1 | Calculus 1 31

UNIT III: THE DERIVATIVE

3.0 Learning Outcomes

At the end of this lesson packet, the students are expected to attain the
following intended learning outcomes:
3.0.1 Identify the different rules of differentiation.
3.0.2 Apply increments method and the rules of differentiation to
find the derivative of a given function.

3.1 Introduction

The Concept of Derivative

The derivative of a function of a real variable measures the

sensitivity to change of the function value (output value) with
respect to a change in its argument (input value).

Derivatives are a fundamental tool of calculus. For example,

the derivative of the position of a moving object with respect
to time is the object's velocity: this measures how quickly the
position of the object changes when time advances.

3.2 Topics and Discussion

3.2.1. Notion of the Derivative

The derivative of a nonlinear function is related to the rate of change of

a linear function, which is the same thing as the slope of a line. We consider
the problem of calculating the slope of the tangent line to a curve, and then we
use the solution to define the derivative. The discussion refers to the figure
below.
 1 | Calculus 1 32

The Slope of a Secant. Suppose we have a function 𝑓(𝑡), and we want

to determine the slope of the tangent line to the function 𝑓 at a point 𝑡 = 𝑎.
This poses a problem because algebraic methods can be used to calculate the
slope of a line only when two points on the line are known. Only one point,
namely the point (𝑎, 𝑓(𝑎)), on the tangent line is known. A secant line to a
curve is a line that connects two points on the graph. Since two points on a
secant line are known, we can always calculate the slope of a secant line. Let’s
consider the secant line that appears in the figure. This line connects the points
(𝑎, 𝑓(𝑎)) and (𝑎 + ℎ, 𝑓(𝑎 + ℎ)). Using the standard calculation for the slope of
a line, we determine the slope of the secant line to be
∆𝑓 𝑓 ( 𝑎 + ℎ ) − 𝑓 (𝑎 )
𝑚 𝑠𝑒𝑐 = =
∆𝑡 ℎ

The Slope of the Tangent. It is clear from the figure that the slope of a
secant is different from the slope of the tangent. However, it is also clear that
a secant slope approximation can always be improved by choosing a smaller
value of ℎ. Notice that the formula for the secant slope works for any non-zero
value of ℎ, no matter how small. However, it does not work for the value ℎ =
0. This means that we cannot directly calculate the tangent slope; however,
we can indirectly calculate it by determining what will happen to the secant
slope as ℎ becomes arbitrarily close to 0. This can sometimes be done exactly,
but it can always be done approximately, simply by selecting a very small vale
of ℎ. Mathematically, the concept is written using the limit notation.

𝑚𝑡𝑎𝑛 = lim 𝑚𝑠𝑒𝑐

ℎ→0

3.2.2. Definition

The terminology of the tangent slope and the notation 𝑚𝑡𝑎𝑛 are very
clumsy and refer neither to the specific function whose tangent slope is being
determined nor to the point at which the slope is being determined. A better
terminology is that of the derivative; specifically, the derivative of 𝑓 at 𝑎 is
defined to be the slope of the tangent to the curve 𝑓(𝑡) at the point t = a, and
is denoted 𝑓′ (𝑎).

𝑓(𝑎+ℎ)−𝑓(𝑎)
𝑓′(𝑎) = 𝑚𝑡𝑎𝑛 = lim 𝑚𝑠𝑒𝑐 = lim
ℎ→0 ℎ→0 ℎ

3.2.3. Determination of the Derivative by Increments

This section we will extend our discussion of limits and examine the
idea of the derivative, the basis of differential calculus. We will assume we
have a particular function of 𝑥, such that
 1 | Calculus 1 33

𝑦 = 𝑥2

If 𝑥 is assigned the value 10, the corresponding value of y will be

(10)2 𝑜𝑟 100. Now, if we increase the value of 𝑥 by 2, making it 12, we may
call this increase of 2 an increment or ∆ 𝑥. This results in an increase in the
value of 𝑦, and we may call this increase an increment or ∆ 𝑦. From this
we write
𝑦 + ∆𝑦 = (𝑥 + ∆ 𝑥)2
= (10 + 2)2
= 144
As x increases from 10 to 12, 𝑦 increases from 100 to 144 so that
∆𝑥 = 2
∆𝑦 = 44
∆𝑦 44
= = 22
∆𝑥 2
∆𝑦
We are interested in the ratio ∆𝑥 because the limit of this ratio as ∆𝑥 approaches
zero is the derivative of 𝑦 = 𝑓(𝑥)

As you recall from the discussion of limits, as ∆𝑥 is made smaller, ∆𝑦 gets

smaller also.

Example: Find the derivative of the given equation using the increment
𝑥−1
method, 𝑦 = 𝑥+1

Solution:

This is the original method in finding the derivative of any equations

using the Increment Method. If you know already the List of Formulas to find
a derivative of a certain equation, then you can use it. If you don't remember
the formulas, then you have to use the Increment Method and it is a long
process to do it.

Well, for the Increment Method, here are the steps

Step 1. You have to substitute x with x + Δx and substitute y with y + Δy,

𝑥−1
𝑦=
𝑥+1
 1 | Calculus 1 34

Step 2. Subtract the given equation from the original equation, we have,

Step 3. Get the Least Common Denominator (LCD) of the two terms of the
right,

Step 4. Simplify the above equation,

Step 5. Divide both sides of the equation by Δx,

Therefore,

The above equation can be written as

3.2.4. Differentiation Rules

The basic rules of differentiation are presented here along with several
examples. Remember that if 𝑦 = 𝑓(𝑥) is a function then the derivative of
𝑑𝑦 𝑑𝑓
𝑦 can be represented by 𝑑𝑥 or 𝑦’ or 𝑓’ or 𝑑𝑥.
 1 | Calculus 1 35

Rule 1. The derivative of a constant is zero.

Rule 2: The General Power Rule. The derivative of 𝑥 𝑛 is 𝑛𝑥 𝑛−1 . (Will discuss
further in Unit VI)

Example: Differentiate = 𝑥 4 .

𝑑𝑦
Solution: If 𝑦 = 𝑥 4 then using the general power rule, = 4𝑥 3 .
𝑑𝑥

Rule 3: The Derivative of a Constant times a Function. The derivative of

𝑘𝑓(𝑥), where 𝑘 is a constant, is 𝑘𝑓’(𝑥).

Example: Differentiate = 3𝑥 2 .

Solution: In this case 𝑓(𝑥) = 𝑥 2 and 𝑘 = 3, therefore the derivative is

3(2𝑥 1 ) = 6𝑥.

Rule 4: The Derivative of a Sum or a Difference.

𝑑𝑓 𝑑ℎ 𝑑𝑔
If 𝑓(𝑥) = ℎ(𝑥) ± 𝑔(𝑥), then 𝑑𝑥 = 𝑑𝑥 ± 𝑑𝑥

Example: Differentiate 𝑓(𝑥) = 3𝑥 2 − 7𝑥.

𝑑ℎ
Solution: In this case ℎ(𝑥 ) = 3𝑥 2 and 𝑔(𝑥) = 7𝑥 and so 𝑑𝑥 = 6𝑥 and
𝑑𝑔
= 7.
𝑑𝑥

𝑑𝑓
Therefore, 𝑑𝑥 = 6𝑥 − 7.

Rule 5: The Product Rule. The derivative of the product 𝑦 = 𝑢(𝑥)𝑣(𝑥), where
𝑑𝑦 𝑑𝑣 𝑑𝑢
𝑢 and v are both functions of 𝑥 is 𝑑𝑥 = 𝑢 𝑑𝑥 + 𝑣 𝑑𝑥

Example: Differentiate 𝑓(𝑥) = (6𝑥 2 + 2𝑥)(𝑥 3 + 1).

Solution: Let 𝑢(𝑥) = 6𝑥 2 + 2𝑥 and 𝑣(𝑥) = 𝑥 3 + 1. Therefore,

𝑑𝑢 𝑑𝑣
Therefore, 𝑑𝑥 = 12𝑥 + 2 and = 3𝑥 2 .
𝑑𝑥

𝑑𝑦 𝑑𝑣 𝑑𝑢
Therefore using the formula for the product rule, 𝑑𝑥 = 𝑢 𝑑𝑥 + 𝑣 𝑑𝑥 .

We get,
 1 | Calculus 1 36

𝑑𝑓
= (6𝑥 2 + 2𝑥 )(3𝑥 2 ) + (𝑥 3 + 1)(12𝑥 + 2)
𝑑𝑥
= 18𝑥 4 + 6𝑥 3 + 12𝑥 4 + 2𝑥 3 + 12𝑥 + 2
= 30𝑥 4 + 8𝑥 3 + 12𝑥 + 2

𝑢(𝑥)
Rule 6: The Quotient Rule. The derivative of the quotient (𝑥 ) = , where 𝑢
𝑣(𝑥)
and 𝑣 are both function of 𝑥 is

𝑑𝑢 𝑑𝑣
𝑑𝑓 𝑣 −𝑢
𝑑𝑥 𝑑𝑥
=
𝑑𝑥 𝑣2

𝑥2 + 7
Example: Differentiate (𝑥) = 3𝑥 − 1 .

Solution: Let 𝑢(𝑥) = 𝑥2 + 7 and 𝑣(𝑥) = 3𝑥 − 1.

𝑑𝑢 𝑑𝑣
Differentiate these to get 𝑑𝑥 = 2𝑥 and 𝑑𝑥 = 3.

Now using the formula for the quotient rule we get,

𝑑𝑓 (3𝑥 – 1)(2𝑥) – (𝑥 2 + 7)(3)

= (3𝑥 – 1)2
𝑑𝑥
6𝑥 2 − 2𝑥 − 3𝑥 2 − 21
=
(3𝑥 − 1)2
2
𝑑𝑓 3𝑥 − 2𝑥 − 21
=
𝑑𝑥 (3𝑥 − 1)2

Rule 7: The Chain Rule. If 𝑦 is a function of 𝑢, meaning 𝑦 = 𝑓(𝑢), and 𝑢 is a

function of 𝑥, meaning 𝑢 = 𝑔(𝑥) then the derivative of 𝑦 with respect
to 𝑥 is
𝑑𝑦 𝑑𝑦 𝑑𝑢
= 𝑑𝑢 . 𝑑𝑥 . (Will discuss further in Unit VI)
𝑑𝑥

Example: Differentiate = (𝑥 2 − 5)4 .

Solution: Let 𝑢 = 𝑥 2 − 5, therefore 𝑦 = 𝑢4 .

𝑑𝑢 𝑑𝑦
= 2𝑥 and 𝑑𝑢 = 4𝑢3 .
𝑑𝑥
Using the chain rule we then get
𝑑𝑦 𝑑𝑦 𝑑𝑢
= . , = (4𝑢3 )(2𝑥 )
𝑑𝑥 𝑑𝑢 𝑑𝑥
= 4(𝑥 2 − 5)3 (2𝑥 )
𝑑𝑦
= 8𝑥 (𝑥 2 − 5)3
𝑑𝑥
 1 | Calculus 1 37

👉 Exercise #1: Determine the derivatives of the following function using the
increments method.

1. 𝑓 (𝑧) = 5𝑧 6. 𝑓 (𝑧) = 2𝑧 -3
1
2. 𝑓 (𝑧) = −8𝑧 7. 𝑓 (𝑧) = 𝑧
−2
3. 𝑓 (𝑧) = 𝑧5 8. 𝑓 (𝑧) = 𝑧
−𝑧
4. 𝑓 (𝑧) = 6𝑧3 9. 𝑓 (𝑧) = 3
𝑧4
5. 𝑓 (𝑧) = −𝑧4 10. 𝑓 (𝑧) = 4

👉 Exercise #2: Complete the table below. Show your solution in a separate sheet.
𝑓(𝑥) 𝑓′(𝑥) 𝑓(𝑥) 𝑓′(𝑥)
1
𝑥
5𝑥 2
3
−4𝑥 + 2 −
𝑥
𝑥 4 − 3𝑥 −𝑥
3𝑥 2
−3𝑥 2 − 5
√𝑥
−3𝑥 −3 3 − 𝑥2

👉 Exercise #3: Use the rules of differentiation to find the derivative of each of the
following.

5 (𝑡+3)3
1. 𝑚 = 2 𝑛3 − 2 4. 𝑘(𝑡) =
√𝑡
9 3
2. 𝑞 = 2 𝑝 5. 𝑓 (𝑥 ) = (2ℎ + 1)5 (ℎ + 2)
(𝑥+3)2
3. 𝑔(ℎ) = 3(ℎ + 1)2 (ℎ − 2) 6. 𝑦 = 𝑥3
 1 | Calculus 1 38

1.3 References

Math Principles in everyday life (2020). Derivative – Increment Method.

http://www.math-principles.com/2012/11/derivative-increment-method.html

EVLM (2020). Rules of Differentiation.

http://evlm.stuba.sk/~partner7/DBfiles/Modules/Differentiation/RulesDifferenti
ation.pdf
Integrated Publishers (2020). Increments and Differentiation.
https://www.tpub.com/math2/40.htm

Glenn Ledder (2002). The Concept of the Derivatives.

https://www.math.unl.edu/~gledder1/Math208/DerivativeConcept.pdf

The ARC (2020). Definition of Derivatives.

https://web.iit.edu/sites/web/files/departments/academic-affairs/academic-
resource-center/pdfs/what_are_Derivative.pdf

Mohamed A. Khamsi (2000). The Derivative.

http://www.sosmath.com/calculus/diff/der00/der00.html

Math is Fun (2017) Introduction to Derivatives.

https://www.mathsisfun.com/calculus/derivatives-introduction.html

Note: Should you have questions about the topics discussed in this module, please feel
free to text or call me thru this cellphone number, 09672200502. I will appreciate if
you’ll ask questions for clarifications to have sufficient learning.