PRACTICE BOOK
DUM20132
ENGINEERING
MATHEMATICS 2
For Second Semester of Diploma
SAJIDAH NUR AMALINA NURHAFIZA FATHIYAH
DUM20132 ENGINEERING MATHEMATICS 2
CONTENT PREFACE
UNIT 1 MATRICES
1.1 Definition
1.2 Operation of Matrices This book is written with the
1.3 Solving System of Linear Equations hopes of sharing the
Revision Unit 1 excitement found in the study
of mathematics. Mathematics
UNIT 2 VECTOR can help us unlock the
2.1 Definition mysteries of our universe, but
2.2 Vector Algebra beyond that, conquering it can
2.3 Component of a vector be personally satisfying.
2.4 Scalar Product
2.5 Vector Product The content is chosen and
Revision Unit 2 developed by MARA Japan
Industrial Institute (MJII)
UNIT 3 DIFFERENTIATION Mathematicians with the goal
3.1 Introduction of Differentiation of helping people achieve that
3.2 Derivatives of Functions feeling of accomplishment.
3.3 Techniques of Differentiation
3.4 Higher Derivatives
Revision Unit 3
MJII MATHEMATICS LECTURER
UNIT 4 INTEGRATION
4.1 Integration as a Reverse Function NORHASLINDA BINTI JAMIL
Head of General Studies Dept. /
of Differentiation Mathematics Lecturer
4.2 Indefinite Integral of Function
4.3 Techniques of Integration SAJIDAH BINTI ABDUL RAHMAN
4.4 Definite Integral Mathematics Lecturer
Revision Unit 4
NUR AMALINA BINTI ZAINAL
Mathematics Lecturer
NURHAFIZA BINTI MOHD HASSIM
Mathematics Lecturer
FATHIYAH BINTI JAMALUDIN
Mathematics Lecturer
DUM20132 ENGINEERING MATHEMATICS 2 1
What is a matrix?
Matrix mathematics applies to several branches of science, as well as different mathematical disciplines.
Let’s start with computer graphics, then touch on science, and return to mathematics. We see the results
of matrix mathematics in every computer-generated image that has a reflection, or distortion effects such
as light passing through rippling water. Before computer graphics, the science of optics used matrix
mathematics to account for reflection and for refraction.
Matrix arithmetic helps us calculate the electrical properties of a circuit, with voltage, amperage,
resistance, etc.
In mathematics, one application of matrix notation supports graph theory. In an adjacency matrix, the
integer values of each element indicate how many connections a particular node has.
The field of probability and statistics may use matrix representations. A probability vector lists the
probabilities of different outcomes of one trial. A stochastic matrix is a square matrix whose rows are
probability vectors. Computers run Markov simulations based on stochastic matrices in order to model
events ranging from gambling through weather forecasting to quantum mechanics.
Matrix mathematics simplifies linear algebra, at least in providing a more compact way to deal with groups
of equations in linear algebra.
Learning Objective:
After completing the topic, students should be able to:
1. Identify matrix and its properties.
2. Identify the various types of matrices.
3. Operate matrices.
4. Solve systems of linear equations using matrix inversion method and Cramer’s rule..
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TYPES OF MATRICES ADD AND SUBSTRACT DETERMINANT
a b m n ab
Row Equal c d p 2 2matrix = c = ad − bc
o d
a b c a b m n a m b n abc
Column c d = p d p
a o = ef df de
b c o
c 3 3matrix = d e f =a −b +c
h ig ig h
Square Identity gh i
1 0 1 0 0 MULTIPLY ADJOINT
0 1 0 0
0 1 1 ef df d e
0 + − +
CONSTANT h i g h
gi
a b Transpose k a b = ka kb Adj ( A) = b c ac a b
c d a bT a c c d kc kd − + −
c d = b d h i gi g h
−a c
BY MATRIX + b c df +a b
e f d
Order : A23B32 e
ORDER 1 d −b
A −c a
same INVERSE A−122 =
ROW x COLUMN 1 Adj
Remember: Rocky Cute A
( )A−133= A
Apply AA−1 = A−1A = I THE INVERSION METHOD CRAMERS’ RULE STEP OF CRAMERS’ RULE
Ax = B Determinant based 1
x = A−1B Ax = B Find A
Find the value of m and n for the following a b x = m mb
simultaneous c d y 2 nd
n
Find x =
2 −3 m = 7 CRAMERS’ RULE A = 2 −3 = 2(4) − −3(1) A
−2
1 4 n 14
INVERSION METHOD 7 −3 = 11 am
3 bn
m = 1 4 3 7 −2 4 = 7(4) − −3(−2) = 22 = 2
−1 2 −2 x= Find y =
− −3(1) A
n 4(2) 11 11 11
1 22 2 27
−1
= 11 −11 = 1 −2 = 2(−2) − 7(1) = −11 = −1
y=
11 11 11
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1.1 Definition
A matrix is a rectangular array of numbers arranged in rows and columns. The array of numbers below is an
example of a matrix.
Column 4
Column 3
Column 2
Column 1
Row 1 ORDER OF MATRIX = 3 X 4
Row 2
Row 3
The number of rows and columns that a matrix has is called its dimension or its order. By convention, rows are
listed first; and columns, second. Thus, we would say that the dimension (or order) of the above matrix is 3 x 4,
meaning that it has 3 rows and 4 columns.
Task A : Identify the row, column and order of each matrix.
Example : Remember !!!!! 1. Matrix
1 0 3 RC 3 1 2 5
5 −2 1
Row 1st then Column Row =
Row = 2 Column =
Column = 3 Order =
Order = 2 3
2. Matrix 1 3 3. Matrix 3 2 3 8
2 9 1 5 9 2
Row = 6 0 Row = 3 2 4 6
Column = Column = 5 1 2 3
Order = 3 −2 7 −5 Order =
5. Matrix 7
4. Matrix 1 9 −1 8 1
Row=
Row = Column=
Column= Order=
Order =
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Task B: Identify the element of a matrix.
Example:
3 −6 2 −1
1
−5 5 1 B = 9 1 8 7 4 −7 0
2 −6 2 5
A = 7 6 8 C = 4 12
8 0 −3 0 9 −3 8
Based on matrix A, find the 1. Based on matrix B, find the 2. Based on matrix C, find the
element of : element of : element of :
Element Answer Element Answer Element Answer
1 a13 a42
a13 a21 a23
a 32 0 a12 a32
a 21
7
1 3 4 5 7 −3
5 2
D = 7 9 E = −7 8 5 4 F = 6 0 2 5
7 1 3 −1
3 −1 −5 2
0 4
3. Based on matrix D, find the 4. Based on matrix E, find the 5. Based on matrix F, find the
element of : element of : element of :
Element Answer Element Answer Element Answer
a13 a13 a14
a32 a12 a23
a21 a34
a 21
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1.1.1 Type of Matrices
1. Row Matrix 2. Column Matrix
If a matrix is having only one row, then that matrix A vector may be represented with a list of
is called to be a row matrix.
The dimension of a row matrix is given as 1 n numbers called a column matrix. A column matrix
where n is the number of columns.
is an ordered list of numbers written in a column.
Example : D = 1 2 0 5
The dimension of a row matrix is given as m1
where m is the number of rows.
1.7
3.1
Example : E =
6.3
4.5
3. Square matrix 4. Identity matrix
A square matrix has the same number of rows as A square matrix which contains ones along the
main diagonal (from the top left to the bottom
columns. right), while all its other entries are zero.
1 3 2 0 2
1 2 , 1 1
Example: F = G = 7 3 5 1 0 0
6
Example: I = 1 0 , I = 0 1 0
0 1 0 0 1
5. Equal matrix 6. Transpose matrix
Two matrices are equal if and only if these A matrix which is formed by turning all the rows of
matrices have the same dimensions and equal a given matrix into columns and vice-versa. The
corresponding elements.
transpose of matrix A is written A T .
Example : H = 2 5 3 , J= 2 5 3 3 7
1 7 4 1 7 4
1 6
Example: M =
Matrix H and J have equal corresponding 2 4
element. 5 1
= 6 1 = M −1 1 then the transpose of matrix M,
2 3 3
If matrix K equal to matrix L 2 3 1 2 5
7 6 4 1 .
MT =
,
Can you guess the value of M?
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Task C: Find the value of a variable.
Example:
11 3 7 S = 5 1 −6 7
−5 , 7 8 9 10 ,
2 −5 2 b Q = 9 −7 1 −6 7
11 a + 3 8 f g + 6
N = 3 P = 4 8 1 4 T = e − 3
7
c − 3 3 7
d+4 −5
If matrix N = P , find the value of R = 9 4
variable: 1
8 2. If matrix S = T , find the
i. a. value of variable:
Ans: 1. If matrix Q = R , find the i. e
a + 4 = 11 value of variable:
i. c
a = 11− 4 = 7
ii. b. ii. d ii. f
Ans: (compare element) iii. g
b = −5
Task D: Find the transpose of the matrix.
Example:
11 3 7 S = 5 1 −6 7 2 0 2
−5 7 8 9 10 G = 1 3 1
F = 1 3 Q = 9 −7 4
1 2 7 6 5
8 1 Find GT .
Find FT . Find QT . Find ST .
Ans:
FT = 1 1
3 2
4 5 7 −3 B = 9 1 8 D = 1 2 0 5 1 3
E = −7 8 2 −6 2 Find DT . H = 2 9
5 4
6 0
3 −1 −5 2 Find HT
Find ET . Find BT .
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1.2 Operation of Matrices
Addition and Subtraction of Matrices
Addition and subtraction of matrices can only be done if the dimensions are exactly the same.
To add/subtract two matrices, just add/subtract the corresponding entries, and place this sum in the
corresponding position in the matrix which results.
Example :
i) 2 −9 + 3 5 = 2 + 3 −9 + 5 = 5 4
Addition 4 2 −7 4 + 2 5 + −7 6 −2
5
ii) Subtraction 12 10 9 − 3 5 4 = 12 − 3 10 − 5 9 − 4 = 9 5 5
Multiplication of Matrices
Matrix multiplication falls into two general categories:
i) Scalar in which a single number is multiplied with every entry of a matrix.
Example:
2 5 = 14 35
7 −3 −1 −21 −7
ii) Multiplication of an entire matrix by another entire matrix. For matrix multiplication to work,
the columns of the second matrix must have the same number of entries as do the rows of
the first matrix.
Example:
2 1 1 = 2(1) + 1(4) = 6
3 5 4 3 (1) + 5(4) 23
The order of first and second matrix is 2 2 and 2 1, respectively. Then, the product is a matrix with the
order of 11 .
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Task A : Find the product of two matrices.
5 −12 2. 10 20 − 7 6 3. 6 −4 2 + 1 7 −5
3 −12 15 1 21 =
1. + 10 =
7 8 =
4. 4 −1 + −6 −5 5. 20 + −11 6. 4 −1 − −6 −5
6 −3 16 6 −3
7 3 5 7 3
= = =
1 2 −7 4 6 0 8. 4 −3 1 + 9 12 7 9. 4 −3 1 − 9 12 7
6 0 1 4 5 6 −7 3 −4 3 5 6 −7 3 −4 3
7. 9 − 3
3 5 4 4 1 5 = =
=
Task B : Simplify the following scalar multiplication.
4 4 3 3. 2k 7 11 6k
1. 8a 3a 2. 9 6 −8
7
4. −5 −5 1 5. 3 −1 3 − 2 −5 1
−9 6 −9
7 0 7
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Task C: Choose the best answer.
1. Which of the following 2. What is the answer for 3. What is the product if
matrices can be multiply? 1
matrix 3 4 multiply with
A. 2 3 1 3 0 6 5 2 4 ? 45?
6
0 9 A. 3 4
B. 6 8 A. 6 20 12 B. 3 5
6 C. 4 4
C. 3 2 D. 4 5
0 5 4 1 6 B. 20
12 4
4 2 6 3. 6 7 3
D. 1 3 3 C. 38
D. −26
Task D : Multiplying matrix with matrix.
1. 8 3 1 −2 1 3 4
0 5 3 2. 2 1 6 2 2 −1
4
4 0 5
0 1 9 1 4 0 3
4. 3 1 2 3 5. 3 6 10 6. 3 −2 1 2
2 5 6 2 7 −5 1 7 2 8 6
−3 −4 3 −1 0 3
7. 8. 2 9.
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1.3 Solving System of Linear Equations
One of the most important applications of matrices is to the solution of linear simultaneous equations.
A system of simultaneous linear equations can be written as
a11 x + a12 y + a13 z = b1
a21 x + a22 y + a23 z = b2
a31 x + a32 y + a33 z = b3
In matrix form, the above simultaneous linear equation can be written as
a11 a12 a13 x = b1 ➔ AX = B
a21 a22 a23 y b2
a31 a32 a33 z b3
There are 2 methods will be introduced here: Matrix Inversion Method and Cramer’s Rule.
1.3.1 Matrix Inversion Method
If AX = B, then X = A-1 B which is A-1 = inverse of matrix A.
The inverse of Matrices
A matrix A-1 is called the inverse of matrix A if AA-1 = A-1A = I where I = identity matrix.
For n x n matrix,
A−1 = 1 adjoint A *A-1 exist if and only if determinant A ≠ 0.
det er minant A
= adjoint A
A
Determinant
The determinant is a scalar quantity. It contains much information about the matrix it came from and is
quite useful in many applications. The determinant is most often used to
• test whether or not a matrix has an inverse
• test for linear dependence of vectors (in certain situations)
• test for existence/uniqueness of solutions of linear systems of equations
Determinant of 2 x 2 Matrices
The product of determinant of 2 2 Matrices can be determined by using the following formula:
Deter minant, A = ab = ad − bc
cd
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Determinant of 3 x 3 Matrices
While for 3 3 Matrices, the following formula should be applied:
n
A = (−1)i+jaijMij Formula of a cofactor
j=1
abc
Det, A = d e f = a e f − b d f + c d e
h i gi gh
gh i
Example of finding determinant:
i) 5 4 = 5(3) − 2(4) = 15 − 8 = 7
23
124
ii) 3 5 6 = 1 5 6 − 2 3 6 + 4 3 5 = 1(45 − 48) − 2(27 − 42) + 4(24 − 35)
89 79 78
789
= 1(−3) − 2(−15) + 4(−11) = −17
Task A: Find the determinant of the following matrices.
2 1 3 − 2 1 2
1. 0 3 2. 1 4 3. 5 3
2 2 5. −3 −2 6. 8 −1
4. 6 4 −4
9 5 1
Task B: Find the determinant for 3 3 Matrices.
1 2 4 5 2 1 1 0 0
1. 3 1 0 2. 3 1 0 3. 2 2 0
5 2 1 1 2 4 1 3 4
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Adjoint of Matrices
The matrix formed by taking the transpose of the cofactor matrix of a given original matrix. The adjoint of
matrix A is often written as adj A.
1 2 3 Adjoint also have another
Example: Find the adjoint of the matrix A = 0 4 5 term. What is that term?
1 0 6 ADJUGATE or ADJUNCT
To find the Adjoint of Matrices A, find the cofactor of each element first.
a11 = ( )−1 1+1 4 5 a12 = ( )−1 1+2 0 5 = (−) − 5 = 5 a13 = ( )−1 1+3 0 4
0 = 24 1 1 = −4
6
6 0
a21 = ( )−1 2+1 2 3 = (−)12 = −12 a22 = ( )−1 2+2 1 3 a23 = ( )−1 2+3 1 2 = (−)(−2) = 2
0 1 = 6−3 = 3 1
6 0
6
a31 = ( )−1 3+1 2 3 a32 = ( )−1 3+2 1 3 = (−1)5 = −5 a33 = ( )−1 3+3 1 2
4 = 10 −12 = −2 0 0 =4
5
5 4
24 5 −4
Hence, the cofactor of matrix A = −12 3
2 .
−2 −5 4
24 −12 −2
3 −5 .
Consequently, the adjoint of matrix A is adj( A ) = 5
−4 2 4
Task A: Find the Adjoint of the following matrices:
1 2 3
i. B = 1 3
5 .
1 5 12
Cofactor , a11 Cofactor , a12 Cofactor , a13
Cofactor , a21 Cofactor , a22 Cofactor , a23
Cofactor , a31 Cofactor , a32 Cofactor , a33
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Cofactor of Matrix B = DUM20132 ENGINEERING MATHEMATICS 2 13
Adjoint of Matrix B =
1 2 3 Cofactor , a12 Cofactor , a13
ii. C = 0 4 5 Cofactor , a22 Cofactor , a23
Cofactor , a32 Cofactor , a33
1 0 6
Cofactor , a11
Cofactor , a21
Cofactor , a31
Cofactor of Matrix B = Adjoint of Matrix B =
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The inverse of 2 2 matrix
The inverse of a 2 2 matrix is denoted by A−1 with the property that AA−1 = A−1A = I , where I is the
1 0
identity matrix 0 1 . That is, multiplying a matrix by its inverse produces an identity matrix. Note that in
this context A−1 1 .
A
Not all 2 2 matrices have an inverse matrix. If the determinant of the matrix is zero, then it will not have
an inverse, the matrix is then said to be singular. Only-non-singular matrices have inverses.
Let A = a b
c d , then the inverse of matrix A:
A−1 = 1 d −b
A −c
a
Task A: Find the inverse of the following matrices.
1. B = 10 4 2. C = 6 2
2 7 3
Example: 6
A = 3 1
4 2
A −1 = 1 2 −1
−4
3(2) −1(4) 3
= 1 2 −1
2 −4
3
1 − 1
2
=
−2 3
2
3. D = 5 4 4. E = 3 1 5. F = 1 2
5 6 2 2 1 3
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The inverse of 3 3 matrix
The formula for the inverse matrix is A−1 = 1 adj (A)
A
Example:
1 2 3
Find the inverse of A = 0 4 5
1 0 6
24 −12 −2
3 −5 , refer previous topic.
We already have that adj( A ) = 5
−4 2 4
We need to find determinant of A,
A = 1 4 5 − 2 0 5 + 3 0 4 = 1(24) − 2 (−5) + 3 (−4) = 24 +10 −12 = 22
06 16 10
Then,
12 −6 − 1
11 11
24 −12 −2 11
A −1 = 1 5 3 −5 = 5 3 − 5
22 2 4 22 22
−4 22 1
11
− 2 2
11 11
Task A : Find the inverse of the following matrices.
2 0 6
i. K = 1 2 0
1 1 3
K =6
1 1 −2
− 1
K −1 = 2 0 1
−1
1 2
− 3
6 3
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1 2 2
ii. L = 0 2 3
1 1 0
L = −1
3 −2 −2
L−1 = −3 2
3
2 −1 −2
3 1 1
iii. 0 −1 2
1 1 1
A = −6
1 0 − 1
2
2
A−1 = − 1 −1 1
3 3
− 1 1 1
6 3 2
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Solution of system of linear equations using matrix inversion method
Example 1 : Consider the simultaneous equations
x + 2y = 4
3x − 5y = 1
Written in matrix form as
1 2 x = 4 ➔ AX = B X = A−1 B
3 −5 y 1
Solve using the matrix inversion method
x = 1 −5 −2 4
y −3 1
1(−5) − 2(3) 1
= 1 −5(4) + −2(1)
−11
−3(4) + 1(1)
= − 1 −22 = 2
11 −11 1
The final answer (x,y) for the system is (2,1)
Example 2 : Solve the system with three variables by using suitable methods.
−2x + y + z = 4
−4x + 2y − z = 8
−6x − 3y + z = 0
Written in matrix form as
−2 1 1 x 4
−4 2 −1 y = 8
−6 −3 1 z 0
−2 1 1 4
Let A = −4 2 −1 and B = 8 ,
−6 −3 1 0
Solve using the matrix inversion method
Step 1: Find Determinant of Matrix A
−2 1 1 2 −1 −4 −1 + 1 −4 2
A = −4 −1 −3 −6 1 −6 −3
−6 2 1 = −2 1 − 1
−3
= −2 (2 − 3) − 1(−4 − 6) + 1(12 − −12)
= −2 (−1) − 1(−10) + 1(24)
= 2 + 10 + 24
= 36
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Step 2: Find Adjoint of Matrix A
.
2 −1 −4 −1 −4 2 T
−
−3 1 −6 1 −6 −3 −1 − (−10) 24 T −1 10 24 T −1 −4 −3
−2 1 4 −12 = 10 4
Adj 1 1 −6 1 − −2 1 = − (4) 4 − (12) = −4 −6 −12 6
A = − −2 1 0 24
−3 1 −1 −6 −3 −3 − (6) 0 −3
− 0
1 1 −4 −2 1
2 −1 −4 2
Step 3: Substitute into formula.
x −1 −4 −3 4
y 1 10 8
z = A−1 B = 36 24 4 6
−12
0 0
−1(4) + −4 (8) + −3 (0)
1
= 36 10 (4) + 4 (8) + 6 (0)
24 (4) + −12 (8) + 0 (0)
1 −36
36
= 72
0
−1
= 2
0
x = −1 , y = 2 , z = 0
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1.3.2 Cramer’s Rule
Cramer’s rule relies on determinants.
Consider the system below with variables x, y and z:
a11 x + a12 y + a13 z = b1 a11 a12 a13 x b1
a21 x + a22 y + a23 z = b2 a21 y b2
a31 x + a32 y + a33 z = b3 a22 a23 =
a31 a32 a33 z b3
b1 a12 a13 a11 b1 a13 a11 a12 b1
A1 = b2 A2 = a21 A3 = a21
a22 a23 b2 a23 a22 b2
b3 a32 a33 a31 b3 a33 a31 a32 b3
x = det A1 @ A1 ; y = det A2 @ A2 ; z = det A3 @ A3
det A A det A A det A A
Example 1 : Consider the simultaneous equations
x + 2y = 4
3x − 5y = 1
Solve using Cramer’s’ Rule
1 2 x = 4 Ax = B
3 −5 y 1
A = 1 2 = 1(−5) − 2(3) = −11
3 −5
To find x
42
1 −5 = 4(−5) − 2(1) = −22 = 2
x= −11 −11 −11
To find y
1 4
y = 3 1 = 1(1) − 4(3) = −11 = 1
−11 −11 −11
The final answer (x,y) for the system is (2,1) .
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Example 2 : Solve the system with three variables by using suitable methods.
−2x + y + z = 4
−4x + 2y − z = 8
−6x − 3y + z = 0
Written in matrix form as
−2 1 1 x 4 −2 1 1 4
−4 2 −1 y = 8 ➔ Let A = −4 2 −1 and B = 8 ,
−6 −3 1 z 0 −6 −3 1 0
Solve using Cramer’s Rule
Step 1: Find Determinant of Matrix .
−2 1 1 2 −1 −4 −1 + 1 −4 2
A = −4 −1 −3 −6 1 −6 −3
−6 2 1 = −2 1 − 1
−3
= −2 (2 − 3) − 1(−4 − 6) + 1(12 − −12)
= −2 (−1) − 1(−10) + 1(24)
= 2 + 10 + 24
= 36
Step 2: Find the value of x .
41 1
8 2 −1
0 −3 1 = 4(2 − 3) −1(8 − 0) + 1(−24 − 0) = 4(−1) −1(8) + 1(−24) = −36 = −1
x=
36 36 36 36
Step 3: Find the value of y .
−2 4 1
−4 8 −1
−6 0 1 = −2(8 − 0) − 4(−4 − 6) + 1(0 − −48) = −2(8) − 4(−10) + 1(48) = 72 = 2
y=
36 36 36 36
Step 4: Find the value of z .
−2 1 4
−4 2 8
−6 −3 0 −2(0 − −24) −1(0 − −48) + 4(12 − −12) −2(24) −1(48) + 4 (24) 0 =0
y=
= ==
36 36 36 36
The final answer (x,y,z) for the system is (−1, 2,0) .
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Equations with no unique solution
If A = 0 , A−1 does not exist and therefore, early on, it is easy to see that the system of linear equations
has no unique solution. But it is not obvious whether this is because the equations are inconsistent or
whether they have infinitely many solutions.
Task A: Find the solution of each variable of the following system by using matrix inversion method.
2x + 3y = 5 5x + y = 13 3x + 2y = −2
1. 2. 3.
x − 2y = −1 3x + 2y = 5 x + 4y = 6
x = 1, y = 1 x = 3, y = −2 x = −2, y = 2
Task B: Find the solution of each variable of the following system by using Cramer’s’ rule.
2x + 3y = 3 2. 8m − 3n = 3 3k + 2g = 3
1. −3m + 9n = 4 3.
5x + 4y = 11 −2k − g = −1
x = 3, y = −1 m = 39 ,n = 41 k = −1,g = 3
63 63
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Task C: Find the solution of each variable of the following system by using matrix inversion method.
x − 2y + z = 3
1. 2x + y − z = 5
3x − y + 2z = 12
A = 10,x = 3,y = 1,z = 2
4x + 5y − 2z = −14
2. 7x − y + 2z = 42
3x + y + 4z = 28
A = −154,x = 4,y = −4,z = 5
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Task D: Find the solution of each variable of the following system by using Cramer’s rule.
x + 2y + 3z = −5
1. 3x + y − 3z = 4
−3x + y + 7z = −7
A = 40,x = −1,y = 1,z = −2
2. − y − 2z = −8
x − 3z = 2
7x + y + z = 0
A = −22,x = −1,y = 6,z = 1
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Revision Unit 1
2 −4 4 −1 4 D = 3 1 E = −3 2 0 .
1. Given A = 1 3 B = 2 0 C = 3 −1 −2
1
Calculate:
a. A + B
b. 3B
c. AC
d. AE
e. B + D
2. Evaluate the following determinants:
24
a.
69
2 2 −3
b. 4 −5 4
3 −1 0
3 10
c. 4 2 3
544
323
d. Show that 2 −1 1 = –12
606
3. Evaluate the determinant by choosing the simplest row or column by which to expand:
10 1
a. −1 0 4
5 2 −2
−2 1 8
b. 4 1 6
0 0 −2
4. Use matrices to solve the following systems of equations:
x−y =1
a.
x+y = 4
y = x−4
b. 2x + y = 6
x + 2y + 3z = −7
c. x + y − 2z = 4
x + z = −1
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Answer:
1. Calculate:
a. 6 −5
3
3
b. 12 −3
6 0
c. −4
13
d. −10 8 8
−1 −6
0
e. Undefined
2. Determinant:
a. −6
b. −1
c. −13
323
−1 1 2 1 2 −1
d. 2 −1 1 = 3 − 2 + 3
0 6 66 6 0
606
= 3(−6 − 0) − 2(12 − 6) + 3(0 + 6)
= −18 − 12 + 18
= −12 (as required)
3. Determinant
a. −10
b. 12
4. Solve:
a. x = 5 ,y = 3
22
b. x = 10 ,y = − 2
33
c. x = 1,y = −1,z = −2
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What are Vector and How Are They Used?
Denoting both direction and magnitude, vectors appear throughout the world of science and
engineering.
Vectors are used in science to describe anything that has both a direction and a magnitude. They are
usually drawn as pointed arrows, the length of which represents the vector's magnitude. A
quarterback's pass is a good example, because it has a direction (usually somewhere downfield) and
a magnitude (how hard the ball is thrown).
Off the field, vectors can be used to represent any number of physical objects or phenomena. Wind,
for instance, is a vectorial quantity, because at any given location it has a direction (such as
northeast) and a magnitude
After completing the unit, students should be able to:
1. Identify the notation of vector
2. State the definition of :
• scalar and vector.
• Zero vector
• Unit vector
• Negative vector
• Equal vector
3. Perform the standard operation with vectors including :
• Addition
• Scalar multiplication
4. Compute the addition of a vector in two and three dimensions.
5. Find the magnitude and unit vectors of two and three dimensions.
6. Perform scalar product in two and three dimensions.
7. Perform vector product in two and three dimensions.
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TYPER OF VECTOR COMPARISON MAGNITUDE
ZERO VECTORS EQUAL VECTOR SCALAR VECTOR (=a1,a2 ) and Q (b1,b2 )
Only Magnitude =Let P
u (a,b) = (m,n) magnitude
and direction (b1 − )a1 2 + (b2 − a2 )2
Eg: speed, =PQ
temperature
Eg: velocity, (=a1,a2 ,a3 ) and Q
UNIT VECTOR momentu=m Let P (b1,b2 ,b3 )
−u PQ = PQ
PQ PQ = (b1 − )a1 2 + (b2 − a2 )2 + (b3 − a3 )2
NEGATIVE VECTOR MULTIPLICATION
NOTATION k (a,=b) (ka,k=b) kai + kbj ADDITION AND SUBSTRACTION
k=u
u
=(a, b) a = k (a,b,c) = (ka,kb,kc) = kai + kbj + kck
PQ =u =ai = ku v
+ bj b
ADDITION AND SUBSTRACTION
PQ =u =ai + bj + ck
a u± v = (a,b) ± (c,d) = (a ± c,b ± d)
= (=a, b, c) b
u± v =
c (a,b,c) ± (d,e, f) = (a ± d,b± e,c ± f) PARALLELOGRAM TRIANGLE
FORMULA FORMULA AREA OF PARALLELOGRAM
(=a1,a2 ) and v (b1,b2 ) ONLY FOR 3D u=× v u v sin θ
=Let u where
altit=ude v sin θ
=a1,a2 ,a3 and v b1,b2 ,b3
i j k
u × v =a1 a2 a3
b1 b2 b3
( ) ( )u ⋅ v= u1v1 + u2v2 =Let u
Let u (=a1,a2 ,a3 ) and v (b1,b2 ,b3 )
u ⋅ v= u1v1+ u 2v2 + u3 v3
AREA OF TRIANGLE
=u ⋅ v u v cos θ
= a2 a3 i − a1 a3 j + a1 a2 k 1×
b2 b3 b1 b3 b1 b2 u× v
cos θ = u ⋅ v
u v = new vector Area of triangle =
2
PARALLEL TYPE OF ANGLE ORTHOGONAL /
(Same d irection) PERPEN DICULAR
u ⋅ v = u v ,θ =0° PARALLEL u ⋅ v= v ⋅ u,=θ 90°
(Opposite direction)
u ⋅ v =− u v ,θ =180°
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2.1 Definition
If there is a movement happened from point A to B, so the vector is defined as AB where A is initial
point and B is terminal point with the existing of a magnitude and direction.
Task A : Identify the notation of vector
Example : 1. Notation=
K
S
R L
Notation = RS
2. Notation= 3. Notation=
G
MN
H
Task B: Give the position vectors in the diagram.
Example: 1. = −7 =−7i + 8 j =57
a 8 2. =n
a= 5 =
−2 5i − 2j
3. m =−6i − 5j 4. k = i + 7j 5. s= 3i − 5j
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Task C: Expressing vector in term of a and b.
Example: 1. If LP = a and LR = b , express in term of
A EB a and b forthe following :
F i. LM
ii. LN
DC iii. LQ
iv. PN
→→
M
If AE = a and AF = b , express the following in PQ
terms of a andb. L RN
i. AD = a + b + b + b − a = 3b
ii. AB = b + a + a − b = 2a
iii. DB = a + a − b − b − b = 2a − 3b
iv. FC = b + b + a + a = 2a + 2b
2.2 Vector Algebra
The ways to find the sum of the vector as following:
a) Triangle Law for addition
There are two vector a and b. to find the sum, vector b is moved to a location such that its
initial point coincides with the end point of vector a. Vector c is resultant vector, that is sum of
a and b, a+b.
c= a + b
b
ab a
b) Parallelogram Law for addition
Let the two vectors be the two sides of a parallelogram, which starts at the same initial point.
Then the sum u+v is the diagonal of the parallelogram.
c=a+b
ab
Multiplication of vector
λ a is a vector of magnitude λ a and in the same direction as aa. It also follows that is − λa
opposite to that of
a vector of magnitude λ a and with direction
a λa
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Task A : Find the sum of two vectors.
Example 2. Sum vector = 3. Sum vector =
MD
1. Sum vector = RP E
P
Q N OF
R
4. Sum vector = 6. Sum vector =
R 5. Sum vector = JG
K H
ST L I
Task B : Find the addition of two vector in ℜ2 and ℜ3 .
Example : 1. Let v = (−10,5) and 2. Let =v (3,−4) and
w= (2,−8) in ℜ2 . Find =w (11,−7) in ℜ2 . Find
Let v = (8,4) and w = (−1,7) in
v+w. v+w.
ℜ2 .
Find v + w
v + w= (8 + −1,4 + 7)
= (7,11)
= 7i + 11j
Example : 3. Let =v (9,−3,5) and 4. Let =v (7,−4,1) and
=w (1,9,−4) in ℜ3 . Find w = (−2,8,11) in ℜ3 . Find
Let v = (−3,8,4) and
v+w. v+w.
w = (5,−1,7) in ℜ3 .
Find v + w
v + w = (−3 + 5,8 + −1,4 + 7)
= ( 2,7,11)
= 2i + 7j + 11k
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2.3 Component of vector
Task A: Find the component form of the vectors.
Example: 1. If a= 3i + 7j , b= 2i − 4j and 2. If a= 3i + 7j , b= 2i − 4j and
If a= 3i + 7j , b= 2i − 4j and
c =−3i − 4j , find c =−3i − 4j , find
c =−3i − 4j , find a) 2a + 3b a) 5c − a
b) 7c b) 4c − 2b
a) 2a
=2(3i + 7j) =6i + 14j
b) b + c
(2i − 4j) + (−3i − 4j)
= −i − 8j
c) a − b
= (3i + 7j) − (2i − 4j)
= 3i + 7j − 2i + 4j
= i + 11j
Direction of the vector
Task B: Find the angle of a vector.
Example: 1. Given the k= 5i − j . Find the 2. Given the G =−2i + 3j . Find
Given the p= 3i + 5j
direction of k. the direction of G.
Find the direction of p.
Solution:
tan θ =5
3
θ =tan−1 5
3
= 59.04°
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Magnitude of vectors
For 2 dimensional
If a has coordinate A = (a1,a2 ) and b has coordinate B = (b1,b2 )
(b1 − a1)2 + (b2 − a2 )2
Then the magnitude of AB =
For 3 dimensional
If a has coordinate A = (a1,a2,a3 ) and b has coordinate B = (b1,b2,b3 )
(b1 − a1)2 + (b2 − a2 )2 + (b3 − a3 )2
Then the magnitude of AB =
Task E: Find the magnitude of each vector.
Example : 1. Given that vector 2. Given that vector
PQ= 8i + 12j , find PQ KL =−5i + 13j , Find KL .
Given that vector PQ= 3i − 4j , find
PQ
SP=oQlution:(3)2 + (−4)2
= 9 + 16 = 25 = 5
Example: 1. Given 2 vector, L= 2i + 6j 2. Given 2 vector,
Given 2 vector, D=5i + 3j and
E =−2i + 4j , find ED and M =−3i + 7j , find =R 10i + 7j and
LM S= 3i − 8j , find RS
SEo=Dlution:(5 − −2)2 + (3 − 4)2
= 72 + (−1)=2 49 +=1 50
Example: 1. Given vector 2. Given the vector
H = 2i + 6j − 5k and
Given vector =L (3,−1,9) and =H (6,7,−5) and
I =i + 3j − 2k . Find the
=K (4,5,−2) , find KL magnitude of IH . =P (3,−6,10) .Find the
magnitude of PH .
K=L
(3 − 4)2 + (−1− 5)2 + (9 − −2)2
= 1+ 36 + 121 = 158
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Unit vector
For 2 dimensional
If =a a1i + a2 j ,
∧ a a1i + a 2 j
Then the unit vector,
a= a (a1)2 + (a2 )2
For 3 dimensional
If a =a1i + a2 j + a3k ,
∧ a1i + a 2 j + a3k
a
Then the unit vector, a= (a1)2 + (a2 )2 + (a3 )2
a
Task A: Find the unit vector of each of the following.
Example: ^^
P ( 4, 5 ) ∧ 1. Find the unit vector of W 2. Find the unit vector of V ,
Given = , find the unit vector, P ^ (−2,3) . given ^ = (5,11) .
. , given W = V
SP=olu(ti4o,n5:=) 4i + 5j
P= 42 + 52 = 41 = 6.403
∧ P= 4i + 5=j 0.625i + 0.781j
P 6.403
=P
Example: 2 2.
G 2 Given Z = 2i + 2j + k , find
1 ∧ 1. Given = , find the
Given R= ∧
−2 , find the unit vector, P 4
the unit vector, P .
4 ∧
. unit vector, G .
Solution:
P =1i − 2j + 4k
P= 12 + (−2)2 + (4)2= 2=1 4.58
∧ P= 1i − 2j + 4k
P 4.58
=P
= 0.218i − 0.437j + 0.873k
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2.4 Scalar vector / Dot Product
If a =a1i + a2 j + a3k and b =b1i + b2 j + b3k
Then,
and if θ is a ⋅=b a1a2 + b1b2 + c1c2
angle between and b
a
Then,
=a ⋅ b a b cos θ
Task A : Find the angle between two vector.
Example: Given m = 8i − 5j + 2k Given a = 9i − 5j + 4k
Given a = 2i − 3j + 4k n =i + 4j + 12k b = 2i + 4j + k
b =i + 3j − 2k Find angle between m and n. Find angle between a and b.
Find angle between a and b.
Solution:
a•b
= (2×1)+ (− 3×3)+ (4)(− 2)
= 2 − 9 − 8 = −15
=a 22 + (−3)2 + 4=2 29
b= 12 + 32 + (−2)2= 14
Als=o, a.b a b cos θ
cos θ = − 15
29. 14
=θ cos−1 − 1=5 138.11°
406
Task B: find the dot product between two vectors.
Example: Find the dot product of vectors Find the dot product of vectors
Find the dot product of vectors v = (1,0,4) and w = (−2,5,3) . v = (2,4,1) and=w (3,2,−8) .
v = (2,0,9) and w = (−3,4,8) .
Solution:
v.w = (2 × −3) + (0 × 4) + (9 × 8)
=−6 + 0 + 56 =50
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Properties of Scalar Product/Dot Product b
a. Parallel Vectors
If a and b are parallel
b
a or a=⋅ b aab cos π
a ⋅ b =ab cos0
a ⋅ b =ab for like parallel vectors, θ = 0° .
a ⋅ b =−ab for unlike parallel vectors, =θ 180° .
In the special case when a = b, a ⋅ b= a.a= a2 (sometimes a.a is written as a2 )
b. Perpendicular/Orthogonal Vectors
If a and b are perpendicular th=en, a.b a=b cos π 0 i.e a ⋅ b =0
2
b
a
In the special case of the Cartesian unit vectors these results give
i⋅i = j⋅ j = k⋅k = 1
i⋅ j = j⋅k = k⋅i = 0
Task C : Determine the properties of two vectors
Example : 1. Show that vector 2. Show that vector
Find the properties of two vector,
b = 2i + 2j − k and c = 5i − 4j + 2k . =r (3,2,−1) is at right =k (4,2,−5) and
b ⋅ c= (2 × 5) + (2 × −4) + (−1× 2) angle to vector l = (2,1,2) are
= 10 + −8 + −2= 0 s = (−3,6,3) . perpendicular.
∴perpendicular since the b ⋅ c =0
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2.5 Vector Product/Cross Product
(a) Defined as a vector having magnitude a b sinθ , and the vector acts in a
direction perpendicular (orthogonal) to both a and b .
(b) If a = x1i + y1j + z1k and b = x2i + y2 j + z2k , then vector product
i jk
a × b =x1 y1 z1
x2 y2 z2
(c) Properties of vector product:
a. Not commutative, a × b ≠ b × a but a × b =−(b × a)
b. a × (b + c) = a × b + a × c
c. α (a × b) = αa × b = a × αb
d. (a × b) × c ≠ a × (b × c)
e. Parallel, if a × b =0 .
f. Perpendicular, if a × b =a b
Task A : Find the vector normal/cross product to the plane containing the following vector.
Example: i. m = 4i − 2j + k and ii. p = 9i + j + 5k and
a =i − 2j + 3k and
n =i + 8j − 2k q = 2i − 5j + 4k
b =−2i + 3j + 2k
Solution:
i jk
a ×=b 1 −2 3
−2 3 2
= i(−4 − 9) − j(2 − −6) + k (3 − 4)
=−13i − 8j − k
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Task B: Applications of Cross Product.
Area of parallelogram 1. Find the area of 2. Find the area of
a=× b a b sin θ where parallelogram determined parallelogram determined
by vectors u = 5i − 2j + k by vectors u =i + 4j + 2k
b sinθ =altitude
Example : and w = 3i + 8j − 2k and w = 5i − 3j + k
Find the area of parallelogram
determined by vectors u = 3i + j + 2k
and w =i − 2j − 2k
Solution:
ijk
u × w =3 1 2
1 −2 −2
= i(−2 − −4) − j(−6 − 2) + k (−6 − 1)
= 2i + 8j − 7k
u ×=w (2)2 + (8)2 + (−7)2
= 4 + 64 + 49
= 10.82
Area of a triangle 3. Find the area of triangle in 4. Find the Area of triangle
Area of triang=le 1 u × v space with vertices ABC given that
2 A (1,−1,3),B(3,1,4) and a = (3,0,1) and
Example:
Find the area of triangle in space with C(−2,4,3) . b= (2,−1,−1)
vertices A (0,1,3),B(4,−1,2) and
C(1,3,−5) .
Solution:
BA =−4i + 2j + k
BC =−3i + 4j − 7k
i j k
BA × BC =−4 2 1
−3 4 −7
= i(−14 − 4) − j(28 − −3) + k (−16 − −6)
=−18i − 31j − 10k
BA × BC = (−18)2 + (−31)2 + (−10)2
= 1385
Area of triangle
( )1 BA=× BC 1 1=385 18.6 unit2
22
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Revision Unit 2
1. Given the vector a = (3,1) and b = (1,2) , find:
a. a + b
b. 2b
c. 3a − 2b
d. 3a − 2b
2. Given the vector m = (−4,5,3)
a. Find a unit vector in the same direction as m
b. Find a vector that has the same direction as m but has length 10
3. Find the component form and magnitude of vector AB from A (4,−7) to B(−1,5) .
4. Express the vector u= 2i − 3j in terms of its magnitude and the angle it makes with x-axis.
5. Let v = (−2,5) and w = (3,2) , find each of the following vector:
a. 2v
b. w − v
c. v + 2w
6. If k = i + 3j and h= 2i − j , calculate k ⋅ h
7. Consider three vectors:
A =−3i + 3j + 2k
B =−2i − 4j + 2k .
C = 2i + 3j + k
Find : ( )
a.
A⋅ B+C
b.
( )
c.
A⋅ B×C
( )
A× B+C
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Answer:
1. Vector:
a. (4,3)
b. (2,4)
c. (7,−1)
d. 50
2. Unit vector:
a. m= 50,u= 4 , 1 , 3
−
50 50 50
b. 10m =10 − 4 , 1 , 3 = − 40 , 10 , 30
50 50 50 50 50 50
3. AB =(−5,12), AB =13
4. u = 13,θ = −56°
5. Vector:
a. (−4,10)
b. (5,−1)
c. (4,13)
6. −1
7. 3 vector:
( )a. B + C = 0i − j + 3k,A ⋅ B + C = 3
( )
b. B × C =−10i + 6j + 2k,A ⋅ B × C =52
( )
c. B + C = 0i − j + 3k,A × B + C = 11i + 9j + 3k
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In mathematics, differential calculus is a subfield of calculus concerned with the study of the rates at
which quantities change. It is one of the two traditional divisions of calculus, the other being integral
calculus, the study of the area beneath a curve.
Differentiation has applications to nearly all quantitative disciplines.
After completing this topic, students should be able to:
1. Find the differential of a function from first principle.
2. Differentiate algebraic functions or polynomials using Basic Rule of Differentiation or formula
3. Differentiate the trigonometric logarithmic by using appropriate methods.
4. Differentiate product of functions using Product Rule.
5. Differentiate quotient of functions using Quotient Rule.
6. Differentiate composite functions using Chain Rule.
7. Differentiate parametric functions.
8. Differentiate implicit functions.
9. Solve the higher order derivatives; differentiate the previous derivative’s functions.
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3.1 Introduction of Differentiation
Differentiation is an important concept in calculus and its applications are widely used in many
areas, especially in the manufacturing industries.
The definitions of the derivative of a function y = f (x) are given below.
3.1.1 Derivatives of First Principle
The derivative or differentiation of a function f (x) at a point x is defined as
f '(x) = lim f (x + h) − f (x) , provided the limit exists.
h→0 h
The step of determine the derivative of a function f (x) is given by following steps:
Step 1 : Find f (x + h)
Step 2 : Find f (x + h) − f (x)
Step 3 :
f (x + h) − f (x)
Step 4 :
Find
h
f (x + h) − f (x)
Find lim
h→0 h
Task A : Find the derivative of the following functions using the first principles.
Example 3.1:
f (=x) 3x2 + 5
Answer:
Step 1 : f (x + h) = 3(x + h)2 + 5
( )f (x + h=) 3 x2 + 2hx + h2 + 5
Step 2 :
( )f (x + h) − f (x) 3 + 5 − 3x2 + 5
=
x2 + 2hx + h2
Step 3 : h h
= 3x2 + 6hx + 3h2 + 5 − 3x2 − 5
h
= 6hx + 3h=2 h(6x + 3h) 6x + 3h
=
hh
f (x + h) − f (x)
lim = lim 6x + 3h
Step 4 : h→0 h h→0
= 6x + 3(0)
= 6x
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1. f (x) = x2
f '(x) = 2x
2. f (x) = 3x3
3. f (=x) 5x2 − 6x f '(x) = 9x2
4. f (x)= 3 + 12x f '(=x) 10x − 6
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3.2 Derivative of standard functions
In this section, we will discuss theorems and techniques that can be used for differentiation.
Notation of differentiation
Type of notation Notation
Leibniz’s notation dy , d2y , d3y
dx dx2 dx3
Lagrange’s notation
f '(x),f ''(x),f '''(x),f 4 (x),...fn (x)
Euler’s notation =Df d=f ,D2f d2f=,...Dnf dnf
Newton’s notation dx dx2 dxn
4n dn y
• •• ••• • • = dtn
y, y, y ,y,...,y
3.2.1 Differentiation of Basic Functions
Theorem f '(x) Example
y = 28
y=a dy = 0 dy = 0
Where a is a constant dx dx
y = ax dy = a y = 16x
Where a is a constant dx dy = 16
dx
y = xn dy = nxn−1 y = x5
Where n is a constant dx dy = 5x4
dx
y = axn dy = naxn−1 y = 2x3
dx dy = 6x2
Where a and n are dx
constants
=y axn ± bxm =dy naxn−1 ± maxm−1 =y 10x4 − 3x5
=dy 40x3 − 15x4
Where a, b, m and n are dx dx
constants
=y (ax + b)n =dy n(ax + b)n−1 (a) =y (8x + 12)4
dy =4(8x + 12)3 (8) =32(8x + 12)3
Known as dx
‘Composite Function’ dx
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DUM20132 ENGINEERING MATHEMATICS 2 44
Task B : Find the derivative of the following functions.
Example 3.2: 1. y = 167x 2. y = πx 3. y = − x
7
y = −125
dy = 0
dx
4. y = 3x6 5. y = 9x−4 6. y = 17 7. y = x5
5 x3 3
( )3 9. y = x5 10. y = 16x−2 11. y = 18x6
x2 x
8. y = x
12. y = 19x 3 + 3x − 100 13. y = 16x + x − 8
x5
( )14=. y 3x2 − 4 5
15.=y (7x + 23)−3
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DUM20132 ENGINEERING MATHEMATICS 2 45
3.2.2 Differentiation involving Trigonometric Functions
Theorem f '(x) Example
sin x y = 2 sin x
cos x cos x dy = 2cos x
tan x dx
ln x − sin x
y = 6 cos x
sec2 x dy = −6 sin x
dx
1
x y = 7 tan x
dy = 7 sec2 x
dx
y = 5ln x
=dy 5= x1 5
dx x
ex ex y = 4ex
sin u dy = 4ex
cos u dx
tan u
ln u =y sin(7x + 5)
cos u du dy = cos(7x + 5) 7 = 7cos(7x + 5)
dx dx
du ( )y = cos 18x2
dx
− sin u ( ) ( )dy
dx
=− sin 18x2 36x =−36x sin 18x2
y = tan(16x)
sec2 u du =dy s=ec2 (16x) 16 16 sec2 (16x)
dx dx
=y ln(3x + 4)
1 du =dy (=3x1+ 4) 3 3
u dx dx
(3x + 4)
y = 2e3x
eu eu du =dy 2=e3x 3 6e3x
dx dx
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DUM20132 ENGINEERING MATHEMATICS 2 46
Task C : Find the derivatives of the following functions.
1. y = cos 5x ( )=2. y tan x2 + 3 ( )3. y = sin x
3
4. =y ln(x + 3) 5. y = ln(5x) ( )6.=y ln x3 − 2
7. y = ex+5 8. y = ex2 9. y = e6x−1
1=0. y 6 sin x + e4x 11. y= cos(5x) + ln(7x − 3) =12. y ln(10x) − e8x
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DUM20132 ENGINEERING MATHEMATICS 2 47
3.1 Techniques of Differentiation
3.3.1 Product Rule
Product of functions means two functions multiplied together.
For example:
y = x2 cos3x
So, we have one function, x2 , multiplied by a second function, cos3x .
Notice that we can write this as y = uv where u = x2 and v = cos3x .
To differentiate this product functions, we can use product rule as below:
If u and v are two differentiable functions and f (x) = u(x) v (x) , then
f '(x) =v (x)u'(x) + u(x) v '(x) =vu'+ uv '
Example 3.3:
( )( )Find the derivative of f (x) =5 + 3x3 x2 + 6
Solution: =v x2 + 6
u= 5 + 3x3 v ' = 2x
u' = 9x2
f '(x=) vu'+ uv '
( )( ) ( )= x2 + 6 9x2 + 5 + 3x3 (2x)
= 9x4 + 54x2 + 10x + 6x4
= 15x4 + 54x2 + 10x
Task D : By using product rule, find the derivative of the following functions.
( )1. y =x2 + 2 (3x + 1) ( )2. y =(2x + 3) x3 + x2
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( )1. y = 2x2 − 1 (3x + 4) DUM20132 ENGINEERING MATHEMATICS 2 48
( )2. y = x3 + 5 (2 − 3x)
( )( )3. y =x2 − 4 2x5 + x =4. y 3x12 (7 − x)
3.3.2 Quotient Rule
Functions often come as quotients, by which we mean one function divided by another
function.
For example:
y = cos x
x2
Notice that we can write this as y = u where u = cos x and v = x2 .
v
To differentiate this quotient of functions, we can use quotient rule as below:
If u and v are two differentiable functions and f (x) = u(x) , then
v(x)
=f '(x) v(x)u'(x) −u(x)v'(x) vu'+ uv '
v2
=
v2
Example 3.4:
Given y = 3x − 1 , find dy .
x2 + 4 dx
Solution: =v x2 + 4
=u 3x − 1 v ' = 2x
u' = 3
MJII MARA JAPAN INDUSTRIAL INSTITUTE
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