Exponential and Logarithm functions :
Exponential functions : If a is a positive number and x is any number ,
we define the exponential function as :
y = ax
with domain : -∞ < x < ∞
Range : y > 0
The properties of the exponential functions are :
1. If a > 0 ↔ ax > 0 .
2. ax . ay = ax + y .
3. ax / ay = ax - y .
4. ( ax )y = ax.y .
5. ( a . b )x = ax . bx .
x
6. a y  a x  ( a ) x .
7. a-x = 1 / ax and ax = 1 / a-x .
8. ax = ay ↔ x = y .
9. a0 = 1 ,
a∞ = ∞ , a-∞ = 0 , where a > 1 .
a∞ = 0 , a-∞ = ∞ , where a < 1 .
The graph of the exponential function y = ax is :
y
y
Logarithm function : If a is any positive number other than 1 , then
the logarithm of x to the base a denoted by :
y = logax
where x > 0
At a = e = 2.7182828… , we get the natural logarithm and denoted by :
y = ln x
Let x , y > 0 then the properties of logarithm functions are :
1. y = ax ↔ x = logay and y = ex ↔ x = ln y .
2. logex = ln x .
3. logax = ln x / ln a .
١
4.
5.
6.
7.
8.
9.
ln (x.y) = ln x + ln y .
ln ( x / y ) = ln x – ln y .
ln xn = n. ln x .
ln e = logaa = 1 and ln 1 = loga1 = 0 .
ax = ex. ln a .
eln x = x .
The graph of the function y = ln x is :
y2
1.5
1
0.5
0
-0.5 0
1
2
e
X
3
4
5
-1
-1.5
-2
Application of exponential and logarithm functions :
We take Newton's law of cooling :
T – TS = ( T0 – TS ) et k
where T is the temperature of the object at time t .
TS is the surrounding temperature .
T0 is the initial temperature of the object .
k is a constant .
EX-1- The temperature of an ingot of metal is 80 oC and the room
temperature is 20 oC . After twenty minutes, it was 70 oC .
a) What is the temperature will the metal be after 30 minutes?
b) What is the temperature will the metal be after two hours?
c) When will the metal be 30 oC?
Sol. :
T  T S  ( T0  T S )e tk  50  60 e 20 k  k 
ln 5  ln 6
  0 .0091
20
a)
T  20  60 e 30 ( 0.0091 )  60 * 0.761  45 .6 o C  T  65.6 o C
b)
T  TS  60 e 120 ( 0.0091 )  60 * 0.335  20.1 o C  T  40 .1 o C
c)
10  60 e 0.0091 t  0.0091 t   ln 6  t  3.3 hrs .
٢
Exponential functions : If u is any differentiable function of x , then :
d u
du
a  a u .ln a .
dx
dx
EX-2 –Find
d u
du
e  eu .
dx
dx
and
dy
for the following functions :
dx
a ) y  2 3x
b) y  2 x .3 x
c) y  ( 2 x ) 2
d ) y  x.2 x
5x
e) y  e (x  e )
f ) ye
2
1 5x 2
Sol.dy
 2 3 x * 3 ln 2
dx
dy
b ) y  2 x .3 x  y  6 x 
 6 x . ln 6
dx
dy
c ) y  (2 x )2  y  2 2 x 
 2 2 x ln 2.2  2 2 x  1 ln 2
dx
2
2
2
2
dy
d ) y  x.2 x 
 x.2 x ln2.2x  2 x  2 x (2x 2 ln2  1)
dx
5x
dy
(x  e 5x )
e) ye

 e (x  e ) ( 1  5 e 5 x )
dx
1
1
1

2 2
2 2 1
2
dy
f ) y  e (1 5x ) 
 e (1 5x ) ( 1  5 x 2 ) 2 .10 x  e 1 5 x
dx
2
a ) y  2 3x 
5x
1  5x2
Logarithm functions : If u is any differentiable function of x , then :
8)
d
1 du
loga u 
.
dx
u .ln a dx
EX-3 – Find
d
1 du
ln u  .
dx
u dx
and
dy
for the following functions :
dx
a ) y  log10 e x
b ) y  log 5 ( x  1 )2
c ) y  log 2 ( 3 x 2  1 )3
d ) y  ln(x 2  2 )2
e ) y  ln(xy)  1
(2x 3  4 ) 3 .( 2 x 2  3 ) 2
f) y 
( 7 x 3  4 x  3 )2

2
3

3
5
Sol. –
dy
 log 10 e  ln e  1
dx
ln 10 ln 10
dy
2
b ) y  log 5 ( x  1 )2  2 log 5 ( x  1 ) 

dx ( x  1 ) ln 5
dy
3
18 x
c ) y  3 log 2 ( 3 x 2  1 ) 
. 6x 

dx 3 x 2  1 ln 2 ( 3 x 2  1 ) ln 2
2
2
dy
48 x ln( x 2  2 )
2
2
d)
 3 2 ln( x  2 )
.2 x 
dx
x2  2
x2  2
a ) y  log 10 e x  y  x log 10 e 



dy 1 1 dy
dy
y
  . 0

dx x y dx
dx
x( y  1 )
2
5
f ) lny  ln( 2 x 3  4 )  ln( 2 x 2  3 )  2 ln( 7 x 3  4 x  3 )
3
2
2
1 dy 2 6 x
5
4x
21 x 2  4
 .  . 3
 . 2
 2. 3
y dx 3 2 x  4 2 2 x  3
7 x  4x  3
2

dy
2x
5x
21 x 2  4 

 2 y 3



2
3
dx
2x  4 2x  3 7 x  4x  3
e ) y  lnx  lny  1 
4

Hyperbolic functions : Hyperbolic functions are used to describe
the motions of waves in elastic solids ; the shapes of electric power lines
; temperature distributions in metal fins that cool pipes …etc.
The hyperbolic sine (Sinh) and hyperbolic cosine (Cosh) are defined
by the following equations :
4.
1 u
1
( e  e  u ) and Coshu  ( e u  e  u )
2
2
Sinhu e u  e  u
Coshu e u  e  u
tanh u 

and Cothu 

Coshu e u  e u
Sinhu e u  e  u
1
2
1
2
Sechu 
 u
and Cschu 
 u
u
Coshu e  e
Sinhu e  e u
2
2
Cosh u  Sinh u  1
5.
tanh 2 u  Sech 2 u  1
1. Sinhu 
2.
3.
and
Coth 2 u  Csch 2 u  1
Coshu  Sinhu  e u
and Coshu  Sinhu  e  u
Cosh(  u )  Coshu
and Sinh(  u )   Sinhu
Cosh0  1
and Sinh0  0
Sinh( x  y )  Sinhx .Coshy  Coshx .Sinhy
Cosh( x  y )  Coshx .Coshy  Sinhx .Sinhy
Sinh 2 x  2.Sinhx .Coshx
Cosh 2 x  Cosh 2 x  Sinh 2 x
Cosh 2 x  1
Cosh 2 x  1
13. Cosh 2 x 
and Sinh 2 x 
2
2
6.
7.
8.
9.
10.
11 .
12.
y=Sinhx
y=Cschx
y=Cschx
0
5
y
y=Coshx
1
1
1
1
1
y=Sechx
0
0
0
0
y=Cothx
1
y=tanhx
0
y=Cothx
y  Sinhx
y  Coshx
y  tanh x
y  Cothx
y  Sechx
y  Cschx
-1
D x : x
and
D x : x
and
D x : x
and
D x : x  0 and
D x : x
and
D x : x  0 and
R y : y
Ry : y  1
R y : 1  y  1
R y : y  1 or y  1
Ry : 0  y  1
R y : y  0
EX-1 - Let tanh u = - 7 / 25 , determine the values of the remaining five
hyperbolic functions .
Sol.-
6
Cothu 
1
25

tanh u
7
tanh 2 u  Sech 2 u  1
1
25

Sechu 24
Sinhu
tanh u 
Coshu

49
24
 Sech 2 u  1  Sechu 
625
25
Coshu 
Cschu 

7
Sinhu
7

 Sinhu  
25
25
24
24
1
24

Sinhu
7
EX-2 - Rewrite the following expressions in terms of exponentials .
Write the final result as simply as you can :
a ) 2Cosh(ln x )
b ) tanh(ln x )
c ) Cosh5 x  Sinh5 x d ) ( Sinhx  Coshx )4
Sol.e ln x  e  ln x
1
a ) 2Cosh(ln x )  2.
 x
2
x
1
x
2
e ln x  e  ln x
x  x 1
b ) tanh(ln x )  ln x

1 x2  1
e  e  ln x
x
x
5 x
5x
e e
e 5 x  e 5 x
c ) Cosh5 x  Sinh5 x 

 e5x
2
2
4
x
x
x
e e
e  ex 
4
  e 4 x
d ) ( Sinhx  Coshx )  

2
2


EX-3 - Solve the equation for x : Cosh x = Sinh x + 1 / 2 .
1
1
Sol. - Coshx  Sinhx   e  x    x  ln 1  ln 2  x  ln 2
2
2
EX-4 – Verify the following identity :
a) Sinh(u+v)=Sinh u. Cosh v + Cosh u.Sinh v
b) then verify Sinh(u-v)=Sinh u. Cosh v - Cosh u.Sinh v
Sol.7
a ) R .H .S .  Sinhu .Coshv  Coshu .Sinhv
e u  e u e v  e v e u  e u e v  e v
.


2
2
2
2
u v
( u v )
e e

 Sinh( u  v )  L .H .S .
2
b ) L .H .S .  Sinh( u  (  v ))  Sinhu .Cosh(  v )  Coshu .Sinh(  v )
 Sinhu .Coshv  Coshu .Sinhv  R .H .S .
EX-5 – Verify the following identities :
1
a)
Sinhu .Coshv  Sinh( u  v )  Sinh( u  v )
2
1
b ) Coshu .Coshv  Cosh( u  v )  Cosh( u  v )
2
3
c ) Sinh 3 u  Sinh u  3Cosh 2 u .Sinhu  3 Sinhu  4 Sinh 3 u
d ) Sinh 2 u  Sinh 2 v  Cosh 2 u  Cosh 2 v
Sol. –
1
a ) R .H .S .
 Sinh( u  v )  Sinh( u  v )
2
1
 Sinhu .Coshv  Coshu .Sinhv  Sinhu .Coshv  Coshu .Sinhv 
2
 Sinhu .Coshv  L .H .S .
1
b ) R .H .S .
 Cosh( u  v )  Cosh( u  v )
2
1
 Coshu .Coshv  Sinhu .Sinhv  Coshu .Coshv  Sinhu .Sinhv 
2
 Coshu .Coshv  L .H .S .
c ) L .H .S .
 Sinh( 2 u  u )  Sinh 2 u .Coshu  Cosh2 u .Sinhu
 2 Sinhu .Coshu .Coshu  ( Cosh 2 u  Sinh 2 u ).Sinhu
 3 Sinhu .Cosh 2 u  Sinh 3 u  R .H .S .( I )
 3 Sinhu .( 1  Sinh 2 u )  Sinh 3 u  3 Sinhu  4 Sinh 3 u  R .H .S .( II )
d ) L .H .S .
 Sinh 2 u  Sinh 2 v  Cosh 2 u  1  ( Cosh 2 v  1 )
 Cosh 2 u  Cosh 2 v  R .H .S .
Inverse hyperbolic functions : All hyperbolic functions have
inverses that are useful in integration and interesting as differentiable
functions in their own right .
8
0
y  Sinh 1 x
1
y  Cosh 1 x
D x : x
R y : y
-١
-1
١
y  tanh1 x Dx : 1  x  1
R y : y
D x : x  1
R y : y  0
١
y  Coth1 x Dx : x  1 or x  1
Ry : y  0
١
y  Sech 1 x
y  Csch1 x Dx : x  0
R y : y  0
Dx : 0  x  1
R y : y  0
9
Some useful identities :
Sinh 1 x  ln( x  x 2  1 )
Cosh 1 x  ln( x  x 2  1 )
1 1 x
3. tanh 1 x  . ln

2  1 x 
1  x  1
1 1
4. Coth 1 x  . ln
  tanh
2  x  1
x
2
1 1 x 
  Cosh 1 1
5. Sech1 x  ln


x
x


1.
2.
1
6. Csch x  ln 
x

1
x 2  1 
1
 Sinh 1

x
x

EX-6 - Derive the formula :
Sinh  1 x  ln( x  x 2  1 )
Sol.e y  e y
e2y  1
Let y  Sinh x  x  Sinhy 
 x
2
2e y
 e 2 y  2 x .e y  1  0
2x  4 x2  4
y
e 
 e y  x  x2  1
2
2
either y  ln( x  x  1 ) neglected sin ce x  x 2  1  0
1
or
y  ln( x 
x2  1 )
10