CAIE A-Level · Mathematics 9709 · Differentiation

The Product and Quotient Rules

9 min readPreviewBy Uzair Khan

What you'll be able to do

Differentiate products and quotients, e.g. (2x − 4)/(3x² + 2), x² ln x, x e^(1 − x²).

Introduction

Many functions encountered in A-Level Mathematics cannot be differentiated by simple rules applied to a single function — they are built by multiplying or dividing two distinct functions together. Examples such as x2ln⁡xx^2 \ln x, xe1−x2x e^{1-x^2}, and 2x−43x2+2\dfrac{2x-4}{3x^2+2} appear regularly in 9709 Paper 3 examinations. The Product Rule and Quotient Rule are the two essential techniques that handle these cases systematically. Mastery of both rules — including knowing which to apply and how to simplify the result — is a core differentiable skill for Paper 3.


Core Concept

The Product Rule

When a function can be written as y=u(x) v(x)y = u(x)\,v(x) — a product of two differentiable functions — its derivative is not simply u′v′u'v'. Instead:

dydx=udvdx+vdudx\frac{\mathrm{d}y}{\mathrm{d}x} = u\frac{\mathrm{d}v}{\mathrm{d}x} + v\frac{\mathrm{d}u}{\mathrm{d}x}

In words: (first × derivative of second) + (second × derivative of first).

Identification: Look for two distinct function types multiplied together — e.g. a power of xx multiplied by exe^x, ln⁡x\ln x, sin⁡x\sin x, etc.

The Quotient Rule

When y=u(x)v(x)y = \dfrac{u(x)}{v(x)} — a quotient of two differentiable functions — the derivative is:

dydx=vdudx−udvdxv2\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{v\dfrac{\mathrm{d}u}{\mathrm{d}x} - u\dfrac{\mathrm{d}v}{\mathrm{d}x}}{v^2}

In words: (bottom × derivative of top) − (top × derivative of bottom), all over (bottom)².

Identification: A fraction in which both numerator and denominator contain xx (and cannot be trivially simplified to a single standard function).

Note: An alternative to the Quotient Rule is to rewrite uv\dfrac{u}{v} as u⋅v−1u \cdot v^{-1} and apply the Product Rule combined with the Chain Rule. Both methods are equally valid in examinations.


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