# 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

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