# Product Rule

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

## Product Rule

We use the product rule to find derivatives of functions which are (funnily enough), products of separate functions – we cannot simply differentiate our terms and multiply them together.

A Level   ## Product Rule Formula

For a function $\textcolor{limegreen}{y} = f(\textcolor{blue}{x}) = u(\textcolor{blue}{x})v(\textcolor{blue}{x})$, we have the derivative (with respect to $\textcolor{blue}{x}$) given by

$\dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}} = \dfrac{df(\textcolor{blue}{x})}{d\textcolor{blue}{x}} = u(\textcolor{blue}{x})\dfrac{dv(\textcolor{blue}{x})}{d\textcolor{blue}{x}} + v(\textcolor{blue}{x})\dfrac{du(\textcolor{blue}{x})}{d\textcolor{blue}{x}}$

A Level   ## Extensions to the Formula

Let’s now say that we have $\textcolor{limegreen}{y} = u(\textcolor{blue}{x})v(\textcolor{blue}{x})w(\textcolor{blue}{x})$, and we want to find derivative with respect to $\textcolor{blue}{x}$.

By setting $u(\textcolor{blue}{x})v(\textcolor{blue}{x}) = a(\textcolor{blue}{x})$, we have

$\dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}} = \dfrac{d(a(\textcolor{blue}{x})w(\textcolor{blue}{x}))}{d\textcolor{blue}{x}}$

$= a(\textcolor{blue}{x})\dfrac{dw(\textcolor{blue}{x})}{d\textcolor{blue}{x}} + \dfrac{da(\textcolor{blue}{x})}{d\textcolor{blue}{x}}w(\textcolor{blue}{x})$

$= \dfrac{du(\textcolor{blue}{x})}{d\textcolor{blue}{x}}v(\textcolor{blue}{x})w(\textcolor{blue}{x})+ u(\textcolor{blue}{x})\dfrac{dv(\textcolor{blue}{x})}{d\textcolor{blue}{x}}w(\textcolor{blue}{x}) + u(\textcolor{blue}{x})v(\textcolor{blue}{x})\dfrac{dw(\textcolor{blue}{x})}{d\textcolor{blue}{x}}$

This technique can be repeated endlessly for $n$ functions, so we have a linear combination of $n$ terms, where each term is the product of one differentiated function and all other functions.

A Level   A Level   ## Example 1: Using the Product Rule

Say we have the function $\textcolor{limegreen}{y} = e^{\textcolor{blue}{x}}\sin \textcolor{blue}{x}$. Find $\dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}}$.

[2 marks]

Let $u(\textcolor{blue}{x}) = e^\textcolor{blue}{x}$ and $v(\textcolor{blue}{x}) = \sin \textcolor{blue}{x}$. Then

$\dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}} = \dfrac{du(\textcolor{blue}{x})}{d\textcolor{blue}{x}}v(\textcolor{blue}{x}) + u(\textcolor{blue}{x})\dfrac{dv(\textcolor{blue}{x})}{d\textcolor{blue}{x}}$

$= e^\textcolor{blue}{x}\sin \textcolor{blue}{x} + e^\textcolor{blue}{x}\cos \textcolor{blue}{x} = e^\textcolor{blue}{x}(\sin \textcolor{blue}{x} + \cos \textcolor{blue}{x})$

A Level   ## Example 2: Using the Product Rule for Larger Functions

Let $\textcolor{limegreen}{y} = (\textcolor{blue}{x}^2 - 1)(\ln \textcolor{blue}{x})\cos \textcolor{blue}{x}$ where $\textcolor{blue}{x}$ is measured in radians. Find $\dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}}$ and verify that there is a stationary point at the point $(1, 0)$.

[4 marks]

Let $u(\textcolor{blue}{x}) = \textcolor{blue}{x}^2 - 1$, $v(\textcolor{blue}{x}) = \ln \textcolor{blue}{x}$ and $w(\textcolor{blue}{x}) = \cos \textcolor{blue}{x}$. Then

$\dfrac{du}{d\textcolor{blue}{x}} = 2\textcolor{blue}{x}$, $\dfrac{dv}{d\textcolor{blue}{x}} = \dfrac{1}{\textcolor{blue}{x}}$ and $\dfrac{dw}{d\textcolor{blue}{x}} = -\sin \textcolor{blue}{x}$

This gives

$\dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}} = (2\textcolor{blue}{x}\ln \textcolor{blue}{x} \cos \textcolor{blue}{x}) + \left( \dfrac{(\textcolor{blue}{x}^2 - 1)\cos \textcolor{blue}{x}}{\textcolor{blue}{x}}\right) - ((\textcolor{blue}{x}^2 - 1)\ln \textcolor{blue}{x}\sin \textcolor{blue}{x})$

When $\textcolor{blue}{x} = 1$,

$\dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}} = (2(1)\ln 1 \cos 1) + \left( \dfrac{0\cos 1}{1}\right) - \left( 0\ln 1 \sin 1\right)$

$= 0$

So, we can confirm that there is a stationary point at $\textcolor{blue}{x} = 1$.

A Level   ## Example Questions

Let $u(x) = x$ and $v(x) = x^2$.

Then

$\dfrac{dy}{dx} = 1x^2 + x2x$

$= x^2 + 2x^2$

$= 3x^2$

Let $u(x) = 2\sin x$ and $v(x) = \cos x$. Then

$\dfrac{df(x)}{dx} = 2\cos x \cos x - 2\sin x \sin x$

$= 2\cos 2x$, by double angle formulae.

We also have

$\dfrac{d(\sin 2x)}{dx} = 2\cos 2x$

Therefore, $\dfrac{d(\sin 2x)}{dx} = \dfrac{d(2\sin x \cos x)}{dx}$

This is an example of the Uniqueness Theorem. You won’t necessarily need this, but it’s an interesting proof, all the same.

Set $u = x^3$, $v = 3^x$ and $w = \tan x$.

Then

$u' = 3x^2$, $v' = 3^x\ln 3$ and $w' = \sec ^2 x$

Using the rule we learned for extended functions, we have

$\dfrac{dy}{dx} = (3x^2 3^x \tan x) + (x^3 3^x\ln 3 \tan x) + (x^3 3^x \sec ^2 x)$

Set $x = 0$ to give

$\dfrac{dy}{dx} = (0 \times 1 \times \tan 0) + (0 \times \ln 3 \times \tan 0) + (0 \times 1 \times \sec ^2 0)$

$= 0$

So, we can confirm there is a stationary point at the origin.

A Level

A Level

## You May Also Like... ### A Level Maths Revision Cards

The best A level maths revision cards for AQA, Edexcel, OCR, MEI and WJEC. Maths Made Easy is here to help you prepare effectively for your A Level maths exams.

£14.99 ### A Level Maths – Cards & Paper Bundle

A level maths revision cards and exam papers for Edexcel. Includes 2022 predicted papers based on the advance information released in February 2022! MME is here to help you study from home with our revision cards and practise papers.

From: £22.99 ### Transition Maths Cards

The transition maths cards are a perfect way to cover the higher level topics from GCSE whilst being introduced to new A level maths topics to help you prepare for year 12. Your ideal guide to getting started with A level maths!

£8.99