dt} = \frac{dA}{dx} \cdot \frac{dx}{dt}$ |

## What's next

Differentiation is the foundation for almost all remaining calculus topics in CIE P3, including further integration techniques, solving differential equations, and analyzing complex curves. Questions on Paper 3 regularly combine differentiation with other topics like inverse trigonometric functions and exponential growth, so mastery of these techniques is critical for a high grade. Next, you will build on these core differentiation skills to find tangents, normals, stationary points, and solve optimisation problems, before moving on to integration, the inverse operation of differentiation that makes up a large portion of the P3 exam.

- [Integration](https://www.owlsprep.com/study/cie-9709-u2-integration/)
- [Numerical methods](https://www.owlsprep.com/study/cie-9709-u2-numerical-methods/)
- [3D Vector Geometry (Lines and Planes)](https://www.owlsprep.com/study/cie-9709-u2-vectors/)

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