BEM 111 – ENGINEERING CALCULUS 1 | 1st Sem.

SY 2022-2023

BEM 111 – ENGINEERING CALCULUS 1

Lecture No. 3

DERIVATIVE, SLOPE AND RATE OF CHANGE

3.1 INTRODUCTION

How fast are you really going?

When you ride your bicycle down a straight bike path and as you proceed, you keep track of your
exact position—so that, at every instant, you know exactly where you are. How fast were you going one
second after you started?

By considering this question, we are led to some important mathematical ideas. First, recall the
∆𝑥
formula for average velocity: 𝒗 = , where ∆𝑥 is the change in your position and ∆𝑡 is the time taken.
∆𝑡
Thus, average velocity is the rate of change of position with respect to time.

Velocity is an important example of a derivative, but this is just one example. The world, as we
know it, is full of quantities which change with respect to each other – and these rates of change can often
be expressed as derivatives.

3.2 INTERPRETATION OF DERIVATIVES BY GEOMETRY

There’s no better way to discuss calculus other than giving the proper illustration, in the
succeeding discussions, the theorem which is fundamental in all applications of the differential calculus to
geometry will be considered.

Consider a function 𝑦 = 𝑓(𝑥) and its graph is a curve in the plane. We wish to find the slope of
this curve at a certain point. It is understood that the slope is defined as the rise over run. Now, given a
curve defined by 𝑦 = 𝑓(𝑥) and a point P on the curve, consider another point P’ on the curve near P, and
draw the line PP’ connecting P and P’. This line is called a secant line.

In the figure, the curve AB represent the graph of a given

∆𝑥
function. The ratio ∆𝑦 is the slope of the line joining the points
P and P’. The delta ∆ is the increment of a value, thus ∆𝑥 is
read as ‘increment of x”. It is also denoted that when x
changes to 𝑥 + ∆𝑥 , y changes to the value of 𝑦 + ∆𝑦.

Figure 1 shows the secant line drawn as described above.

Thus, it can be interpreted using the four-step rule below, given the function:
𝑦 = 𝑓(𝑥)

FIRST STEP. 𝑦 + ∆𝑦 = 𝑓(𝑥 + ∆𝑥)

SECOND STEP. ∆𝑦 = 𝑓 (𝑥 + ∆𝑥 ) − 𝑦 → transpose y but since y = f(x)

Then ∆𝑦 = 𝑓 (𝑥 + ∆𝑥 ) − 𝑓(𝑥)

THIRD STEP. By dividing both sides by ∆𝑥, we have

∆𝑦 𝑓(𝑥+∆𝑥)−𝑓(𝑥)
= → the slope of the secant line PP’
∆𝑥 ∆𝑥

1 | E n g r . J U N I E V I N R A M I L L A N O | INSTRUCTOR 1
 BEM 111 – ENGINEERING CALCULUS 1 | 1st Sem. SY 2022-2023

FOURTH STEP. As ∆𝑥 approaches to zero, ∆𝑦 also approaches to zero and P’ approaches P

along the curve. In this case, the line PP’ approaches a certain line PT, as
∆𝑦
the limiting position. That is, the ratio ∆𝑥 approaches a limit, is actually also
the slope of the line PT. This limit is called the derivative of y with respect
to x. In other words, the derivative of y with respect x is the limit of the
∆𝑦
ration ∆𝑥 when ∆𝑥 approaches to zero. That is,
𝒅𝒚 ∆𝒚 𝒇(𝒙 + ∆𝒙) − 𝒇(𝒙)
= 𝐥𝐢𝐦 = 𝐥𝐢𝐦
𝒅𝒙 ∆𝒙→𝟎 ∆𝒙 ∆𝒙→𝟎 ∆𝒙
𝑑𝑦
Other symbols of the derivative are 𝑦 ′, 𝑓 ′(𝑥 ) 𝑜𝑟 .
𝑑𝑥
Example 3.1
Differentiate the function 𝑦 = 𝑥 3 − 2𝑥 using the four-step rule.

Solution
𝑦 = 𝑥 3 − 2𝑥
𝑦 + ∆𝑦 = (𝑥 + ∆𝑥)3 − 2(𝑥 + ∆𝑥)
∆𝑦 = (𝑥 + ∆𝑥)3 − 2(𝑥 + ∆𝑥 ) − 𝑦
∆𝑦 = (𝑥 + ∆𝑥)3 − 2(𝑥 + ∆𝑥 ) − (𝑥 3 − 2𝑥 )
∆𝑦 = 𝑥 3 + 3𝑥 2 ∆𝑥 + 3𝑥(∆𝑥)2 + (∆𝑥 )3 − 2𝑥 − 2∆𝑥 − 𝑥 3 + 2𝑥
∆𝑦 = 3𝑥 2 ∆𝑥 + 3𝑥(∆𝑥)2 + (∆𝑥 )3 − 2∆𝑥
1 1
( ) [ ∆𝑦 = 3𝑥 2 ∆𝑥 + 3𝑥(∆𝑥)2 + (∆𝑥 )3 − 2∆𝑥 ] ( )
∆𝑥 ∆𝑥
∆𝑦 3𝑥 2 ∆𝑥 3𝑥(∆𝑥)2 (∆𝑥 )3 2∆𝑥
= + + −
∆𝑥 ∆𝑥 ∆𝑥 ∆𝑥 ∆𝑥
∆𝑦
= 3𝑥 2 + 3𝑥∆𝑥 + (∆𝑥 )2 − 2
∆𝑥
As ∆x approaches to zero
𝑑𝑦 ∆𝑦
= lim = 3𝑥 2 + 3𝑥(0) + (0)2 − 2
𝑑𝑥 ∆𝑥→0 ∆𝑥
𝒅𝒚
= 𝟑𝒙𝟐 − 𝟐
𝒅𝒙

Example 3.2
1
Differentiate the function 𝑦 = 𝑡 using the four-step rule.

Solution
1
𝑦=
𝑡
1
𝑦 + ∆𝑦 =
𝑡 + ∆𝑡
1 1 1 𝑡 − (𝑡 + ∆𝑡)
∆𝑦 = −𝑦= − =
𝑡 + ∆𝑡 𝑡 + ∆𝑡 𝑡 𝑡 (𝑡 + ∆𝑡)
𝑡 − 𝑡 − ∆𝑡 −∆𝑡
∆𝑦 = =
𝑡(𝑡 + ∆𝑡) 𝑡(𝑡 + ∆𝑡)

1 −∆𝑡 1
( ) [∆𝑦 = ]( )
∆𝑡 𝑡(𝑡 + ∆𝑡) ∆𝑡
∆𝑦 −∆𝑡 −1
= =
∆𝑡 𝑡(𝑡 + ∆𝑡)∆𝑡 𝑡(𝑡 + ∆𝑡)
As ∆t approaches to zero
𝑑𝑦 ∆𝑦 −1 𝟏
= lim = =− 𝟐
𝑑𝑡 ∆𝑡→0 ∆𝑡 𝑡(𝑡 + 0) 𝒕

2 | E n g r . J U N I E V I N R A M I L L A N O | INSTRUCTOR 1
 BEM 111 – ENGINEERING CALCULUS 1 | 1st Sem. SY 2022-2023

Example 3.3
Differentiate the function 𝑦 = √𝑥 using the four-step rule.

Solution 𝑦 = √𝑥
𝑦 + ∆𝑦 = √𝑥 + ∆𝑥
∆𝑦 = √𝑥 + ∆𝑥 − 𝑦 = √𝑥 + ∆𝑥 − √𝑥
(√𝑥 + ∆𝑥 + √𝑥)
∆𝑦 = √𝑥 + ∆𝑥 − √𝑥 ∙
(√𝑥 + ∆𝑥 + √𝑥)
𝑥 + ∆𝑥 − 𝑥 ∆𝑥
∆𝑦 = =
(√𝑥 + ∆𝑥 + √𝑥) (√𝑥 + ∆𝑥 + √𝑥)
1 ∆𝑥 1
( ) ( ∆𝑦 = )( )
∆𝑥 (√𝑥 + ∆𝑥 + √𝑥) ∆𝑥
∆𝑦 ∆𝑥 1
= =
∆𝑥 ∆𝑥(√𝑥 + ∆𝑥 + √𝑥) √𝑥 + ∆𝑥 + √𝑥

As ∆x approaches to zero
𝑑𝑦 ∆𝑦 1 𝟏
= lim = =
𝑑𝑥 ∆𝑥→0 ∆𝑥 √𝑥 + 0 + √𝑥 𝟐√𝒙

Example 3.4
Differentiate the function 𝑦 = sin 𝑥 using the four-step rule.

Solution
𝑦 + ∆𝑦 = sin(𝑥 + ∆𝑥)
∆𝑦 = sin(𝑥 + ∆𝑥) − 𝑦
∆𝑦 = sin(𝑥 + ∆𝑥) − 𝑦
Since:
∆𝑦 = sin 𝑥 cos ∆𝑥 + sin ∆𝑥 cos 𝑥 − sin 𝑥 sin(𝑎 + 𝑏) = sin(𝑎) cos(𝑏) + sin(𝑏) cos(𝑎)
if: ∆x approaches zero
sin 𝑥 cos ∆𝑥
= 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
∆𝑥

Simplify further to have a form of

∆𝑦 = sin ∆𝑥 cos 𝑥 − sin 𝑥 (1 − cos ∆𝑥 )
Identities:
1
sin2 𝐴 = (1 − cos 2𝐴) or 2sin2 𝐴 = (1 − cos 2𝐴)
2

1
𝐿𝑒𝑡, 2𝐴 = ∆𝑥 ; 𝐴= ∆𝑥
2
Therefore:
1
2sin2 (2 ∆𝑥) = (1 − cos ∆𝑥 )

∆𝑥
∆𝑦 = sin ∆𝑥 cos 𝑥 − sin 𝑥 (2 sin2 )
2

1 ∆𝑥 1
( ) (∆𝑦 = sin ∆𝑥 cos 𝑥 − 2sin 𝑥 sin2 ) ( )
∆𝑥 2 ∆𝑥

∆𝑥
∆𝑦 2sin 𝑥 sin2 2
= sin ∆𝑥 cos 𝑥 −
∆𝑥 ∆𝑥

2 ∆𝑥
∆𝑦 sin ∆𝑥 cos 𝑥 sin 𝑥 sin 2
= −
∆𝑥 ∆𝑥 ∆𝑥
2
∆𝑥
∆𝑦 sin ∆𝑥 cos 𝑥 ∆𝑥 sin 2
= − sin 𝑥 ∙ sin ∙
∆𝑥 ∆𝑥 2 ∆𝑥
2

3 | E n g r . J U N I E V I N R A M I L L A N O | INSTRUCTOR 1
 BEM 111 – ENGINEERING CALCULUS 1 | 1st Sem. SY 2022-2023

∆𝑦 ∆𝑥
= cos 𝑥 − sin 𝑥 ∙ sin ∙
∆𝑥 2

As ∆x approaches to zero

𝑑𝑦 ∆𝑦 0
= lim = cos 𝑥 − sin 𝑥 ∙ sin = 𝐜𝐨𝐬 𝒙
𝑑𝑥 ∆𝑥→0 ∆𝑥 2

Example 3.5
𝟏
Differentiate the function 𝒚 = 𝟐 (𝟑𝒙𝟐 + 𝟏)𝟐 using the four-step rule.

Solution
1
𝑦= (3𝑥 2 + 1)2
2
1
𝑦 + ∆𝑦 = (3(𝑥 + ∆𝑥)2 + 1)2
2
1
∆𝑦 = (3(𝑥 + ∆𝑥)2 + 1)2 − 𝑦
2
1 1
∆𝑦 = (3(𝑥 + ∆𝑥)2 + 1)2 − (3𝑥 2 + 1)2
2 2
1 1
∆𝑦 = (3𝑥 2 + 6𝑥∆𝑥 + 3(∆𝑥)2 + 1)2 − (9𝑥 4 + 6𝑥 2 + 1)
2 2
1 1
∆𝑦 = (3𝑥 2 + 6𝑥∆𝑥 + 3(∆𝑥)2 + 1)2 − (9𝑥 4 + 6𝑥 2 + 1)
2 2

1 1
∆𝑦 = (9𝑥 4 + 9(∆𝑥)4 + 36𝑥 3 ∆𝑥 + 54𝑥 2 (∆𝑥)2 + 36𝑥(∆𝑥)3 + 6𝑥 2 + 12𝑥∆𝑥 + 3(∆𝑥)2 + 1) − (9𝑥 4 + 6𝑥 2 + 1)
2 2
1
∆𝑦 = (9(∆𝑥)4 + 36𝑥 3 ∆𝑥 + 54𝑥 2 (∆𝑥)2 + 36𝑥(∆𝑥)3 + 12𝑥∆𝑥 + 3(∆𝑥)2 )
2
Divide by ∆𝑥
∆𝑦 9(∆𝑥)4 + 36𝑥 3 ∆𝑥 + 54𝑥 2 (∆𝑥)2 + 36𝑥(∆𝑥)3 + 12𝑥∆𝑥 + 3(∆𝑥)2
=
∆𝑥 2∆𝑥
∆𝑦 9(∆𝑥)3
= + 18𝑥 3 + 27𝑥 2 ∆𝑥 + 18𝑥(∆𝑥)2 + 6𝑥 + 3∆𝑥
∆𝑥 2

As ∆x approaches to zero

𝑑𝑦 ∆𝑦
= lim
𝑑𝑥 ∆𝑥→0 ∆𝑥
9(0)3
= + 18𝑥 3 + 27𝑥 2 (0) + 18𝑥(0)2 + 6𝑥 + 3(0)
2

𝒅𝒚
= 𝒚′ = 𝟏𝟖𝒙𝟑 + 𝟔𝒙
𝒅𝒙

4 | E n g r . J U N I E V I N R A M I L L A N O | INSTRUCTOR 1
 BEM 111 – ENGINEERING CALCULUS 1 | 1st Sem. SY 2022-2023

3.2 DERIVATIVES INTERPRETED AS SLOPE

Recall that in the fourth step of the four-step rule, the value of the derivative at any point of a
curve is equal to the slope of the tangent line to the curve at that point. This will enable us to determine
slopes of the tangent line of any curve by using only the concept of derivatives.

Let’s start by determining the slope of a straight line. Though we know already that if the equation
of the line is in the form of 𝑦 = 𝑚𝑥 + 𝑏, the slope is ‘m’, we are going to determine it using derivatives.

Given the function: 𝑦 = 𝑚𝑥 + 𝑏

FIRST STEP. 𝑦 + ∆𝑦 = 𝑚(𝑥 + ∆𝑥) + 𝑏

SECOND STEP ∆𝑦 = 𝑚(𝑥 + ∆𝑥) + 𝑏 − (𝑚𝑥 + 𝑏)

∆𝑦 = 𝑚𝑥 + 𝑚∆𝑥 + 𝑏 − 𝑚𝑥 − 𝑏
THIRD STEP
∆𝑦 𝑚∆𝑥
= =𝑚
∆𝑥 ∆𝑥
As ∆x approaches to zero
FOURTH STEP
𝒅𝒚 ∆𝒚
= 𝐥𝐢𝐦 = 𝒚′ = 𝒎
𝒅𝒙 ∆𝒙→𝟎 ∆𝒙
Example 3.6
Find the slope of the curve 1 at 𝑥 = 2.
𝑦 = 𝑥 3 − 2𝑥
4
Solution
1 3
𝑦= 𝑥 − 2𝑥
4
1
𝑦 + ∆𝑦 = (𝑥 + ∆𝑥)3 − 2(𝑥 + ∆𝑥)
4
1
∆𝑦 = (𝑥 + ∆𝑥)3 − 2(𝑥 + ∆𝑥) − 𝑦
4
1 1
∆𝑦 = ( (𝑥 + ∆𝑥)3 − 2(𝑥 + ∆𝑥)) − ( 𝑥 3 − 2𝑥)
4 4
1 1
∆𝑦 = ( [𝑥 3 + 3𝑥 3 ∆𝑥 + 3𝑥(∆𝑥)2 + (∆𝑥)3 ] − 2𝑥 − 2𝑥∆𝑥) − ( 𝑥 3 − 2𝑥)
4 4
1 3 3 3 3 1 1
∆𝑦 = 𝑥 + 𝑥 ∆𝑥 + 𝑥(∆𝑥)2 + (∆𝑥)3 − 2𝑥 − 2𝑥∆𝑥 − 𝑥 3 + 2𝑥
4 4 4 4 4
3 3 3 1
∆𝑦 = 𝑥 ∆𝑥 + 𝑥(∆𝑥)2 + (∆𝑥)3 − 2𝑥∆𝑥
4 4 4
Divide by ∆𝑥,

∆𝑦 3 𝑥 3 ∆𝑥 3 𝑥(∆𝑥)2 1 (∆𝑥)3 2𝑥∆𝑥

= + + −
∆𝑥 4 ∆𝑥 4 ∆𝑥 4 ∆𝑥 ∆𝑥
∆𝑦 3 3 3 1
= 𝑥 + 𝑥∆𝑥 + (∆𝑥)2 − 2𝑥
∆𝑥 4 4 4

As ∆x approaches to zero,
𝑑𝑦 ∆𝑦 3 3 3 1
= lim = 𝑥 + 𝑥(0) + 02 − 2𝑥
𝑑𝑥 ∆𝑥→0 ∆𝑥 4 4 4
𝑑𝑦 3
= 𝑦′ = 𝑥 3 − 2𝑥
𝑑𝑥 4
at x = 2
𝑑𝑦 3
= 𝑦 ′ = (2)3 − 2(2) = 𝟏 ; 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑢𝑟𝑣𝑒 𝑎𝑡 𝑃(2, −2)
𝑑𝑥 4

NOTE: More examples will be given in the application of derivatives in the succeeding lessons.

5 | E n g r . J U N I E V I N R A M I L L A N O | INSTRUCTOR 1
 BEM 111 – ENGINEERING CALCULUS 1 | 1st Sem. SY 2022-2023

3.2 RATE OF CHANGE

Although the succeeding discussion is purely application, the highlight is still the use of the
∆𝑦
fundamental concept of the four-step rule. While the ratio ∆𝑥 in the preceding lessons is denoted as the
slope of a tangent line, it can also denote as the average rate of change over the interval ∆𝑥. As ∆𝑥
∆𝑦
approaches to zero, the ratio approaches a limiting value. It is also called the instantaneous rate or
∆𝑥
instant rate in a certain point.

Example 3.7
Find the rate at which the reciprocal of a number changes as the number increases.

Solution
n = independent (number)
1
The key formula is 𝑟 = r = dependent (reciprocal)
𝑛
𝑑𝑟
1 We want to solve
𝑟 + ∆𝑟 = 𝑑𝑛
𝑛 + ∆𝑛
1 1 1 𝑛 − (𝑛 + ∆𝑛)
∆𝑟 = −𝑟 = − =
𝑛 + ∆𝑛 𝑛 + ∆𝑛 𝑛 𝑛(𝑛 + ∆𝑛)
𝑛 − 𝑛 − ∆𝑛 ∆𝑛
∆𝑟 = =
(
𝑛 𝑛 + ∆𝑛 ) 𝑛 𝑛 + ∆𝑛)
(

Divide by ∆𝑛,
∆𝑟 ∆𝑛 1
= =
∆𝑛 𝑛(𝑛 + ∆𝑛)∆𝑛 𝑛(𝑛 + ∆𝑛)
As ∆n approaches to zero,

𝑑𝑟 1 𝟏
= = 𝟐
𝑑𝑛 𝑛(𝑛 + 0) 𝒏

Example 3.8
The surface area of a sphere, initially zero increases uniformly at the rate of 4 sq.in per sec. Find the rate at which
the radius is increasing at the end of 2 sec.

Solution
Let:
t = time (sec) → independent variable
r = radius of sphere (in) → dependent variable
S = surface area (𝑖𝑛2 )

𝑑𝑟
We want to solve 𝑑𝑡 at t = 2 sec

At a constant rate (4 sq. in per sec.)

Surface Area = rate ∙ time
𝑆 =4∙𝑡

To solve for the rate of change of the radius with respect to time, we need equation/s in terms of r and t.

Surface area of sphere : 𝑆 = 4𝜋𝑟 2 and 𝑆=4 ∙ 𝑡

Equate the two equations

𝑆=𝑆
4𝜋𝑟 2 = 4𝑡 𝑡
𝑟=√
𝜋
𝑡
𝑟2 =
𝜋

6 | E n g r . J U N I E V I N R A M I L L A N O | INSTRUCTOR 1
 BEM 111 – ENGINEERING CALCULUS 1 | 1st Sem. SY 2022-2023

Differentiate the function,

𝑡 √𝑡
𝑟=√ =
𝜋 √𝜋

√𝑡 + ∆𝑡
𝑟 + ∆𝑟 =
√𝜋
√𝑡 + ∆𝑡 √𝑡 + ∆𝑡 √𝑡
∆𝑟 = −𝑟= −
√𝜋 √𝜋 √𝜋
Rationalizing:
√𝑡 + ∆𝑡 − √𝑡
∆𝑟 = (√𝑎 − √𝑏)(√𝑎 + √𝑏) = 𝑎 − 𝑏
√𝜋
√𝑡 + ∆𝑡 − √𝑡 (√𝑡 + ∆𝑡 + √𝑡)
∆𝑟 = ∙
√𝜋 (√𝑡 + ∆𝑡 + √𝑡)
𝑡 + ∆𝑡 − 𝑡 ∆𝑡
∆𝑟 = =
√𝜋(√𝑡 + ∆𝑡 + √𝑡) √𝜋(√𝑡 + ∆𝑡 + √𝑡)

Divide by ∆𝑡,
∆𝑟 ∆𝑡 1
= =
∆𝑡 √𝜋(√𝑡 + ∆𝑡 + √𝑡)∆𝑡 √𝜋(√𝑡 + ∆𝑡 + √𝑡)

As ∆t approaches to zero,

𝑑𝑟 1 1 𝟏
= = =
𝑑𝑡 √𝜋(√𝑡 + 0 + √𝑡) √𝜋(√𝑡 + √𝑡) 𝟐√𝝅√𝒕

when t = 2 sec
𝑑𝑟 1 1
= = = 𝟎. 𝟐𝟎 𝒔𝒆𝒄
𝑑𝑡 2√𝜋√𝑡 2√𝜋√2

The rate at which the radius is increasing at the end of 2.0 seconds.

NOTE: More examples will be given in the application of derivatives in the succeeding lessons.

SEAT WO RK #3

Using the four-step rule, differentiate the following functions:

1. 𝑦 = 𝑥 4 − 2𝑥 2 + 3𝑥 − 4
𝜃
2. 𝜌 = 𝜃+2
3. 𝑦 = cos 𝑥
4. 𝑠 = 𝑎𝑡 2 + 𝑏𝑡 + 𝑐 where a, b and c are constant
𝑥2
5. Find the slope of the function 𝑦 = 2𝑥 − at 𝑥 = 3
2

7 | E n g r . J U N I E V I N R A M I L L A N O | INSTRUCTOR 1