Differentiation of Products and Quotients (2.4.2) | CIE A-Level Maths Notes

Differentiation enables the calculation of the instantaneous rate of change of functions. In particular, the product and quotient rules allow us to tackle functions expressed as the product or quotient of two other functions.

Introduction

In various situations, functions are intertwined through operations of multiplication or division. Understanding and applying the product and quotient rules is crucial for differentiating such compound expressions.

Product and Quotient Rules

Product Rule:

When two functions $u$ and $v$ are multiplied together:

\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}

Example:

Differentiate $x^2\ln(x)$ .

Solution:

u = x^2, \quad \frac{du}{dx} = 2x

v = \ln(x), \quad \frac{dv}{dx} = \frac{1}{x}

\frac{d}{dx}(x^2\ln(x)) = x^2 \cdot \frac{1}{x} + \ln(x) \cdot 2x = x + 2x\ln(x)

Quotient Rule:

When a function $u$ is divided by another function $v$ :

\frac{d}{dx} \left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}

Example:

Differentiate $\frac{x}{e^{1-x^2}}.$

Solution:

u = x, \quad \frac{du}{dx} = 1

v = e^{1-x^2}, \quad \frac{dv}{dx} = -2xe^{1-x^2}

\frac{d}{dx} \left(\frac{x}{e^{1-x^2}}\right) = \frac{e^{1-x^2} \cdot 1 - x \cdot (-2xe^{1-x^2})}{(e^{1-x^2})^2} = \frac{1 + 2x^2}{e^{x^2 - 1}}

Written by:

Dr Rahil Sachak-Patwa

Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

Oxford University - PhD Mathematics

CIE A-Level Maths Study Notes

2.4.2 Differentiation of Products and Quotients

Introduction

Product and Quotient Rules

Product Rule:

Example:

Quotient Rule:

Example:

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