# Quotient Rule

## Quotient Rule Revision

**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.

**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}

**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}

**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.

## Quotient Rule Example Questions

**Question 1:** Using the quotient rule, show that the function y = \dfrac{1}{x^3} has derivative \dfrac{dy}{dx} = \dfrac{-3}{x^4}.

**[3 marks]**

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

Then

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

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

**Question 2:** For f(x) = \dfrac{2\sin x}{\cos x}, use the quotient rule to find its derivative with respect to x, and prove that \dfrac{2\sin x}{\cos x} = 2\tan x.

**[5 marks]**

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).

**Question 3: **Find the derivative (w.r.t x), of the function y = \dfrac{x^3}{3^x}. Verify that there is a stationary point at the origin.

**[5 marks]**

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.

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