“Teach A Level Maths”
Vol. 2: A2 Core Modules
12: The Quotient Rule
© Christine Crisp
The Quotient Rule
Module C3
AQA
Edexcel
MEI/OCR
"Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with
permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"
The Quotient Rule
The following are examples of quotients:
x
x
(a) y 
sin x
e
(b) y 
1 x
1  x2
sin x
(d) y 
cos x
(c) y 
x2
(c) can be divided out to form a simple function as
there is a single polynomial term in the denominator.
y
1  x2
x
2
 y
1
2

x2
x
x2
dy
2
 y  x 1 
 2 x  3
dx
For the others we use the quotient rule.
The Quotient Rule
The quotient rule gives us a way of differentiating
functions which are divided.
The rule is similar to the product rule.
u
y
v

du
dv
v
 u
dy
 dx 2 dx
dx
v
where u and v are functions of x.
This rule can be derived from the product rule but
it is complicated. If you want to go straight to the
examples, click on the box below.
Examples
The Quotient Rule
We can develop the quotient rule by using the product
rule!
u
y
v

1
yu
v
 y  u  v 1
The problem now is that this v is not the same as the
v of the product rule. That v is replaced by v 1.
1
dy
du
dv
dy
du
d
(
v
)
1
So,
becomes
v
u
v
u
dx
dx
dx
dx
dx
dx
1
dy
1
du
d
(
v
)
Simplifying 

u
dx v dx
dx
1
d
(
v
) , is the derivative of v 1
Part of the
term,
dx
but with respect to x not v.
2nd
The Quotient Rule
1
d
(
v
)
We use the chain rule:
dx
1
d
(
v
)
So,
dx
1
dy
1
du
d
(
v
)
Then,

u

dx v dx
dx
d (v 1 ) dv


dv
dx
1
dv
  2 
dx
v
dy 1 du
u dv

 2
dx v dx v dx
Make the denominators the same by multiplying the
numerator and denominator of the 1st term by v.
dy
v du
u dv
 2
 2
du
dv
dx v dx v dx
v
 u
dy
Write with a common denominator: 
 dx 2 dx
dx
v
The Quotient Rule
x2
e.g. 1 Differentiate y 
to find dy .
x 1
dx
u
Solution: y 
 u  x 2 and v  x  1
v
dv
du
1
 2x
dx
dx
du
dv
v
 u
u
dy
y

 dx 2 dx
v
dx
v
dy
2 x( x  1)  x 2


dx
( x  1) 2
We now need to simplify.
The Quotient Rule
dy
2 x( x  1)  x 2

dx
( x  1) 2
We could simplify the numerator by taking out the
common factor x, but it’s easier to multiply out the
brackets. We don’t touch the denominator.
2
2
dy
2
x

2
x

x
Multiplying out numerator:

dx
( x  1) 2
dy
x2  2x

Now collect like terms:
dx
( x  1) 2
dy
x ( x  2)

and factorise:
dx
( x  1) 2
We leave the brackets in the denominator as the
factorised form is simpler.
The Quotient Rule
Quotients can always be turned into products.
e.g.
ex
1 x
2
can be written as e x (1  x 2 ) 1
However, differentiation is usually more awkward
if we do this.
In the quotient above, u  e x
and v  1  x 2
( both simple functions )
In the product ,
u  e x and
v  (1  x 2 ) 1
( v needs the chain rule )
SUMMARY
The Quotient Rule
To differentiate a quotient:
 Check if it is possible to divide out. If so, do it
and differentiate each term.
 Otherwise use the quotient rule:
If
u
y
v
, where u and v are both functions of x
du
dv
v
 u
dy
 dx 2 dx
dx
v
The Quotient Rule
Exercise
Use the quotient rule, where appropriate, to
differentiate the following. Try to simplify your
answers:
1.
y
x3
e
x
2.
y
4.
y
2
3.
x
y
2 x
cos x
x4
2 x
x2
The Quotient Rule
1.
y
x3
e
Solution:
x
u
y
v

du
dv
v
 u
dy
 dx 2 dx
dx
v
u
y   u  x3
v
du
 3x 2
dx
and
v  ex
dv
 ex
dx
dy 3 x 2 e x  x 3 e x


x 2
dx
e
 
2
dy x 2 e x ( 3  x )
dy x ( 3  x )




2
x
x
dx
dx
e
e
 
2.
y
cos x
x
Solution:
The Quotient Rule
4
u
y
v

du
dv
v
 u
dy
 dx 2 dx
dx
v
u
y   u  cos x and v  x 4
v
dv
du
3

4x
  sin x
dx
dx
dy  x 4 sin x  4 x 3 cos x


4 2
dx
x
 
 ( x sin x  4 cos x )
dy  x 3 ( x sin x  4 cos x )



dx
x5
x8 5
x
The Quotient Rule
3.
x2
y
2 x
Solution:
u
y
v

du
dv
v
 u
dy
 dx 2 dx
dx
v
u
y   u  x 2 and v  2  x
v
dv
du
 1
 2x
dx
dx
2
dy 2 x( 2  x )  x ( 1)


2
dx
(2  x )
2
2
2
dy 4 x  2 x  x
4x  x
x(4  x )




2
2
dx
(2  x )
(2  x )
(2  x ) 2
4.
y
The Quotient Rule
2 x
x2
Solution:
y
Divide out:
2 x
x
2
 y
2
2

1x
x
x2
 y  2 x 2  x  1
dy
3
2

 4 x  x
dx
4
1
 3  2
x
x
The Quotient Rule
We can now differentiate the trig function
by writing
y  tan x
sin x
y  tan x 
cos x
u
v  cos x
u  sin x
y

v
dv
du
  sin x
 cos x
dx
dx
dy (cos x )(cos x )  (sin x )( sin x )


dx
(cos x )(cos x )

cos2 x  sin2 x
cos2 x
The Quotient Rule
So,
y  tan x 
dy cos2 x  sin2 x

dx
cos2 x
This answer can be simplified:
dy
1

cos x  sin x  1 
dx cos2 x
1
Also,
is defined as secx
cos x
2
So,
2
y  tan x 
dy
 sec2 x
dx
The Quotient Rule
Those of you taking the OCR/MEI spec can omit the
next ( final exercise ).
The Quotient Rule
Exercise
Use the quotient rule ( or, for (a) and (b), the
chain rule ) to find the derivatives with respect to x
of
(a) y  cosec x
(b) y  sec x
(c) y  cot x
Before you check the solutions, look in your formula
books to see the forms used for the answers. Try
to get your answers into these forms.
The Quotient Rule
(a) Solution:
1
y  cosec x 
sin x
dy sinx(0)  1(cos x )

dx
(sinx ) 2
cos x

sin x sin x
cos x
1


sin x sin x
  cot x cosec x
(   cose cx cot x )
The Quotient Rule
(b) Solution:
1
y  sec x 
cos x
dy cos x(0)  1( sinx )

2
dx
(cos x )
sin x

cos x cos x
sin x
1


cos x cos x
 tan x sec x
(  se cx tan x )
The Quotient Rule
(c) Solution:
cos x
y  cot x 
sin x
dy sin x(  sin x )  cos x cos x

2
dx
(sinx )


 (sin2 x  cos2 x )
sin2 x
1
2
si n x
  cosec2 x
The Quotient Rule
The Quotient Rule
The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
For most purposes the slides can be printed
as “Handouts” with up to 6 slides per sheet.
The Quotient Rule
The following are examples of quotients:
x
x
(a) y 
sin x
e
(b) y 
1 x
1  x2
sin x
(d) y 
cos x
(c) y 
x2
(c) can be divided out to form a simple function as
there is a single polynomial term in the denominator.
y
1  x2
x
2
 y
1
2

x2
x
x2
dy

2
 y  x 1 
 2 x  3
dx
For the others we use the quotient rule.
The Quotient Rule
SUMMARY
To differentiate a quotient:
 Check if it is possible to divide out. If so, do it
and differentiate each term.
 Otherwise use the quotient rule:
If
u
y
v
, where u and v are both functions of x
du
dv
v
 u
dy
 dx 2 dx
dx
v
The Quotient Rule
x2
e.g. 1 Differentiate y 
to find dy .
x 1
dx
u
Solution: y 
 u  x 2 and v  x  1
v
dv
du
1
 2x
dx
dx
du
dv
v
 u
u
dy
y

 dx 2 dx
v
dx
v
dy
2 x( x  1)  x 2


dx
( x  1) 2
We now need to simplify.
The Quotient Rule
dy
2 x( x  1)  x 2

dx
( x  1) 2
We could simplify the numerator by taking out the
common factor x, but it’s easier to multiply out the
brackets. We don’t touch the denominator.
2
2
dy
2
x

2
x

x
Multiplying out numerator:

dx
( x  1) 2
dy
x2  2x

Now collect like terms:
dx
( x  1) 2
dy
x ( x  2)

and factorise:
dx
( x  1) 2
We leave the brackets in the denominator.
( A factorised form is considered to be simpler. )
The Quotient Rule
We can now differentiate the trig function
by writing
y  tan x
sin x
y  tan x 
cos x
u
v  cos x
u  sin x
y

v
dv
du
  sin x
 cos x
dx
dx
dy (cos x )(cos x )  (sin x )( sin x )


dx
(cos x )(cos x )

cos2 x  sin2 x
cos2 x
The Quotient Rule
So,
y  tan x 
dy cos2 x  sin2 x

dx
cos2 x
This answer can be simplified:
dy
1

cos x  sin x  1 
dx cos2 x
1
Also,
is defined as secx
cos x
2
So,
2
y  tan x 
dy
 sec2 x
dx