“Teach A Level Maths”
Vol. 2: A2 Core Modules
8: The Product Rule
© Christine Crisp
The Product Rule
Module C3
AQA
Edexcel
MEI/OCR
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The Product Rule
The product rule gives us a way of differentiating
functions which are multiplied together.
Consider (a) y
 x ( x  1)
10
2
and (b)
y  x ( x  1)
2
10
(a) can be differentiated by multiplying out the
brackets . . . y  x 12  x 10
but, we need an easier way of doing (b) since
( x  1)10 has 11 terms!
However, doing (a) gives us a clue for a new method.
The Product Rule
(a)
Multiplying out:
y  x ( x  1)
10
yx
12
2
x
10
dy
11
9

 12 x  10 x
dx
Now suppose we differentiate the 2 functions in
x 10 ( x 2  1) without multiplying them out.
u  x 10 and v  x 2  1
du
dv
9
 10x
 2x
dx
dx
du
Multiplying
BUT
v  these
 2 answers does NOT
dx
Let
give
dy
dx
The Product Rule
(a)
Multiplying out:
y  x ( x  1)
10
yx
12
2
x
10
dy
11
9

 12 x  10 x
dx
Now suppose we differentiate the 2 functions in
x 10 ( x 2  1) without multiplying them out.
u  x 10 and v  x 2  1
du
dv
9
 10x
 2x
dx
dx
du
v
 10 x 11  10 x 9 and u  dv 
dx
dx
Let
BUT
The Product Rule
(a)
Multiplying out:
y  x ( x  1)
10
yx
12
2
x
10
dy
11
9

 12 x  10 x
dx
Now suppose we differentiate the 2 functions in
x 10 ( x 2  1) without multiplying them out.
u  x 10 and v  x 2  1
du
dv
9
 10x
 2x
dx
dx
du
v
 10 x 11  10 x 9 and u  dv  2x 11
dx
dx
Let
BUT
Adding these gives the answer we want.
The Product Rule
So, if
y  uv
where u and v are
e.g.
y  x ( x  1)
10
2
u  x 10
v  x2  1
both functions of x,
du
dv
9
 10x
 2x
dx
dx
dy
du
dv
v
 u
dy
10
2
9
dx
dx
dx
x
(2 x )
(
x

1
)
10x


dx
11
9
11
Multiply out the brackets:

10
x

10
x

2
x
We need to simplify the answer:
Collect like terms:
 12 x 11  10 x 9
The Product Rule
Now we can do (b) in the same way.
y  x 2 ( x  1)10
Let u  x 2
and
v  ( x  1)10
dv
du
9

10
(
x

1
)
 2x
dx
dx
dy
du
dv
v
 but
u since the derivative
v is a function of a function
dx
dx
dx
of the inner function is 1, we can ignore it.
dy
10
x
( x complicated
 will meet
 2more
1) 
However, we
functions of
dx
a function later!
du
10
v
 ( x  1)  2 x , but I’ve changed the order.
dx
The standard order is to have constants first, then
powers of x and finally bracketed factors.
The Product Rule
Now we can do (b) in the same way.
Let
y  x 2 ( x  1)10
v  ( x  1)10
u  x 2 and
dv
du
9

10
(
x

1
)
 2x
dx
dx
dy Tip: The
du crossdv
v
 u ( multiply! ) acts as
dx a reminder
dx
ofdxthe product rule!
dy
10
2
9

x
x
10 ( x  1 )

 2 ( x  1)
dx
Don’t be tempted to try to multiply out. Think how
many terms there will be!
There are common factors.
The Product Rule
Now we can do (b) in the same way.
Let
y  x 2 ( x  1)10
v  ( x  1)10
u  x 2 and
dv
du
9

10
(
x

1
)
 2x
dx
dx
dy
du
dv
v
 u
dx
dx
dx
dy
10
2
9

x
x
10 ( x  1 )

 2 ( x  1)
dx
 2 x ( x  1 )10 2  5 x 2 ( x  1 ) 9
How many
factors
 (2x x (1x) 
 1)common?
 5 x
1) 9 ( xare
The common factors.
The Product Rule
Now we can do (b) in the same way.
Let
y  x 2 ( x  1)10
v  ( x  1)10
u  x 2 and
dv
du
9

10
(
x

1
)
 2x
dx
dx
dy
du
dv
v
 u
dx
dx
dx
dy
10
2
9

x
x
10 ( x  1 )

 2 ( x  1)
dx
 2 x ( x  1 )10 2  5 x 2 ( x  1 ) 9
 2 x ( x  1) 9 ( x  1)  5 x 
 2 x( x  1) (6 x  1)
9
SUMMARY
The Product Rule
To differentiate a product:
 Check if it is possible to multiply out. If so, do it
and differentiate each term.
 Otherwise use the product rule:
If y  uv , where u and v are both functions of x
dy
du
dv
v
 u
dx
dx
dx
The product rule says:
• multiply the 2nd factor by the derivative of
the 1st.
• Then add the 1st factor multiplied by the
derivative of the 2nd.
The Product Rule
N.B. You may, at first, find it difficult to simplify
the answers to look the same as those given in
textbooks. Don’t worry about this but keep trying
as it gets easier with practice.
The Product Rule
Reminder:
A function such as y  3 sin x is a product, BUT we
don’t need the product rule.
When we differentiate, a constant factor just “tags
along” multiplying the answer to the 2nd factor.
e.g.
y  3 sin x

dy
 3 cos x
dx
However, the product rule will work even though you
shouldn’t use it
u  3 and
du
0
dx
N.B.
v  sin x
The Product Rule
Reminder:
A function such as y  3 sin x is a product, BUT we
don’t need the product rule.
When we differentiate, a constant factor just “tags
along” multiplying the answer to the 2nd factor.
e.g.
y  3 sin x

dy
 3 cos x
dx
However, the product rule will work even though you
shouldn’t use it
so,
u  3 and v  sin x
du
dv
0
 cos x
dx
dx
dy
dy
du
dv

 3 cos x
v
u
dx
dx
dx
dx
as before.
The Product Rule
Exercise
Use the product rule, where appropriate, to
differentiate the following. Try to simplify your
answers by removing common factors:
1.
3 x
yx e
2.
y  x 4 sin x
3.
y  x (2  x )
4.
y  x (2  x )
2
2
4
The Product Rule
Solutions:
1.
Let
dy
du
dv
v
u
dx
dx
dx
yx e
3 and
u x
v  ex
du
dv
2
x
 3x
e
dx
dx
3 x
dy

 3 x 2e x  x 3e x
dx
order within each
Remove Notice
commonthe
factors:
x 2 e xterm:
(3  x )
constants, powers of x, then exponentials.
The Product Rule
2.
y  x 4 sin x
Let
u  x and v  sin x
du
dv
3
 4x
 cos x
dx
dx
4
dy

 4 x 3 sin x  x 4 cos x
dx
3
factors:
 x (4 sin x  x cos x )
dy
du
dv
v
u
dx
dx
dx
Remove common
3.
y  x 2 (2  x )
 2x  x
2
3
No need for the product
rule: just multiply out.
dy
dy
2
 4x  3x 
 x(4  3 x )
dx
dx
The Product Rule
4.
y  x 2 (2  x )4
Let
u x
du
 2x
dx
2
and
v  (2  x )
dv
3
 4( 2  x )
dx
4
dy you notice that
dy
du
dv  Did
4 v was
2 a
3

2
x
(
2

x
)

4
x
(
2

x
)
v
u
dx
function
of a function?
dx
dx
dx
Remove common factors:  2 x( 2  x ) 3 ( 2  x )  2 x 
 2 x( 2  x ) 3 2  3 x 
Product Rule or Chain Rule?
We can now differentiate all of the following:
simple functions,
products and
compound functions ( functions of a function ).
A simple function could be like any of the following:
y  3x  7x  1
y  2 sin x  3 cos x
2
y
1
x2
y
2ex
3
We differentiate them term by term using the
4 rules for
x n , e x , sin x and cos x
The multiplying constants just “tag along”.
Product Rule or Chain Rule?
For products we use the product rule and for
functions of a function we use the chain rule.
Decide how you would differentiate each of the
following ( but don’t do them ):
(c)
y  sin 2 x
y  x sin x
y  2 sin x
This is a simple function
(d)
y  sin x
Chain rule
(a)
(b)
3
Chain rule
Product rule
 y  (sin x) 
3
Product Rule or Chain Rule?
Exercise
Decide with a partner how you would differentiate the
following ( then do them if you need the practice ):
Write C for the Chain rule and P for the Product Rule
1.
y  x sin x
3.
x2 C
e
y
P
5.
y  sin x
2
2.
y  sin x 2
4.
y  e x (1  x )
C or P
C
P
Product Rule or Chain Rule?
Solutions
1. y  x sin x P
u x
v  sin x
du
dv
1
 cos x
dx
dx
dy
du
dv
dy
v
 u
 sin x  x cos x

dx
dx
dx
dx
2.
y  sin x 2
C
u  x 2  y  sin u
du
dy
 2x
 cos u  cos x 2
dx
du
dy
dy dy du



 2 x cos x 2
dx du dx
dx
Product Rule or Chain Rule?
3.
ye
x2
C
u  x2 
y  eu
du
dy
u
x2
 2x
 e e
dx
du
dy
dy dy du
x2



 2x e
dx du dx
dx
x
x
u

e
v  1 x
4. y  e (1  x ) P
du
dv
x
e
 1
dy
du
dv
dx
dx
v
 u
dx
dx
dx
dy
dy
x
x
 e (1  x )  (  e ) 

  xe  x
dx
dx
Product Rule or Chain Rule?
5.
y  sin 2 x
u  sin x  y  u 2
Either C
du
dy
 cos x
 2u  2 sin x
dx
du
dy
dy dy du



 2 sin x cos x
dx du dx
dx
Or P
u  sin x
v  sin x
du
dv
 cos x
 cos x
dx
dx
dy
dy
du
dv
v
 u
 sin x cos x  sin x cos x

dx
dx
dx
dx
dy

 2 sin x cos x
dx
The Product Rule
The Product 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 Product Rule
SUMMARY
To differentiate a product:
 Check if it is possible to multiply out. If so, do it
and differentiate each term.
 Otherwise use the product rule:
If y  uv , where u and v are both functions of x
dy
du
dv
v
 u
dx
dx
dx
The Product Rule
e.g.
y  x ( x  1)
v  ( x  1)10
u  x 2 and
dv
du
9

10
(
x

1
)
 2x
dx
dx
dy
du
dv
v
 u
dx
dx
dx
dy
10
2
9

 2 x ( x  1 )  10 x ( x  1 )
dx
10
2
9
 2 x( x  1)  2  5 x ( x  1)
2
Let
10
Remove common factors:
 2 x( x  1) 9 ( x  1)  5 x 
 2 x( x  1) 9 (6 x  1)
The Product Rule
We can now differentiate all of the following:
simple functions,
products and
A simple function could be like any of the following:
y  3x  7x  1
y  2 sin x  3 cos x
2
y
1
x2
y
2ex
3
We differentiate them term by term using the
4 rules for
x n , e x , sin x and cos x
The multiplying constants just “tag along”.
Product Rule or Chain Rule?
For products we use the product rule and for
functions of a function we use the chain rule.
Decide how you would differentiate each of the
following ( but don’t do them ):
(c)
y  sin 2 x
y  x sin x
y  2 sin x
This is a simple function
(d)
y  sin x
Chain rule
(a)
(b)
3
Chain rule
Product rule
 y  (sin x) 
3
Product Rule or Chain Rule?
It can be tricky to distinguish between products and
functions of a function.
If you do have difficulty try the following:
•
Assume you have a product.
Then, you must have 2 factors ( at least ).
•
Try to split up the function into one
function multiplied by another.
If you can’t then it isn’t a product.