The Quotient Rule
Master the quotient rule for Queensland Year 11 Mathematical Methods (QCAA). It differentiates one function divided by another using a set formula built from the top, the bottom and their derivatives.
You will learn to apply the rule to ratios of power and polynomial functions, keep the numerator in order, give the derivative in simplest and factorised form, and use it to find where a curve has a horizontal tangent.
Every question with a fully worked solution.
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Theory
In Year 11 Mathematical Methods (QCAA, Unit 2), the quotient rule differentiates a ratio of two functions: \(\left(\dfrac{u}{v}\right)'=\dfrac{v\,u'-u\,v'}{v^{2}}\). This page applies it to ratios of power and polynomial functions, shows how to keep the numerator in the correct order, simplify it, and use the derivative to find gradients and horizontal tangents.
The quotient rule gives the derivative of a ratio \(y=\dfrac{u}{v}\), where \(u\) (the numerator) and \(v\) (the denominator) are both functions of \(x\). It is \(\dfrac{dy}{dx}=\dfrac{v\,u'-u\,v'}{v^{2}}\).
The order in the numerator matters: it is \(v\,u'\) minus \(u\,v'\), not the other way round. Reversing it is the single most common error. The denominator is always \(v^{2}\), the square of the original denominator.
To use it, identify \(u\) and \(v\), write \(u'\) and \(v'\), substitute, then simplify the numerator. Setting the numerator to zero (with \(v\neq 0\)) locates any horizontal tangents.
The quotient rule, with \(u\) and \(v\) functions of \(x\):
A horizontal tangent occurs where the numerator is zero (with \(v\neq0\)):
How to use the quotient rule
- Split: write the function as \(\dfrac{u}{v}\) and state \(u\) and \(v\).
- Differentiate each part: find \(u'\) and \(v'\).
- Substitute in order: put them into \(\dfrac{v\,u'-u\,v'}{v^{2}}\), keeping \(v\,u'\) first.
- Simplify / use: expand and simplify the numerator, then evaluate at a point or set the numerator to zero as required.
Identify \(u,\,v\) and their derivatives:
| \(u\) | \(=\) | \(x,\quad u'=1\) |
| \(v\) | \(=\) | \(x+1,\quad v'=1\) |
Substitute into \(\dfrac{v\,u'-u\,v'}{v^{2}}\):
| \(\dfrac{dy}{dx}\) | \(=\) | \(\dfrac{(x+1)(1)-x(1)}{(x+1)^{2}}\) |
Simplify the numerator:
| \(=\) | \(\dfrac{x+1-x}{(x+1)^{2}}\) | |
| \(=\) | \(\dfrac{1}{(x+1)^{2}}\) |
\(\dfrac{dy}{dx}=\dfrac{1}{(x+1)^{2}}\).
Identify \(u,\,v\) and their derivatives:
| \(u\) | \(=\) | \(2x-1,\quad u'=2\) |
| \(v\) | \(=\) | \(x+3,\quad v'=1\) |
Substitute in the correct order:
| \(\dfrac{dy}{dx}\) | \(=\) | \(\dfrac{(x+3)(2)-(2x-1)(1)}{(x+3)^{2}}\) |
Expand and simplify the numerator:
| \(=\) | \(\dfrac{2x+6-2x+1}{(x+3)^{2}}\) | |
| \(=\) | \(\dfrac{7}{(x+3)^{2}}\) |
\(\dfrac{dy}{dx}=\dfrac{7}{(x+3)^{2}}\).
Identify \(u,\,v\) and their derivatives:
| \(u\) | \(=\) | \(x^{2},\quad u'=2x\) |
| \(v\) | \(=\) | \(x-1,\quad v'=1\) |
Substitute into the quotient rule:
| \(\dfrac{dy}{dx}\) | \(=\) | \(\dfrac{(x-1)(2x)-x^{2}(1)}{(x-1)^{2}}\) |
Expand, simplify and factorise the numerator:
| \(=\) | \(\dfrac{2x^{2}-2x-x^{2}}{(x-1)^{2}}\) | |
| \(=\) | \(\dfrac{x^{2}-2x}{(x-1)^{2}}\) | |
| \(=\) | \(\dfrac{x(x-2)}{(x-1)^{2}}\) |
\(\dfrac{dy}{dx}=\dfrac{x(x-2)}{(x-1)^{2}}\).
Identify \(u,\,v\) and their derivatives:
| \(u\) | \(=\) | \(x,\quad u'=1\) |
| \(v\) | \(=\) | \(x^{2}+1,\quad v'=2x\) |
Substitute and simplify the numerator:
| \(\dfrac{dy}{dx}\) | \(=\) | \(\dfrac{(x^{2}+1)(1)-x(2x)}{(x^{2}+1)^{2}}\) |
| \(=\) | \(\dfrac{1-x^{2}}{(x^{2}+1)^{2}}\) |
Set the numerator to \(0\) (denominator is never zero):
| \(1-x^{2}\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\pm 1\) |
\(y\)-coordinates from \(y=\dfrac{x}{x^{2}+1}\):
| \(y\) | \(=\) | \(\dfrac{1}{2}\ \text{ or }\ -\dfrac{1}{2}\) |
Horizontal tangents at \(\left(1,\tfrac12\right)\) and \(\left(-1,-\tfrac12\right)\).
Common pitfalls
Frequently asked questions
What is the quotient rule?
For \(y=\dfrac{u}{v}\), the derivative is \(\dfrac{v\,u'-u\,v'}{v^{2}}\): denominator times derivative of the numerator, minus numerator times derivative of the denominator, all over the denominator squared.
What order does the numerator go in?
It is \(v\,u'-u\,v'\), in that order. Reversing it to \(u\,v'-v\,u'\) is the most common mistake and changes the sign.
Why is the denominator squared?
The quotient rule always divides by \(v^{2}\), the square of the original denominator; it comes from differentiating \(u\,v^{-1}\).
How do I find where a rational function has a horizontal tangent?
Differentiate with the quotient rule, then set the numerator equal to zero (the denominator cannot be zero) and solve for \(x\).
Do I need the second derivative to classify the stationary point?
Not in Year 11 for this topic — you locate where \(f'(x)=0\); classifying its nature by \(f''\) is beyond this scope.