Differential Calculus

BWMA01
Business Mathematics
Chapter 3:
Differential Calculus
Prof. Dr. Silke Jütte und Dr. Mehmet Evrim

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(18th ed.) Springer.
List of Sources

QM220 BUSINESS MATHEMATICS – CHAPTER 4 – DIFFERENTIAL CALCULUS
3
Chapter Outline 3: Differential Calculus
3
3.1 Introduction
3.2 Rules of Differentiation
3.3 First and Higher Derivatives
Chapter 3 Differential Calculus
3.4 Curve Sketching

BACKGROUND
4
Introduction (1/7)
The slope of a linear function
“What is the change in y per unit of change in x?”
Take the following linear function:
𝑓 𝑥 = 2𝑥 + 4
The slope of a linear function is always the coefficient of x
Let us see why: Make a table of values and take the difference quotient ∆𝑦
∆𝑥
(= change of y, per change of unit of x)
x y ∆𝑦
∆𝑥
1 6
2 8 From x = 1 to x = 2: 2
1 = 2
3 10 From x = 1 to x = 3: 4
2 = 2
4 12 From x = 1 to x = 4: 6
3 = 2

BACKGROUND
5
Introduction (2/7)
Δx = 4
Δy = 8

BACKGROUND
6
Introduction (3/7)
The slope of a non-linear function
Determine the slope of the following non-linear function at the point A:
A
B
Δx = 0.2
Δy = 0.2

BACKGROUND
7
Introduction (4/7)
A
B
Δx ≈ 0.14
Δy ≈ 0.19

BACKGROUND
8
Introduction (5/7)
A
B
Δx ≈ 0.10
Δy ≈ 0.15

BACKGROUND
9
Introduction (6/7)
The slope of a non-linear function can be found if the difference between A and B becomes very
small (marginal), i.e., we have to determine 𝐥𝐢𝐦
∆𝒙→𝟎
∆𝒚
∆𝒙
We call this the
1st derivative or 𝒚′, or
𝒅𝒚
𝒅𝒙
. A

BACKGROUND
10
Introduction (7/7)
Example:
𝑦 = 𝑥2
Build the limit of the difference quotient:
𝑦′
= lim
∆𝑥→0
∆𝑦
∆𝑥
𝑦′
= lim
∆𝑥→0
(𝑥 + ∆𝑥)2−𝑥2
∆𝑥
𝑦′
= lim
∆𝑥→0
𝑥2
+ 2∆𝑥 ∙ 𝑥 + ∆𝑥2
− 𝑥2
∆𝑥
= lim
∆𝑥→0
2∆𝑥 ∙ 𝑥 + ∆𝑥2
∆𝑥
= lim
∆𝑥→0
∆𝑥(2𝑥 + ∆𝑥)
∆𝑥
𝑦′
= lim
∆𝑥→0
2𝑥 + ∆𝑥 = 2𝑥
y: The change in y, if x
changes by x.

11
Class Exercise 3.1: Rules of Differentiation (1/2)
CLASS
EXERCISE
Task 3.1.1
Differential calculus is a field that studies
a) change tendencies.
b) functional tendencies.
c) calculation tendencies.
d) mathematical tendencies.
Task 3.1.2
Which of the following statements is correct?
a) The derivative indicates the slope at a point and
thus corresponds to the difference quotient.
b) The derivative indicates the slope between two
points and thus corresponds to the differential
quotient.
c) The derivative indicates the slope at a point and
thus corresponds to the differential quotient.
d) The derivative indicates the slope between two
points and thus corresponds to the difference
quotient.

12
Class Exercise 3.1: Rules of Differentiation (1/2)
CLASS
EXERCISE
Task 3.1.3
Given that the continuous
function f holds F'(x) = ƒ(x), what is
function F called?
a) primitive function to ƒ
b) primitive function to ƒ'
c) integrand to ƒ
d) integrand to ƒ'
Task 3.1.4
With which of the following can you solve
an optimization problem?
a) the distributive property
b) the associative property
c) functional calculus
d) differential calculus

13
13
3.1 Introduction
3.4 Curve Sketching

14
Luckily, one must not always built the limit of the difference quotient, but rather apply rules of
differentiation.
Basic rules: 𝒇 𝒙 𝒇′ 𝒙
𝒚 = 𝒙𝒏
𝒚′ = 𝒏 ∙ 𝒙𝒏−𝟏
𝑦 = 𝑥2 𝑦′ = 2𝑥
𝑦 = 𝑥3
𝑦′
= 3𝑥2
𝑦 =
3
𝑥2 = 𝑥
2
3
𝑦′ =
2
3
𝑥
2
3
−1
=
2
3
𝑥−
1
3
𝑦 =
1
𝑥2
= 𝑥−2 𝑦′ = −2𝑥−3
Basic Rules of Differentiation (1/2)
If possible, reformulate
roots and quotients as
powers
HOW TO

15
Luckily, one must not always built the limit of the difference quotient, but rather apply rules of
differentiation.
Basic rules: 𝒇 𝒙 𝒇′ 𝒙
𝑦 = 𝑒𝑥 𝑦′ = 𝑒𝑥
𝑦 = ln x
𝑦′
=
1
𝑥
𝑦 = sin 𝑥 𝑦′
= cos 𝑥
𝑦 = cos 𝑥 𝑦′
= − sin 𝑥
𝑦 = 𝐶 𝑦′ = 0
𝑦 = 𝐶 ∙ f(x) 𝑦 = 𝐶 ∙ f′(x)
Basic Rules of Differentiation (2/2)
Taking the derivative,
remove constant
summands, but keep
constant factors
TROUBLE SPOT

16
Sum or Difference rule:
𝑦 = 𝑓 𝑥 ± 𝑔(𝑥) 𝑦′
= 𝑓′
𝑥 ± 𝑔′
𝑥
Example:
𝑦 = 3𝑥2 + 𝑙𝑛𝑥 − 𝑒𝑥 + 𝑥 + 2
𝑦′
= 6𝑥 +
1
𝑥
− 𝑒𝑥
+
1
2
𝑥−
1
2
Sum or Difference Rule

17
Product rule:
𝑦 = 𝑓 𝑥 ∙ 𝑔(𝑥) 𝑦′
= 𝑓′
𝑥 ∙ 𝑔 𝑥 + 𝑓 𝑥 ∙ 𝑔′
𝑥
Often written as follows:
𝑦 = 𝑢 ∙ 𝑣 𝑦′
= 𝑢′
∙ 𝑣 + 𝑢 ∙ 𝑣′
Example:
𝑦 = 4𝑥5
∙ 𝑙𝑛𝑥
Product Rule
𝑢 = 4𝑥5 𝑣 = 𝑙𝑛𝑥
𝑢′ = 20𝑥4 𝑣′ =
1
𝑥
𝑦′
= 20𝑥4
∙ 𝑙𝑛𝑥 + 4𝑥5
∙
1
𝑥
= 20𝑥4
∙ 𝑙𝑛𝑥 + 4x4

18
Quotient rule:
𝑦 =
𝑓 𝑥
𝑔(𝑥)
𝑦′ =
𝑓′ 𝑥 ∙𝑔 𝑥 −𝑓 𝑥 ∙𝑔′ 𝑥
𝑔 𝑥
2
Often written as follows:
𝑦 =
𝑢
𝑣
𝑦′
=
𝑢′∙ 𝑣 − 𝑢 ∙ 𝑣′
𝑣2
Example:
𝑦 =
5𝑥3−4𝑥2+1
2𝑥2−𝑥
Quotient Rule
𝑢 = 5𝑥3 − 4𝑥2 + 1 𝑣 = 2𝑥2 − 𝑥
𝑢′ = 15𝑥2 − 8𝑥 𝑣′ = 4𝑥 − 1
𝑦′
=
15𝑥2
− 8𝑥 2𝑥2
− 𝑥 − 5𝑥3
− 4𝑥2
+ 1 4𝑥 − 1
2𝑥2 − 𝑥 2

19
Chain rule:
𝑦 = 𝑓 𝑔 𝑥 set 𝑦 = 𝑓 𝑧 with 𝑧 = 𝑔 𝑥
Then: 𝑦′ = 𝑓′(𝑧) ∙ 𝑧′
Example 1:
𝑦 =
3
𝑥4 − 3𝑥2 + 1
Chain Rule (1/2)
set 𝑦 = 3
𝑧 with 𝑧 = 𝑥4
− 3𝑥2
+ 1
𝑧′ = 4𝑥3 − 6𝑥
𝑦′
= 𝑓′(𝑧) ∙ 𝑧′ =
1
3
𝑧
1
3
−1
∙ 𝑧′ =
1
3
𝑧−
2
3 ∙ 𝑧′
Transform this back:
𝑦′
=
1
3
𝑥4
− 3𝑥2
+ 1 −
2
3 ∙ (4𝑥3
− 6𝑥)

20
Chain rule:
𝑦 = 𝑓 𝑔 𝑥 set 𝑦 = 𝑓 𝑧 with 𝑧 = 𝑔 𝑥
Then: 𝑦′ = 𝑓′(𝑧) ∙ 𝑧′
Example 2:
𝑦 = 𝑒𝑥2−3𝑥
Chain Rule (2/2)
set 𝑦 = 𝑒𝑧 with 𝑧 = 𝑥2 − 3𝑥
𝑧′ = 2𝑥 − 3
𝑦′ = 𝑓′(𝑧) ∙ 𝑧′ = 𝑒𝑧 ∙ 𝑧′
Transform this back:
𝑦′ = 𝑒𝑥2−3𝑥 ∙ (2𝑥 − 3)

21
Class Exercise 3.2: Rules of Differentiation
CLASS
EXERCISE
Task 3.2.1
Find the derivative of the following function:
𝑓 𝑥 =
𝑥
2𝑥 − 5
Task 3.2.2
𝑓 𝑥 = (5𝑥 + 1) ∙ 𝑒𝑥2+1

22
CLASS
EXERCISE
Task 3.2.3
𝑓 𝑥 = ln(4𝑥2 + 7)
Task 3.2.4
𝑓 𝑥 = 2𝑥2 + 5

23
CLASS
EXERCISE
Task 3.2.5
𝑓 𝑥 = (7 − 3𝑥)3
Task 3.2.6
Whic of the following statement is not correct for
differentiation:
a) If 𝑦 = 𝑓 𝑥 ∙ 𝑔(𝑥), then
𝑦′
= 𝑓′
𝑥 ∙ 𝑔 𝑥 + 𝑓 𝑥 ∙ 𝑔′
𝑥
b) If 𝑦 = 𝑓 𝑥 ∙ 𝑔(𝑥), then
𝑦′
= 𝑓′
𝑥 ∙ 𝑔 𝑥 − 𝑓 𝑥 ∙ 𝑔′
𝑥
c) If If 𝑦 = 𝑓 𝑥 + 𝑔(𝑥), then
𝑦′
= 𝑓′
𝑥 + 𝑔′
𝑥
d) If If 𝑦 = 𝑓 𝑥 − 𝑔(𝑥), then
𝑦′
= 𝑓′
𝑥 − 𝑔′
𝑥

24
24
3.1 Introduction
3.4 Curve Sketching

25
Take the following function:
𝑓 𝑥 =
1
3
𝑥3
− 3𝑥2
+ 8𝑥
𝑓′
𝑥 = 𝑥2
− 6𝑥 + 8
𝑓′′ 𝑥 = 2𝑥 − 6
𝑓′′′ 𝑥 = 2
𝑓4
𝑥 = 0
Note: Many times derivatives are written using the following notation:
𝒇′ 𝒙 =
𝒅𝒇
𝒅𝒙
First derivative, f prime
Second derivative, f double prime
Third derivative, f triple prime
Fourth derivative
First and Higher Derivatives

26
𝑓 𝑥 =
1
3
𝑥3
− 3𝑥2
+ 8𝑥, 𝑓′
𝑥 = 𝑥2
− 6𝑥 + 8
𝒇′ gives information on the slope of f.
𝑓′ < 0: negative slope
𝑓′
> 0: positive slope
𝑓′
= 0: slope is zero
First Derivative: Slope

27
𝑓 𝑥 =
1
3
𝑥3
− 3𝑥2
+ 8𝑥, 𝑓′
𝑥 = 𝑥2
− 6𝑥 + 8, 𝑓′′
𝑥 = 2𝑥 − 6
𝒇′′ gives information on the curvature of f.
𝑓′′ < 0: concave curvature
𝑓′′
> 0: convex curvature
𝑓′′
= 0: curvature might change
Second Derivative: Curvature

A function f is
 strictly increasing and concave if 𝑓′ 𝑥 > 0 and 𝑓′′ 𝑥 < 0 ∀ 𝑥 ∈ 𝐷.
 strictly increasing and convex if 𝑓′ 𝑥 > 0 and 𝑓′′ 𝑥 > 0 ∀ 𝑥 ∈ 𝐷.
 strictly decreasing and concave if 𝑓′ 𝑥 < 0 and 𝑓′′ 𝑥 < 0 ∀ 𝑥 ∈ 𝐷.
 strictly decreasing and convex if 𝑓′ 𝑥 < 0 and 𝑓′′ 𝑥 > 0 ∀ 𝑥 ∈ 𝐷.
If 𝑓′′ 𝑥 = 0 and 𝑓′′′ 𝑥 ≠ 0 for a value 𝑥 ∈ 𝐷, then the graph has an inflection
point at x.
Classification of Derivatives

29
Find the maximum of the following function f(x):
𝑓 𝑥 = −𝑥2 + 3𝑥, 𝑓′ 𝑥 = −2𝑥 + 3 𝑓′′ 𝑥 = −2
Conditions for a maximum:
Necessary condition: 𝑓′ 𝑥 = 0
Sufficient condition: 𝑓′′ 𝑥 < 0
Finding a Maximum

30
Find the minimum of the following function f(x):
𝑓 𝑥 = 𝑥2 − 3𝑥 + 2, 𝑓′ 𝑥 = 2𝑥 − 3 𝑓′′ 𝑥 = 2
Conditions for a minimum:
Necessary condition: 𝑓′ 𝑥 = 0
Sufficient condition: 𝑓′′ 𝑥 > 0
Finding a Minimum

31
Find the inflection point of the following function f(x):
𝑓 𝑥 =
1
3
𝑥3
− 3𝑥2
+ 8𝑥, 𝑓′
𝑥 = 𝑥2
− 6𝑥 + 8, 𝑓′′
𝑥 = 2𝑥 − 6, 𝑓′′′
𝑥 = 2
Conditions for an inflection point:
Necessary condition: 𝑓′′ 𝑥 = 0
Sufficient condition: 𝑓′′′ 𝑥 ≠ 0
Finding an Inflection Point

32
Class Exercise 3.3: First and Higher Derivatives
CLASS
EXERCISE
Task 3.3.1
Given the function f(x) = x³ − 15x²,
which of the following statements is
correct?
a) The function has an extreme point at x
= 5.
b) The function has an inflection point
at x = 5.
c) The function has an inflection point
at x = 1.
d) The function has an extreme point at x
= 1.
Task 3.3.2
The second derivative (or second-
order derivative) gives information
about ...
a) ... the straight-line slope of the
tangent.
b) ... the straight-line slope of the
secant.
c) ... the curvature of a curve.
d) ... the slope of a curve.

33
CLASS
EXERCISE
Task 3.3.1
Consider the following function:
𝑓 𝑥 = 𝑥3 − 3𝑥 + 5.
Find the extreme points of the function,
determine the kind of the extreme point
(minimum or maximum) and find whether
there is an inflection point.
Task 3.3.2
Consider the following function:
𝑓 𝑥 = 𝑥4 − 4𝑥3 + 4𝑥2.
Find the extreme points of the function,
determine the kind of the extreme point
(minimum or maximum) and find whether
there is an inflection point.

34
CLASS
EXERCISE
Task 3.3.3
For a function ƒ, the first-order derivative at the
point x = 1 is ƒ'(1) = 0. Furthermore, ƒ''(x) >
0 holds for all x from the domain of definition.
Which of the following statements is correct?
a) At the point x = 1, the curve of the function
has a global maximum.
b) At the point x = 1, the curve of the function
has a local minimum.
c) At the point x = 1, the curve of the function
has a global minimum.
d) At the point x = 1, the curve of the function
has a local maximum.
Task 3.3.4
The solution of optimization problems
always involves the determination of …
a) extreme points.
b) inflection points.
c) zeros.
d) saddle points.

35
CLASS
EXERCISE
Task 3.3.5
Given the function ƒ(x) = 10 + 5x2, the
slope of the graph is …
a) negative as long as x is a positive
integer.
b) negative as long as x is a positive real
number.
c) positive as long as x is a positive real
number.
d) positive as long as x is a negative real
number.
Task 3.3.6
Given the function ƒ(x) = 4x2 + 8x + 4,
a. determine the zero(s) of this function;
b. determine whether the function has a
minimum or maximum and motivate
your answer; and
c. calculate the extreme point.

36
CLASS
EXERCISE
Task 3.3.7
If the slope increases continuously along
a curve in the first quadrant of the
coordinate system, then …
a) the curve is concave.
b) the curve is a straight line.
c) the curve is a tangent.
d) the curve is convex.
Task 4

37
37
3.1 Introduction
3.4 Curve Sketching

38
Task: Sketch the curve of the following function:
𝑓 𝑥 = 𝑥 +
1
4𝑥2 =
4𝑥3
+ 1
4𝑥2
Step 1: Determine roots
Set the numerator to zero:
4𝑥3
+ 1 = 0
4𝑥3
= −1
𝑥3 = −
1
4
𝑥 =
3
−
1
4
Root: 𝑥1 =
3
−
1
4
Step 2: Determine domain and vertical asymptotes
Set the denominator to zero:
4𝑥2
= 0
𝑥 = 0
Domain: 𝐷 = ℝ 0
Vertical Asymptote: 𝑥2 = 0 𝐸𝑣𝑒𝑛 𝑉𝐴
Curve Sketching (1/4)

39
𝑓 𝑥 = 𝑥 +
1
4𝑥2 =
4𝑥3
+ 1
4𝑥2 = 𝑥 +
1
4
𝑥−2
Step 3: Find extreme points
𝑓′ 𝑥 = 1 +
(−2)
4
𝑥−3 = 1 −
1
2
𝑥−3
𝑓′′
𝑥 =
3
2
𝑥−4
Necessary Condition: 𝑓′ 𝑥 = 0
1 −
1
2
𝑥−3
= 0 ⟹ 1 =
1
2
𝑥−3
⟹ 2 = 𝑥−3
⟹
1
2
= 𝑥3
⟹ 𝑥 =
3 1
2
Sufficient Condition: 𝑓′′ 𝑥 > 0 for minimum, 𝑓′′ 𝑥 < 0 for maximum
𝑓′′ 3 1
2
= 3.78 > 0
Hence, sufficient condition is met for a minimum at 𝑥3 =
3 1
2

40
𝑓 𝑥 = 𝑥 +
1
4𝑥2 =
4𝑥3
+ 1
4𝑥2
Step 4: Determine inflection points
Set the second derivative to zero:
𝑓′′
𝑥 = 0 ⟹
3
2
𝑥−4
= 0 ⟹
3
2𝑥4
= 0
No solution, hence no inflection point
Step 5: Determine horizontal or oblique asymptote
Find limit of the function for x approaching infinity:
𝑓 𝑥 = 𝑥 +
1
4𝑥2
Thus, the whole functions approaches y = x,
as x approaches infinity
y = x is the oblique asymptote
Approaches zero

41
𝑓 𝑥 = 𝑥 +
1
4𝑥2 =
4𝑥3
+ 1
4𝑥2

42
TEST YOURSELF: Curve Sketching (1/5)
TEST
YOURSELF
Sketch the curve of the following function:
𝑓 𝑥 =
𝑥2 − 4𝑥 + 4
𝑥2
Your analysis should include
• domain
• roots
• vertical/horizontal/oblique asymptotes
• extreme points
• inflection points

43
TEST
YOURSELF
𝑓 𝑥 =
𝑥2 − 4𝑥 + 4
𝑥2
Domain & roots:

44
TEST
YOURSELF
𝑓 𝑥 =
𝑥2 − 4𝑥 + 4
𝑥2
Vertical/horizontal/oblique asymptotes:

45
TEST
YOURSELF
𝑓 𝑥 =
𝑥2 − 4𝑥 + 4
𝑥2
Extreme points:

46
TEST
YOURSELF
𝑓 𝑥 =
𝑥2 − 4𝑥 + 4
𝑥2
Inflection points:

Differential Calculus

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Differential Calculus