Quotient Rule
Quotient Rule
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}
A LevelTransition 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!
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 LevelExample 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 LevelExample 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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