Integral Calculus

1. God of grace,
We open our hearts, minds
and souls to worship you.
Thank you that today we dwell in your kingdom
and live in your presence.
Thank you that as we gather together
we join with all Christians across the world
to glorify your holy name.
Come be with us, inspire us and lead us in our time together.
We ask all this in the beautiful name of Jesus. Amen

3. LEARNING COMPETENCIES:
 Identify and define the different rules in finding the
derivatives of functions.
 Apply the different rules in order to find the derivatives of the
given polynomial functions.
CORE VALUE: CHARISM
As a Paulinian, this lesson will develop your sense of
responsibility and discipline as you will be dealing with
problems with different difficulties.

4. Now that we've seen many different functions, it's time to
turn our attention to the central notion of this course---the
derivative. This is the key to deepening our discussion about
functions. The derivative is one of the greatest ideas that we as
humans have created to describe our natural world. It took
thousands of years for the idea to be formulated in a way which is
really useful (so don't be discouraged if it takes you a day or two!).
As with any great idea, the derivative may be considered from
several different perspectives. However, they all boil down to this:
The derivative tells us how to approximate a graph, near some base
point, by a straight line.
INTRODUC
TION

5. Raymond Barnett Michael Ziegler

6. FIRST,
Slope of the
tangent
SECON
D,
Instantaneous
rate of change
DERIVATIVE
DERIVATIVE

7. Derivative Notation
derivative of a function 𝒚 = 𝒇(𝒙)
𝑓(𝑥), 𝑦,
𝑑𝑦
𝑑𝑥
, or
𝑑𝑓(𝑥)
𝑑𝑥
derivative of y with respect to 𝒙” or for
short, “dee 𝑦 over dee 𝑥”

8. DIFFERENTIA
TION
- process of finding the derivative of a function.
That is, the derivative of a function is obtained by
differentiating the function. (Barnett, et.al, 2003)

9. Rules of Differentiation:
constant
function
identity
function
power
constant multiple
sum and difference
product
quotient

10. THE CONSTANT FUNCTION RULE
Definition. If 𝑦 = 𝐶, where 𝐶 is a constant, then 𝒚’ = 𝟎.
Illustrative Examples
1. Find the derivative of 𝑓(𝑥) = 1
. 2. Find the derivative of 𝑦 = −23
3. Find the derivative of 𝑦 = 𝜋
Answer: 𝒇’(𝒙) = 𝟎
Answer: 𝒚′ = 𝟎
Answer: 𝒚′ = 𝟎
REMEMBER:
The derivative of any
constant is always equal to
zero.

11. THE IDENTITY FUNCTION RULE
Definition. If 𝑦 = 𝑥, where 𝑥 is a differentiable function, then 𝒚’ = 𝟏.
Illustrative Examples
1. Find the derivative of 𝑦 = 𝑥
. 2. Find the derivative of 𝑓(𝑥) = −2𝑥
3. Find the derivative of𝑦 = 6𝑥
Answer: 𝒚’ = 𝟏
Answer: 𝒇’(𝒙) = −2
Answer: 𝒚′ = 𝟔
REMEMBER:
The derivative of identity function is
equal to 1. Likewise, the derivative
of a function on the first degree is
equal to its numerical coefficient.

12. THE POWER RULE
Definition. If 𝑦 = 𝑥𝑛, where 𝑥 is a differentiable function and n
is a real number, then 𝒚’ = 𝒏𝒙𝒏−𝟏.
Illustrative Examples
1. Find the derivative of 𝑦 = 𝑥10
. 2. Find the derivative of 𝑓(𝑥) = 𝑥−5
Answer: 𝒚’ = 𝟏𝟎𝒙𝟏𝟎−𝟏
Answer: 𝒇’ 𝒙 = −𝟓𝒙−𝟓−𝟏
REMEMBER:
In finding the derivative using
the Power Rule, just make the
exponent as the numerical
coefficient and subtract one
(1) from the given exponent
of the variable x. simplify the
derivatives that are not in
their simplified forms.
𝒚’ = 𝟏𝟎𝒙𝟗
𝒇’ 𝒙 = −𝟓𝒙−𝟔
𝒇’ 𝒙 =
−𝟓
𝒙𝟔

13. THE CONSTANT MULTIPLE RULE
Definition. If 𝑦 = 𝐶. 𝑓(𝑥), where 𝑥 is a differentiable function, then 𝒚’ = 𝑪. 𝒇′(𝒙).
Illustrative Examples
1. Find the derivative of 𝑦 = 3𝑥2 . 2. Find the derivative of𝑦 = −6𝑥−4
Answer: 𝑦′ = (3)(2)𝑥2−1 Answer: 𝑦′ = (−6)(−4)𝑥−4−1
𝒚′ = 𝟔𝒙 𝑦′
= 24𝑥−5
𝒚′ =
𝟐𝟒
𝒙𝟓

14. REMEMBER:
In finding the derivative using the Constant
Multiple Rule, the following steps should be
followed:
1.Multiply the exponent by the numerical
coefficient of the given function.
2.Subtract one from the exponent of the given
function.
3.Simplify the resulting derivative.

15. THE SUM AND DIFFERENCE RULE
Definition. If 𝑦 = 𝑢 𝑥 ± 𝑣 𝑥 , where 𝑢 and 𝑣 are differentiable functions, then
𝒚’ = 𝒖′(𝒙) ± 𝒗′(𝒙)
Illustrative Examples
1. 𝑦 = 3𝑥2 + 2𝑥 − 5
Answer: 𝑦′ = 3 2 𝑥2−1 + 2 𝑥1−1 − 0
𝒚′ = 𝟔𝒙 + 𝟐
2. 𝑦 = 4𝑥6 − 9𝑥3 + 3𝑥2
Answer: 𝑦′
= 4 6𝑥5
− (9) 3𝑥2
+ (3)(2𝑥)
𝒚′ = 𝟐𝟒𝒙𝟓 − 𝟐𝟕𝒙𝟐 + 𝟔𝒙

16. REMEMBER:
In finding the derivative using the Sum and
Difference Rules, the following steps should be
followed:
1. Get the derivative of each term by applying
the previously discussed rule in finding the
derivatives.
2. Simplify the resulting derivatives.

17. THE PRODUCT RULE
Definition. If 𝑦 = 𝑢 𝑥 . 𝑣 𝑥 , where 𝑢 and 𝑣 are differentiable functions, then
𝒚’ = 𝒖 𝒙 . 𝒗′ 𝒙 + 𝒗 𝒙 . 𝒖′(𝒙)
Illustrative Examples
1. Find the derivative of 𝒚 = 𝟑𝒙𝟐 𝟐𝒙𝟐 − 𝟑
Solution:
Let 𝒖 𝒙 = 𝟑𝒙𝟐 and 𝒗 𝒙 = (𝟐𝒙𝟐 − 𝟑)
𝑦′ = 3𝑥2 2𝑥2 − 3 ′ + 2𝑥2 − 3 (3𝑥2)′
)
𝑦′
= 3𝑥2
(4𝑥) + 2𝑥2
− 3 (6𝑥
𝑦′ = 12𝑥3 + 12𝑥3 − 18𝑥
𝒚′
= 24𝒙3
− 18𝒙
𝑪𝒉𝒆𝒄𝒌𝒊𝒏𝒈:
)
𝑦 = 3𝑥2
(2𝑥2
− 3
𝑦 = 6𝑥4 − 9𝑥2
𝒚′
= 24𝒙3
− 18𝒙
𝑦′ = 6(4)𝑥4−1
− 9(2)𝑥2−1

18. THE PRODUCT RULE
Illustrative Examples
𝑪𝒉𝒆𝒄𝒌𝒊𝒏𝒈:
2. Find the derivative of 𝒚 = (𝟑𝒙𝟐 − 𝟓)(𝟐𝒙 + 𝟑).
Let 𝑢 𝑥 = 3𝑥2 − 5 and 𝑣 𝑥 = 2𝑥 − 3
Solution:
𝑦′ = 3𝑥2 − 5 2𝑥 − 3 ′ + 2𝑥 − 3 (3𝑥2 − 5 )′
)
𝑦′ = 3𝑥2 − 5 (2) + 2𝑥 − 3 (6𝑥
𝑦′ = 6𝑥2 − 10 + 12𝑥2 + 18𝑥
𝒚′
= 18𝒙2
+ 18𝒙 − 10
)
𝑦 = (3𝑥2 − 5)(2𝑥 + 3
𝑦 = 6𝑥3 + 9𝑥2 − 10𝑥 − 15
𝒚′ = 18𝒙2 + 18𝒙 − 10
𝑦 = 6(3)𝑥3−1
+ 9(2)𝑥2−1
− 10 − 0

19. REMEMBER:
In finding the derivative of the products of a function, the following methods can be
used:
1. By applying the Product Rule; and
2. By getting first the product, then, taking the derivative by applying the Sum
and Difference Rules.
In using the Product Rule, the following steps will be followed:
1. Multiply the first factor by the derivative of the second factor, and then add to
the product of the second factor and the derivative of the first factor.
2. Simplify the resulting derivatives.
In using getting the first product, the following steps will be followed:
1. Get the product of the first factor and the second factor.
2. Find the derivatives of the resulting products.
3. Simplify the final answer.

20. THE QUOTIENT RULE
2 Methods
1. By using the Quotient Rule
2. By splitting the fractions into two or more
fractions, or finding the quotient if possible,
then, taking the derivative using any of the
previous rules and simplify.

21. Definition. If 𝑦 =
𝑢(𝑥)
𝑣(𝑥)
where 𝑢 and 𝑣 are differentiable functions, then
𝒚’ =
𝒗 𝒙 . 𝒖′
𝒙 − 𝒖 𝒙 . 𝒗′(𝒙)
𝒗(𝒙) 𝟐
Illustrative Examples
1.Find the derivative of 𝒚 =
𝒙𝟐
𝒙−𝟏
.
Solution:
Let 𝑢 𝑥 = 𝑥2
and 𝑣 𝑥 = 𝑥 − 1
𝑦′
=
)
𝑥 − 1 2𝑥 − 𝑥2
(1
𝑥 − 1 2
𝑦′
=
2𝑥2
− 2𝑥 − 𝑥2
𝑥2 − 2𝑥 + 1
𝒚′
=
𝒙2
− 2𝒙
𝒙2 − 2𝒙 + 1
𝑪𝒉𝒆𝒄𝒌𝒊𝒏𝒈:
𝑦 =
𝑥2
𝑥 − 1
𝑥 − 1 ) 𝑥2
𝑥 − 1 +
1
𝑥−1
𝑥2
− 𝑥
𝑥
𝑥 − 1
1
𝑦 = 𝑥 + 1 + 𝑥 − 1 −1
𝑦′ = 1 + 0 + (−1) 𝑥 − 1 −2
𝑦′ = 1 −
1
𝑥 − 1 2
𝒚′
=
𝒙2
− 2𝒙
𝒙2 − 2𝒙 + 1

22. Illustrative Examples
2. Find the derivative of 𝒚 =
𝒙𝟑−𝟐
𝒙𝟐
Solution: 𝑪𝒉𝒆𝒄𝒌𝒊𝒏𝒈:
Let 𝑢 𝑥 = 𝑥3 − 2 and 𝑣 𝑥 = 𝑥2
𝑦′ =
)
𝑥2
3𝑥2
− (𝑥3
− 2)(2𝑥
𝑥4
𝑦′
=
3𝑥4
− 2𝑥4
+ 4𝑥
𝑥4
𝒚′ =
𝒙4
+ 4𝒙
𝑥4
𝒚′ =
)
𝒙(𝒙3
+ 4
𝑥4
𝒚′
=
𝒙3 + 4
𝑥3
𝑦 =
𝑥3
− 2
𝑥2
𝑦 = 𝑥 −
2
𝑥2
𝑦 = 𝑥 − 2𝑥−2
𝑦 = 1 + 4𝑥−3
𝑦 = 1 +
4
𝑥3
𝒚′
=
𝒙3 + 4
𝑥3

23. REMEMBER:
In finding the derivative of the quotient of a function, the following methods can
be used:
1. By applying the Quotient Rule; and
2. By splitting the quotient into two or more fractions or finding the quotient if
possible, then, take the derivative using any of the previous rules and
simplify.
In using the Quotient Rule, the following steps will be followed:
1. Subtract the product of the derivative of the numerator and the denominator
from the product of the derivative of the denominator and the numerator all
over the square of the denominator.
2. Simplify the resulting derivatives.
In finding the derivatives by splitting the quotient into two or more fractions, or
finding the quotient if possible, the following steps will be followed:
1. Express the given quotient of a function into two or more fractions, or take
the quotient of the given function.
2. Take the derivative of each term of the resulting quotient.
3. Simplify the resulting derivatives.

24. THE CHAIN RULE
Definition. If 𝑦 = 𝑢(𝑥) is a differentiable function, 𝑛 is is any real
number, and 𝒚 = 𝒖(𝒙) 𝒏, then 𝒚′ = 𝒏 𝒖 𝒙 𝒏−𝟏. 𝒖′(𝒙)
Illustrative Examples
1. Find the derivative of 𝑦 = (5𝑥 + 2)3.
Answer: 𝑦′
= 3(5𝑥 + 2)3−1
. 5
𝒚′
= 15(5𝒙 + 2)2
2. Find the derivative of 𝑦 = (3𝑥 − 5)
1
2.
Answer: 𝑦′ =
1
2
(3𝑥 − 5)
1
2
−1
. 3
𝑦′ =
3
2
3𝑥 − 5 −
1
2
𝒚′ =
3
2 3𝒙 − 5

25. REMEMBER:
In finding the derivative using Chain Rule, the following
methods will be followed:
1. Take the derivative of the given function by applying the
Constant Multiple or Power Rule.
2. Multiple the resulting derivative in step 1 to the derivative
of the function inside the parenthesis or grouping symbol.
3. Simplify the resulting derivative.

26. HIGHER ORDER DERIVATIVES
Illustrative Examples
1. If y=3𝑥5 − 2𝑥, 𝑓𝑖𝑛𝑑 𝑦′′′′
Solution:
𝑦′ = 15𝑥4 − 2
𝑦′′
= 60𝑥3
𝑦′′′
= 180𝑥2
𝒚′′′′ = 𝟑𝟔𝟎𝒙
2. If y = 2𝑥4
− 3𝑥3
+ 2𝑥 − 5, 𝑓𝑖𝑛𝑑 𝑦𝑉
𝑦′=8𝑥3 − 9𝑥2 + 2
𝑦′′
= 24𝑥2
− 18𝑥
𝑦′′′
= 48𝑥 − 18
𝑦′′′′
= 48
𝒚𝑽
= 𝟎
Solution:

27. Thank You!

28. May God, the Father protect us,
Jesus continue to teach us
and the Holy Spirit guide us lovingly
to live well the life God has given us.
We ask this through Christ, our Lord.
Amen.