Learning Enhancement Team
Steps into Calculus
The Product Rule
This guide describes how the product rule is used to differentiate
functions which are two or more basic functions multiplied together. It
also shows you how to use your page in an efficient way when performing
lengthy calculations.
Introduction
The most important skill when differentiating is being able to identify the form of the function
you have to differentiate. Once you know the form of the function you can choose an
appropriate rule to differentiate it. Many complicated functions are created by combining
basic functions in some way, usually by addition/subtraction, multiplication, division or
composition (see study guide: More Complicated Functions and the factsheet: Five Basic
Functions). This guide is concerned with differentiating complicated functions which are
made by multiplying together more basic functions. In numeracy, the result of
multiplying numbers is called their product. Extending this idea, the rule to differentiate a
function which is the result of multiplying simpler functions is called the product rule.
The product rule is given by:
If
y  uv
then
dy
dv
du
u
v
dx
dx
dx
Here the function you are trying to differentiate y is made by multiplying together the
functions u and v. Importantly, u and v are basic functions of x and so they take the form of
one of the functions in the table given on the next page of this guide.
The derivative
dy
dv
du
is dependent on u and v and their respective derivatives
and
.
dx
dx
dx
To use the product rule you need to be able to differentiate basic functions. Methods to do
this are discussed in the study guides: Differentiating Using the Power Rule and
Differentiating Basic Functions. You should read these guides and familiarise yourself with
the methods discussed in them before you continue with this guide. The results are
summarised in the following table:
rule
function
derivative
1
k
0
2
ax
a
3
ax n
anx n 1
4
a sin kx
ak cos kx
5
a cos kx
 ak sin kx
6
ae
kx
akekx
a
x
aln kx
7
Throughout this guide these results will be referred to by the rule number in this table.
The product rule is a two-stage rule, in a similar manner to the power rule. First you must
ensure that the function you are dealing with fits the pattern – that it is a product of two basic
functions. If this is the case you can proceed to the second part which gives the derivative.
Example:
Which of the following functions can be differentiated using the product rule?
(a) y  16 x 2 sin x
(b)
y  sin16 x 2 
(c)
y  16 sin x
(d)
y
16 sin x
x2
(a)
This function can be made by multiplying the basic functions 16x 2 and sin x ; 16x 2
can be differentiated using rule 3 and sin x using rule 4. If you set u  16x 2 and
v  sin x then y  uv and you can use the product rule to find the derivative.
(b)
This function is made by composition as it is the sine of 16x 2 . As it is not the product
of two elementary functions you cannot use the product rule here. In fact you can
use the chain rule to differentiate this function (see study guide: The Chain Rule).
Although this function can be differentiated by the product rule, with u  16 (which
can be differentiated by rule 1) multiplied by v  sin x (which can be differentiated by
rule 4), it can be directly differentiated by rule 4 in the table with a  16 and k  1
with no need for the product rule. Both methods give the same answer but the latter
is easier.
(c)
(d)
Upon first inspection this function looks like 16 sin x divided by x 2 which would lead
to use of the quotient rule (see study guide: The Quotient Rule). However, you can
think of the function as 16 sin x multiplied by x 2 using the laws of indices. 16 sin x
can be differentiated using rule 4 and x 2 using rule 3. If you set u  16 sin x and
v  x 2 then y  uv and you can use the product rule to find the derivative.
Using the product rule
Once you have decided that your function is the product of two basic functions and therefore
it is appropriate to use the product rule to differentiate the function, the next step is to assign
u and v. In the product rule it does not matter which function you assign as u and
which you assign as v, unlike the quotient rule (see study guide: The Quotient Rule).
When this is done you can proceed to the second part of the rule. Looking carefully:
dy
dv
du
u
v
dx
dx
dx
The derivative is made of four separate parts and you can think of the product rule as:
u multiplied by the derivative of v
plus v multiplied by the derivative of u
As you have already defined u and v, all you need to do is to find the derivative of each of
them. Next you substitute u and v and their derivatives into the product rule formula and
simplify if necessary. You may find it useful to divide your page vertically into an answer
space (where your answer will be written) and an exploring space (where you can do any
working and thinking which informs your answer). Usually in product rule calculations the
exploring space is simply a list of the mathematics that you need to calculate in order to
perform the substitution and any algebra you need to think about (such as factorisation).
However if you progress to more complicated mathematics the exploring space offers a vital
place to play with ideas as you work towards a solution.
Differentiate y  16 x 2 sin x .
Example:
As mentioned in example (b) above the function can be differentiated by the product rule
where u  16x 2 and v  sin x .
Answer Space
Exploring Space
y  16 x sin x
u  16x 2
y  uv
v  sin x
2
so
dy
dv
du
u
v
dx
dx
dx
 16 x 2  cos x   sin x   32x 
du
 32 x
dx
(rule 3)
 16 x 2 cos x  32x sin x
dv
 cos x
dx
(rule 4)
 16 x x cos x  2 sin x 
So the derivative of y  16 x 2 sin x is
dy
 16 x x cos x  2 sin x  .
dx
By writing u, v and so on in the exploring space to the right of the line you keep them apart
from the flow of the answer to the left. It is also useful for further reference and checking
when substituting, especially so when examples become more complicated.
Example: What is the gradient of y  5e x cos x when x   ?
The function can be thought of as 5e x multiplied by cos x so by setting u  5e x and
v  cos x you can use the product rule.
Answer Space
y  5e cos x
y  uv
so
Exploring Space
u  5e x
v  cos x
x
dy
dv
du
u
v
dx
dx
dx
 5e x   sin x   cos x   5e x 
du
 5e x
dx
(rule 6)
 5e x sin x  5e x cos x
dv
  sin x
dx
(rule 5)
 5e x cos x  sin x 
dy
 5e x cos x  sin x  .
dx
Remember that the derivative and the gradient are equivalent so when x   the gradient is:
So the derivative of y  5e x cos x is
dy
dx
 5e cos   sin    5e  115.703 to 3 d.p.
x 
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