# Quotient Rule

A LevelAQAEdexcelOCR

## Quotient Rule

You could use the Product Rule here, but that might get a little messy and a bit laborious. Here’s another rule which saves us a lot of time and effort.

A LevelAQAEdexcelOCR

## Quotient Rule Formula

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

$\dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}} = \dfrac{df(\textcolor{blue}{x})}{d\textcolor{blue}{x}} = \dfrac{v(\textcolor{blue}{x})u'(\textcolor{blue}{x}) - u(\textcolor{blue}{x})v'(\textcolor{blue}{x})}{(v(\textcolor{blue}{x}))^2}$

A LevelAQAEdexcelOCR
A LevelAQAEdexcelOCR

## Example 1: Using the Quotient Rule

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

[3 marks]

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

\begin{aligned}\dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}} &= \dfrac{e^\textcolor{blue}{x}\sin \textcolor{blue}{x} - e^\textcolor{blue}{x}\cos \textcolor{blue}{x}}{\sin ^2 \textcolor{blue}{x}}\\[1.2em]&=\dfrac{e^\textcolor{blue}{x}(\sin \textcolor{blue}{x} - \cos \textcolor{blue}{x})}{\sin ^2 \textcolor{blue}{x}}\\[1.2em]&=e^\textcolor{blue}{x}(\sin \textcolor{blue}{x} - \cos \textcolor{blue}{x}) \cosec^2 \textcolor{blue}{x} \end{aligned}

A LevelAQAEdexcelOCR

## Example 2: Using the Quotient Rule (with the Product Rule)

Let $\textcolor{limegreen}{y} = \dfrac{(\textcolor{blue}{x}^2 - 1)\ln \textcolor{blue}{x}}{\cos \textcolor{blue}{x}}$ 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 $(1, 0)$.

[5 marks]

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

$\dfrac{da}{d\textcolor{blue}{x}} = 2\textcolor{blue}{x}$ and $\dfrac{db}{d\textcolor{blue}{x}} = \dfrac{1}{\textcolor{blue}{x}}$

so

$\dfrac{du(\textcolor{blue}{x})}{d\textcolor{blue}{x}} = 2\textcolor{blue}{x}\ln \textcolor{blue}{x} + \dfrac{\textcolor{blue}{x}^2 - 1}{\textcolor{blue}{x}}$ and $\dfrac{dv(\textcolor{blue}{x})}{d\textcolor{blue}{x}} = -\sin \textcolor{blue}{x}$

This gives

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

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

$\dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}} = \dfrac{\cos 1((2 \times 0) + 0) + \sin 1 \times 0 \times 0}{\cos ^2 1}$

$= 0$

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

A LevelAQAEdexcelOCR

## Quotient Rule Example Questions

Let $u(x) = 1$ and $v(x) = x^3$.

Then

$\dfrac{dy}{dx} = \dfrac{0 - 3x^2}{x^6}$

$= \dfrac{-3}{x^4}$

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

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

$= 2\sec ^2 x$, by the identity $\sin ^2 x + \cos ^2 x \equiv 1$.

We also have

$d\dfrac{(2\tan x)}{dx} = 2\sec ^2 x$

Therefore, $\dfrac{df(x)}{dx} = \dfrac{d\tan x}{dx}$

This is an example of the Uniqueness Theorem. (See Product Rule, Q2).

Set $u = x^3$ and $v = 3^x$.

Then

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

Using the quotient rule, we have

$\dfrac{dy}{dx} = \dfrac{3^{x + 1}x^2 - x^3 3^x\ln 3}{3^{2x}}$

Set $x = 0$ to give

$\dfrac{dy}{dx} = \dfrac{(3 \times 0) - (0 \times 1 \times \ln 3)}{1}$

$= 0$

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

A Level

A Level

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