ENGINEERING
MATHEMATICS
GEC 210
By
Miss Victoria Aina
(Department of Petroleum Engineering)
08/09/2015
COVENANT UNIVERSITY GEC 210
1
CONTENT
• Concept of continuity and differentiability
• Mean Value Theorem
• Taylor’s Series Expansion
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2
EXPECTATIONS
By the end of the Lecture you would have
known:
 The Concept of Continuity
differentiability
 The mean value theorem
 Taylor’s Series Expansion
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COVENANT UNIVERSITY GEC 210
and
3
1.1 Concept of Continuity
Continuity at a point
A function f is continuous at x = a when
the three following conditions are
satisfied:
i. 𝑓 𝑎 𝑖𝑠 𝑑𝑒𝑓𝑖𝑛𝑒𝑑;
ii. lim 𝑓 𝑥 𝑒𝑥𝑖𝑠𝑡𝑠
𝑥→𝑎
iii. lim 𝑓 𝑥 = 𝑓(𝑎)
𝑥→𝑎
Otherwise f is said to be discontinuous
at x = a
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210
4
Concept of Continuity...
The concept of continuity in simple terms
says a function is continuous at a point
(x = a ) when you can compute its limit at
that point by substituting in x = a in the
function.
Example: check for continuity at x=3 for
the following functions
1. f(x)  x 3 - 5
2.
3.
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f ( x )  x13
2 3 x  2
x
f( x) 2
x 5 x  6
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5
Concept of Continuity...
Solution:
1. f(x)  x - 5
3
i . at x  3, f(3)  3 - 5  22 (defined)
3
ii . lim f(x)  3 - 5  22
3
(exist)
x 3
iii. lim f(x)  22  f(3)
x 3
Hence, f(x)  x 3 - 5 is continuous at x=3
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6
Concept of Continuity...
Solution:
2. f(x)  x-13
i . at x  3, f(3)  31- 3  1
0 (undefined )
 f(x)  x-13
3.
is not continuous at x=3
2 3 x  2
x
f( x) 2
x 5 x  6
i . at x  3, f(3) 
2 3 x  2
x
 f(x)  2
x 5 x  6
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32  3(3)  2
32 5(3)  6
 02 (undefined)
is not continuous at x=3
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7
Class Work 1 (Continuity at a Point)
Check for continuity of the following
functions at x=2
1
1. f(x)  2 x5
2. f(x) 
x 1
3. f(x)  2 x 2
x 3 x  2
4. f ( x ) 
x 2  3 x 10
x 7
5. f(x) 
x
x 4 16
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8
Continuity on an Interval
A function f(x) is said to be continuous
on an interval x (a,b) if f(x) is continuous
at every point within the interval (a,b).
e.g. for the interval (-5,7) the functions;
1. f(x)  x 3 - 5
is continuous everywhere on the interval (-5,7)
2. f ( x )  x1
3
is discontinuous at x=3 hence it is not
continuous on the interval
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(-5,7)
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9
Continuity on an Interval…
Question: is the function
2 3 x  2
x
f( x) 2
x 5 x  6
continuous on the interval (-5,7)?
Why?
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10
Solved Examples: Interval of
Continuity
Determine interval of continuity if
i . f(x) 
x5
ii . f(x) 
Solution:
i . f(x) 
x 2 4
x 2 4
x5
Inspection shows that for this root
function to be continuous
x50
 x  5
That is the function is continuous on the
5 x  
interval
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Solved Examples (contd.)
Solution:
ii . f(x) 
x 2 4
x 2 4
Now we have a root in the denominator
so that this time for continuity:
x 4  0
2
 x 4
 x  4
2
 x  -2 or x  2
That is f(x) is continuous on the double
interval
   x  2  2  x  
Which can be written alternatively as
x     2  x  2 
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Class Work II
(Interval of Continuity)
1. Determine the interval of continuity
x 1
i . f(x) 
x 2 6 x  9
ii . f(x)  2 x 5
x 6 x  9
iii. f(x) 
x  6x  9
2
iv . f(x)  x  6 x  9
2
2. What can you say about the domain of a
function and its interval of continuity?
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1.2 Concept of Differentiability
A function f(x) is said to be differentiable
at a point x = a in its domain if for a small
difference h  Δx the limit given by
f(a  h) f ( a )
lim
h
h0
exists.
This gives the derivative of f(x) at x = a as
f(a  h) f(a)
f ' (a)  lim
h
h0
Hence, if f(x) is differentiable at x = a then f(x)
is continuous at x = a.
Question: Can you recall the conditions for the
continuity of the function f(x) at the point x = a?
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1.2.1 Differentiation: The Derivative of a
function
The process of computing the derivative
of a function is known as Differentiation.
The derivative of the differentiable
function f(x) is given by
f(x  h) f ( x )
f ' ( x )  lim
h
h0
This is known as the first principle of
differentiation. In another notation we write
dy
f(x  δx) f(x)
 lim
δx
dx δx 0
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1.2.2 Differentiation using First
Principle
Practice: Find the derivative of the
following functions using the first
principle of differentiation.
i . f ( x )  5x  2
ii . f ( x )  3 x  2 x  8
2
iii. f ( x )  1x
iv . f ( x )  x
v. f ( x )  1
x
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1.2.2 Differentiation using First
Principle
Solution: i . f ( x )  5 x  2
f(x

h)

f
(
x
)
from f ' ( x )  lim
h
h0
5(x  h)  2 5 x  2 

 lim
h0
h
 lim 5 x 5hh25 x 2
h0
 lim 5 x 5hh25 x  2  lim 5hh
h0
h0
 f ' ( x )  lim 5  5
h0
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17
1.2.2 Differentiation using First
Principle
Solution: ii . f ( x )  3 x 2  2 x  8 thus,
f ' ( x)  lim

 

 3( x  h ) 2  2( x  h ) 8   3 x 2  2 x 8 

 

h
h 0
 lim

 

 3( x 2  2 xh  h 2 )  2 x  2 h 8    3 x 2  2 x 8 

 

h
h 0
2 2h  lim 6 x  h  2  6 x  2
6
xh

h
 lim
h0
h
h0
hence; f ( x )  3 x 2  2 x  8
 f ' ( x )  6x  2
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1.2.2 Differentiation using First
Principle
iii. f ( x )  1x
Solution:
then,



1   1
xh  x
f ' ( x )  lim
h
h0
x  x  h 
1
 lim h  x  hx  lim 1  x  x h 
h0
h0 h  x 2  xh 



h
1

 lim h  2   lim 21 
h0  x  xh  h0  x  xh 
 12
x
 f ' ( x )  12
x
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1.2.2 Differentiation using First
Principle
Solution:
For the remaining solutions, we will try to
use the differential notation:
dy
f(x  δx) f(x)
 lim
dx δx 0
δx
Comparing this with the previous notation,
we see that for the function y = f(x)
dy
 f' ( x )
dx
also,
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δx  h
Let us now apply the new notation to the
remaining solutions.
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20
1.2.2 Differentiation using First
Principle
iv . y  x
Solution:
dy
dx
for which
x  δx  x
δx
 lim
δx 0
like we' re 

δx  x
x  δx  x  rationalising 
 lim x  δx



δx 0
surds!
x  δx  x 

x  δx  x
δx
 lim
 lim
δx 0
 lim
δx
δx 0
 x  δx  x 
1
x  δx  x

δx 0
δx
 x  δx  x 
dy
1


x x
dx 2 x
1
We see that the approach is just the same!
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1.2.2 Differentiation using First
Principle
Solution:
v. y  1
x

1
 1
x  δx
x
δx
dy
 lim
dx δx 0
 x  x  δx 
 lim δx 

δx 0
 x  δx x 
1

 
 x  x  δx
x  x  δx 
 lim δx 


δx 0
x  x  δx 
 x  δx x

x

x

δx
1 
 lim δx 

3
2
3
2
2
δx 0
 x  x x  x  2 x x  xx 

1
 


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1
x  x
3
3


dy
dx

1
2 x
3
 21
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x
3
2
22
1.2.2 First Principle Derivatives:
Results Summary
Using the first principle of differentiation
we have obtained the derivatives of the
given functions as follows
i . f ( x )  5x  2
 f' ( x )  5
ii . f ( x )  3 x  2 x  8  f ' (x)  6 x  2
2
iii. f ( x )  1x
 f' ( x ) 
1
x2
iv . f ( x )  x
v. f ( x )  1
 f' ( x ) 
1
x
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 f' ( x ) 
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2 x
1
2
x3
23
1.2.3 Class Work III: First Principle
Differentiation
Obtain the derivatives of the following
functions by using first principle
i . f ( x )  11 x  3
ii . f ( x )  x  5 x  1
iii. f ( x )  x12
2
iv . f ( x )  2 x  1
v. f ( x )  1
x 1
vi . f ( x )  7
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1.2.4 Differentiation of Polynomial
functions
For any polynomial expression of the form
y  f(x)  ax
n
constant a, and exponent n; the
derivative is given by
dy
n1

f
'
(x)

nax
dx
e.g.
f ( x )  5x; f ' ( x )  1  5 x11  5x 0  5
1
21
 6x

6x
f
'
(
x
)

2

3
x
f ( x )  3x ;
2
1
f ( x )  x  x ; f ' ( x )  1 x
1
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210
11
2
 1 x 
1
x2
25
1.2.5 Derivatives of functions with
Fractional Exponents
Fractional exponents are not left out since
n thus,
dy
n1
y  f(x)  ax ;

f
'
(x)

nax
dx
even when n is a fractional exponent.
1
e.g. i . f ( x )  x  x 2 ;
1
1 1

1
1
1
2
2 
f' ( x )  x
 x
n
2
ii . f ( x )  1 
x
f' ( x )   x
1
2
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 12 1
2
1
2 x
 12
x x ;
1
2

1
2
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x
 32

1
2 x3
26
1.2.6 Class Work IV: Polynomial
Differentiation
Obtain f ’(x) for the following functions
i . f ( x )  5x 3
ii . f ( x )  x
5
iii. f ( x )  12
x
iv . f ( x )  x
3
v . f ( x )  41
x
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1.3 Standard Derivatives
Next, we present, in tabular form, the
standard derivatives of commonly
encountered functions.
It is worth noting that while many of the
standard derivatives are obtainable
from first principle some of them can
also be obtained from the applications
of the rules of differentiation, presented
later, to the definitions of the functions
concerned; e.g (as shall be shown later)
tanx can be obtained from sinx and cosx.
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1.3.1 Standard Derivatives
Table
S/No
1.
2.
3.
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y f(x)
y' 
n
4.
x
x
e
x
a
ln x
5.
loga x
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dy
dx
n 1
nx
x
e
x
a  ln a
1
x
1
x ln a
29
1.3.1 Standard Derivatives
Table
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S/No
y f(x)
6.
7.
8.
9.
10.
11.
sin x
cos x
tan x
cot x
sec x
csc x
dy
dx
y' 
cos x
 sin x
2
sec x
2
 csc x
sec x  tan x
 csc x  cot x
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1.3.1 Standard Derivatives
Table
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S/No
y f(x)
12.
13.
14.
15.
sinh x
cosh x
tanh x
coth x
16.
sec hx
17.
csc hx
y' 
dy
dx
cosh x
sinh x
2
sec h x
2
 csc h x
 sec hx  tanh x
 csc hx  coth x
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1.3.2 Memory Test 1: Standard
Derivatives
Try to write down the derivatives of the
following functions without looking at
the table of standard derivatives.
x
iii
.
y

log
x
7
3
i
i
.
y

5
i. y  x
v . y  cos x
vi . y  csc hx
viii. y  sec x
ix. y  sinh x
x . y  tan x
xi. y  sin x
xiii . y  cosh x
xvi . y  tanh x
xiv. y  sec hx
xvii. y  cot x
xii. y  csc x
xv . y  coth x
iv. y  ln x
vii . y  e
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x
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xviii. y  x
32
1.4 Rules of Differentiation
Let u = f(x) and v = g(x) be two
differentiable
functions
then
the
following rules hold:
dy du dv
i . y  u  v  y'  dx  dx  dx
ii . y  u  v  y'  dy  du  dv
dx dx dx
iii. y  uv

iv . y  uv

v . y  u( x )
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
sum
rule 
difference

rule
dy
dv  v du product 
y'  dx  u dx
rule
dx
du  u dv
v
quotient
dy
dx
dx

y'  dx 
rule 
v
composite


y' 

rule
2
dy
dx
dy du

du dx
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33
1.4 Rules of Differentiation
Applied Examples: Differentiate the
following functions with respect to x
i . y  3x2  2x
ii . y  x  cos x
3 x
2
iii. y  x e
3 5 x 1
x
iv . y 
3 x 2 5
v . y  tan x
vi . y  ( x 3  2 x  1)8
vii .
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y  cos x  ln( 2 x 3  1)
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1.4.1 The Sum Rule
i . y  3x2  2x
Solution:
This is a sum of two functions
u  3x 
2
du
dx
 6 x ; and
then y  u  v
dv  2
v  2 x  dx
 
dy du dv
sum
 dx  dx  dx
rule
dy d ( 3 x 2 ) d ( 2 x )
 dx  dx  dx
dy
 dx  6 x  2
Sum Rule: The derivative of sums of functions
equals the sum of derivatives of each functions
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35
1.4.2 The Difference Rule
Solution: ii . y  x  cos x
2
This is a difference of two functions
dv   sin x
u  x 2  du

2
x
;
and
v

cos
x

dx
dx
dy du dv
y

u

v
 dx  dx  dx
then

difference

rule
dy d ( x 2 ) d ( cos x )


dx
dx
dx
dy
 dx  2 x  (  sin x )  2 x  sin x
Difference Rule: The derivative of differences of
functions equals the differences of derivatives
of each functions
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36
1.4.3 The Product Rule
Solution:
iii. y  x e
3
x
This is a product of two functions
2
x
x
dv
u  x 3  du

3x
;
and v  e   e
dx
dx
then
y  uv  dy  u dv  v du
dx
dx
dx
product
rule 
x
3
dy
3 d (e )
x d( x )
 dx  x dx  e dx
dy
 dx  x 3e x  3 x 2e x
Product Rule: The derivative of product of two
functions = first (function) X differential second
+ second (function) X differential first.
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37
1.4.4 The Quotient Rule
3 5 x 1
x
iv . y 
3 x 2 5
Solution:
This is a function with quotient
2
2
dv  6 x
u  x 3  5 x  1  du

3
x

5;
v

3
x

5

dx
dx
then y  uv

v2
quotient
rule 
3  5 x  1)
2  5)
d
(
x
d
(
3
x
( 3 x 2  5)
 ( x 3  5 x  1)
dy
dx
dx

dx
( 3 x 2  5) 2
dy
 dx 
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dy
 dx 
dv
v du

u
dx
dx
( 3 x 2 5 )(3x 2  5 )( x 3  5 x 1)6 x
( 3 x 5 )
2
2
COVENANT UNIVERSITY GEC 210

3 x 4  30 x 2  6 x  25
( 3 x 2 5 )2
38
1.4.5 The Composite Rule
Solution: vii . y  cos x  ln( 2 x 3  1)
Combines product and composite functions
u  cos x  du   sin x; v  ln(2 x 3  1);
dx
2
3
2
dv
dv
dw
1
6
x
let w  2 x  1  v  ln w ;    w  6x 
dx dw dx
2 x 3 1
product
thus y  uv  dy  u dv  v du
rule
dx
dx
dx
2
dy
6
x
 dx  cos x( 3 )  ln( 2 x 3  1)(  sin x )
2 x 1
2
dy
3
6
x
 dx  3 cos x  sin x ln( 2 x  1)
2 x 1

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COVENANT UNIVERSITY GEC 210

39
1.4 Class Work V: Applying Rules of
Differentiation
Differentiate the following functions with
respect
to
x (using rules of
differentiation as appropriate).
i . y  cot x
ii . y  x 2 ln x
iii. y  x  2 sin x
iv . y  cos x
3
v. y  e  5
x
vi . y  log10 x  ln x
x
vii . y  ( x 2  3 x  1)17
ix.
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y  3 x
2
 2 x 1
4
  tan x
viii. y 
COVENANT UNIVERSITY GEC 210
2x
x 2 2
x . y  sec x
40
1.5 Derivative Theorems
Two derivative theorems lead to the
Mean Value Theorem.
From the Mean Value Theorem, itself,
we can draw three corollaries.
We hereby state the derivative
theorems including the Mean Value
Theorem and its three corollaries.
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41
1.5.1 Interior Extremum Theorem
If the function f(x) defined, at least, on an
open interval (a,b) takes an extreme
value (maximum or minimum) at a
number c  (a,b) and if f'(c) exist then
f '(c ) 0
Warnings:
1. Take note that the open interval (a,b) does
not include the numbers x = a and x = b
2. That f  (w)=0, w(a,b) may not imply there is
an extremum at w; this turns the theorem.
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42
1.5.2 Rolle’s Theorem
Let f(x) be continuous on the (closed)
interval [a,b], and differentiable on the
(open) interval (a,b). If f(a) = f(b) then
there is at least one number c  (a,b)
such that f'(c)=0
e.g. f ( x )  x 2  7 x  10
6
y
f (b= 6)
f (a= 1)
4
2
c=3.5
0
-2 0
-4
1
2
3
4
x
5
6
7
8
9
f '(c )=0
f ( 1 )  f ( 6 )  4 ;  c=3.5 :1<3.5<6 where f ' ( c )  0
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COVENANT UNIVERSITY GEC 210
43
1.5.3 Mean Value Theorem
If f(x) is continuous on the (closed)
interval [a,b], and differentiable on the
(open) interval (a,b); then there exist a
number c  (a,b) such that
f (b) f (a )
f '(c ) 
ba
Example:
60
y
f'(c)
40
f(b)
c=1.8206
20
0
-20
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x
0
f(a)1
2
3
4
COVENANT UNIVERSITY GEC 210
5
6
7
8
9
44
1.5.4 Corollary I of Mean Value
Theorem
If the derivative of a function is 0 (zero)
throughout an interval I then the
function is constant throughout that
interval.
Mathematically:
f ' ( x )  0 , x  I  f(x)  C on I
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COVENANT UNIVERSITY GEC 210
45
1.5.5 Corollary II of Mean Value
Theorem
If two functions f(x) and g(x) has the same
derivative for all x in an open interval I
then the two derivatives differs by a
constant.
Mathematically:
f ' ( x )  g ' ( x ) , x  I  f(x)  g(x)  C
This theorem has significant implications in
the reversal of derivatives i.e. Integration.
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46
1.5.6 Corollary III of Mean Value
Theorem
If f(x) is continuous on (a,b) and has a
positive derivative on (a,b) then f(x) is
increasing on the interval [a,b]. If f(x) is
continuous on (a,b) and has a negative
derivative on (a,b) then
f(x) is
decreasing on the interval [a,b].
Mathematically:
if a  x1  x2  b  f ' ( x1 )  f ' ( x2 ) then
f (a ) f (b)
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COVENANT UNIVERSITY GEC 210
47
QUESTIONS???
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48