lecture note on mathematics for social scientists ii (mts 108)
TRANSCRIPT
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LECTURE NOTE
ON
MATHEMATICS FOR SOCIAL SCIENTISTS II
(MTS 108)
BY
ADEOSUN SAKIRU ABIODUN
E-mail: [email protected]
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Course Outline
Functions and equations: Definition and types of functions, definition and types of
equations. Matrices. Introduction to statistical distribution and density functions,
especially the binomial, Poisson and normal. Introduction to calculus functions of the
variable and their continuity. Techniques of differentiation; logarithmic, trigonometric
and exponential functions. Integral calculus; Optimization of functions, maximal,
minimal and in flexional points, and the application of these concepts in Business
and Economics.
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§ 1.0 FUNCTIONS
Mathematical modeling is an attempt to describe some part of the real world
in mathematical terms. Our models will be functions that show the relationship
between two or more variables. These variables will represent quantities that we
wish to understand or describe. We will begin by reviewing the basic concept of
functions. In short, we call any rule that assigns or corresponds to each element in
one set precisely one element in another set a function.
Definition of a function: A function ð from ð· to ð is a rule that assigns to each
element ð¥ in ð· one and only one (unique) element ðŠ = ð(ð¥) in ð .
The set ð· in the definition is called the domain of f. We might think of the
domain as the set of inputs. We then think of the values ð(ð¥) as outputs. The set of
all the possible outputs (set of functional), ð is called the range of f. The letter
representing elements in the domain is called the independent variable and the letter
representing elements in the range is called the dependent variable. Thus, if
ðŠ = ð(ð¥), ð¥ is the independent variable while ðŠ is the dependent variable. Note that
ðŠ = ð(ð¥) is read as âðŠ is a function of ð¥â.
Example 1: A restaurant serves a steak special for N500. Write a function that
models the amount of revenue made from selling these specials. How much revenue
will 15 steak specials earn?
Solution: We first need to decide if the independent variable is the price of the
steak specials, the number of specials sold, or the amount of revenue earned. Since
the price is fixed at N500 per special and revenue depends on the number of special
sold, we choose the independent variable, ð¥, to be the number of specials sold and
the dependent variable, ð = ð(ð¥) to be the amount of revenue.
⎠ð = ð ð¥ = 500ð¥. We note that ð¥ must be a whole number, so the domain is
= 0,1,2, ⊠. Hence, when selling 15 steak specials, ð = ð 15 = 500 15 = 7500. So
the revenue is N7500.
Example 2: If ð ð¥ = ð¥3 â 3ð¥ + 4, find the values of : ð(0), ð(â1), ð(ð), ð(3ð¥).
Solution: ð 0 = 03 â 3 0 + 4 = 4
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ð â1 = â1 3 â 3 â1 + 4 = 6
ð ð = ð3 â 3ð + 4
ð 3ð¥ = 3ð¥ 3 â 3 3ð¥ + 4 = 27ð¥3 â 9ð¥ + 4
Example 3: If ð ð¥ =ð¥+1
ð¥2+ð¥â2, determine
(a) the image of â1
(b) ð(ð¥ â 1)
(c) What values of ð¥ will have no images for this function?
Solution: (a) The image of ð¥ = â1 is ð(â1):
ð â1 =â1+1
â1 2+ â1 â2=
0
â2= 0. (Note ð¥ = â1 is the zero for this function)
(b) ð ð¥ â 1 =ð¥â1+1
ð¥â1 2+ ð¥â1 â2=
ð¥
ð¥2âð¥â2
(c) ð ð¥ =ð¥+1
ð¥2+ð¥â2=
ð¥+1
ð¥+2 ð¥â1 .
When ð¥ = â2, ð â2 =â1
0 and when ð¥ = 1, ð 1 =
2
0. And since division by 0 is
meaningless (undefined), therefore ð(â2) and ð(1) do not exist. Hence the values
ð¥ = â2 and ð¥ = 1 have no images for this function.
Example 4: Given the functions ð ð¥ = ð¥2 â 1, ð ð¥ = 2ð¥ + 3 and ð ð¥ =1
ð¥, find
(i) the domain and range of ð(ð¥)
(ii) the range of ð(ð¥)
(iii) ð â ð ð¥ (called the composition function)
Solution: (i) The domain and range of ð(ð¥) is the set of all real numbers i.e â.
(ii) Note that one way of determining the range is to express ð¥ as a function of
ðŠ; ð¥ = ð(ðŠ), then the range is the set of all real values of ðŠ for which the function ð
is defined.
Solve for ð¥:
ð¥ =1
ðŠ define for all real numbers except when ðŠ = 0. Therefore the range of ð(ð¥)
is the set of real numbers when zero is deleted i.e. Rng h = â\ 0 .
(iii) ð â ð ð¥ = ð ð(ð¥) = 2ð¥ + 3 2 â 1 = 4ð¥2 + 12ð¥ + 8 .
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Different between a Function and an Equation
Function Equation
Let ðŠ = ð ð¥ = ð¥2 â 3ð¥ â 10 = ð¥ â 5 ð¥ + 2 .
This states the function. A function cannot be
solved but is used to find the images of ð¥. The
values of ð¥ which make ð ð¥ = 0 are called the
zeros of ð. The zeros of this function are
ð¥ = 5 and ð¥ = â2.
For example ð¥2 â 3ð¥ â 10 = 0. An
equation is solved to find its roots
which are the values of ð¥ which
satisfy the equation. The roots of
this equation are ð¥ = 5 or ð¥ = â2.
Some important types of functions:
1. Linear function ðŠ = 2ð¥ â 4
2. Quadratic function ðŠ = ð¥2 â ð¥ + 12
3. Polynomial function ðŠ = ððð¥ð + ððâ1ð¥ðâ1 + ⯠+ ð1ð¥ + ð0
4. Trigonometric function ðŠ = sin ð¥
5. Logarithmic function ðŠ = log ð¥
6. Exponential function ðŠ = 3ð¥ or ðð¥
Many other functions will be defined by formula. For example, area, A, of a circle is a
function of radius r, as ðŽ = ðð2, also Total value (T) = Price(P) à Quantity (Q) and
so on.
Exercise 1
1. Find the values stated for these functions:
(a) ð ð¥ = 3ð¥ â 1 ; ð(0), ð(â1), ð(1)
(b) ð ð¥ = ð¥2 + ð¥ + 1 ; ð(â1), ð(0), ð(ð¥ + 1)
(c) ð ð¥ = ð¥2 â ð¥ â 2 ; ð â1 , ð â1
2 , ð 0 , ð 1
2
(d) ð ð¥ =ð¥
ð¥+1; ð 0 , ð 1 , ð(ð¥ â 1)
2. Find the zeros of the following functions:
(a) ð ð¥ = ð¥ â 2 (b) ð ð¥ = ð¥ + 3 2
(c) ð ð¥ =ð¥â1
ð¥+6 (d) ð ð¥ =
ð¥2+2ð¥â15
ð¥2â1
3. If ð ð¥ = 3ð¥ + 2, what is the value of ð¥ whose image is 5? If also given that
ð ð¥ = 2ð¥ â 3 and ð ð¥ + ð ð¥ = 4, find the value of ð¥.
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4. Given the function ð ð¥ = 2ð¥2 â ð¥ + 1, express ð ð¥+ð âð(ð¥)
ð in its simplest form.
§ 2.0 MATRICES AND DETERMINANTS
2.1 Basic Concepts
A matrix is an array (enclosed between brackets) of real numbers arranged in
ð rows and ð columns. The matrix is then referred to as an ð by ð matrix or
(ð Ã ð) matrix.
For example,
(a) ðŽ = 1 2 3 is a 1 by 3 matrix.
(b) ðµ = 2 4 12 0 3
is a 2 by 3 matrix
(c) ð¶ = 8 24 50 1
is a 3 by 2 matrix
(d) ð· = 1 2 34 5 67 8 9
is a 3 by 3 matrix.
Any ð by ð matrix is called a square matrix.
2.2 Types of Matrices
(I) Equal matrices: Two matrices ðŽ and ðµ are said to be equal when
(i) The number of rows of ðŽ = the number of rows of ðµ and
(ii) The number of columns of ðŽ = the number of columns of B and
(iii) The entries in corresponding position are the same in both ðŽ and ðµ.
For example:
1. If ðŽ = 3 12 2
and ðµ = 3 1 22 2 0
, then ðŽ â ðµ.
2. ð = 2 0 33 1 00 0 6
, ð =1
2 4 0 66 2 00 0 12
, then ð = ð since entries in
corresponding positions in both matrices are equal
(II) .Identity Matrix: A square matrix ðŽ is said to be an identity matrix when each
entry on the leading diagonal of ðŽ is unity and other entries off the leading
diagonal are zeros. For example, the following are identity matrices:
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I2 = 1 00 1
, ðŒ3 = 1 0 00 1 00 0 1
, ðŒ4 =
1000
0100
0010
0001
(III)Diagonal Matrix: A square matrix with zero entries off the leading diagonal is a
diagonal matrix provided the entries on the leading diagonal are non-zero. For
example,
ðŽ = 3 0 00 6 00 0 9
, ðµ =
4000
0500
0090
0002
are diagonal matrices.
(IV) Triangular Matrix
A matrix whose entries above or below the leading diagonal are zeros is
referred to as a triangular matrix. It is said to be upper triangular where all
entries below the leading diagonal are zeros. It is lower triangular when all
entries above the leading diagonal are zeros. Following are the two examples:
ð = 1 4 30 7 50 0 8
, ð = 1 0 03 6 04 5 7
.
(V) The Transpose of a Matrix
A matrix ðµ is the transpose of ðŽ if when the rows of ðµ are turned into columns,
the resulting matrix is the same as ðŽ. For example, let ðŽ = 1 3 59 6 32 1 4
, if the
rows of ðŽ are now put into the columns of another matrix ðµ, we have
ðµ = 1 3 59 6 32 1 4
. Then ðµ is the transpose of ðŽ. This is written as ðµ = ðŽð .
(VI) Symmetric Matrix
A matrix ð is said to be symmetric if ð = ðð. In other words, a matrix is
symmetric if the matrix and its transpose are the same. For example,
ð = 2 8 18 9 21 2 3
is such that ð = ðð. Hence ð is a symmetric matrix.
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(VII) Skew Symmetric Matrix
A matrix ð is said to be skew symmetric if ðð = âð. We noted that this could
only be possible if all the leading diagonal entries of ð are zeros. For
example, ð = 0 1 â4
â1 0 64 â6 0
, ðð = 0 â1 41 0 â6
â4 6 0 , âð =
0 â1 41 0 â6
â4 6 0 .
Observed that ðð = âð. Therefore, ð is a skew symmetric matrix.
(VIII) Zero Matrix or Null Matrix
A null matrix is a matrix with all its entries zero. For example, ð = 0 0 00 0 00 0 0
.
2.3 Basic Operations on Matrices
(a) Multiplication of a Matrix by a Constant (Scalar)
Let ð be a constant and ð a matrix. The matrix ð = ðð, is obtained from ð by
multiplying each entry of ð by ð.
Example: If ðŽ = 3 2 11 4 5
, then ððŽ = 3ð 2ð ðð 4ð 5ð
= 12 8 44 16 20
when
ð = 4.
(b) Addition of Two Matrices
Let ðŽ and ðµ be two matrices. The sum ðŽ + ðµ exists only when the number
of rows of both matrices are the same, and the number of columns of both
matrices are also the same. That is, both matrices have the same order.
When ðŽ + ðµ exists, then the sum is obtained by adding corresponding entries
of ðŽ and ðµ and putting the sum in the corresponding positions of a new matrix
ðŽ + ðµ.
Example: Let ðŽ = 2 14 79 1
, ðµ = 5 44 12 3
. Then
ðŽ + ðµ = 2 + 5 1 + 44 + 4 7 + 19 + 2 1 + 3
= 7 58 8
11 4 .
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(c) Difference of Two Matrices
The difference of two matrices ðŽ and ðµ which are of the same order is defined
as the sum of the two matrices ðŽ and (âðµ) where âðµ = â1 â ðµ. Thus if
ðŽ =
ð11 ⯠ð1ð
â® â± â®ðð1 ⯠ððð
and ðµ = ð11 ⯠ð1ð
â® â± â®ðð1 ⯠ððð
then
ðŽ â ðµ = ð11 â ð11 ⯠ð1ð â ð1ð
â® â± â®ðð1 â ðð1 ⯠ððð â ððð
.
Example: If ðŽ = 5 2 13 8 7
and ðµ = 6 1 â39 7 5
, evaluate (i) ðŽ â ðµ (ii)
4ðŽ â 3ðµ.
Solution:
ðŽ â ðµ = 5 â 6 2 â 1 1 â (â3)3 â 9 8 â 7 7 â 5
= â1 1 4â6 1 2
.
4ðŽ â 3ðµ = 20 â 18 8 â 3 4 + 912 â 27 32 â 21 28 â 15
= 2 5 13
â15 11 13 .
Note: ðŽ + ðµ = ðµ + ðŽ, and ðµ â ðŽ = â ðŽ â ðµ â ðŽ â ðµ.
(d) Product of Two Matrices
Let ðŽ and ðµ be two matrices. The condition that the product ðŽðµ (in that order)
exists is that the number of columns of ðŽ must be equal to the number of
rows of ðµ. When the product ðŽðµ exists, then the entry in row ð and column ð
of ðŽðµ is obtained as the scalar product of the entries in row ð of ðŽ and those
in the column ð of ðµ. For instance, if ðŽ = ð11 ð12
ð21 ð22 , ðµ =
ð11 ð12 ð21 ð22
ð13
ð23 .
Let ðŽðµ = ð¶ = ð11 ð12 ð13
ð21 ð22 ð23 then the matrices ðŽ and ðµ are comfortable for
product where ð11 = ð11 ð12 ð11
ð21 = ð11ð11 + ð12ð21
ð12 = ð11 ð12 ð12
ð22 = ð11ð12 + ð12ð22
ð13 = ð11 ð12 ð13
ð23 = ð11ð13 + ð12ð23
ð21 = ð21 ð22 ð11
ð21 = ð21ð11 + ð22ð21
ð22 = ð21 ð22 ð12
ð22 = ð21ð12 + ð22ð22
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ð23 = ð21 ð22 ð13
ð23 = ð21ð13 + ð22ð23.
Note: If ðŽ is an ð à ð matrix and ðµ is an ð à ð matrix, then the product ðŽðµ is
an ð Ã ð matrix.
Examples
1. If ðŽ = 6 1 45 3 27 0 5
and ðµ = 2 48 61 7
, evaluate ðŽðµ.
Solution
ðŽðµ =
6 2 + 1 8 + 4 (1) 6 4 + 1 6 + 4 (7) 5 2 + 3 8 + 2 (1) 5 4 + 3 6 + 2 (7) 7 2 + 0 8 + 5 (1) 7 4 + 0 6 + 5 (7)
= 24 5836 5219 63
.
2. Find the value of ð and ð so that 2 7
â3 0
ð ð1 5
= 11 29â6 9
.
Solution
2 7
â3 0
ð ð1 5
= 2ð + 7 2ð + 35â3ð â3ð
= 11 29â6 9
â 2ð + 7 = 11, 2ð + 35 = 29, â 3ð = â6, â 3ð = 9
â ð = 2 and ð = â3.
2.4 Determinants
The determinant of a square matrix ð Ã ð is a specific number associated
with the matrix, and is usually denoted by ðŽ or ððð¡ ðŽ . The determinant may be
positive, negative or zero. A square matrix ðŽ is called singular matrix if its
determinant ðŽ = 0.
Given matrix ðŽ = ð11 ð12
ð21 ð22 , ðŽ can be obtained as
ðŽ = ð11 ð12
ð21 ð22 = ð11ð22 â ð12ð21. Also given a 3 à 3 matrix ðŽ =
ð11 ð12 ð13
ð21 ð22 ð23
ð31 ð32 ð33
.
The determinant of ðŽ is given by
ððð¡ ðŽ =
ð11 ð12 ð13
ð21 ð22 ð23
ð31 ð32 ð33
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= ð11 ð22 ð23
ð32 ð33 â ð12
ð21 ð23
ð31 ð33 + ð13
ð21 ð22
ð31 ð32
= ð11ð22ð33 + ð12ð23ð31 + ð13ð21ð32 â ð11ð23ð32 â ð12ð21ð33 â ð13ð22ð31.
Examples
1. If ðŽ = 2 â35 4
then ðŽ = 2 4 â â3 5 = 23.
2. If ð = â2 28 4
then ð = â2 4 â 2 8 = â24.
3. Given that ð = 1 0 â24 6 12 â3 1
, find ð .
Solution ð = +1 6 1
â3 1 â 0
4 12 1
+ â2 4 62 â3
= 6 â â3 â 0 â 2 â12 â 12
= 9 + 48 = 57
4. Given that ð¥ + 1 1â1 ð¥ â 1
= 4, find the value of ð¥.
Solution
ð¥ + 1 ð¥ â 1 â â1 1 = 4
â ð¥2 â ð¥ + ð¥ â 1 + 1 = 4
â ð¥2 = 4
â ð¥ = ±2
2.5 Minor, Cofactor and Adjoint of a Matrix
I. A minor of a matrix ðŽ is any square submatrix of ðŽ. For example, if
ðŽ = 1 2 34 5 67 8 9
, then 1 24 5
, 1 34 6
, 5 68 9
, 4 57 8
are submatrices of ðŽ.
Each of them is therefore a minor of ðŽ. Some particular types of minors of a
matrix ðŽ are obtained by deleting from ðŽ its ðth row and the ð th column. The
determinant of such a minor is denoted by ðŽðð .
Example 1: Given the matrix ðŽ = 1 2 34 5 67 8 9
, find the determinant of all 2 Ã 2
minors of ðŽ.
Solution: The (1, 1) minor is obtained by deleting from ðŽ the first row and the
first column. The determinants are as follows:
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ðŽ11 = 5 68 9
= 45 â 48 = â3
ðŽ12 = 4 67 9
= 36 â 42 = â6
ðŽ13 = 4 57 8
= 32 â 35 = â3
ðŽ21 = 2 38 9
= 18 â 24 = â6
ðŽ22 = 1 37 9
= 9 â 21 = â12
ðŽ23 = 1 27 8
= 8 â 14 = â6
ðŽ31 = 2 35 6
= 12 â 15 = â3
ðŽ32 = 1 34 6
= 6 â 12 = â6
ðŽ33 = 1 24 5
= 5 â 8 = â3.
II. Cofactor of a Matrix
The (ð, ð ) cofactor of a matrix is the determinant of the matrix obtained by
deleting the ðth and the ð th column of ðŽ and then multiplying the result by
â1 ð+ð . Let ð¶ðð denote the (ð, ð ) cofactor. Then ð¶ðð = â1 ð+ð ðŽðð .
Example 2: Find the cofactor of ðŽ in the example 1 above.
Solution: Using ð¶ðð = â1 ð+ð ðŽðð , we have the following.
ð¶11 = â1 1+1ðŽ11 = â3
ð¶12 = â1 1+2ðŽ12 = 6
ð¶13 = â1 1+3ðŽ13 = â3
ð¶21 = â1 2+1ðŽ21 = 6
ð¶22 = â1 2+2ðŽ22 = â12
ð¶23 = â1 2+3ðŽ22 = 6
ð¶31 = â1 3+1ðŽ31 = â3
ð¶32 = â1 3+2ðŽ32 = 6
ð¶33 = â1 3+3ðŽ33 = â3.
Therefore, the cofactor of ðŽ is given by
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ð¶ = â3 6 â36 â12 6
â3 6 â3 .
III. The Adjoint of a Matrix
The Adjoint of a matrix ðŽ is the transpose of its matrix of cofactors. Let ðŽ be a
matrix whose matrix of cofactor is ð¶. Then ðŽðð ðŽ = ð¶ð. Thus in the example
two above,
ðŽðð ðŽ = â3 6 â36 â12 6
â3 6 â3
ð
= â3 6 â36 â12 6
â3 6 â3 .
2.6 The Inverse of a Matrix ðš ðšâð
Let ðŽ be an ð à ð. Let ðŒð be the ð à ð identity matrix. Then ðŽðŒ = ðŒðŽ = ðŽ.
Furthermore, ðŽâ1ðŽ = ðŽðŽâ1 = ðŒ. Note that we do not write ðŽâ1 as 1
ðŽ since
1
ðŽ has no
meaning in the theory of matrices.
Let ðŽ be an ð à ð matrix such that ðŽâ1 exist. Then, the expression for the
inverse of matrix ðŽ is given as:
ðŽâ1 =ðŽðð ðŽ
ðŽ provided ðŽ â 0.
Matrix ðŽ is said to be singular when ðŽ = 0 and so inverse not exist. While matrix ðŽ
is non singular when ðŽ â 0.
Examples
1. Find ðâ1 if ð = 8 43 1
.
Solution:
ð = 8 43 1
= 8 â 12 = â4.
Cofactors are:
ð¶11 = â1 1+1ðŽ11 = 1 ð¶12 = â1 1+2ðŽ12 = â3
ð¶21 = â1 2+1ðŽ21 = â4 ð¶22 = â1 2+2ðŽ22 = 8.
Then ð¶ = 1 â3
â4 8
ðŽðð ð = 1 â3
â4 8
ð
= 1 â4
â3 8
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⎠ðâ1 =ðŽðð ð
ð =
1 â4
â3 8
â4=
â1
41
3
4â2
.
Check: ð¿ð¿âð = ð°ð. As exercise.
2. If ðŽ = 2 1 01 3 23 1 â1
, find ðŽâ1.
Solution:
ðŽ = 2 3 21 â1
â 1 1 23 â1
+ 0 1 33 1
= 2 â3 â 2 â â1 â 6 + 0 = â3
Cofactor of matrix ðŽ:
ð¶ =
+
3 21 â1
â 1 23 â1
+ 1 33 1
â 1 01 â1
+ 2 03 â1
â 2 13 1
+ 1 03 2
â 2 01 â1
+ 2 11 3
= â5 7 â81 â2 12 â4 5
ðŽðð ðŽ = ð¶ð = â5 1 27 â2 â4
â8 1 5
⎠ðŽâ1 =ðŽðð ðŽ
ðŽ =
â5 1 27 â2 â4
â8 1 5
â3=
1
3
5 â1 â2â7 2 48 â1 â5
.
Check that ðŽðŽâ1 = ðŒ3
2.7 SYSTEMS OF LINEAR EQUATIONS
A system of linear equations is of the form:
ð11ð¥1 + ð12ð¥2 + ⯠+ ð1ðð¥ð = ð1
ð21ð¥1 + ð22ð¥2 + ⯠+ ð2ðð¥ð = ð2
â®
ðð1ð¥1 + ðð2ð¥2 + ⯠+ ðððð¥ð = ðð
The method of solution could be
(a) Solution by direct algebraic method (substitution method or elimination
method)
(b) Solution using the matrix inverse
(c) Solution using Cramerâs rule.
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For solution by direct algebraic method, it has been discussed in MTS 107 and so
we omit it here.
(b) Solution using Inverse Method
In matrix form:
ð11 ⯠ð1ð
â® â± â®ðð1 ⯠ððð
ð¥1
â®ð¥ð
= ð1
â®ðð
ðŽð = ðµ
ðŽâ1ðŽð = ðŽâ1ðµ
ðŒð = ðŽâ1ðµ
â ð = ðŽâ1ðµ provided ðŽ â 0
Example: Solve the following equations by inverse method:
1. ð¥1 + 4ð¥2 = 22
ð¥1 + ð¥2 = 7
2. 3ð¥1 â 2ð¥2 + ð¥3 = 2
ð¥1 + 3ð¥2 + 4ð¥3 = 19
5ð¥1 + 4ð¥2 â 3ð¥3 = 4.
Solution:
1. In matrix form:
1 41 1
ð¥1
ð¥2 =
227
Where ðŽ = 1 41 1
, ð = ð¥1
ð¥2 and ðµ =
227
.
The solution of which is
ð¥1
ð¥2 =
1 41 1
â1
227
= â1
3
1 â4â1 1
227
= â1
3
â6â15
= 25
That is, ð¥1 = 2 and ð¥2 = 5.
2. This can be written in the form
3 â2 11 3 45 4 â3
ð¥1
ð¥2
ð¥3
= 2
194
Or ðŽð = ðµ
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Hence, ð = ðŽâ1ðµ or
ð¥1
ð¥2
ð¥3
= 3 â2 11 3 45 4 â3
â1
2
194
thus we need ðŽâ1.
ðŽ = 3ðŽ11 â â2 ðŽ12 + (1)ðŽ13
= 3 â25 + 2 â23 + (â11)
= â132.
Cof ðŽ =
+
3 44 â3
â 1 45 â3
+ 1 35 4
â â2 14 â3
+ 3 15 â3
â 3 â25 4
+ â2 13 4
â 3 11 4
+ 3 â21 3
= â25 23 â11â2 â14 â22â11 â11 11
ðŽðð ðŽ = ð¶ð = â25 â2 â1123 â14 â11
â11 â22 11
ðŽâ1 =ðŽðð ðŽ
ðŽ =
â25 â2 â1123 â14 â11
â11 â22 11
â132=
1
132
25 2 11â23 14 1111 22 â11
Therefore,
ð¥1
ð¥2
ð¥3
= ðŽâ1 2
194
=1
132
25 2 11â23 14 1111 22 â11
2
194
=1
132
50 + 38 + 44â46 + 266 + 4422 + 418 â 44
=1
132 132264392
= 123 .
(c) Solution by Cramerâs rule
1. The 2 Ã 2 matrix case:
ð¥1 + 4ð¥2 = 22
ð¥1 + ð¥2 = 7 Or 1 41 1
ð¥1
ð¥2 =
227
i.e ðŽð = ðµ.
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By Cramerâs rule,
ð¥1 = 22 47 1
1 41 1
, ð¥2 =
1 221 7
1 41 1
ð¥1 =â1
â , ð¥2 =
â2
â
Where ââ ðŽ , â1 is the determinant obtained by replacing the first column of ðŽ by
the matrix 227
and â2 is the determinant obtained by replacing the second column
of ðŽ by the matrix 227
. Thus,
â= 1 41 1
= 1 â 4 = â3
â1= 22 47 1
= 22 â 28 = â6
â2= 1 221 7
= 7 â 22 = â15
Hence,
ð¥1 =â1
â=
â6
â3= 2 and ð¥2 =
â2
â=
â15
â3= 5 .
Therefore, ð¥1
ð¥2 =
25 .
2. The (3 Ã 3) matrix case:
3 â2 11 3 45 4 â3
ð¥1
ð¥2
ð¥3
= 2
194
â= 3 â2 11 3 45 4 â3
= 132
â1= 2 â2 1
19 3 44 4 â3
= â132
â2= 3 2 11 19 45 4 â3
= â264
â3= 3 â2 21 3 195 4 4
= â396
Hence,
ð¥1 =â1
â=
â132
â132= 1
ð¥2 =â2
â=
â264
â132= 2
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ð¥3 =â3
â=
â396
â132= 3.
2.8 BUSINESS APPLICATION
Example: There are 30 secondary schools and 60 primary schools in a local
government area in a certain state. Each of the secondary schools and the primary
schools has 1 Messenger (ME), 5 Clerks (CL) and 1 Cashier (CA). Each secondary
school in addition has 1 Accountant (AC) and 1 Head Clerk (HC). The monthly salary
(Nâ000) of each of them is as follows:
Messenger â N20; Clerk â N40, Cashier â N35, Accountant â N50, and Head clerk â
N60. Use matrices to find
(i) The total number of posts of each kind of primary and secondary schools
taken together.
(ii) The total monthly salary of each primary school and each secondary
school separately.
(iii) The total monthly salary bill of all secondary and primary schools.
Solution: Consider the row matrix of the two types of schools:
ð = 30 60
Representing the number of secondary schools and primary schools respectively,
we have
Let ME CL CA AC HC
ð = 1 5 11 5 1
10
10
This is the matrix of the number of person in each position for each type of
school. Consider the product matrix:
ðð = 30 60 1 5 11 5 1
10
10
= 90 450 90 30 30 .
Then, there are in the two types of schools, 90 Messengers, 450 Clerks, 90
Cashiers, 30 Accountants and 30 Head Clerks.
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(ii) The monthly salaries (Nâ000) can be put as a column matrix ð =
2040355060
.
Total monthly bill of each kind of school is
ðð = 1 5 11 5 1
10
10
2040355060
= 20 + 200 + 35 + 50 + 60
20 + 200 + 35 + 0 + 0
= 365255
Total monthly salary bill for secondary schools is N365,000 while total monthly for
primary schools is N255,000.
(iii) ð ðð gives the total monthly salary bill for all secondary and primary schools.
ð ðð = 30 60 365255
= ð26,250 (â000)
= ð26,250,000.
Exercise 2
1. Use the following information to answer question (i) â (iv):
ðŽ = 2 4 13 â5 0
and ðµ = 6 7 10 0 2
.
(i) Find ðŽ + ðµ
(ii) Find ðŽðµ
(iii) ðŽððµ
(iv) ðŽððµ .
2. Evaluate 3 5 18 0 24 0 6
, â10 15
2 2 ,
1 â2 3â4 5 67 8 â9
.
3. Use the following information to answer question (a) â (c):
2 34 5
ð¥ðŠ =
2543
is written as ðŽð = ðµ.
(a) Write down ðŽ and ðµ.
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(b) Find ðŽâ1
(c) Find the matrix ð.
(d) Find ðŽ â ððŒ
4. A woman invested different amounts at 8%, 83
4% and 9%, all at simple
interest. Altogether she invested N40,000 and earns N3,455 per year. How
much does she have invested at each rate if she has N4,000 more invested at
9% than 8%? Solve by using matrices.
5. A salesman has the following record of sales during three months for three
items ðŽ, ðµ and ð¶ which have different rates of commission.
Months Sales units Total commission drawn
(in £) A B C
May 90 100 20 800
June 130 50 40 900
July 60 100 30 850
Find out the rates of commission on the items A, B and C. Solve by Cramerâs
rule (determination method)
§ 3.0 STATISTICAL DISTRIBUTION
3.1 BINOMIAL DISTRIBUTION
An experiment consisting of ð repeated trials such that
a) the trials are independent and identical
b) each trial result in only one or two possible outcomes
c) the probability of success ð remains constant
d) the random variable of interest is the total number of success.
The binomial distribution is one of the widely used in statistics and it used to
find the probability that an outcome would occur ð¥ times in ð performances of an
experiment and its probability density function is given by
ð ð¥; ð, ð = ðð¥ ðð¥ 1 â ð ðâð¥ , ð¥ = 0,1, ⊠, ð
Where ð = Total number of trials
ð = Probability of success
1 â ð = Probability of failure
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ð â ð¥ = Number of failure in ð â trials
To find the probability of ð¥ success in ð trials, the only values we need are that of ð
and ð.
Properties of Binomial Distribution
Mean ð = ðð
Variance ð2 = ððð ; ð = 1 â ð
Standard deviation ð = ððð
Note: ð~ðµ(ð, ð) reads as ð follows binomial distribution.
Examples:
1. Observation over a long period of time has shown that a particular sales man
can make a sale on a single contact with the probability of 20%. Suppose the
same person contact four prospects,
(a) What is the probability that exactly 2 prospects purchase the product?
(b) What is the probability that at least 2 prospects purchase the product?
(c) What is the expected value (mean) of the prospects that would purchase the
product?
Solution: Let ð denote the number of prospect.
Let ð denote the probability of (success) purchase = 0.2
Then ð~ðµ(4,0.2).
ð ð¥ = ðð¥ ðð¥ððâð¥ ð¥ = 0,1,2,3,4
= 0, ððð ðð€ðððð
ð ð¥ = 4
ð¥ 0.2 ð¥ 0.8 4âð¥
(a) ð ð¥ = 2 = 42 0.2 2 0.8 4â2 = 0.1536
(b) ð ð¥ ⥠2 = ð 2 + ð 3 + ð(4) ( OR 1 â ð 𥠆1 = 1 â {ð 0 + ð(1)})
= 0.1536 + 43 0.2 3 0.8 4â3 + 4
4 0.2 4 0.8 4â4
= 0.1536 + 0.0256 + 0.0016
= 0.1808
(c) Expected value = ðð
= 4 Ã 0.2
= 0.8
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2. The probability that a student is accepted to a prestigious college is 0.3. If 5
students from the same school apply, what is the probability that at most 2 are
accepted?
Solution:
ð 𥠆2 = ð 0 + ð 1 + ð(2)
= 5C0 0.3 0 0.7 5â0 + 5C1 0.3 1 0.7 5â1 + 5C2 0.3 2 0.7 5â2
= 0.1681 + 0.3602 + 0.3087
= 0.837
3.2 POISSON DISTRIBUTION
When the size of the sample (ð) is very large and the probability of obtaining
success in any one trial very small, then Poisson distribution is adopted. Given an
interval of real numbers, assumed counts of occur at random throughout interval, if
the interval can be partition into sub interval of small enough length such that
a) The probability of more than one count sub interval is 0
b) The probability of one count in a sub interval is the same for all sub intervals
and proportional to the length of the sub interval
c) The count in each of the sub interval is independent of all other sub intervals.
A random experiment of this type is called a Poisson Process. If the mean
number of count in an interval is ð > 0, the random variable ð¥ that equals the number
of count in an interval has a Poisson distribution with parameter ð and the probability
density function is given by
ð ð¥; ð =ðð¥ðâð
ð¥ !; ð¥ = 0,1, ⊠, ð.
For a Poisson distribution, the mean and the variance is given by ðž ð¥ = ð, ð ð¥ = ð.
Poisson distribution probability density function can also be used to approximate
binomial distribution when ð is large (ð ⥠30) and ðð < 5.
Examples
1. Attendance in a factory shows 7 absences. What is the probability that on a
given day there will be more than 8 people absent?
Solution
ð ð > 8 = 1 â ð ð †8
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= 1 â ð 1 + ð 2 + ð 3 + ð 4 + ð 5 + ð 6 + ð 7 + ð(8)
= 1 â 71ðâ7
1!+
72ðâ7
2!+
73ðâ7
3!+
74ðâ7
4!+
75ðâ7
5!+
76ðâ7
6!+
77ðâ7
7!+
78ðâ7
8!
= 1 â 0.0064 + 0.0223 + 0.0521 + 0.0912 + 0.1277 + 0.149
+0.149 + 0.1304
= 0.2709
2. An automatic production line breaks down every 2 hours. Special production
requires uninterrupted operation for 8 hours. What is the probability that this
can be achieved?
Solution ð =8
2= 4, ð¥ = 0 (no breaks down)
ð ð¥ = 0 =40ðâ4
0! = 0.0183 = 1.83%
3.3 NORMAL DISTRIBUTION
The normal distribution is the most important and the most widely used
among all continuous distribution in the statistics. The graph of a normal distribution
is a bell â shaped curved that extends indefinitely in both direction.
Normal curve
1 or 100%
ð
3.3.1 Features (Properties) of Normal Curve
1. The curve is symmetrical about the vertical axis through the mean ð.
2. The mode is the highest point on the horizontal axis where the curve is
maximum and occurs where ð¥ = ð.
3. The normal curve approaches the horizontal axis asymptotically.
4. The total area under the curve is one (1) or 100%.
It is clear from these properties that a knowledge of the population means and
standard deviation gives a complete picture of the distribution of all the values.
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Notation: Instead of saying that the values of a variable ð¥ are normally distributed
with mean ð and standard deviation ð, we simply say that ð¥ has an ð ð, ð2 or ð¥ is
ð ð, ð2 or ð~ð ð, ð2 .
A random variable ð¥ is said to have a normal distribution if its probability
density function is given by
ð ð¥; ð, ð =1
2ðð ðâ1
2 ð¥âð
ð
2
, ââ < ð¥ < â
Where ð and ð are the parameters of the distribution (standard deviation and mean
respectively).
3.3.2 The Standard Normal Curve
The standard normal distribution is a special case. If ð~ð 0,1 , then ð¥ is
called the standard normal variable with probability density function ð ð§ =
1
2ððâð§2
2 , ââ < ð§ < â.
The table usually used to determine the probability that a random variable ð¥
drawn from a normal population with no mean and standard deviation 1 is the
standard normal distribution table (or ð§- scores table).
We shall note the following point when using the table:
1. ð 𧠆0 = 0.5
2. ð 𧠆âð§ = 1 â ð 𧠆ð§
= ð(ð§ ⥠ð§)
3. ð âð§1 < ð§ < ð§2 = ð ð§ < ð§2 â ð ð§ > âð§1
= ð ð§ < ð§2 â [1 â ð ð§ < ð§1 ]
= ð ð§ < ð§2 + ð ð§ < ð§1 â 1
Example 1: Find the probability that a random variable having the standard normal
distribution will take a value
(a) Less than 1.53 (b) Less than â0.82
(c) Between 1.25 and 2.34 (d) between â0.35 and 1.26
Solution:
(a) ð ð§ < 1.53 = 0.9370 (using the table)
(b) ð ð§ < â0.82 = 0.2061
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(c) ð 1.25 †ð§ †2.34 = ð 𧠆2.34 â ð ð§ ⥠1.25
= ð 𧠆2.34 â 1 â ð(ð§ < 1.25)
= 0.9904 â 1 + 0.8944
= 0.8848
(d) ð â0.35 †ð§ †1.26 = ð 𧠆1.26 â ð ð§ ⥠â0.35
= ð 𧠆1.26 â 1 â ð(ð§ < 0.35)
= ð 𧠆1.26 + ð ð§ < 0.35 â 1
= 0.8962 + 0.6368 â 1
= 0.533
3.3.3 STANDARDIZED NORMAL VARIABLE
In real world application, the given continuous random variable may have a
normal distribution in value of a mean and standard deviation different from 0 and 1.
To overcome this difficulty, we obtain a new variable denoted by ð§ and this is given
by
ð§ =Value ð¥âMean
Standard deviation =
ðâð
ð (for ð â 0 and ð â 1)
Example 2: If ð~ð(510ð, 6.25ð), calculate ð(ð = 507.5ð)
Solution: ð§ =ðâð
ð
=507.5â510
2.5
= â1
⎠ð ð = 507.5ð = ð 𧠆â1
= 0.1587
Example 3: It is known from the previous examination results that the marks of
candidates have a normal distribution with mean 55 and standard deviation 10. If the
pass mark in a new examination is set at 40, what percentages of the candidates will
be expected to fail?
Solution: We have ð¥~ð(55,100) and the required proportion of ð¥ âvalues that are
below 40 is ð(ð¥ < 40).
ð ð¥ < 40 = ð(ð§ <ðâð
ð)
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= ð ð§ <40â55
10
= ð(𧠆â1.5)
= 0.0668
⎠6.68% of the candidates are expected to fail.
Exercise 3
1. A surgery is successful for 75% patients. What is the probability of its success
in at least 7 cases out of randomly selected 9 patients?
2. If the 5% of the electric bulbs manufactured by a company are defective. Find
the probability that in the sample of 130 bulbs, at most 3 doors are defective.
a) Use binomial to solve the problem
b) Use Poisson distribution and compare your results.
3. It is known that the marks in a University direct entry examination are normally
distributed with mean 70 and standard deviation 8. Given that your score is
69, what percentage of all the candidates will be expected to score more than
you?
§ 4.0 LIMIT AND CONTINUITYOF FUNCTIONS
4.1 Definition
A given function ð(ð¥) is said to have a limit ð¿ as ð¥ approaches ð (in symbol
limð¥âð ð(ð¥) = ð¿) if and only if, ð(ð¥) is as near to ð¿ as we please for all values of ð¥ â ð
but sufficiently near to ð.
We say that ð(ð¥) approaches the limit in real number as ð¥ approaches a fixed
number ð, if and only if, limð¥âðâ ð(ð¥) = ð¿ = limð¥âð+ ð(ð¥). When this is so, we
write limð¥âð ð(ð¥) = ð¿. Alternatively, this simply means that ð(ð¥) is arbitrary close to ð¿
whenever the variable ð¥ is sufficiently near to the number ð.
Remark: A function ð(ð¥) cannot have more than one limit.
4.2 Rules for evaluating limits
If limð¥âð ð(ð¥) and limð¥âð ð(ð¥) exist, then the following rules hold.
(i) limð¥âð ð = ð where ð is a constant.
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(ii) limð¥âð ðð ð¥ = ð limð¥âð ð(ð¥) for any ð real number
(iii) limð¥âð ð(ð¥) ± ð(ð¥) = limð¥âð ð(ð¥) ± limð¥âð ð(ð¥)
(iv) limð¥âð ð(ð¥)
ð(ð¥) =
lim ð¥âð ð(ð¥)
lim ð¥âð ð(ð¥) , provided limð¥âð ð(ð¥) â 0.
Example: Evaluate the following limits:
(1) limð¥â0(ð¥3 â 5ð¥2 + 2) (2) limð¥â0 9+ð¥â3
ð¥ (3) limð¥â2
ð¥2âð¥â2
ð¥2â4 (4) limð¥ââ
4ð¥3
ð¥3+3 .
Solution
(1) limð¥â0(ð¥3 â 5ð¥2 + 2) = limð¥â0
ð¥3 â limð¥â0
5ð¥2 + limð¥â0
2
= 0 2 â 5 02 + 2
= 2
(2) limð¥â0 9+ð¥â3
ð¥= lim
ð¥â0
9+ð¥â3
ð¥Ã
9+ð¥+3
9+ð¥+3
= limð¥â0
ð¥
ð¥ 9+ð¥+3
= limð¥â0
1
9+ð¥+3
=1
9+3
=1
6
(3) limð¥â2ð¥2âð¥â2
ð¥2â4= limð¥â2
ð¥+1 ð¥â2
ð¥+2 ð¥â2
= limð¥â2
ð¥+1
ð¥+2
=limð¥â2
(ð¥+1)
limð¥â2
(ð¥+2)
=3
4
(4) limð¥ââ4ð¥3
ð¥3+3= lim
ð¥ââ
4ð¥3
ð¥3
ð¥3
ð¥3+3
ð¥3
= limð¥ââ
4
1+3
ð¥3
=limð¥ââ
4
limð¥ââ
1+lim3
ð¥3ð¥ââ
= 4
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4.3 CONTINUITY OF FUNCTION
We say that the function ð is continuous at ð¥ = ð if limð¥âð ð(ð¥) = ð(ð). We say
that ð is continuous if it is continuous at each point of its domain.
Remark:
(1) The continuity definition requires that the following conditions be met if ð is to
be continuous at ð (a point): (a) ð(ð) is defined as a finite real number, (b)
limð¥âðâ ð(ð¥) exists and equals ð(ð), (c) limð¥âðâ ð(ð¥) = f(c) = limð¥âð+ ð(ð¥) .
When a function ð is not continuous at ð, one or more, of these conditions are
not met.
(2) All polynomials, sin ð¥, cos ð¥ , ðð¥ , sinh ð¥, cosh ð¥, ðð¥ , ð â 1 are continuous for all
real values of ð¥. All logarithmic functions, logð ð¥ , ð > 0, ð â 1 are continuous
for all ð¥ > 0. Each rational function, ð(ð¥)/ð(ð¥), is continuous where ð(ð¥) â 0.
Examples
1. Determine whether the function ð ð¥ = 3ð¥2 + 4ð¥ â 10 is continuous at ð¥ = 3
or not.
Solution:
ð 3 = 3 32 + 4 3 â 10 = 29.
limð¥â3 ð ð¥ = limð¥â3 3ð¥2 + limð¥â3 4ð¥ â limð¥â3 10
= 27 + 12 â 10
= 29
⎠limð¥â3
ð ð¥ = ð 3 = 29
ð ð¥ is continuous at ð¥ = 3.
2. Verify the continuity of the function ð ð¥ =ð¥â2
ð¥2â5ð¥+6 at ð¥ = 2.
Solution:
ð ð¥ =ð¥â2
ð¥2â5ð¥+6=
ð¥â2
ð¥â2 ð¥â3 =
1
ð¥â3
ð 2 =1
2â3= â1
limð¥â2 ð ð¥ = limð¥â31
ð¥â3
=lim ð¥â2 1
lim ð¥â2 ð¥â3
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=1
2â3= â1
⎠limð¥â2
ð ð¥ = ð 2 = â1
ð ð¥ is continuous at ð¥ = 2.
3. The function ð ð¥ =ð¥2âð¥â6
ð¥â3 is not continuous at ð¥ = 3 since ð 3 is not
defined.
Exercise 4
1. Evaluate each of the following limits.
(a) limð¥â1ð¥2â1
ð¥3â1
(b) limð¥ââð¥2âð¥
ð¥3+2ð¥
(c) limð¥â4 ð¥â2
ð¥â4
2. Is function ð ð¥ =2ð¥2+3
2ð¥+1 continuous at the point ð¥ = â
1
2?
§ 5.0 BASIC CONCEPT OF DIFFERENTIATION
Differentiation is a mathematical concept dealing with the rate of change. This
rate of change is generally termed the derivative, slope, gradient or marginal
measure in various areas of usage. That is, the process of finding the derivative of a
function is known as differentiation. Formally, differentiation (the differential calculus)
is the process of finding the derivative of a function. And it has a lot of applications in
many areas of human endeavours such as the field of engineering, science,
economics, business, and so on.
5.1 Determination of Derivative from First Principle
There is need to apply the limits theory to the incremental principle of the
variables involved in the function. For instance, let us consider the function ðŠ = ð(ð¥).
A small increment in ð¥ (denoted by âð¥) will cause a small increment in ðŠ (also
denoted by âðŠ). Then for
ðŠ = ð(ð¥) âŠâŠâŠâŠâŠâŠâŠâŠ. (i)
We have ðŠ + âðŠ = ð ð¥ + âð¥ âŠâŠâŠâŠ (ii)
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Subtract equation (i) from equation (ii), gives
ðŠ + âðŠ â ðŠ = ð ð¥ + âð¥ â ð(ð¥)
âðŠ = ð ð¥ + âð¥ â ð(ð¥)
Divide all through by âð¥, we have
ÎðŠ
Îð¥=
ð ð¥+âð¥ âð(ð¥)
âð¥ âŠâŠâŠâŠâŠ. (iii)
As âð¥ â 0, we have
limâð¥â0ÎðŠ
Îð¥= limâð¥â0
ð ð¥+âð¥ âð(ð¥)
âð¥ âŠ. (iv)
By the limitâs theory, we have
ððŠ
ðð¥= limâð¥â0
ð ð¥+âð¥ âð(ð¥)
âð¥ âŠâŠâŠ.. (v) (i.e.
ððŠ
ðð¥â¡ limâð¥â0
ÎðŠ
Îð¥)
Note that ððŠ
ðð¥ is also represented by ð â² (ð¥) which is called the derivative of ð(ð¥).
Others are ð
ðð¥ ð ,
ðð
ðð¥, ðŠâ² .
If the left hand side of the equation (v) exists, then ððŠ
ðð¥ is called the differential
coefficient of ðŠ with respect to ð¥.
Example: Find the derivative of (i) ðŠ = ð¥3 + 2 (ii) ðŠ = ðð¥ð from the first principle.
Solution:
(i) ðŠ = ð¥3 + 2 ------------- (*)
ðŠ + âðŠ = ð¥ + âð¥ 3 + 2 -------------- (**)
Then subtracting equation (*) from equation (**), we have
ðŠ + âðŠ â ðŠ = ð¥ + âð¥ 3 + 2 â ð¥3 + 2
âðŠ = ð¥3 + 3ð¥2 âð¥ + 3ð¥ âð¥ 2 + âð¥ 3 + 2 â ð¥3 â 2
âðŠ = 3ð¥2 âð¥ + 3ð¥ âð¥ 2 + âð¥ 3
Divide through by âð¥, we have
ÎðŠ
Îð¥= 3ð¥2 + 3ð¥ âð¥ + âð¥ 2
limÎð¥â0
ÎðŠ
Îð¥= lim
Îð¥â0 3ð¥2 + 3ð¥ âð¥ + âð¥ 2
âŽððŠ
ðð¥= 3ð¥2 + 0 + 0
= 3ð¥2
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(ii) Given ðŠ = ðð¥ð .
Let ðŠ + âðŠ = ð ð¥ + âð¥ ð
Then
ðŠ + âðŠ â ðŠ = ð ð¥ + âð¥ ð â ðð¥ð
âðŠ = ðð¥ð 1 +âð¥
ð¥
ð
â 1 Using 1 + ð¥ ð = 1 + ð ðâ1 ðâ2 âŠ(ðâð+1)
ð !âð=1 ð¥ð
âðŠ = ðð¥ð 1 + ð âð¥
ð¥ +
ð(ð â 1)
2! âð¥
ð¥
2
+ ⯠â 1
= ð ðð¥ðâ1 âð¥ +ð(ðâ1)
2!ð¥ðâ2 âð¥ 2 + â¯
âŽÎðŠ
Îð¥= ð ðð¥ðâ1 + terms containing higher power of âð¥
limÎð¥â0
ÎðŠ
Îð¥= lim
âð¥â0 ððð¥ðâ1 + terms containing higher power of âð¥
âŽððŠ
ðð¥= ð ðð¥ðâ1 âŠâŠâŠâŠâŠâŠâŠâŠ. (vi)
We note that equation (vi) is the derivative of ð = ððð and it shall be applied as
a rule when the derivative from first principle is not required.
Example: (a) Obtain the derivative for the following. (i) ðŠ = ð¥4 (ii) ðŠ = 4ð¥5.
(b) Obtain the value of the derivative for the following functions at ð¥ = 2: (i) ðŠ = 2ð¥3
(ii) ðŠ = 3ð¥6.
Solution:
(a) (i) ððŠ
ðð¥= 4 ð¥4â1 = 4ð¥3
(ii) ððŠ
ðð¥= 4 5 ð¥5â1 = 20ð¥4
(b) (i) ððŠ
ðð¥= 6ð¥2
At ð¥ = 2, ððŠ
ðð¥= 6 22 = 24
(ii) ððŠ
ðð¥= 18ð¥5
At ð¥ = 2,ððŠ
ðð¥= 18 25 = 576
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32 | P a g e
5.2 Rules of Differentiation
The following rules shall be applied to various functions:
Rule 1: If a polynomial ðŠ = 𢠱 𣠱 ð€ ± ⯠where ð¢, ð£, ð€ are functions of ð¥, then
ððŠ
ðð¥=
ð
ðð¥ 𢠱 𣠱 ð€ ± â¯
=ðð¢
ðð¥Â±
ðð£
ðð¥Â±
ðð€
ðð¥Â± â¯
Hence, the derivative of a sum is the of the derivatives and the derivative of a
difference is the difference of the derivatives.
Example: Find the derivative of each of the following functions.
(i) ðŠ = 2ð¥5 â 3ð¥6 + 6ð¥2; (ii) ð ð¥ = 2ð¥3 + 3ð¥4 â2
ð¥ .
Solution:
(i) ððŠ
ðð¥= 2 5 ð¥5â1 â 3 6 ð¥6â1 + 6(2) ð¥2â1
= 10ð¥4 â 18ð¥5 + 12ð¥
(ii) ðð
ðð¥= 2 3 ð¥3â1 + 3 4 ð¥4â1 â 2 â1 ð¥â1â1
= 6ð¥2 + 12ð¥3 +2
ð¥2 .
Rule 2: If ðŠ = ð¶, where ð¶ is a constant, then ððŠ
ðð¥= 0. E.g. (a) ðŠ = 2,
ððŠ
ðð¥= 0.
(b) ðŠ = ð2 , ððŠ
ðð¥= 0 since ð is a constant and has no degree of ð¥ from 1 and above.
Rule 3: If a product ðŠ = ð¢ð£, where ð¢ and ð£ are functions of ð¥, then
ððŠ
ðð¥= ð£
ðð¢
ðð¥+ ð¢
ðð£
ðð¥ .
By extension, if ðŠ = ð¢ð£ð€, where ð¢, ð£ and ð€ are functions of ð¥, then
ððŠ
ðð¥= ð£ð€
ðð¢
ðð¥+ ð¢ð€
ðð£
ðð¥+ ð£ð¢
ðð€
ðð¥ .
Example: Find the derivative of the following functions:
(a) ðŠ = 2ð¥ + 2 3ð¥2 + 2ð¥ (b) ðŠ = 2ð¥ + 4 ð¥2 + 3ð¥ 2ð¥3 + 5ð¥ .
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Solution:
(a) Let ð¢ = 2ð¥ + 2, ðð¢
ðð¥= 2
ð£ = 3ð¥2 + 2ð¥, ðð£
ðð¥= 6ð¥ + 2 .
By product rule,
ððŠ
ðð¥= ð£
ðð¢
ðð¥+ ð¢
ðð£
ðð¥
= 3ð¥2 + 2ð¥ 2 + 2ð¥ + 2 6ð¥ + 2
= 6ð¥2 + 4ð¥ + 12ð¥2 + 4ð¥ + 12ð¥ + 4
= 18ð¥2 + 20ð¥ + 4 .
(b) Let ð¢ = 2ð¥ + 4, ðð¢
ðð¥= 2
ð£ = ð¥2 + 3ð¥, ðð£
ðð¥= 2ð¥ + 3
ð€ = 2ð¥3 + 5ð¥, ðð€
ðð¥= 6ð¥2 + 5
But ððŠ
ðð¥= ð£ð€
ðð¢
ðð¥+ ð¢ð€
ðð£
ðð¥+ ð£ð¢
ðð€
ðð¥
= ð¥2 + 3ð¥ 2ð¥3 + 5ð¥ 2 + 2ð¥ + 4 2ð¥3 + 5ð¥ 2ð¥ + 3
+ ð¥2 + 3ð¥ 2ð¥ + 4 6ð¥2 + 5
Rule 4: If quotient ðŠ =ð¢
ð£, where ð¢ and ð£ are functions of ð¥, then
ððŠ
ðð¥=
ð£ ðð¢
ðð¥ â ð¢
ðð£
ðð¥
ð£2 .
Example: Differentiate the following with respect to ð¥.
(i) ðŠ =1+ð¥2
3ð¥ (ii) ðŠ =
2+3ð¥
ð¥2+3ð¥+5
Solution:
(i) Let ð¢ = 1 + ð¥2 , ðð¢
ðð¥= 2ð¥
ð£ = 3ð¥, ðð£
ðð¥= 3
By Quotient Rule:
ððŠ
ðð¥=
ð£ ðð¢
ðð¥ â ð¢
ðð£
ðð¥
ð£2
=3ð¥ 2ð¥ â 1+ð¥2 (3)
3ð¥ 2
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34 | P a g e
=6ð¥2â3â3ð¥2
9ð¥2
=ð¥2â1
3ð¥2 .
(ii) Let ð¢ = 2 + 3ð¥, ðð¢
ðð¥= 3
ð£ = ð¥2 + 3ð¥ + 5, ðð£
ðð¥= 2ð¥ + 3
But
ððŠ
ðð¥=
ð£ ðð¢
ðð¥ â ð¢
ðð£
ðð¥
ð£2
= ð¥2+3ð¥+5 (3)â 2+3ð¥ (2ð¥+3)
ð¥2+3ð¥+5 2
=3ð¥2+9ð¥+15â4ð¥â6â6ð¥2â9ð¥
ð¥2+3ð¥+5 2
=â3ð¥2â4ð¥+9
ð¥2+3ð¥+5 2 .
Rule 5: If a function of function ðŠ = ðð¥ + ð ð where ð and ð are constants and ð is
the index, then ððŠ
ðð¥=
ð
ðð¥ ðð¥ + ð ð = ð ðð¥ + ð ðâ1
ð
ðð¥(ðð¥ + ð) .
This rule is also called the composite or chain rule. In another way, this can
be written as: ððŠ
ðð¥=
ððŠ
ðð¢
ðð¢
ðð¥
Where ðŠ is a function of ð¢ and ð¢ is a function of ð¥. It is called function of a
function.
Example: Differentiate the following: (i) ðŠ = ð¥ + 3 4 (ii) ðŠ = 3ð¥2 â 5 6 .
Solution:
(i) Let ð¢ = ð¥ + 3; ðð¢
ðð¥= 1 then ðŠ = ð¢4 ,
ððŠ
ðð¢= 4ð¢3 .
âŽððŠ
ðð¥=
ððŠ
ðð¢Ã
ðð¢
ðð¥
= 4ð¢3(1)
= 4ð¢3
= 4 ð¥ + 3 3 .
(ii) Let ð¢ = 3ð¥2 â 5; ðð¢
ðð¥= 6ð¥ then ðŠ = ð¢6 ,
ððŠ
ðð¢= 6ð¢5 .
âŽððŠ
ðð¥=
ððŠ
ðð¢Ã
ðð¢
ðð¥
= 6ð¢5 (6ð¥)
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35 | P a g e
= 36ð¥ð¢5
= 36ð¥ 3ð¥2 â 5 5 .
Rule 6: For the implicit function, the relationship between ðŠ and ð¥ variables may not
be expressed explicitly (e.g. ð¥ðŠ2, ð¥ðŠ + ð¥2 = 3ð¥2ðŠ5). In this case, treat ðŠ as if
it is a function of ð¥ and the rules of differentiation discussed above are
applied as appropriate.
Example: Differentiate the following implicitly:
(i) ð¥3 + ðŠ4 = 7 (ii) 6ðŠ2ð¥ â 4ð¥2ðŠ3 + 2ðŠ = 0
Solution:
(i) Differentiate the function term by term with respect to ð¥
3ð¥2 ðð¥
ðð¥ + 4ðŠ3
ððŠ
ðð¥= 0
3ð¥2 + 4ðŠ3 ððŠ
ðð¥= 0
â 4ðŠ3 ððŠ
ðð¥= â3ð¥2
âŽððŠ
ðð¥=
â3ð¥2
4ðŠ3 .
(ii) 6 ðŠ2 ðð¥
ðð¥+ 2ðŠð¥
ððŠ
ðð¥ â 4 2ð¥
ðð¥
ðð¥ðŠ3 + 3ð¥2ðŠ2 ððŠ
ðð¥ + 2
ððŠ
ðð¥= 0
6ðŠ2 + 12ð¥ðŠððŠ
ðð¥â 8ð¥ðŠ3 â 12ð¥2ðŠ2
ððŠ
ðð¥+ 2
ððŠ
ðð¥= 0
â 12ð¥ðŠ â 12ð¥2ðŠ2 + 2 ððŠ
ðð¥= 8ð¥ðŠ3 â 6ðŠ2
âŽððŠ
ðð¥=
8ð¥ðŠ3 â 6ðŠ2
12ð¥ðŠ â 12ð¥2ðŠ2 + 2
=4ð¥ðŠ3â3ðŠ2
6ð¥ðŠâ6ð¥2ðŠ2+1 .
Rule 7: If the exponential function ðŠ = ððð¥ +ð , where ð and ð are constants, then
ððŠ
ðð¥=
ð
ðð¥ ððð¥+ð
=ð
ðð¥ ðð¥ + ð ððð¥ +ð
= ðððð¥ +ð
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36 | P a g e
Example: Differentiate the following with respect to ð¥. (i) ðŠ = 6ð2ð¥2+ð¥ (ii) ðŠ = ð¥3ðð¥3.
Solution:
(i) ððŠ
ðð¥=
ð
ðð¥ 2ð¥2 + ð¥ â 6ð2ð¥2+ð¥
= 4ð¥ + 1 â 6ð2ð¥2+ð¥
= 6 4ð¥ + 1 ð2ð¥2+ð¥ .
(ii) Applying product rule:
Let ð¢ = ð¥3; ðð¢
ðð¥= 3ð¥2
ð£ = ðð¥3;
ðð£
ðð¥= 3ð¥2ðð¥3
But ððŠ
ðð¥= ð£
ðð¢
ðð¥+ ð¢
ðð£
ðð¥
= ðð¥3 3ð¥2 + ð¥3 3ð¥2ðð¥3
= 3ð¥2ðð¥3+ 3ð¥5ðð¥3
= ðð¥3 3ð¥2 + 3ð¥5 .
Rule 8: If the logarithmic function ðŠ = logð ðð¥ + ð = ln ðð¥ + ð where ð and ð are
constants, then ððŠ
ðð¥=
1
ðð¥ +ðâ
ð
ðð¥ ðð¥ + ð =
ð
ðð¥ +ð .
Example: Differentiate the following with respect to ð¥.
(i) ðŠ = ln ð¥ (ii) ðŠ = logð 5ð¥ + 2 (iii) ðŠ = ln 6ð¥ â 1 2 .
Solution:
(i) ððŠ
ðð¥=
1
ð¥
(ii) ððŠ
ðð¥=
1
5ð¥+2â
ð
ðð¥ 5ð¥ + 2
=5
5ð¥+2
(iii) ððŠ
ðð¥=
1
6ð¥â1 2 âð
ðð¥ 6ð¥ â 1 2
=1
6ð¥â1 2 6 2 (6ð¥ â 1)
=12
6ð¥â1 .
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37 | P a g e
Rule 9: Differentiation of Trigonometric functions:
(a) If ðŠ = sin ð¥ then ððŠ
ðð¥= cos ð¥
(b) If ðŠ = cos ð¥ then ððŠ
ðð¥= â sin ð¥
(c) If ðŠ = tan ð¥ then ððŠ
ðð¥= sec2 ð¥
Examples:
1. ðŠ = cos 7ð¥ , ððŠ
ðð¥= â7 sin 7ð¥
2. ðŠ = sin 3ð¥ â 4 , ððŠ
ðð¥= 3 cos 3ð¥ â 4
3. ðŠ = cos ð
4â 2ð¥ ,
ððŠ
ðð¥= 2 sin
ð
4â 2ð¥
5.3 SECOND ORDER DERIVATIVE
Second order derivative is one of higher derivatives where successive
differentiations are carried out. The process of differentiating a function more that
one is called successive differentiation.
If ðŠ = ð ð¥ ,ððŠ
ðð¥ is also a function of ð¥. The derivative of
ððŠ
ðð¥ with respect to ð¥
ð
ðð¥
ððŠ
ðð¥ . The expression
ð
ðð¥
ððŠ
ðð¥ is called the second order derivative of ðŠ with respect
to ð¥ and it is denoted by ð2ðŠ
ðð¥2 or ðŠâ²â² or ð â²â² (ð¥). It is useful when determining the turning
point of a function.
Example: Find the first, second and third derivatives of the following.
(i) ðŠ = 4ð¥5 (ii) ðŠ = 3ð¥6 â 2ð¥5 + ðð¥ + 3ð¥2 â 8 .
Solution:
(i) ððŠ
ðð¥= 20ð¥4
ð2ðŠ
ðð¥2= 80ð¥3
ð3ðŠ
ðð¥3 = 240ð¥2 .
(ii) ððŠ
ðð¥= 18ð¥5 â 10ð¥4 + ðð¥ + 6ð¥
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38 | P a g e
ð2ðŠ
ðð¥2= 90ð¥4 â 40ð¥3 + ðð¥ + 6
ð3ðŠ
ðð¥3 = 360ð¥3 â 120ð¥2 + ðð¥ .
5.4 APPLICATION OF DIFFERENTIATION
5.4.1 Application to maximum and minimum values
The concept of maximum and minimum values is termed the zero slope
analysis. The focus here is to find a point where the slope of the function to be
optimized is zero. It should be noted that the concept is not restricted to plotting of
curve alone, but more importantly to determine the maximum and minimum values of
a function such as maximizing profit or minimizing cost, etc.
We have the following procedures for the determination of maximum and
minimum values of a function ðŠ = ð(ð¥) if it exists:
Step 1: Obtain the first derivative and set it equal to zero i.e. ððŠ
ðð¥= 0. This occur at
turning point.
Step 2: From the ððŠ
ðð¥= 0, determine the stationary point.
Step 3: Compute ð2ðŠ
ðð¥2 at these stationary points.
If ð2ðŠ
ðð¥2 < 0, the stationary point is maximum.
If ð2ðŠ
ðð¥2 > 0, the stationary point is minimum.
If ð2ðŠ
ðð¥2 = 0, (that is point of inflection) higher derivative (than second) can be
used to decide.
Example 1: ADAS company planning to have a new product in the market came up
with a total sales function ð = â1000ð2 + 10000ð and the total cost function
ð¶ = â2000ð + 2500, where ð and ð¶ are respectively the price and cost (in US dollar)
of the new product. Find the optimal price for the new product and the maximum
profit expected for the company.
Solution: The profit (ð) is obtained by the difference in total sales and total cost.
⎠ð = ð â ð¶
= â1000ð2 + 10000ð â â2000ð + 2500
= â1000ð2 + 10000ð + 2000ð â 2500
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39 | P a g e
= â1000ð2 + 12000ð â 2500
ðð
ðð= â2000ð + 12000
But at turning point, ðð
ðð= 0, then
0 = â2000ð + 12000
â ð =12000
2000= 6.
But ð2ð
ðð2= â2000 (which is less than zero, indicating maximum)
Hence, ð = 6, gives a maximum point. Therefore, the optimal price is $6. And the
maximum profit ð = â1000 6 2 + 12000(6) â 2500
= $33500 .
Example 2: ð articles are produced at a total cost of N 2ð2 + 30ð + 20 , and each
one is sold for N ð
3+ 100 . Determine the value of ð which gives the greatest profit,
and find this profit.
Solution: Total cost of articles = N 2ð2 + 30ð + 20
Total money received for ð articles sold = Nð ð
3+ 100 .
Then the profit Nð = N ð ð
3+ 100 â 2ð2 + 30ð + 20 .
So, ð = 70ð â5
3ð2 â 20.
The minimum value of ð will be given by ðð
ðð= 0
ðð
ðð= 70 â
10
3ð = 0 â ð = 21.
ð2ð
ðð2= â
10
3< 0.
Therefore, the maximum profit = N 70 Ã 21 â5
3 21 2
= N715 .
5.4.2 Application to Marginal Cost and Revenue
For the cost analysis (that is, relationship between average and marginal cost)
Let ð = Total cost
ð = Quantity demand
Then, the Average Cost ðŽð¶ =ð
ð
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40 | P a g e
And the Marginal Cost ðð¶ =ðð
ðð, i.e.
ð
ðð ð
Also the slope of average cost is obtained by:
ð
ðð
ð
ð =
ð ðð
ðð âð
ðð
ðð
ð2 (Quotient Rule)
=
ðð
ððâ
ð
ð
ð
=1
ð
ðð
ððâ
ð
ð
=1
ððð¶ â
1
ððŽð¶
It is established that if the cost curve is U â shape and the ðŽð¶ has a sloping
downward curve, then ð
ðð
ð
ð < 0 which means that ðð¶ < ðŽð¶. When ðŽð¶ decreases
ðð¶ < ðŽð¶ and on the lowest point of ðŽð¶ curve, the tangent will be horizontal.
Therefore ð
ðð
ð
ð = 0 i.e. ðð¶ = ðŽð¶. Also when ðŽð¶ increasing, ðð¶ > ðŽð¶ and
ð
ðð
ð
ð > 0.
Relationship between Average and Marginal revenue
Let Total revenue = ð and we know that ð = ððððð à ðð¢ððð¡ðð¡ðŠ i.e. ð = ðð.
Then, Average Revenue ðŽð =ð
ð
And, the Marginal Revenue ðð =ð
ðð ð = ð + ð
ðð
ðð. (Product Rule)
Example 1: Determine the minimum average cost if the cost function is given by:
ð = 36ð â 20ð2 + 4ð3. Also obtain the marginal cost at the point of minimum
average cost.
Solution:
Given ð = 36ð â 10ð2 + 2ð3
⎠ðŽð¶ =ð
ð= 36 â 10ð + 2ð2
For maximum or minimum, ð
ðð
ð
ð = 0
i.e. â10 + 4ð = 0
â ð =5
2.
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41 | P a g e
More so, ð2
ðð2 ð
ð at ð = 5
2 = 4, which is greater than zero.
Hence, ð =5
2 is at minimum point.
⎠The Minimum Average Cost = 36 â 10 5
2 + 2
5
2
2
= 23.5
For the Marginal Cost (ðð¶), we have
ðð¶ =ðð
ðð= 36 â 20ð + 6ð2
ðð¶ at ð =5
2 is equal to 36 â 20
5
2 + 6
5
2
2
= 23.5 .
Example 2: Find the maximum profit of a company with revenue function
ð = 200ð â 2ð2 and the cost function ð = 2ð3 â 57ð2.
Solution:
Profit ð = ð â ð
= 200ð + 55ð2 â 2ð3
At turning point, ðð
ðð= 0
ðð
ðð= 200 + 110ð â 6ð2 = 0
â 100 + 55ð â 3ð2 = 0
3ð2 â 55ð â 100 = 0
3ð2 â 60ð + 5ð â 100 = 0
3ð ð â 20 + 5 ð â 20 = 0
3ð + 5 ð â 20 = 0
â ð = â5
3 or 20
But ð2ð
ðð2= 110 â 12ð
At ð = â5
3,
ð2ð
ðð2 = 110 â 12 5
3 = 90 > 0 indicating minimum point.
At ð = 20, ð2ð
ðð2 = 110 â 12 20 = â130 < 0 indicating maximum point.
Hence, at ð = 20, we have the maximum profit.
⎠The maximum profit = 200 20 + 55 20 2 â 2 20 3
= 10,000
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42 | P a g e
5.4.3 Application of differentiation to the elasticity of demand
The price elasticity of demand is the rate of change in response to quantity
demanded to the change in price. Let ð = ð(ð) be the demand function, where ð is
the demand and ð is the price. Then the elasticity of demand is
ðð = âð
ðâ
ðð
ðð When ðð > 1, the demand is elastic and, when ðð < 1, the demand is
inelastic.
For elasticity of supply, Let ð¥ = ð ð be the supply function, where ð¥ is the
supply and ð is the price. The elasticity of supply is defined as ðð =ð
ð¥â
ðð¥
ðð .
Example: Given a demand function ð· = 60 â ð â ð2 , determine its elasticity of
demand when ð = 4
Solution: ðð·
ðð= â1 â 2ð
But ðð = âð
ð·
ðð·
ðð
= âð
60âðâð2 â1 â 2ð
=2ð2+ð
60âðâð2 .
When ð = 4 we get
ðð = 0.9
Exercise 5
1. A firm produces ð¥ tonnes of output at a total cost ð¶ =1
10ð¥3 â 5ð¥2 + 10ð¥ â 32.
At what level of output will the marginal cost and the average variable cost
attain their respective minimum?
2. A certain manufacturing company has total cost function
ð¶ = 15 + 9ð¥ â 6ð¥2 + ð¥3 . Find ð¥, when the total cost is minimum.
3. The relationship between profit ð and advertising cost ð¥ is given by
ð =4000ð¥
500+ð¥â ð¥. Find ð¥ which maximizes ð.
4. The total cost and total revenue of a firm are given by
ð¶ = ð¥3 â 12ð¥2 + 48ð¥ + 11 and ð = 83ð¥ â 4ð¥2 â 21.
Find the output (i) when the revenue is maximum (ii) when profit is maximum.
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43 | P a g e
5. A GSM company has a profit of N2 per SIM when the number of SIM in the
exchanges is not over 10,000. The profit per SIM decreases by 0.01 kobo for
each SIM over 10,000. What is the maximum profit?
6. The total cost function of a firm is ð¶ =1
3ð¥3 â 5ð¥2 + 28ð¥ + 10 where ð¥, is the
output. A tax at N2 per unit of output is imposed and the producer adds it to
his cost. If the market demand function is given by ð = 2530 â 5ð¥, where Nð
is the price per unit of output. Determine the profit maximizing output and
price.
7. Investigate the maxima and minima of the function 2ð¥3 â 3ð¥2 â 36ð¥ + 10 .
8. For the cost function ð¶ = 2000 + 1800ð¥ â 75ð¥2 + ð¥3 find when the total cost
(ð¶) is increasing and when it is decreasing.
9. Find the equation of the tangent and normal to the demand curve
ðŠ = 10 â 3ð¥2 at (1, 7).
10. ADAS produces ð¥ tonnes of output at a total cost
ð¶ ð¥ = 1
10ð¥3 â 4ð¥2 + 20ð¥ + 5 . Find (i) Average Cost (ii) Average Variable
Cost (iii) Average Fixed Cost (iv) Marginal Cost and (v) Marginal Average
Cost.
11. The total cost ð of making ð units of product is
ð = 0.00005ð3 â 0.06ð2 + 10ð + 79020. Find the marginal cost at 1000 units
of output.
12. Find the elasticity of the following functions.
(a) ð¥ = 2ð2 + 8ð + 10
(b) ðŠ = 4ð¥ â 8 when ð¥ = 5.
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44 | P a g e
§ 6.0 INTEGRAL CALCULUS
Integration is the reverse of differentiation. If ð¹(ð¥) is a function whose
derivative ð¹â² ð¥ = ð(ð¥), then ð¹(ð¥) is called an integral of ð(ð¥). For example,
ð¹ ð¥ = ð¥3 is an integral of ð ð¥ = 3ð¥2 since ð¹â² ð¥ = 3ð¥2 = ð(ð¥).
6.1 Indefinite Integral
Let ðŠ = ð(ð¥) be a function of ð¥. Suppose that ððŠ
ðð¥= ð(ð¥) where ð(ð¥) is a
known function of ð¥. Then ðŠ = ð(ð¥) ðð¥ is called an indefinite integral of ð(ð¥) with
respect to ð¥.
Remark
1. Let ð be an arbitrary constant. Since ð
ðð¥ ð = 0, it follows that ð = 0 ðð¥ .
2. If ð(ð¥) is an indefinite integral of a function ð(ð¥) and ð is an arbitrary constant,
then all positive indefinite integrals of ð(ð¥) are of the form
ð(ð¥) ðð¥ = ð ð¥ + ð. In this case, ð is called a constant of integration.
For all rational values of ð except when ð = â1, ð
ðð¥ ð¥ð+1
ð+1 =
1
ð+1â ð + 1 ð¥ð .
Therefore, if ð â â1, the indefinite integral of ð¥ð is ð¥ð ðð¥ =ð¥ð+1
ð+1+ ð . Note also that
constant different from zero, the integral is obtained by simply multiply the constant
by the independent variable and add the constant of integration. E.g. 2 ðð¥ = 2ð¥ + ð.
Example
1. If ððŠ
ðð¥= 4ð¥2, find ðŠ.
Solution
ððŠ = 4ð¥2ðð¥
ððŠ = 4ð¥2 ðð¥
â ðŠ =4ð¥2+1
2+1+ ð =
4
3ð¥3 + ð
2. Integrate 6ð¡3 + 6 with respect to ð¡.
Solution
6ð¡3 + 6 ðð¡ =6
4ð¡4 + 6ð¡ + ð
=3
2ð¡4 + 6ð¡ + ð .
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6.2 Some Standard form of Integration
(i) ðð¥ ðð¥ = ðð¥ + ð
(ii) ðð¥
ð¥= logð ð¥ + ð
(iii) sin ð¥ ðð¥ = â cos ð¥ + ð
(iv) cos ð¥ ðð¥ = sin ð¥ + ð
(v) ðð¥
ð¥2+ð2 =1
ðtanâ1 ð¥
ð+ ð
(vi) ðð¥
ð2âð¥2= sinâ1 ð¥
ð+ ð
Example: Integrate the following with respect to ð¥.
(a) ðð¥
16âð¥2=
ðð¥
42âð¥2= sinâ1 ð¥
4+ ð
(b) ðð¥
ð¥2+25=
1
5tanâ1 ð¥
5+ ð
6.3 Rules of Integration
Rule 1: The integration of a sum (difference) of a finite number of functions is the
sum (difference) of their separate integrals. E.g.
ð¥2 â ð¥ + 5 ðð¥ = ð¥2 ðð¥ â ð¥ ðð¥ + 5 ðð¥
=ð¥3
3â
ð¥2
2+ 5ð¥ + ð .
Rule 2: A constant factor may be brought outside the integral sign. E.g.
18ð¥3 ðð¥ = 18 ð¥3 ðð¥ = 18 ð¥4
4 + ð =
9
2ð¥4 + ð .
Rule 3: The addition of a constant to the variable makes no difference to the form of
the result. E.g.
(i) ðð¥
ð¥â3 2+32=
1
3tanâ1 ð¥â3
3+ ð
(ii) ðð¥
ð¥+7= logð ð¥ + 6 + ð
Rule 4: Multiplying the variables by the constant also makes no difference to the
form of the result, but we have to divide by the constant. (Note integration by
substitution can be used to verify this claim)
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Examples
1. cos 3ð¥ ðð¥ =1
3sin 3ð¥ + ð
2. ð4ð¥ ðð¥ =1
4ð4ð¥ + ð
3. ð2ð¥â5 ðð¥ =1
2ð2ð¥â5 + ð
4. 2ð¥ + 3 12 ðð¥ =
1
2 2ð¥+3
32
32
+ ð =1
3 2ð¥ + 3
32 + ð
Rule 5: The integration of a fraction whose numerator is derivative of the
denominator is obtained by the logarithm of the denominator. i.e.
ð â² ð¥
ð ð¥ ðð¥ = logð ð ð¥ + ð
Examples
(a) 2ð¥
ð¥2+12ðð¥ = ln ð¥2 + 12 + ð
(b) ð¥2
ð¥3+2ðð¥ =
1
3
3ð¥2
ð¥3+2ðð¥ =
1
3ln ð¥3 + 2 + ð
(c) ð¥ðð¥2
ðð¥2+3
ðð¥ =1
2
2ð¥ðð¥2
ðð¥2+3
ðð¥ =1
2ln ðð¥2
+ 3 + ð
(d) ðð¥
ð¥ log ð ð¥=
1
ð¥ðð¥
log ð ð¥= logð logð ð¥ + ð
Rule 6: (Integration by Substitution)
Consider the indefinite integral ð(ð¥) ðð¥ of an arbitrary function ðŠ = ð(ð¥). Suppose
that ð(ð¥) can be written as ð ð¥ = ð(ð¢)ðð¢
ðð¥ for some function ð(ð¢) of suitably chosen
variable ð¢ = ð¢(ð¥), then ð(ð¥) ðð¥ = ð(ð¢) ðð¢. That is, we get ð(ð¥) ðð¥ by
evaluating ð(ð¢) ðð¢ as a function of ð¢ and then substituting ð¢ = ð¢(ð¥) to get a
function of the original variable ð¥.
Example 1: Evaluate ð¥
1â2ð¥2ðð¥.
Solution: Set ð¢ = 1 â 2ð¥2; ðð¢
ðð¥= â4ð¥ â ðð¥ =
1
â4ð¥ðð¢.
Therefore, ð¥
1â2ð¥2ðð¥ =
ð¥
ð¢
1
â4ð¥ðð¢
= â1
4
1
ð¢ðð¢
= â1
4 ð¢â1
2 ðð¢
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= â1
4 2 ð¢
12 + ð
= â1
2 1 â 2ð¥2 + ð
Example 2: Evaluate ð2â5ð¥ ðð¥
Solution: Let ð¢ = 2 â 5ð¥; ðð¢
ðð¥= â5 â ðð¥ =
ðð¢
â5.
Hence, ð2â5ð¥ ðð¥ = ðð¢ ðð¢
â5
= â1
5 ðð¢ ðð¢
= â1
5ðð¢ + ð
= â1
5ð2â5ð¥ + ð
Rule 7: (Integration by Parts)
If ð and ð are two differentiable functions of ð¥. Then ð
ðð¥ ðð = ð ð¥
ðð
ðð¥+ ð(ð¥)
ðð
ðð¥
which implies that ð ð¥ ðð
ðð¥=
ð
ðð¥ ðð â ð(ð¥)
ðð
ðð¥ . Using the formula for the integral of a
sum which is sum of the integrals, we obtain
ð ð¥ ðð
ðð¥ðð¥ = ð ð¥ ð ð¥ â ð(ð¥)
ðð
ðð¥ðð¥
This is called the formula for integration by parts. If ð¢ = ð(ð¥) and ð£ = ð(ð¥), the
formula can be written as ð¢ ðð£ = ð¢ð£ â ð£ ðð¢ .
Example: Evaluate the following integrals:
1. 2ð¥ sin ð¥ ðð¥
Solution: Let ð¢ = 2ð¥, ðð¢
ðð¥= 2 â ðð¢ = 2ðð¥
ðð£ = sin ð¥ ðð¥ â ð£ = cos ð¥
Hence, by integration by parts, we have
2ð¥ sin ð¥ ðð¥ = 2ð¥ cos ð¥ â cos ð¥ 2ðð¥
= 2ð¥ cos ð¥ â 2 cos ð¥ ðð¥
= 2ð¥ cos ð¥ â 2 sin ð¥ + ð
2. ð¥ðð¥ ðð¥ = ð¥ðð¥ â ðð¥ ðð¥ = ð¥ðð¥ â ðð¥ + ð
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3. ð¥ logð ð¥ ðð¥
Solution: Let ð¢ = logð ð¥ , ðð¢
ðð¥=
1
ð¥ â ðð¢ =
1
ð¥ðð¥
ðð£ = ð¥ ðð¥ â ð£ =ð¥2
2
Hence, by integration by parts, we have
ð¥ logð ð¥ ðð¥ =ð¥2
2logð ð¥ â
ð¥2
2
1
ð¥ðð¥
=ð¥2
2logð ð¥ â
1
2 ð¥ ðð¥
=ð¥2
2logð ð¥ â
1
4ð¥2 + ð
PP: (a) ð¥2ðð¥ ðð¥ (b) ðð¥ sin ð¥ ðð¥ .
6.4 Definite Integral
Let ð be a function of ð¥ and ð, ð be an interval, ð(ð¥)ð
ððð¥ is called definite
integral with respect to ð¥ within the limit ð and ð.
THEOREM (Fundamental Theorem of Calculus): Suppose that ð is a strong
continuous function over a closed interval ð, ð and that ð¹ is anti â derivative of ð.
Then ð(ð¥)ð
ððð¥ = ð¹(ð¥)
ðð
= ð¹ ð â ð¹(ð).
Examples
1. ð¥23
1ðð¥ =
ð¥3
3
31
= 33
3 â
13
3 = 8
2
3
2. 4ð¥5
0ðð¥ = 2ð¥2 5
0= 50
PP: (a) ð2â5ð¥1
0ðð¥ (b) ð¥ ln ð¥
1
0ðð¥
6.5 APPLICATIONS OF INTEGRATION TO ECONOMICS AND COMMERCE
We learnt already that the marginal function is obtained by differentiating the
total function. We were given the total cost, total revenue or demand function and
we obtained the marginal cost, marginal revenue or elasticity of demand. Now we
shall obtain the total function when marginal function is given.
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6.5.1 The cost function and average cost function from marginal cost
function:
If ð¶ is the cost of producing an output ð¥, then marginal cost function,
ðð¶ =ðð¶
ðð¥. Using integration as reverse process of differentiation, we obtain,
Cost function, ð¶ = ðð¶ ðð¥ + ð
where ð is the constant of integration which can be evaluated if the fixed cost is
known. If the fixed cost is not known, then ð = 0.
Average cost function, ðŽð¶ =ð¶
ð¥, ð¥ â 0
Example 1: The marginal cost function of manufacturing ð¥ units of a commodity is
8 + 16ð¥ â 9ð¥2. Find the total cost and average cost, given that the total cost of
producing 1 unit is 20.
Solution: Given that,
ðð¶ = 8 + 16ð¥ â 9ð¥2
ð¶ = ðð¶ ðð¥ + ð
= 8 + 16ð¥ â 9ð¥2 ðð¥ + ð
= 8ð¥ + 8ð¥2 â 3ð¥3 + ð ------------------ (*)
Given, when ð¥ = 1, ð¶ = 20
⎠â â 20 = 8 1 + 8 12 â 3 13 + ð â ð = 13
⎠Total Cost function, ð¶ = 8ð¥ + 8ð¥2 â 3ð¥3 + 13
Average Cost function, ðŽð¶ =ð¶
ð¥, ð¥ â 0
= 8 + 8ð¥ â 23 +13
ð¥ .
Example 2: The marginal cost function of manufacturing ð¥ units of a commodity is
3ð¥2 â 2ð¥ + 32. If there is no fixed cost, find the total cost and average cost
functions. Solution: Given that,
ðð¶ = 3ð¥2 â 2ð¥ + 32
ð¶ = ðð¶ ðð¥ + ð
= 3ð¥2 â 2ð¥ + 32 ðð¥ + ð
= ð¥3 â ð¥2 + 32ð¥ + ð
No fixed cost â ð = 0
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⎠Total cost, ð¶ = ð¥3 â ð¥2 + 32ð¥
Average Cost, ðŽð¶ =ð¶
ð¥, ð¥ â 0
= ð¥2 â ð¥ + 32 .
6.5.2 The revenue function and demand function from marginal revenue
function:
If ð is the total revenue function when the output is ð¥, then marginal revenue
ðð =ðð
ðð¥.
Integrating this with respect to ð¥ we have
Revenue function, ð = ðð ðð¥ + ð
where ð is the constant of integration which can be evaluated under given
conditions.
If the total revenue ð = 0, when ð¥ = 0,
Demand function, ð =ð
ð¥, ð¥ â 0
Example 1: If the marginal revenue for a commodity is ðð = 10 â 6ð¥2 + 2ð¥, find
the total revenue and demand function.
Solution: Given that,
ðð = 10 â 6ð¥2 + 2ð¥
ð = ðð ðð¥ + ð
= 10 â 6ð¥2 + 2ð¥ ðð¥ + ð
= 10ð¥ â 2ð¥3 + ð¥2 + ð
Since ð = 0 when ð¥ = 0, then ð = 0.
⎠ð = 10ð¥ â 2ð¥3 + ð¥2
ð =ð
ð¥, ð¥ â 0
â ð = 10 â 2ð¥2 + ð¥.
Example 2: For the marginal revenue function ðð = 5 + sin 4ð¥ â 2ð¥ â ð¥2, find the
revenue function and demand function.
Solution: Given that,
ðð = 5 + sin 4ð¥ â 2ð¥ â ð¥2
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ð = ðð ðð¥ + ð
= 5 + sin 4ð¥ â 2ð¥ â ð¥2 ðð¥ + ð
= 5ð¥ â1
4coss 4ð¥ + ð¥2 +
ð¥3
3+ ð
Since ð = 0 when ð¥ = 0, then ð = â1
4.
⎠ð = 5ð¥ â1
4coss 4ð¥ + ð¥2 +
ð¥3
3â
1
4
ð =ð
ð¥, ð¥ â 0
â ð = 5 â1
4ð¥coss 4ð¥ + ð¥2 +
ð¥3
3â
1
4ð¥.
6.5.3 The demand function when the elasticity of demand is given
We know that,
Elasticity of demand ðð = âð
ð¥â
ðð¥
ðð
â âðð
ð=
ðð¥
ð¥
1
ðð
Integrating both sides
â ðð
ð=
1
ðð
ðð¥
ð¥
This equation yields the demand function ð as a function of ð¥.
The revenue function can be found out by using the relation, ð = ðð¥.
Example 1: The elasticity of demand with respect to price ð for a commodity is
ð¥â6
ð¥, ð¥ > 6 when the demand is ð¥. Find the demand function if the price is 2 when
demand is 8. Also find the revenue function.
Solution: Given that,
Elasticity of demand, ðð = âð
ð¥â
ðð¥
ðð=
ð¥â6
ð¥
â ðð¥
ð¥â6= â
ðð
ð
Integrating both sides,
ðð¥
ð¥â6= â
ðð
ð+ ð
â logð ð¥ â 6 = â logð ð + logð ð (since c is a constant, ð â¡ logð ð)
â logð ð¥ â 6 + logð ð = logð ð
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â logðð ð¥ â 6 = logð ð
â ð ð¥ â 6 = ð -------------- (1)
When ð = 2, ð¥ = 8, from (1) we get
ð = 4.
⎠The demand function is,
ð =4
ð¥â6, ð¥ > 6.
Revenue, ð = ðð¥ or 4ð¥
ð¥â6, ð¥ > 6.
Example 2: The elasticity of demand with respect to price for a commodity is a
constant and is equal to 2. Find the demand function and hence the total revenue
function, given that when the price is 1, the demand is 4.
Solution: Given that,
Elasticity of demand, ðð = 2
â âð
ð¥â
ðð¥
ðð= 2
â ðð¥
ð¥= â2
ðð
ð
Integrating both sides,
â ðð¥
ð¥= â2
ðð
ð
ln ð¥ = â2 ln ð + ln ð
ln ð¥ + ln ð2 = ln ð
ln ð¥ð2 = ln ð
ð¥ð2 = ð --------------- (1)
Given, when ð¥ = 4, ð = 1
From (1) we get ð = 4
⎠(1) â ð¥ð2 = 4 or ð =2
ð¥, ð¥ > 0
Demand function ð =2
ð¥, ð¥ > 0 ;
Revenue ð = ðð¥ =2ð¥
ð¥= 2 ð¥, ð¥ > 0.
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Exercise 6
1. The marginal cost function of manufacturing ð¥ units of a commodity is
3 â 2ð¥ â ð¥2. If the fixed cost is 200, find the total cost and average cost
function.
2. If the marginal revenue for a commodity is ðð =ð2ð¥
100+ ð¥ + ð¥2, find the
revenue function.
3. The marginal cost and marginal revenue with respect to a commodity of a firm
are given by ðð¶ = 4 + 0.8ð¥ and ðð = 12. Find the total profit, given that the
total cost at zero output is zero. Hence, determine the profit when the output
is 50.
4. The marginal revenue function (in thousands of naira) of a commodity is
7 + ðâ0.05ð¥ where ð¥ is the number of units sold. Find the total revenue from
the sale of 100 units ðâ5 = 0.0067
5. The marginal cost and marginal revenue are given by ðð¶ = 20 +ð¥
20 and
ðð = 30. The fixed cost is N200. Find the maximum profit.
6. ADAS company determines that the marginal cost of producing ð¥ units is
ðð¶ = 10.6ð¥. the fixed cost is N50. The selling price per unit is N5. Find
(i) Total Cost Function
(ii) Total Revenue Function
(iii) Profit Function
7. Find the cost of producing 3000 units of commodity if the marginal cost in
naira per unit is ð¶â² ð¥ =ð¥
3000+ 2.50.
8. The marginal cost of a production level of ð¥ units is given by ð¶â² ð¥ = 85 +375
ð¥2 .
Find the cost of producing 10 incremental units after 15 units have been
produced.