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Differentiation
1
Differentiation
Today’s Objective:
When you have completed this lecture you will be able to:
• Differentiate by using a list of standard derivatives
• Apply the chain rule
• Apply the product and quotient rules
• Perform logarithmic differentiation
• Differentiate parametric equations
2
Here is a revision list of the standard derivatives which you have no
doubt used many times before. Memorize those with which you
are less familiar.
3
Standard derivatives
No.
1
2
3
4
5
6
7
x
sin

No.
8
9
10
11
12
13
14
)
(x
f
y 
dx
dy
n
x 1

n
nx
x
e x
e
x
a a
ax
ln
.
x
ln
x
1
x
a
log
a
x ln
.
1
x
sin x
cos
)
(x
f
y 
dx
dy
x
2
sec
x
cot x
ec2
cos

x
sec x
x tan
.
sec
ecx
cos x
ecx cot
.
cos

x
sinh x
cosh
x
cosh x
sinh
x
tan
x
cos x
sin

kx
e kx
ke
4
Example 1: write down the derivatives for the following.
Solution :
x
x
x
x
e
x
x 2
.
6
tan
.
5
ln
.
4
.
3
sin
.
2
.
1 3
5
x
e
x
x
x x
cos
.
11
.
10
log
.
9
cosh
.
8
sec
.
7 10
x
a
x
a
ecx
x .
16
cot
.
15
.
14
cos
.
13
sinh
.
12 3
2
4
.
20
.
19
log
.
18
.
17
x
a e
x
x
x
2
ln
.
2
.
6
sec
.
5
1
.
4
3
.
3
cos
.
2
5
.
1 2
3
4 x
x
x
x
e
x
x
x
e
x
x
x
x x
sin
.
11
.
10
10
ln
.
1
.
9
sinh
.
8
tan
.
sec
.
7 
x
ec
anx
ecx
x 2
cos
.
15
0
.
14
cot
.
cos
.
13
cosh
.
12 

2
2
1
5
.
20
2
1
.
19
ln
1
.
18
4
.
17
ln
.
.
16
x
x
e
x
a
x
x
a
a 

5
Function of a function
Is a function of since the value of depends on the value of
angle . Similarly is a function of the angle since
the value of the sine depends on the value of this angle.
i.e. Is a function of
But it itself is a function of , since its value depends on .
For example :
the derivative of the function
x
sin x x
sin
x )
5
2
sin( 
x )
5
2
( 
x
)
5
2
sin( 
x
)
5
2
( 
x
)
5
2
( 
x
x x
5
)
3
4
( 
 x
y


 4
)
3
4
(
5 x
dx
dy
)
3
4
( 
x
4
)
3
4
(
5 4



 x
dx
dy
4
)
3
4
(
20 
 x
6
Example 2 : write down the derivatives for the following.
Solution :
)
cos
4
3
ln(
.
5
)
cos(
.
4
2
sin
.
3
.
2
)
5
4
(
.
1 2
3
6
x
y
x
y
x
y
e
y
x
y x






 
x
y
x
y
x
y
e
y
x
y x
3
cos
ln
.
10
)
3
(
cos
.
9
sin
.
8
.
7
)
1
2
(
log
.
6 3
2
2
sin
10 





x
x
y
x
x
y
x
dx
dy
e
dx
dy
x
dx
dy x
cos
4
3
sin
4
.
5
)
sin(
2
.
4
2
cos
2
.
3
.
2
)
5
4
(
24
.
1
2
3
5










 
x
x
x
y
x
x
y
x
x
y
e
x
y
x
y x
3
tan
3
3
cos
3
sin
3
.
10
)
3
(
cos
)
3
sin(
9
.
9
sin
cos
2
.
8
.
2
cos
2
.
7
10
ln
)
1
2
(
2
.
6
2
2
sin















7
Differentiate functions which are products or quotients of two of the functions
dx
du
v
dx
dv
u
dx
dy


1. Products : if y=uv where u and v are functions, of x, then
you already know that:
2
v
dx
dv
u
dx
du
v
dx
dy


2. Quotients : if y=u/v where u and v are functions, of x, then
you already know that:
8
Example 3: write down the derivatives for the following.
Solution :
2
2
2
3
5
2
2
cos
.
8
ln
.
7
1
3
sin
.
6
sinh
ln
.
5
5
sin
.
4
2
cos
.
3
)
1
3
(
.
2
tan
.
1
x
x
y
e
x
y
x
x
y
x
x
y
x
x
y
x
x
y
x
e
y
x
x
y
x
x










)
1
3
(
5
3
.
2
tan
2
sec
.
1
5
5
2
2





x
e
e
dx
dy
x
x
x
x
dx
dy
x
x
)
5
15
(
3 5
5


 x
e
e x
x
)
8
15
(
)
5
15
3
( 5
5




 x
e
x
e x
x
x
x
x
dx
dy
2
cos
2
sin
2
.
3 


x
x
x
x
dx
dy
5
sin
3
5
cos
5
.
4 2
3


)
5
sin
3
5
cos
5
(
2
x
x
x
x 

x
x
x
x
x
dx
dy
sinh
2
sinh
cosh
.
5
2


)
sinh
2
coth
( x
x
x
x 

2
)
1
(
3
sin
3
cos
)
1
(
3
.
6




x
x
x
x
dx
dy
x
x
x
x
e
x
x
e
x
e
x
e
dx
dy
2
4
2
2
ln
2
1
ln
2
.
7




4
2
2
cos
2
2
sin
2
.
8
x
x
x
x
x
dx
dy 


3
)
2
cos
2
sin
(
2
x
x
x
x 


9
Logarithmic differentiation.
The rules of differentiating a product or a quotient that we are
revised are used when there are just two-factor functions, i.e. uv
or u/v. where there are more than two functions in any
arrangement top or bottom, the derivative is best found by what
is known as logarithmic differentiation. For example;
if where u, v, and w -and also y- are all functions
of x.
First take logs of base e
w
uv
y 
w
v
u
y ln
ln
ln
ln 


dx
dw
w
dx
dv
v
dx
du
u
dx
dy
y
1
1
1
1












dx
dw
w
dx
dv
v
dx
du
u
y
dx
dy 1
1
1
10
Example 4: find dy/dx for the following.
Solution :
x
x
e
y
x
e
x
y
x
x
x
y
x
x
2
cosh
.
3
tan
.
2
2
cos
sin
.
1 3
4
3
4
2

















 















x
x
x
x
x
x
dx
dy
x
x
x
x
x
x
x
x
x
dx
dy
dx
dw
w
dx
dv
v
dx
du
u
y
dx
dy
x
w
x
v
x
u
let
2
tan
2
cot
2
2
cos
sin
2
cos
)
sin
2
(
sin
cos
2
2
cos
sin
1
1
1
2
cos
,
sin
,
.
1
2
2
2
2






























x
x
x
x
e
x
dx
dy
x
x
e
e
x
x
x
e
x
dx
dy
dx
dw
w
dx
dv
v
dx
du
u
y
dx
dy
x
w
e
v
x
u
let
x
x
x
x
x
tan
sec
3
4
tan
tan
sec
3
4
tan
1
1
1
tan
,
,
.
2
2
3
4
2
3
3
4
3
3
4
3
4






























x
x
x
x
e
x
x
x
x
e
e
x
x
e
dx
dy
dx
dw
w
dx
dv
v
dx
du
u
y
dx
dy
x
w
x
v
e
u
let
x
x
x
x
x
2
tanh
2
3
4
2
cos
2
cosh
2
sinh
2
3
4
2
cos
1
1
1
2
cosh
,
,
.
3
3
4
3
2
4
4
3
4
3
4
11
Parametric equations
In same cases, it is convenient to represent a function by
expressing x and y separately in terms of a third independent
variable. E.g. y= cos2t, x=sint.
The third variable is called parameter, and the two expression
of x and y are called parametric equation.
Example 5: find and for the following.
Solution :
1.
t
t
y
t
t
x
x
y
t
x
t
y











1
2
3
,
1
3
2
.
3
cos
,
sin
sin
3
.
2
sin
,
2
cos
.
1 3
3



dx
dy
2
2
dx
y
d
t
dt
dx
t
dt
dy
cos
2
sin
2



dx
dt
dt
dy
dt
dt
dx
dy
dx
dy
.
. 

t
t
dx
dy
cos
1
.
2
sin
2


12
dx
dt
t
t
dt
d
dx
y
d
dt
dt
t
t
dx
d
dx
y
d
t
t
dx
d
dx
y
d
.
cos
2
sin
2
.
cos
2
sin
2
cos
2
sin
2
2
2
2
2
2
2























t
t
t
t
t
t
dx
y
d
cos
1
.
)
(cos
sin
2
sin
2
2
cos
cos
4
2
2
2





 







 



t
t
t
t
t
t
dx
y
d
3
2
2
2
2
2
cos
cos
sin
4
)
sin
(cos
cos
4
4
cos
cos
sin
4
sin
cos
4
cos
4
3
2
2
3
2
2







 



t
t
t
t
t
t
dx
y
d
Solution :
2.







sin
cos
3
cos
sin
3
cos
3
2
2




d
dx
d
dy
dx
d
d
dy
dx
dy 

.

13









sin
cos
3
)
sin
1
(
cos
3
sin
cos
3
cos
sin
3
cos
3
2
2
2
2






dx
dy





cot
sin
cos
3
cos
cos
3
2
2




     
dx
dt
dt
d
dx
y
d
dt
dt
dx
d
dx
y
d
dx
d
dx
y
d
.
cot
.
cot
cot 2
2
2
2
2
2


 













sin
cos
3
cos
sin
cos
3
1
).
cos
( 2
2
2
2
2
2
ec
ec
dx
y
d 





Solution :
3.
2
2
2
2
2
2
)
1
(
5
)
1
(
3
2
3
3
)
1
(
)
3
2
(
)
1
(
3
)
1
(
1
)
1
(
2
3
2
2
)
1
(
)
2
3
(
)
1
(
2
t
t
t
t
t
t
t
dt
dx
t
t
t
t
t
t
t
dt
dy






























dx
dt
dt
dy
dx
dy
.
5
1
5
)
1
(
.
)
1
(
1 2
2






t
t
dx
dy
0
5
1
2
2








dx
d
dx
y
d
Exponential and logarithmic
functions
14
Exponential and logarithmic functions
• Today’s Objective:
When you have completed this lecture you will be able to:
• Solve indicial and logarithmic equations.
• Recognize that the exponential function and the natural
logarithmic function are mutual inverses
• Construct the hyperbolic functions from the odd and
even parts of the exponential function.
• Application of logarithms and exponential function.
15
Introduction to logarithms
If a number y can be written in the form , then the
index x is called the ‘logarithm of y to the base of a’,
Logarithms having a base of 10 are called common
logarithms and is usually abbreviated to .
16
10
log
x
a
y 
y
x
then
a
y
if a
x
log


lg
Laws of logarithms
There are three laws of logarithms, which apply to any base:
B
A
A
n
A
B
A
B
A
B
A
n
log
log
)
log(
.
3
log
)
log(
.
2
log
log
)
log(
.
1






An indicial equation is an equation where the variable appears as an
index and the solution of such an equation requires the application of
logarithms.
17
Indicial equations
Example 1:
Find the value of X, give that 4
.
35
122

x
Solution :
7177
.
0
0792
.
1
2
5490
.
1
5490
.
1
0792
.
1
)
2
(
4
.
35
log
12
log
)
2
(
4
.
35
log
12
log 2








x
x
x
x
A
n
An
log
)
log( 
18
Example 2:
Find the value of X, give that
Solution :
694
.
6
3913
.
0
6192
.
2
2042
.
1
4150
.
1
)
4150
.
1
8063
.
1
(
4150
.
1
4150
.
1
2042
.
1
8063
.
1
4150
.
1
)
1
(
6021
.
0
)
2
3
(
26
log
)
1
(
4
log
)
2
3
(
26
log
4
log 1
2
3















 

x
x
x
x
x
x
x
x
x
x
log(An
)= nlogA
1
2
3
26
4 

 x
x
19
Example 3:
Find the value of X, give that
Solution :
4853
.
2
5164
.
0
2834
.
1
2834
.
1
)
0436
.
2
5600
.
2
(
0436
.
2
2834
.
1
5600
.
2
0436
.
2
9138
.
0
8276
.
1
19726
.
2
7324
.
0
0436
.
2
9138
.
0
)
1
2
(
7324
.
0
)
3
(
)
6812
.
0
(
3
2
.
8
log
)
1
2
(
4
.
5
log
)
3
(
8
.
4
log
3
2
.
8
log
4
.
5
log
8
.
4
log
)
2
.
8
4
.
5
log(
1
2
3
3
1
2
3






























x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
B
A
B
A log
log
)
log( 


x
x
x 3
1
2
3
8
.
4
2
.
8
4
.
5 
 

20
Example 4: Find the value of X, give that
Solution :
5006
.
1
7676
.
2
1533
.
4
1533
.
4
7676
.
2
4683
.
2
6123
.
1
7766
.
5
1553
.
1
8450
.
0
294
log
4
.
6
log
2
3
.
14
log
)
5
(
7
log
294
log
}
4
.
6
)
3
.
14
(
7
log{ 2
5


















x
x
x
x
x
x
x
x
294
4
.
6
)
3
.
14
(
7 2
5


 x
x
Example 5: Solve the equation
Solution :
2
1
4
2
2
2
4
0
)
4
(
2
0
)
2
(
0
)
4
)(
2
(
0
8
6
2
2
























x
x
z
or
z
z
z
or
z
z
z
z
z
z
z
let
x
x
x
0
8
2
6
22



 x
x
Graphs of logarithmic functions
21
A graph of is shown in bellow
x
y 10
log

0
1
log 
a


0
loga
1
log 
a
a
In general, with a logarithm
to any base a, it is noted
that:
Exponential functions
The exponential function is expressed by the equation:
or
where e is the exponential number 2.7182818 _ _ _ . The
graph of this function lies entirely above the x-axis as does
the graph of its reciprocal , as can be seen in the
diagram:
22
)
exp(x
y 
x
e
y 
x
e
y 

The value of can be found to any level of precision desired
from the series expansion:
In practice a calculator is used.
Logarithms with base exponential called Napierian logarithms which is
written as:
So,
23
x
e
x
x
e ln
log 








!
5
!
4
!
3
!
2
1
5
4
3
2
x
x
x
x
x
ex
x
e
x
e x
x
e 

 ln
log
.
..........
ln 3

 x
e
24
Example 6: Solve for X, given that
Solution :
)
1
(
60
20 2
x
e


810
.
0
405
.
0
2
405
.
0
2
2
3
ln
ln
2
3
3
2
6
2
1
)
1
(
60
20
2
2
2
2
2















 


x
x
x
e
e
e
e
e
x
x
x
x
x
Example 7: Solve for X, given that
Solution :
0
2
3
2


 x
x
e
e
0
)
2
)(
1
(
0
2
3
2








 y
y
y
y
e
y
let x
0
1
ln
1
ln
ln
1
1
0
)
1
(








x
x
e
e
y
y
x
x
693
.
0
2
ln
2
ln
ln
2
2
0
)
2
(








x
x
e
e
y
y
x
x
25
Example 8:
Solution : Ans. 0.5545s
The current amperes flowing in a capacitor at time seconds is
given by
where and
Determine :
(a) the time for the current to reach 6.0 A.
(b) Sketch the graph of current against time.
)
1
(
8 CR
t
e
i



i t


 3
10
25
R f
C 6
10
16 


t
26
Odd and even parts
Not every function is either even or odd but many can be written as
the sum of an even part and an odd part. If, given f(x) where f( -x) is
also defined then:
is even and is odd
Furthermore is called the even part of f(x) and is called
the odd part of f(x).
For example, if then
So that the even and odd parts of are:
2
)
(
)
(
)
(
x
f
x
f
x
fe



2
)
(
)
(
)
(
x
f
x
f
x
fo



)
(x
fe )
(x
fo
1
2
3
)
( 2


 x
x
x
f 1
2
3
1
)
(
2
)
(
3
)
( 2
2








 x
x
x
x
x
f
)
(x
f
x
x
x
x
x
x
f
x
x
x
x
x
x
f
o
e
2
2
)
1
2
3
(
)
1
2
3
(
)
(
1
3
2
)
1
2
3
(
)
1
2
3
(
)
(
2
2
2
2
2
















So, the even and odd parts of the odd part of
are………………
4
3
2
)
( 2
3



 x
x
x
x
f
27
Odd and even parts of the exponential function
The exponential function is neither odd nor even but it can be written
as a sum of an odd part and an even part.
That is, and . These
two functions are known as the hyperbolic cosine and the hyperbolic
sine respectively:
and
Using these two functions the hyperbolic tangent can also be defined:
The logarithmic function is neither odd nor even and indeed
does not possess even and odd parts because is not defined.
2
)
exp(
)
exp(
)
(
exp
x
x
x
e



2
)
exp(
)
exp(
)
(
exp
x
x
x
o



2
cosh
x
x
e
e
x



2
sinh
x
x
e
e
x



x
x
x
x
e
e
e
e
x 




tanh
x
y a
log

)
(
log x
a 

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lecture (8)derivatives.pptxdddddddddddddff

  • 2. Differentiation Today’s Objective: When you have completed this lecture you will be able to: • Differentiate by using a list of standard derivatives • Apply the chain rule • Apply the product and quotient rules • Perform logarithmic differentiation • Differentiate parametric equations 2
  • 3. Here is a revision list of the standard derivatives which you have no doubt used many times before. Memorize those with which you are less familiar. 3 Standard derivatives No. 1 2 3 4 5 6 7 x sin  No. 8 9 10 11 12 13 14 ) (x f y  dx dy n x 1  n nx x e x e x a a ax ln . x ln x 1 x a log a x ln . 1 x sin x cos ) (x f y  dx dy x 2 sec x cot x ec2 cos  x sec x x tan . sec ecx cos x ecx cot . cos  x sinh x cosh x cosh x sinh x tan x cos x sin  kx e kx ke
  • 4. 4 Example 1: write down the derivatives for the following. Solution : x x x x e x x 2 . 6 tan . 5 ln . 4 . 3 sin . 2 . 1 3 5 x e x x x x cos . 11 . 10 log . 9 cosh . 8 sec . 7 10 x a x a ecx x . 16 cot . 15 . 14 cos . 13 sinh . 12 3 2 4 . 20 . 19 log . 18 . 17 x a e x x x 2 ln . 2 . 6 sec . 5 1 . 4 3 . 3 cos . 2 5 . 1 2 3 4 x x x x e x x x e x x x x x sin . 11 . 10 10 ln . 1 . 9 sinh . 8 tan . sec . 7  x ec anx ecx x 2 cos . 15 0 . 14 cot . cos . 13 cosh . 12   2 2 1 5 . 20 2 1 . 19 ln 1 . 18 4 . 17 ln . . 16 x x e x a x x a a  
  • 5. 5 Function of a function Is a function of since the value of depends on the value of angle . Similarly is a function of the angle since the value of the sine depends on the value of this angle. i.e. Is a function of But it itself is a function of , since its value depends on . For example : the derivative of the function x sin x x sin x ) 5 2 sin(  x ) 5 2 (  x ) 5 2 sin(  x ) 5 2 (  x ) 5 2 (  x x x 5 ) 3 4 (   x y    4 ) 3 4 ( 5 x dx dy ) 3 4 (  x 4 ) 3 4 ( 5 4     x dx dy 4 ) 3 4 ( 20   x
  • 6. 6 Example 2 : write down the derivatives for the following. Solution : ) cos 4 3 ln( . 5 ) cos( . 4 2 sin . 3 . 2 ) 5 4 ( . 1 2 3 6 x y x y x y e y x y x         x y x y x y e y x y x 3 cos ln . 10 ) 3 ( cos . 9 sin . 8 . 7 ) 1 2 ( log . 6 3 2 2 sin 10       x x y x x y x dx dy e dx dy x dx dy x cos 4 3 sin 4 . 5 ) sin( 2 . 4 2 cos 2 . 3 . 2 ) 5 4 ( 24 . 1 2 3 5             x x x y x x y x x y e x y x y x 3 tan 3 3 cos 3 sin 3 . 10 ) 3 ( cos ) 3 sin( 9 . 9 sin cos 2 . 8 . 2 cos 2 . 7 10 ln ) 1 2 ( 2 . 6 2 2 sin               
  • 7. 7 Differentiate functions which are products or quotients of two of the functions dx du v dx dv u dx dy   1. Products : if y=uv where u and v are functions, of x, then you already know that: 2 v dx dv u dx du v dx dy   2. Quotients : if y=u/v where u and v are functions, of x, then you already know that:
  • 8. 8 Example 3: write down the derivatives for the following. Solution : 2 2 2 3 5 2 2 cos . 8 ln . 7 1 3 sin . 6 sinh ln . 5 5 sin . 4 2 cos . 3 ) 1 3 ( . 2 tan . 1 x x y e x y x x y x x y x x y x x y x e y x x y x x           ) 1 3 ( 5 3 . 2 tan 2 sec . 1 5 5 2 2      x e e dx dy x x x x dx dy x x ) 5 15 ( 3 5 5    x e e x x ) 8 15 ( ) 5 15 3 ( 5 5      x e x e x x x x x dx dy 2 cos 2 sin 2 . 3    x x x x dx dy 5 sin 3 5 cos 5 . 4 2 3   ) 5 sin 3 5 cos 5 ( 2 x x x x   x x x x x dx dy sinh 2 sinh cosh . 5 2   ) sinh 2 coth ( x x x x   2 ) 1 ( 3 sin 3 cos ) 1 ( 3 . 6     x x x x dx dy x x x x e x x e x e x e dx dy 2 4 2 2 ln 2 1 ln 2 . 7     4 2 2 cos 2 2 sin 2 . 8 x x x x x dx dy    3 ) 2 cos 2 sin ( 2 x x x x   
  • 9. 9 Logarithmic differentiation. The rules of differentiating a product or a quotient that we are revised are used when there are just two-factor functions, i.e. uv or u/v. where there are more than two functions in any arrangement top or bottom, the derivative is best found by what is known as logarithmic differentiation. For example; if where u, v, and w -and also y- are all functions of x. First take logs of base e w uv y  w v u y ln ln ln ln    dx dw w dx dv v dx du u dx dy y 1 1 1 1             dx dw w dx dv v dx du u y dx dy 1 1 1
  • 10. 10 Example 4: find dy/dx for the following. Solution : x x e y x e x y x x x y x x 2 cosh . 3 tan . 2 2 cos sin . 1 3 4 3 4 2                                   x x x x x x dx dy x x x x x x x x x dx dy dx dw w dx dv v dx du u y dx dy x w x v x u let 2 tan 2 cot 2 2 cos sin 2 cos ) sin 2 ( sin cos 2 2 cos sin 1 1 1 2 cos , sin , . 1 2 2 2 2                               x x x x e x dx dy x x e e x x x e x dx dy dx dw w dx dv v dx du u y dx dy x w e v x u let x x x x x tan sec 3 4 tan tan sec 3 4 tan 1 1 1 tan , , . 2 2 3 4 2 3 3 4 3 3 4 3 4                               x x x x e x x x x e e x x e dx dy dx dw w dx dv v dx du u y dx dy x w x v e u let x x x x x 2 tanh 2 3 4 2 cos 2 cosh 2 sinh 2 3 4 2 cos 1 1 1 2 cosh , , . 3 3 4 3 2 4 4 3 4 3 4
  • 11. 11 Parametric equations In same cases, it is convenient to represent a function by expressing x and y separately in terms of a third independent variable. E.g. y= cos2t, x=sint. The third variable is called parameter, and the two expression of x and y are called parametric equation. Example 5: find and for the following. Solution : 1. t t y t t x x y t x t y            1 2 3 , 1 3 2 . 3 cos , sin sin 3 . 2 sin , 2 cos . 1 3 3    dx dy 2 2 dx y d t dt dx t dt dy cos 2 sin 2    dx dt dt dy dt dt dx dy dx dy . .   t t dx dy cos 1 . 2 sin 2  
  • 13. 13          sin cos 3 ) sin 1 ( cos 3 sin cos 3 cos sin 3 cos 3 2 2 2 2       dx dy      cot sin cos 3 cos cos 3 2 2           dx dt dt d dx y d dt dt dx d dx y d dx d dx y d . cot . cot cot 2 2 2 2 2 2                  sin cos 3 cos sin cos 3 1 ). cos ( 2 2 2 2 2 2 ec ec dx y d       Solution : 3. 2 2 2 2 2 2 ) 1 ( 5 ) 1 ( 3 2 3 3 ) 1 ( ) 3 2 ( ) 1 ( 3 ) 1 ( 1 ) 1 ( 2 3 2 2 ) 1 ( ) 2 3 ( ) 1 ( 2 t t t t t t t dt dx t t t t t t t dt dy                               dx dt dt dy dx dy . 5 1 5 ) 1 ( . ) 1 ( 1 2 2       t t dx dy 0 5 1 2 2         dx d dx y d
  • 15. Exponential and logarithmic functions • Today’s Objective: When you have completed this lecture you will be able to: • Solve indicial and logarithmic equations. • Recognize that the exponential function and the natural logarithmic function are mutual inverses • Construct the hyperbolic functions from the odd and even parts of the exponential function. • Application of logarithms and exponential function. 15
  • 16. Introduction to logarithms If a number y can be written in the form , then the index x is called the ‘logarithm of y to the base of a’, Logarithms having a base of 10 are called common logarithms and is usually abbreviated to . 16 10 log x a y  y x then a y if a x log   lg Laws of logarithms There are three laws of logarithms, which apply to any base: B A A n A B A B A B A n log log ) log( . 3 log ) log( . 2 log log ) log( . 1      
  • 17. An indicial equation is an equation where the variable appears as an index and the solution of such an equation requires the application of logarithms. 17 Indicial equations Example 1: Find the value of X, give that 4 . 35 122  x Solution : 7177 . 0 0792 . 1 2 5490 . 1 5490 . 1 0792 . 1 ) 2 ( 4 . 35 log 12 log ) 2 ( 4 . 35 log 12 log 2         x x x x A n An log ) log( 
  • 18. 18 Example 2: Find the value of X, give that Solution : 694 . 6 3913 . 0 6192 . 2 2042 . 1 4150 . 1 ) 4150 . 1 8063 . 1 ( 4150 . 1 4150 . 1 2042 . 1 8063 . 1 4150 . 1 ) 1 ( 6021 . 0 ) 2 3 ( 26 log ) 1 ( 4 log ) 2 3 ( 26 log 4 log 1 2 3                   x x x x x x x x x x log(An )= nlogA 1 2 3 26 4    x x
  • 19. 19 Example 3: Find the value of X, give that Solution : 4853 . 2 5164 . 0 2834 . 1 2834 . 1 ) 0436 . 2 5600 . 2 ( 0436 . 2 2834 . 1 5600 . 2 0436 . 2 9138 . 0 8276 . 1 19726 . 2 7324 . 0 0436 . 2 9138 . 0 ) 1 2 ( 7324 . 0 ) 3 ( ) 6812 . 0 ( 3 2 . 8 log ) 1 2 ( 4 . 5 log ) 3 ( 8 . 4 log 3 2 . 8 log 4 . 5 log 8 . 4 log ) 2 . 8 4 . 5 log( 1 2 3 3 1 2 3                               x x x x x x x x x x x x x x x x x x x B A B A log log ) log(    x x x 3 1 2 3 8 . 4 2 . 8 4 . 5    
  • 20. 20 Example 4: Find the value of X, give that Solution : 5006 . 1 7676 . 2 1533 . 4 1533 . 4 7676 . 2 4683 . 2 6123 . 1 7766 . 5 1553 . 1 8450 . 0 294 log 4 . 6 log 2 3 . 14 log ) 5 ( 7 log 294 log } 4 . 6 ) 3 . 14 ( 7 log{ 2 5                   x x x x x x x x 294 4 . 6 ) 3 . 14 ( 7 2 5    x x Example 5: Solve the equation Solution : 2 1 4 2 2 2 4 0 ) 4 ( 2 0 ) 2 ( 0 ) 4 )( 2 ( 0 8 6 2 2                         x x z or z z z or z z z z z z z let x x x 0 8 2 6 22     x x
  • 21. Graphs of logarithmic functions 21 A graph of is shown in bellow x y 10 log  0 1 log  a   0 loga 1 log  a a In general, with a logarithm to any base a, it is noted that:
  • 22. Exponential functions The exponential function is expressed by the equation: or where e is the exponential number 2.7182818 _ _ _ . The graph of this function lies entirely above the x-axis as does the graph of its reciprocal , as can be seen in the diagram: 22 ) exp(x y  x e y  x e y  
  • 23. The value of can be found to any level of precision desired from the series expansion: In practice a calculator is used. Logarithms with base exponential called Napierian logarithms which is written as: So, 23 x e x x e ln log          ! 5 ! 4 ! 3 ! 2 1 5 4 3 2 x x x x x ex x e x e x x e    ln log . .......... ln 3   x e
  • 24. 24 Example 6: Solve for X, given that Solution : ) 1 ( 60 20 2 x e   810 . 0 405 . 0 2 405 . 0 2 2 3 ln ln 2 3 3 2 6 2 1 ) 1 ( 60 20 2 2 2 2 2                    x x x e e e e e x x x x x Example 7: Solve for X, given that Solution : 0 2 3 2    x x e e 0 ) 2 )( 1 ( 0 2 3 2          y y y y e y let x 0 1 ln 1 ln ln 1 1 0 ) 1 (         x x e e y y x x 693 . 0 2 ln 2 ln ln 2 2 0 ) 2 (         x x e e y y x x
  • 25. 25 Example 8: Solution : Ans. 0.5545s The current amperes flowing in a capacitor at time seconds is given by where and Determine : (a) the time for the current to reach 6.0 A. (b) Sketch the graph of current against time. ) 1 ( 8 CR t e i    i t    3 10 25 R f C 6 10 16    t
  • 26. 26 Odd and even parts Not every function is either even or odd but many can be written as the sum of an even part and an odd part. If, given f(x) where f( -x) is also defined then: is even and is odd Furthermore is called the even part of f(x) and is called the odd part of f(x). For example, if then So that the even and odd parts of are: 2 ) ( ) ( ) ( x f x f x fe    2 ) ( ) ( ) ( x f x f x fo    ) (x fe ) (x fo 1 2 3 ) ( 2    x x x f 1 2 3 1 ) ( 2 ) ( 3 ) ( 2 2          x x x x x f ) (x f x x x x x x f x x x x x x f o e 2 2 ) 1 2 3 ( ) 1 2 3 ( ) ( 1 3 2 ) 1 2 3 ( ) 1 2 3 ( ) ( 2 2 2 2 2                 So, the even and odd parts of the odd part of are……………… 4 3 2 ) ( 2 3     x x x x f
  • 27. 27 Odd and even parts of the exponential function The exponential function is neither odd nor even but it can be written as a sum of an odd part and an even part. That is, and . These two functions are known as the hyperbolic cosine and the hyperbolic sine respectively: and Using these two functions the hyperbolic tangent can also be defined: The logarithmic function is neither odd nor even and indeed does not possess even and odd parts because is not defined. 2 ) exp( ) exp( ) ( exp x x x e    2 ) exp( ) exp( ) ( exp x x x o    2 cosh x x e e x    2 sinh x x e e x    x x x x e e e e x      tanh x y a log  ) ( log x a 

Editor's Notes

  1. cos2x=cos↑2x-sin↑2x sin2x=2sinxcosx