The Quotient Rule

Differentiating one function divided by another

SPEC 7.4 • DIFFERENTIATION

Why We Need the Quotient Rule

When we have a function divided by another function, we cannot simply differentiate the top and bottom separately.

Common Mistake

If

y = \frac{f ( x )}{g ( x )}

, then

\frac{d y}{d x} \neq = \frac{f ^{'} ( x )}{g ^{'} ( x )}

For example:

y = \frac{x ^{2}}{x} = x

, so

\frac{d y}{d x} = 1

But

\frac{2 x}{1} = 2 x \neq = 1

— completely wrong!

Key Insight

The quotient rule handles the interaction between the numerator and denominator when both are changing.

The Quotient Rule Formula

Quotient Rule

If

y = \frac{u}{v}

where

u

and

v

are functions of

x

, then:

\frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v ^{2}}

Or using prime notation:

Alternative Form

(\frac{u}{v})^{'} = \frac{v \cdot u ^{'} - u \cdot v ^{'}}{v ^{2}}

Critical Warning

The subtraction order matters: v × u' first, then minus u × v'

Getting this wrong gives the negative of the correct answer.

Worked Example01

Differentiate $y = \frac{2 x + 1}{x - 3}$

1

Identify u and v:

u = 2 x + 1

,

v = x - 3

2

Find derivatives:

\frac{d u}{d x} = 2

,

\frac{d v}{d x} = 1

3

Apply quotient rule:

\frac{d y}{d x} = \frac{( x - 3 ) ( 2 ) - ( 2 x + 1 ) ( 1 )}{( x - 3 ) ^{2}}

4

Expand numerator:

= \frac{2 x - 6 - 2 x - 1}{( x - 3 ) ^{2}}

5

Simplify:

= \frac{- 7}{( x - 3 ) ^{2}}

Answer

\frac{d y}{d x} = \frac{- 7}{( x - 3 ) ^{2}}

Worked Example02

Differentiate $y = \frac{x ^{2} + 1}{x - 2}$

1

Identify:

u = x^{2} + 1

,

v = x - 2

2

Derivatives:

\frac{d u}{d x} = 2 x

,

\frac{d v}{d x} = 1

3

Apply formula:

\frac{d y}{d x} = \frac{( x - 2 ) ( 2 x ) - ( x ^{2} + 1 ) ( 1 )}{( x - 2 ) ^{2}}

4

Expand:

= \frac{2 x ^{2} - 4 x - x ^{2} - 1}{( x - 2 ) ^{2}}

5

Simplify:

= \frac{x ^{2} - 4 x - 1}{( x - 2 ) ^{2}}

Answer

\frac{d y}{d x} = \frac{x ^{2} - 4 x - 1}{( x - 2 ) ^{2}}

Worked Example03

Differentiate $y = \frac{s i n x}{x}$

1

Identify:

u = sin x

,

v = x

2

Derivatives:

\frac{d u}{d x} = cos x

,

\frac{d v}{d x} = 1

3

Apply formula:

\frac{d y}{d x} = \frac{x \cdot c o s x - s i n x \cdot 1}{x ^{2}}

4

Simplify:

= \frac{x c o s x - s i n x}{x ^{2}}

Answer

\frac{d y}{d x} = \frac{x c o s x - s i n x}{x ^{2}}

Worked Example04

Differentiate $y = \frac{e ^{x}}{x ^{2} + 1}$

1

Identify:

u = e^{x}

,

v = x^{2} + 1

2

Derivatives:

\frac{d u}{d x} = e^{x}

,

\frac{d v}{d x} = 2 x

3

Apply formula:

\frac{d y}{d x} = \frac{( x ^{2} + 1 ) ( e ^{x} ) - ( e ^{x} ) ( 2 x )}{( x ^{2} + 1 ) ^{2}}

4

Factor out

e^{x}

:

= \frac{e ^{x} ( x ^{2} + 1 - 2 x )}{( x ^{2} + 1 ) ^{2}}

5

Simplify:

= \frac{e ^{x} ( x ^{2} - 2 x + 1 )}{( x ^{2} + 1 ) ^{2}} = \frac{e ^{x} ( x - 1 ) ^{2}}{( x ^{2} + 1 ) ^{2}}

Answer

\frac{d y}{d x} = \frac{e ^{x} ( x - 1 ) ^{2}}{( x ^{2} + 1 ) ^{2}}

Worked Example05

Differentiate $y = \frac{l n x}{x ^{2}}$

1

Identify:

u = ln x

,

v = x^{2}

2

Derivatives:

\frac{d u}{d x} = \frac{1}{x}

,

\frac{d v}{d x} = 2 x

3

Apply formula:

\frac{d y}{d x} = \frac{x ^{2} \cdot \frac{1}{x} - l n x \cdot 2 x}{( x ^{2} ) ^{2}}

4

Simplify numerator:

= \frac{x - 2 x l n x}{x ^{4}}

5

Factor and simplify:

= \frac{x ( 1 - 2 l n x )}{x ^{4}} = \frac{1 - 2 l n x}{x ^{3}}

Answer

\frac{d y}{d x} = \frac{1 - 2 l n x}{x ^{3}}

Worked Example06

Find the stationary points of $y = \frac{x ^{2}}{x + 2}$

1

Find derivative:

u = x^{2}

,

v = x + 2

, so

u^{'} = 2 x

,

v^{'} = 1

2

\frac{d y}{d x} = \frac{( x + 2 ) ( 2 x ) - x ^{2} ( 1 )}{( x + 2 ) ^{2}} = \frac{2 x ^{2} + 4 x - x ^{2}}{( x + 2 ) ^{2}} = \frac{x ^{2} + 4 x}{( x + 2 ) ^{2}}

3

Set

\frac{d y}{d x} = 0

: numerator = 0, so

x^{2} + 4 x = 0

4

Factorise:

x (x + 4) = 0

, so

x = 0

or

x = - 4

5

Find y-values: When

x = 0

:

y = 0

. When

x = - 4

:

y = \frac{16}{- 2} = - 8

AnswerStationary points at

(0, 0)

and

(- 4, - 8)

Worked Example07

Differentiate $y = \frac{2 x + 1}{x ^{2} + 3}$ and find the gradient when $x = 1$ .

1

Identify:

u = 2 x + 1

,

v = x^{2} + 3

2

Derivatives:

\frac{d u}{d x} = 2

,

\frac{d v}{d x} = 2 x

3

Apply formula:

\frac{d y}{d x} = \frac{( x ^{2} + 3 ) ( 2 ) - ( 2 x + 1 ) ( 2 x )}{( x ^{2} + 3 ) ^{2}}

4

Expand:

= \frac{2 x ^{2} + 6 - 4 x ^{2} - 2 x}{( x ^{2} + 3 ) ^{2}} = \frac{- 2 x ^{2} - 2 x + 6}{( x ^{2} + 3 ) ^{2}}

5

At

x = 1

:

\frac{d y}{d x} = \frac{- 2 - 2 + 6}{( 1 + 3 ) ^{2}} = \frac{2}{16} = \frac{1}{8}

Answer

\frac{d y}{d x} = \frac{- 2 x ^{2} - 2 x + 6}{( x ^{2} + 3 ) ^{2}}

, gradient at

x = 1

is

\frac{1}{8}

Practice

Question 1: Differentiate

y = \frac{3 x - 2}{x + 5}

u = 3 x - 2

,

v = x + 5

,

u^{'} = 3

,

v^{'} = 1

\frac{d y}{d x} = \frac{( x + 5 ) ( 3 ) - ( 3 x - 2 ) ( 1 )}{( x + 5 ) ^{2}} = \frac{3 x + 15 - 3 x + 2}{( x + 5 ) ^{2}}

Answer

\frac{d y}{d x} = \frac{17}{( x + 5 ) ^{2}}

Practice

Question 2: Differentiate

y = \frac{c o s x}{x ^{2}}

u = cos x

,

v = x^{2}

,

u^{'} = - sin x

,

v^{'} = 2 x

\frac{d y}{d x} = \frac{x ^{2} ( - s i n x ) - c o s x ( 2 x )}{x ^{4}} = \frac{- x ^{2} s i n x - 2 x c o s x}{x ^{4}}

Factor out

- x

:

= \frac{- x ( x s i n x + 2 c o s x )}{x ^{4}}

Answer

\frac{d y}{d x} = \frac{- ( x s i n x + 2 c o s x )}{x ^{3}}

Practice

Question 3: Given

y = \frac{x ^{2} - 4}{x + 3}

, find the x-coordinates of the stationary points.

\frac{d y}{d x} = \frac{( x + 3 ) ( 2 x ) - ( x ^{2} - 4 ) ( 1 )}{( x + 3 ) ^{2}} = \frac{2 x ^{2} + 6 x - x ^{2} + 4}{( x + 3 ) ^{2}} = \frac{x ^{2} + 6 x + 4}{( x + 3 ) ^{2}}

Set

\frac{d y}{d x} = 0

:

x^{2} + 6 x + 4 = 0

Using quadratic formula:

x = \frac{- 6 \pm 36 - 16}{2} = \frac{- 6 \pm 20}{2} = - 3 \pm 5

Answer

x = - 3 + 5

and

x = - 3 - 5

Summary

The Quotient Rule

\frac{d y}{d x} = \frac{v \cdot \frac{d u}{d x} - u \cdot \frac{d v}{d x}}{v ^{2}}

Method

Identify

u

(numerator) and

v

(denominator)

Find

\frac{d u}{d x}

and

\frac{d v}{d x}

Apply the formula:

\frac{v \cdot u ^{'} - u \cdot v ^{'}}{v ^{2}}

Simplify the result

Exam Tips

Always clearly identify u, v, u', v' before applying the formula

Remember: subtraction order is v·u' − u·v' (not the other way)

Simplify your final answer where possible

For stationary points, set the numerator equal to zero