It is about differential calculas made by Emon Kundu.

1. DIFFERENTIAL
CALCULUS

2. GROUP PROFILE – 05
SHAMSUN NAHAR TANIA
ID:26-150
MOSAMMOT FYZA AKTER
ID:26-082
REPLICA MADDEN SARKAR
ID:26-048
EMON KUNDU
ID:26-030
MD IMDADUL ISLAM
ID:26-054

3. 7.1:THE INTRODUCTORY PART
We study calculus because many real world problems can be solved
using calculus. Chapter 7 introduces the basic concept of
differential calculus including the definition of the derivative, limit,
continuity and fundamental rules of taking derivatives.

4. DELTA NOTATION
GENERAL EXPRESSION
Calculus is used to solve the problems involving rate of change. And delta notation
finds the change in the value of function for change in the value of independent
variables.
∆𝒇 = 𝒇 𝒙 + 𝒉 − 𝒇(𝒙

5. MATHEMETICAL PROBLEM & SOLUTION
THE SOLUTION
THE PROBLEM
f(x)=2x²-10x+8,find Δf
If x changes from 0 by 0.1
According to general expression:
Δf= f(x+h)-f(x)
=f(0+0.1)-f(0)
=[2(0.1)²-10(0.1)+8]-[2(0)²-10(0)+8]
=7.02-8
=-0.98

6. 7.2:LIMIT
GENERAL EXPRESSION
A limit is the value that a function(or sequences) approaches as the input(or index)
approaches some value. Limits are essential to calculus and mathematical analysis
and are used to define continuity, derivatives and integrals.
𝒍𝒊𝒎
𝒙→𝟎
𝒇(𝒙 = 𝟎
Where the symbol “x 0” means x approach's 0 or gets closer and closer to 0 without ever
equaling zero

7. MATHEMETICAL PROBLEM & SOLUTION
THE SOLUTION
THE PROBLEM
𝒍𝒊𝒎
𝒙→𝟐
𝒙𝟐 + 𝒙 − 𝟔
𝒙𝟐 − 𝟒
𝒍𝒊𝒎
𝒙→𝟐
𝒙𝟐 + 𝒙 − 𝟔
𝒙𝟐 − 𝟒
=𝑙𝑖𝑚
𝑥→2
𝑥+3 (𝑥−2
𝑥+2 (𝑥−2
= 𝒍𝒊𝒎
𝒙→𝟐
𝒙 + 𝟑
𝒙 + 𝟐
=
𝟐+𝟑
𝟐+𝟐
=
𝟓
𝟒

8. 7.3 :CONTINUITY:
A function is said to be continuous if it can be drawn without lifting the pencil
from the paper.
Function is said to be continuous at X = a IF,
𝒇(𝒙 has a definite limit as x → 𝒂 𝒇rom
either side , 𝒍𝒊𝒎
𝒙→𝒂
− 𝒇(𝒙 = 𝒇(𝒙 = 𝒍𝒊𝒎
𝒙→𝒂
+ 𝒇(𝒙

9. MATHEMETICAL PROBLEM & SOLUTION
THE SOLUTION
THE PROBLEM
f(x)= 3x2+2x-1
Show that the function is
continuous at x=2
f(2)= 3*22+2*2-1
=12+4-1 =15
L.H.L. 𝒍𝒊𝒎
𝒙→𝟐−
𝟑𝒙𝟐 + 𝟐𝒙 − 𝟏
=𝒍𝒊𝒎
𝒉→𝟎
𝟑 𝟐 − 𝒉 𝟐 + 𝟐 𝟐 − 𝒉 − 𝟏
=3(2-0)2+2(2-0)-1
=15
R.H.L. 𝒍𝒊𝒎
𝒙→𝟐+
𝟑𝒙𝟐
+ 𝟐𝒙 − 𝟏
=𝒍𝒊𝒎
𝒉→𝟎
𝟑 𝟐 + 𝟎 𝟐 + 𝟐 𝟐 + 𝟎 − 𝟏
=15
Since L.H.S =f(x)=R.H.S at x=2, the function is
continuous at x=2.

10. 7.4:THE DIFFERENT QUOTIENT
• It is basically the calculation of slope .
Step-1: 𝒇 𝒙 + 𝒉
Step-2: Ms=
𝒇 𝒙+𝒉 −𝒇(𝒙
𝒉
• Differential calculus of derivative deals with determining the slope of a line
tangent to a curve at a point on that curve.
• When f(x) tends to the first point Ms tends to tangent slope.
Step-3: Slope f’(x =𝒍𝒊𝒎
𝒉→𝟎
𝒇 𝒙+𝒉 −𝒇(𝒙
𝒉

11. MATHEMETICAL PROBLEM & SOLUTION
THE SOLUTION
THE PROBLEM
Find the slope of f(x)= x3 at the
point where x=0.5
Step 1 find f(x+h)=(x+h)3
=x3 +3x2h +3xh2 +h3
Step 2 find ms,
Ms=
x3+3x2h+3xh2+h3−x3
ℎ
=3x2+3xh+h2
Step 3 find f’(x)
f’(x)=lim
ℎ→0
3𝑥2 + 3𝑥ℎ + ℎ2
= 3x2
slope = 3*(0.5)2
= 0.75

12. 7.5:THE SIMPLE POWER RULE
GENERAL EXPRESSION
The simple power rule is applied to find the derivative of any power of x.
If f(x) = xn ,
then f’(x = nx n-1

13. MATHEMETICAL PROBLEM & SOLUTION
THE SOLUTION
THE PROBLEM
𝑭𝒊𝒏𝒅 𝒇′
𝒙 𝒐𝒇 𝒇 𝒙 =
𝟐
𝟑
− 𝟑
𝒙
𝟐
𝒇′ 𝒙 =
𝒅𝒚
𝒅𝒙
𝟐
𝟑
− 𝟑
𝒙
𝟐
=
𝒅𝒚
𝒅𝒙
2
3
−
3
2
𝑥1
= 0 −
3
2
1 𝑥1−1
=-
3
2

14. 7.6:Function power rule
GENERAL EXPRESSION
Function power rule is applied to find the derivative of any power of x, but this rule is
applied when the function is in [f(x)]n this form.
if f(x)= [f(x)]n ,then f’(x)= n[f(x)]n-1 *
𝒅𝒚
𝒅𝒙
f(x)
= n[f(x)]n-1 * f’(x)

15. MATHEMETICAL PROBLEM & SOLUTION
THE SOLUTION
THE PROBLEM
𝒇 𝒙 =
𝟒
𝟐𝒙 − 𝟑
= 𝟒 𝟐𝒙 − 𝟑 −𝟏
𝒇′(𝒙 = 𝟒 𝟐𝒙 − 𝟑 −𝟏
= 𝟒 −𝟏 𝟐𝒙 − 𝟑 −𝟏−𝟏
∗
𝒅𝒚
𝒅𝒙
𝟐𝒙 − 𝟑
= −𝟒 𝟐𝒙 − 𝟑 −𝟐 ∗ 𝟐
∴ 𝒇′(𝒙 = −𝟖 𝟐𝒙 − 𝟑 −𝟐

16. 7.7:PRODUCT AND QUOTIENT RULES
we now extend our list of rules to include the derivative of product of two
function and quotient of two functions.
𝒅𝒚
𝒅𝒙
𝒙 − 𝟏 𝒙𝟐
+ 𝟐
PRODUCT:
QUOTIENT:
𝒅𝒚
𝒅𝒙
𝟐𝒙𝟐
+ 𝟑
𝒙 + 𝟏

17. THE PRODUCT RULE
MATHEMETICAL EXPRESSION
The derivative of the product of two function is the first times the derivative
of the 2nd, plus the second times the derivative of the first.
CALCULATION
𝒇′ 𝒇 𝒙 𝒈 𝒙
= 𝒇 𝒙
𝒅𝒚
𝒅𝒙
𝒈 𝒙 + 𝒈 𝒙
𝒅𝒚
𝒅𝒙
𝒇 𝒙

18. MATHEMETICAL PROBLEM & SOLUTION
THE SOLUTION
𝒇 𝒙 = 𝟑 − 𝒙𝟐 (𝟓𝒙 + 𝟔
𝒇′(𝒙 = 𝟑 − 𝒙𝟐
𝒅𝒚
𝒅𝒙
𝟓𝒙 + 𝟔 + 𝟓𝒙 + 𝟔
𝒅𝒚
𝒅𝒙
𝟑 − 𝒙𝟐
THE PROBLEM
= 𝟑 − 𝒙𝟐 ∗ 𝟓 + (𝟓𝒙 + 𝟔 *(-2x)
= 𝟏𝟓 − 𝟓𝒙𝟐 − 𝟏𝟎𝒙𝟐 − 𝟏𝟐𝒙
= 15 − 15𝑥2 − 12𝑥

19. THE QUOTIENT RULE
MATHEMETICAL EXPRESSION
The derivative of the quotient of two function is the denominator times the derivative
the nominator minus the nominator times the derivative of the denominator and the
denominator Squared.
CALCULATION
𝒇
′
𝒇 𝒙
𝒈(𝒙 =
𝒈 𝒙
𝒅𝒚
𝒅𝒙
𝒇 𝒙 − 𝒇 𝒙
𝒅𝒚
𝒅𝒙
𝒈 𝒙
𝐠 𝒙 𝟐

20. MATHEMETICAL PROBLEM & SOLUTION
THE SOLUTION
THE PROBLEM
𝒇 𝒙 =
𝒙 − 𝟐
𝒙 + 𝟏
𝒇′(𝒙 =
𝒙 + 𝟏
𝒅𝒚
𝒅𝒙
𝒙 − 𝟐 − 𝒙 − 𝟐
𝒅𝒚
𝒅𝒙
(𝒙 + 𝟏
𝒙 + 𝟏 𝟐
=
(𝒙 + 𝟏 − 𝒙 − 𝟐
𝒙 + 𝟏 𝟐
=
𝑥 + 1 − 𝑥 + 2
𝑥 + 1 2
=
𝟑
𝒙 + 𝟏 𝟐

21. 8.1:MAXIMA AND MINIMA OF FUNCTION
The first derivative test
• Calculate the first derivative f‘(x)
• set f‘(x) = 0, to find stationary point and solve the value of x.
• Test a point left and right of stationary point to find local minimum or Local
maximum.
STATIONARY POINT: A stationary point occurs when the first derivative is 0.
To find local maximum or local minimum of function using 1st Derivative test, we need
to follow following steps:

22. DECISION CRITERIA to find local minimum or local maximum
1
if f’(x) is increasing in the left and decreasing
in the right, then stationary Point is a local
maximum.
2
if f’(x) is decreasing in the left and increasing in
the right, then stationary Point is a local
minimum.
𝒇′
𝒙 = 𝟎
𝒇′(𝒙𝒍 ↑→↓ 𝒇′(𝒙𝒓
𝒇′ 𝒙 = 𝟎
𝒇′(𝒙𝒍 ↓→↑ 𝒇′(𝒙𝒓

23. DECISION CRITERIA to find point of inflection
3
When the left and right point of stationary point is both increasing or both
decreasing then the stationary point is point of inflection.
𝒇′
𝒙 = 𝟎
𝒇′(𝒙𝒍 ↓→↓ 𝒇′(𝒙𝒓
𝒇′
𝒙 = 𝟎
𝒇′(𝒙𝒍 ↑→↑ 𝒇′(𝒙𝒓

24. MATHEMETICAL PROBLEM & SOLUTION
Testing a point just left and right of x=.5
When,
X=0
𝒇′ 𝒙 = 𝟐 ∗ 𝟎 − 𝟏
= −𝟏 ↓
When,
X=1
𝒇′ 𝒙 = 𝟐 ∗ 𝟏 − 𝟏
= 𝟏 ↑
The stationary point is local minimum at x=.5 and co-
ordinate (.75,.5)
0 .5 1
THE SOLUTION
THE PROBLEM
𝒇 𝒙 = 𝒙𝟐 − 𝒙 + 𝟏
𝒇′(𝒙 = 𝟐𝒙 − 𝟏
For stationary point,
𝒇′(𝒙 = 𝟎
⇒ 𝟐𝒙 − 𝟏 = 𝟎
⇒ 𝟐𝒙 = 𝟏
∴ 𝒙 =. 𝟓

25. DECISION CRITERIA to find local minimum or local maximum of END POINT
1 Test a point just to the right of a.
2 Test a point just to the left of b.
a) if 𝑓′(𝑥𝑟) ↑ then End point minimum.
b) if 𝑓′(𝑥𝑟) ↓ then End point maximum.
a) if 𝑓′(𝒙𝒍) ↑ then End point maximum.
b) if 𝑓′(𝒙𝒍) ↓ then End point minimum.
If f(x) is restricted to the interval [a,b], then

26. MATHEMETICAL PROBLEM & SOLUTION
THE SOLUTION
THE PROBLEM
4 5
Testing a point just right to x= -1
When,
X=0
𝒇′ 𝟎 = 𝟏𝟎𝟖 − 𝟒 𝟎 𝟑
= 𝟏𝟎𝟖 ↑
So, Endpoint Minimum at x= -1 and co-ordinate (-109, -1)
-1 0
Testing a point just left to x= 5
When,
X=4
𝒇′ 𝟒 = 𝟏𝟎𝟖 − 𝟒 𝟒 𝟑
= −𝟏𝟒𝟖 ↓
So, Endpoint Minimum at x= -1 and co-ordinate (-85 , 5)
𝒇 𝒙 = 𝟏𝟎𝟖𝒙 − 𝒙𝟒𝒐𝒏 −𝟏, 𝟓
𝒇′ 𝒙 = 𝟏𝟎𝟖 − 𝟒 𝒙𝟑

27. 8.2:THE 2ND DERIVATIVE TEST
The second derivative tells whether the curve is concave up or concave down at that
point. If the second derivative is positive at a point, the graph is bending upwards at that
point. Similarly, if the second derivative is negative, the graph is concave down. The
derivative of the derivative is
𝑑𝑦
𝑑𝑥
f’(x) = f” (x), where f” (x) is called the second derivative.

28. The Second Derivative Test for Maximum and Minimum
The following sequence of steps facilitates the second derivative test, to find the
local maxima and local minima of the real-valued function.
• Finding the first derivative f'(x) of the function f(x)
• Equalizing the first derivative to zero f'(x) = 0 and finding the limiting points.
• Finding the second derivative of the function f''(x).
• Substituting the limiting points in the second derivative.
• If the second derivative is greater than zero, then the limiting point is the local
minima.
• If the second derivative is lesser than zero, then the limiting point is the local
maxima.

29. Third Derivative Test for Inflection Points
Setting f’’(x) = 0 and, solving for x.
a) If f’’’(x) ≠ 0, then we have an inflection point.
b) If f’’’(x) = 0, then we need to test the second derivative just to the left
and right. If the second derivative changes the sign, then we have an
inflection point.

30. MATHEMETICAL PROBLEM & SOLUTION
THE SOLUTION
THE PROBLEM
𝒇 𝒙 = 𝒙3
+ 3𝒙2
− 9𝒙 − 3
𝑓′
𝑥 = 3𝑥2
+ 6𝑥 − 9
𝑓"(𝑥 = 6𝑥 + 6
Setting, f’(x) = 0
3𝑥2
+ 6𝑥 − 9 = 0
 3 𝑥2 + 2𝑥 − 3 = 0
 𝑥2 + 3𝑥 − 𝑥 − 3 = 0
 𝑥 𝑥 + 3 − 1 𝑥 + 3 = 0
𝑥 + 3 𝑥 − 1 = 0
𝑥 + 3 = 0 or 𝑥 − 1 = 0
𝑥 = −3 𝑥 = 1
When 𝑥 = −3,
𝑓"(−3 = 6 ∗ (−3 + 6
= −12
𝑓’ −3 = (−3 3
+ 3(−3 2
− 9 ∗ −3 − 3
= 27 + 27 + 27 − 3
= 24
When 𝑥 = 1,
𝑓"(1 = 6 ∗ 1 + 6
= 12
𝑓′(1 = 13
+ 3 ∗ 12
− 9 ∗ 1 − 3
= −8
We have local maxima point at 𝑥 = −3,24 , and local minima
point at 𝑥 = 1, −8

31. MATHEMETICAL PROBLEM & SOLUTION cont.
THE SOLUTION
Point of inflection check
6𝑥 + 6 = 0
 𝑥 + 1 = 0
 𝑥 = 1
𝑓′′′
𝑥 = 6 > 0
𝑓 −1 = (−1 3
+ 3(−1 2
+ 9 ∗ −1 − 3
= −1 + 3 − 12
= −10
𝑓"(−1 = 6 ∗ (−1 = 6
= 0
Inflection point −1, −10 .

32. Maxima and Minima Applications
There are numerous practical applications in which it is desired
to find the maximum or minimum value of a particular quantity.
Such applications exist in economics, business, and engineering.

33. MATHEMETICAL PROBLEM & SOLUTION
THE PROBLEM
A rectangular shaped manufacturing plant with a floor area of 600,000 square
feet is to be built in a location where zoning regulations require buffer strips 50
feet wide front and back, and 30 feet wide at either side. What plot
dimensions will lead to minimum local area for plant and buffer strips? What is
the minimum total area? If the plot dimensions were made 1500 feet by 400
feet rather than the dimensions leading to minimum area, by how much would
the total plot area exceed the minimum area?

34. MATHEMETICAL PROBLEM & SOLUTION
THE SOLUTION
Area = 𝑥𝑦
Here, A = 𝑥 + 100 𝑦 + 60
𝑥𝑦 = 600000
 𝑦 =
600000
𝑥
A 𝑥 = (𝑥 + 100 (
600000
𝑥
+ 60
= 600000 + 60𝑥 +
60000000
𝑥
+ 6000
A’ 𝑥 = 60 −
60000000
𝑥2
A”=
120000000
𝑥3
Now, A’ 𝑥 = 0
60 −
60000000
𝑥2
= 0
𝑥 = 1000
When 𝑥 = 1000,
A” 1000 =
120000000
10003
= 0.12
So, 𝑦 =
600000
1000
= 600 feet

35. MATHEMETICAL PROBLEM & SOLUTION cont.
THE SOLUTION
Again, if dimensions of plant are 1500 feet by 400 feet,
𝑥 + 100
= 1500 + 100
= 1600 feet
𝑦 + 60
= 400 + 60
= 460 feet
Area = 1600 ∗ 460
= 736000 sq feet
So, total plot area exceeds = 736000 − 726000 sq feet
= 10000 sq feet
Dimensions would be 1000 feet by 600 feet
Total land, 𝑥 + 100
= 1000 + 100
= 1100 feet
𝑦 + 60
= 600 + 60
= 660 feet
Total minimum area, A = 1100 ∗ 660
= 726000 sq feet

36. SPECIAL
THANKS
FROM GROUP 05