Applying the Product, Quotient, and Power of a Function
Rules in Calculus I
Prof. Lee Townsend
Fall 2015
Complicated math rules can be difficult for some students, especially concrete thinkers. Here
is a method that seems to work fairly well for them. I call it the template method. The list of
rules is on the last page. We will apply the template method for rules 5, 6, and 7.
I. The Product Rule
y = ( 3x 2 + 2 ) ( 4x − 1)
(1)
1) Determine the pattern of y so you know what rule to use. This equation is a product.
2) To find its derivative we use the product rule:
d (u ⋅ v )
dv
du
=u +v
dx
dx
dx
(2)
3) The product rule says we need to define u and v , then take their derivatives. I recommend
setting up the equations as follows for taking the derivatives. Keep them tidy and separate.
u = ( 3x 2 + 2 )
du d
= ( 3x 2 + 2 )
dx dx
By rule 4
d (2)
du d
= ( 3x 2 ) +
dx dx
dx
By rule 1, the second term is zero
du d
= ( 3x 2 )
dx dx
By rule 3
du
d
= 3 ( x2 )
dx
dx
By rule 2
du
= 3( 2x )
dx
By algebra
du
= 6x
dx
v = ( 4x − 1)
(3a)
(3b)
dv d
= ( 4x − 1)
dx dx
By rule 4
d (1)
dv d
= ( 4x ) −
dx dx
dx
By rule 1, the second term is zero
dv d
= ( 4x )
dx dx
By rule 3
dv
d
= 4 ( x)
dx
dx
By rule 2
dv
= 4 (1)
dx
By algebra
(4a)
Page 1 of 11
dv
=4
dx
(4b)
Note that I have put boxes around the four formulas I need. I also wrote them in a different
color. That way when I am applying the product rule, the four relevant equations are easy to
find among all those steps. When doing the problems by hand, you may have difficulty
finding the equations you need unless you are very tidy with your work. So, I recommend
circling the four important formulas for the problem.
At this point I find it comforting to check both derivatives using the TI-89.
d ( 3x 2 + 2, x )
give
d ( 4x − 1, x )
and
6x
4
We’re good.
4) Now it’s time to apply the Product Rule. The individual derivatives have been found so it
is just a question of putting the terms in the right place. For this, we create a template.
d (u ⋅ v )
dv
du
=u +v
dx
dx
dx
Recall the rule itself:
The template is therefore
dv
dx
u
v
du
dx
d (u ⋅ v )
=(
)⋅ (
)+(
)(
)
dx
where I have written what is supposed to go in the blank sets of parentheses right above the
parentheses.
5) Now we fill it in with the data from the table above.
dy
4
= ( 3x 2 + 2 ) ⋅ (
) + ( 4x − 1 )(
dx
6x )
(5)
Note that all I have done is fill in the template. I have not done any math in this step.
6) All that remains is algebra – i.e. simplify equation (5) algebraically. I find it less confusing
to keep parentheses so I can tell where terms came from.
{
}
dy
= 4 ⋅ ( 3x 2 + 2 ) + {6x ( 4x − 1)}
dx
dy
= {12x 2 + 8} + {24x 2 − 6x }
dx
Page 2 of 11
Before proceeding, make sure you have done the math correctly so far while the two terms are
still separated. Now combine them by first putting like terms next to each other. Since the
TI-89 gives results by highest power of x down to lowest power of x, I will too.
dy
= (12x 2 + 24x 2 ) − 6x + 8
dx
Finally we get the solution to the problem.
dy
= 36x 2 − 6x + 8
dx
7) Check the final answer with the TI-89. You already know the derivatives,
du
dv
and
, are
dx
dx
correct but did you do the algebra correctly?
Done. They agree.
II. The Quotient Rule
( 3x
y=
2
+ 2)
(6)
( 4x − 1)
1) Determine the pattern of y so you know what rule to use. This equation is a quotient.
2) To find its derivative we use the Quotient rule:
du
dv
d ⎛ u ⎞ v dx − u dx
⎜ ⎟=
dx ⎝ v ⎠
v2
(7)
3) The Quotient rule says we need to define u and v , then take their derivatives. I used the
same u and v that I used in the product rule so we know the derivatives already.
u = 3x 2 + 2
v = ( 4x − 1)
(8)
(
du
= 6x
dx
)
dv
=4
dx
Page 3 of 11
(9)
4) Now it’s time to apply the Quotient Rule. Since the individual derivatives have been
found, it is just a question of putting the terms in the right place. For this, we create a
template.
du
dv
d ⎛ u ⎞ v dx − u dx
⎜ ⎟=
dx ⎝ v ⎠
v2
Recall the rule itself:
The template is therefore
v
du
dx
u
dv
dx
d (u ⋅ v )
=
dx
v
where, again, I have written what is supposed to go in the blank parentheses right above or
below the empty parentheses. The results from before are
du
dv
u = 3x 2 + 2 ,
v = ( 4x − 1) ,
= 6x ,
=4
dx
dx
(
)
5) Now we fill it in with the data.
2
dy ( 4x − 1) ⋅ ( 6x ) − ( 3x + 2 ) ( 4 )
=
dx
( 4x − 1)2
(10)
Note that, once again, all I have done is fill in the template. I have not done any math in this
step.
6) All that remains is algebra – i.e. simplify equation (10) algebraically. I find it less
confusing to keep parentheses so I can tell where terms came from. Then I can do a quick
check by comparing the below with equation (10).
2
2
dy ( 24x − 6x ) − (12x + 8 )
=
dx
( 4x − 1)2
Before proceeding, make sure you have done the math correctly so far while the two terms are
still separated. Now combine them by first putting like terms next to each other. Since the
TI-89 gives results by highest power of x down to lowest power of x, I will too.
dy 12x 2 − 6x − 8
=
dx
( 4x − 1)2
which is the solution to the problem.
Page 4 of 11
7) Check the final answer with the TI-89. You already know the derivatives,
du
dv
and
, are
dx
dx
correct but did you do the algebra correctly?
dy 12x 2 − 6x − 8
, has no
=
dx
( 4x − 1)2
parentheses in the numerator while the TI answer does. We have two choices to see if they
are the same. I am including detail for those of you who are new to the TI-89.
But they don’t exactly agree. Note that our answer above,
A) Factor our numerator
1) Get the factor command
2) Press Enter.
`
Page 5 of 11
3) Type in our numerator.
4) Close the parentheses then press Enter.
The answer is the same as the one the TI-89 got so our algebra was correct.
B) Expand the TI numerator
1) Get the numerator from your TI derivative result.
a) Press Enter to get getNum(.
b) Arrow up to select the fraction.
c) Press Enter.
d) Put on the closing parenthesis then press Enter again.
Page 6 of 11
2) Expand the numerator, now extracted from the fraction.
3) Apply the expand command.
a) Press Enter to get expand(.
b) Arrow up to select 2 ⋅ 6 ⋅ x 2 − 3⋅ x − 4 .
(
)
c) Press Enter.
d) Put on the closing parenthesis then press Enter again.
The numerator is the same as the one we got by hand so our algebra was
correct.
III. The Power Rule
y = ( 3x 2 + 2 )
7
(11)
1) Determine the pattern of y so you know what rule to use. This equation is in the form of a
power of a function.
2) To find y’s derivative we use the Power rule:
du n
du
= nu n−1
dx
dx
(12)
Page 7 of 11
3) The Power rule says we need to define u and take its derivative. I used the same u that I
used above so we know its derivative already.
du
u = 3x 2 + 2
(13)
= 6x
dx
(
)
4) Now it’s time to apply the Power Rule. For this, again, we create a template.
5) Now we fill it in with the data. We use the results from before:
u = 3x 2 + 2
n=7
(
du
= 6x
dx
)
n −1 = 6
(14)
Note that, once again, all I have done is fill in the template. I have not done any math in this
step.
7) Check the final answer with the TI-89.
Once again, the answers differ.
Our answer
6
dy
= 7 ⋅ ( 3⋅ x 2 + 2 ) ⋅ ( 6 ⋅ x )
dx
TI Answer
6
dy
= 42 ⋅ x ⋅ ( 3⋅ x 2 + 2 )
dx
How do they differ? Ours has a 7 in front and a ( 6 ⋅ x ) at the end. The TI has 42 ⋅ x in
front with nothing at the end. Where did the ( 6 ⋅ x ) go? The TI put it in front then
multiplied, as shown below.
Page 8 of 11
6
dy
= 7 ⋅ ( 3⋅ x 2 + 2 ) ⋅ ( 6 ⋅ x )
dx
Our answer:
Move the ( 6 ⋅ x ) in front since
multiplication is commutative:
6
dy
= 7 ⋅ ( 6 ⋅ x ) ⋅ ( 3⋅ x 2 + 2 )
dx
Multiply the 7 and the ( 6 ⋅ x ) :
6
dy
= 42 ⋅ x ⋅ ( 3⋅ x 2 + 2 )
dx
The answers agree.
Please feel free to ask questions. Since this is the first draft, I may have made some typos.
Please let me know of any errors or anything that is unclear.
Page 9 of 11
Method Summary
1) Look at the overall pattern of the problem to decide which rule to use first.
2) Write down the rule.
3) What information does the rule need? Get the information.
4) Create the annotated template for the rule.
5) Put the appropriate information inside the template.
6) Do the algebra.
7) Check your work by putting the original function in the calculator. The form for the TI
dy
derivatives, , is d(y,x), where y is the dependent variable and x is the independent
dx
variable.
Page 10 of 11
The Derivative Rules
Notation:
c is a constant.
u and v are any functions.
dc
=0
dx
dx n
= nx n−1
2)
dx
1)
3)
d ( c ⋅u )
du
= c⋅
dx
dx
4)
d ( u + v ) du dv
=
+
dx
dx dx
d (u ⋅ v )
dv
du
=u +v
dx
dx
dx
du
dv
d ⎛ u ⎞ v dx − u dx
6)
⎜ ⎟=
dx ⎝ v ⎠
v2
5)
with the associated rule
dx
=1
dx
The Product Rule
The Quotient Rule
⎛ u⎞
d⎜ ⎟
⎝ v⎠
d ⎛ u⎞
Note that sometimes it is less confusing to write
as
⎜ ⎟ . They are equivalent.
dx ⎝ v ⎠
dx
7)
du n
du
= nu n−1
dx
dx
The Power of a Function Rule
Note that it is a specific example of the more general Chain Rule:
Page 11 of 11
dy ( u ) dy du
=
dx
du dx