Uni-Sine
Mathematics
+25XP
MathematicsCalculus

The quotient rule for differentiation

Published 7 min read

Quotient rule differentiation is how you differentiate one function divided by another. If y = u ÷ v, where u and v are both functions of x, then dy/dx = (v du/dx − u dv/dx) ÷ v². In words: bottom times the derivative of the top, minus top times the derivative of the bottom, all over the bottom squared. For example, y = x² ÷ (x − 2) gives dy/dx = ((x − 2)(2x) − x²(1)) ÷ (x − 2)² = (x² − 4x) ÷ (x − 2)².

The quotient rule formula

\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}The quotient rule

The same rule in function notation:

\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{\left(g(x)\right)^2}

AQA's AS and A-level formulae booklet gives the quotient rule in this function form, and Pearson Edexcel's specification expects skill in differentiating functions built from standard forms using products, quotients and composition.

How to remember the quotient rule

The order matters, because of the minus sign. A common memory aid is "low d-high minus high d-low, over low squared": the bottom (low) times the derivative of the top (high), minus the top times the derivative of the bottom, over the bottom squared. If you swap the two terms on top, you get the right answer with the wrong sign.

How to use the quotient rule, step by step

  1. Write the top as u and the bottom as v.
  2. Differentiate each: find du/dx and dv/dx.
  3. Substitute into (v du/dx − u dv/dx) ÷ v².
  4. Expand and simplify the top. Usually leave the bottom squared, not multiplied out.

Quotient rule examples

Example 1: a linear over a linear

Differentiate y = (2x + 1) ÷ (x − 3).

u = 2x + 1, so du/dx = 2. v = x − 3, so dv/dx = 1.

\frac{dy}{dx} = \frac{(x - 3)(2) - (2x + 1)(1)}{(x - 3)^2} = \frac{2x - 6 - 2x - 1}{(x - 3)^2} = \frac{-7}{(x - 3)^2}

The gradient is always negative, so the curve is always decreasing on each side of x = 3.

Example 2: a quadratic over a linear

Differentiate y = x² ÷ (x − 2).

u = x², du/dx = 2x. v = x − 2, dv/dx = 1.

\frac{dy}{dx} = \frac{(x - 2)(2x) - x^2}{(x - 2)^2} = \frac{x^2 - 4x}{(x - 2)^2}

Example 3: an exponential

Differentiate y = eˣ ÷ x.

u = eˣ, du/dx = eˣ. v = x, dv/dx = 1.

\frac{dy}{dx} = \frac{xe^x - e^x}{x^2} = \frac{e^x(x - 1)}{x^2}

Factorising eˣ out of the top shows straight away that the gradient is zero when x = 1.

Example 4: the derivative of tan x

Write tan x = sin x ÷ cos x. u = sin x, du/dx = cos x. v = cos x, dv/dx = −sin x.

\frac{d}{dx}(\tan x) = \frac{\cos x \cos x - \sin x(-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x

The same method gives the derivatives of sec x, cosec x and cot x.

Example 5: a gradient at a point

Find the gradient of y = (3x − 2) ÷ (x² + 1) at x = 1.

u = 3x − 2, du/dx = 3. v = x² + 1, dv/dx = 2x. At x = 1: u = 1, v = 2, du/dx = 3 and dv/dx = 2.

\frac{dy}{dx} = \frac{2 \times 3 - 1 \times 2}{2^2} = \frac{4}{4} = 1

Substituting the numbers before simplifying is quicker when you only need the gradient at one point.

Example 6: stationary points

Find the stationary points of y = x ÷ (x² + 4).

u = x, du/dx = 1. v = x² + 4, dv/dx = 2x.

\frac{dy}{dx} = \frac{(x^2 + 4) - x(2x)}{(x^2 + 4)^2} = \frac{4 - x^2}{(x^2 + 4)^2}

A fraction is zero when its top is zero, so 4 − x² = 0 and x = ±2. The stationary points are (2, ¼) and (−2, −¼). The bottom of the fraction is always positive, so only the top decides the sign of the gradient.

Now try a question in the course. Choose the right rule and differentiate, then press Mark to check it there.

Differentiate using the chain, product, or quotient rule.

Press Mark to have your answer marked in the Calculus course.

The Derivative Calculator differentiates any quotient and shows which rule it used at each step, so you can check your working.

Try it

Derivative Calculator

Differentiate any function and see each rule applied

Go to its page

Quotient rule vs product rule

Any quotient u ÷ v can be rewritten as a product, u × v⁻¹, and differentiated with the product rule and the chain rule. Both routes give the same answer. For y = x² ÷ (x − 2), the product rule gives 2x(x − 2)⁻¹ − x²(x − 2)⁻², which simplifies to the quotient rule answer once you put it over a common denominator.

UseWhen
Quotient ruleThe function is clearly one expression divided by another
Product ruleTwo expressions are multiplied together
Chain ruleOne function is inside another, such as (3x + 1)⁵
Simplify firstThe bottom is a single term, such as (x³ + 2x) ÷ x = x² + 2

Our guides to the product rule and the chain rule cover the other two.

Common mistakes

  • The wrong order on top. It is v du/dx − u dv/dx. Swapping gives the negative of the right answer.
  • Forgetting to square the bottom. The denominator is v², not v.
  • Expanding the denominator. (x + 1)² is usually best left as it is.
  • Losing brackets on the minus. In Example 1, −(2x + 1) becomes −2x − 1, not −2x + 1.
  • Using the quotient rule when you can simplify. (x³ + 2x) ÷ x is just x² + 2.

Quotient rule questions with answers

1. Differentiate y = x ÷ (x + 2).
((x + 2)(1) − x(1)) ÷ (x + 2)² = 2 ÷ (x + 2)²
2. Differentiate y = (x + 5) ÷ (x − 1).
((x − 1) − (x + 5)) ÷ (x − 1)² = −6 ÷ (x − 1)²
3. Differentiate y = sin x ÷ x.
(x cos x − sin x) ÷ x²
4. Find the gradient of y = (x + 1) ÷ x² at x = 1.
dy/dx = (x² − 2x(x + 1)) ÷ x⁴ = (−x² − 2x) ÷ x⁴. At x = 1: −3
5. Show that the derivative of cot x is −cosec² x.
cot x = cos x ÷ sin x, so dy/dx = (sin x(−sin x) − cos x cos x) ÷ sin² x = −1 ÷ sin² x = −cosec² x

Practise calculus in a short timed paper

A-level Maths: Calculus gives you a short timed paper every day, with 8 questions worth 25 to 35 marks across the course, so differentiation rules turn up regularly but not in every paper. The Calculus course builds up the rules one at a time.

Paper resets in

Common questions

What is the quotient rule?
A rule for differentiating a fraction of two functions: d/dx(u ÷ v) = (v du/dx − u dv/dx) ÷ v².
How do you remember the quotient rule?
"Low d-high minus high d-low, over low squared." Low is the bottom, high is the top, and d means "the derivative of".
When should you use the quotient rule?
When the function is one expression divided by another and you cannot simplify it into separate terms first.
Is the quotient rule in the formula booklet?
In AQA's AS and A-level formulae booklet, yes. Check your own board's booklet.
Can you use the product rule instead of the quotient rule?
Yes. Write u ÷ v as u × v⁻¹ and use the product and chain rules. The answer is the same after simplifying.
What is the derivative of tan x?
sec² x, found with the quotient rule by writing tan x as sin x ÷ cos x.

Revision cards

Say the answer first, then flip the card to check it.

Where to go next

  • Solve it

    Derivative Calculator

    Differentiate any function and see each rule applied

    Open the tool
  • Learn it

    Calculus

    Differentiation, integration and the mathematics that drives physics and engineering

    Start the course
  • Test it

    A-level Maths: Calculus

    A new exam-style paper every day, with instant marking and a mark scheme

    Start today's paper
Read next
Maths · CalculusProduct rule: formula, examples and how to use itThe product rule formula, how to differentiate a product in four steps, 6 worked examples, the proof, practice questions and what the booklet gives.