MATH 206 JAMES STEWART

CALCULUS
METRIC VERSION, 8 E
Early Transcendentals
SECTION 3.11
HYPERBOLIC FUNCTIONS

Examples: 1 – 5
Exercises:10, 12, 16, 32, 34, 37, 40
Figures: 1, 2, 3, 8, 9 and 10.
Introduction

• Certain combinations of the exponential functions ex and

e–x arise so frequently in mathematics and its
applications that they deserve to be given special names.

• In many ways they are analogous to the trigonometric

functions, and they have the same relationship to the
hyperbola that the trigonometric functions have to the
circle. For this reason they are collectively called
hyperbolic functions and individually called
hyperbolic sine, hyperbolic cosine, and so on.

P3
Definition of the Hyperbolic Functions

e x  e x 1
sinh x  csc h x 
2 sinh x
e x  e x 1
cosh x  sec h x 
2 cosh x
sinh x cosh x
tanh x  coth x 
cosh x sinh x

P4
Figures: 1, 2, 3, 8, 9 and 10.

P5
SUMMARY:HYPERBOLIC FUNCTIONS

P6
Hyperbolic Identities

The hyperbolic functions satisfy a number of

identities that most are similar to well-known
trigonometric identities.

sinh(  x )   sinh x
cosh(  x )  cosh x
cosh 2 x  sinh 2 x  1
1  tanh 2 x  sech 2 x
sinh( x  y )  sinh x cosh y  cosh x sinh y
cosh( x  y )  cosh x cosh y  sinh x sinh y
P7
Example 1
• Prove that

• (a) cosh2x – sinh2x = 1

• (b) 1 – tanh2x = sech2x

• Solution (a)
2 2
e e  e e 
x x x x
cosh x  sinh x  
2 2
  
 2   2 
e 2 x  2  e 2 x e 2 x  2  e 2 x
 
4 4
4
 1
P8
4
Example 1(b)

Solution (b)

2
sinh x
1  tanh x  1 
2
2
cosh x
cosh 2 x  sinh 2 x

cosh 2 x
1

cosh 2 x
 sech 2 x
P9
 Derivatives of Hyperbolic Functions

• The derivatives of the hyperbolic functions are easily

computed.
 For example,

d d  e x  e x  e x  e x
(sinh x )     cosh x
dx dx  2  2
d d e e x x
 e e
x x
(cosh x )     sinh x
dx dx  2  2
P10
 Derivatives of Hyperbolic Functions

d d
(sinh x )  cosh x (csc h x )   csc h x coth x
dx dx

d d
(cosh x )  sinh x (sec h x )   sec h x tanh x
dx dx

d d
(tanh x )  sec h 2 x (coth x )   csc h 2 x
dx dx P11
 Recall: (Chain Rule) d
f ( g (x )  f ( g (x )) g (x )
•

dx

Example 2
Find
d 
dx 
cosh  x 
• Solution:

d d
(cosh x )  sinh x x
dx dx
1  12
 sinh x x
2
d sinh x
(cosh x ) 
dx 2 x

P12
Inverse Hyperbolic Functions

y  sinh 1 x  sinh y  x
y  cosh 1 x  cosh y  x and y  0
1
y  tanh x  tanh y  x


sinh 1 x  ln x  x 2  1  x

cosh 1
x  ln  x  x 2
 1 x 1

1
1 1 x 
tanh x  ln    1 x 1
2 1 x  P13
INVERSE HYPERBOLIC FUNCTIONS

P14
 Derivatives of Inverse Hyperbolic Functions
d 1 1 d 1 1
(sinh x )  (csc h x )  
dx 1 x 2 dx x x 12

d
 cosh x  
1 1 d 1
(sec h x ) 
1
dx x 1
2 dx x 1 x 2

d 1 1 d 1 1
(tanh x )  (coth x ) 
dx 1 x 2
dx 1 x 2

P15
Example 3
•
1
Show that sinh x  ln x  x 1
2
 
Let y  sinh 1 x , then x  sinh y .Thus
• Solution e y e y
x 
2
2xe y  e 2 y  1
e 2 y  2xe y  1  0

e  y 2
 
 2x e y  1  0

solve the above equation using quadratic formula for e y

here, a  1, b  2x , c  1
b  b 2  4ac 2x  4x 2  4
e y
 
2a 2
e y  x  x 2  1,
e y  x  x 2 1 x - x 2  1 <0 but e y  0

y  ln x  x 2  1  P16
Example 4
• Prove that
d
 sinh 1
x  
l
dx 1 x 2

• Solution

Let y  sinh 1 x
x  sinh  y 
differentiating implicitly with respect to x , we get
dy
1  cosh  y 
dx
dy 1

dx cosh  y 
dy 1

dx 1 sinh 2 y
dy 1

dx 1 x 2 P17
 Example 5:
d
• Find  tanh 1  sin x  
dx

• Solution

d 1 d
 tanh (sin x )  
1
(sin x )
dx 1  (sin x ) dx
2

1
 cos x
1  sin x
2

cos x
 2
 sec x
cos x
P17
 Exercise 10:
Prove the identity cosh x  sinh x  e
x
•

• Solution

e x  e x e x  e x
cosh x  sinh x  
2 2
e x  e x  e x  e x

2
2 e x

2
 e x

P18
 Exercise 12:
• Prove the identity cosh  x  y   cosh x cosh y  sinh x sinh y

• Solution

cosh x cosh y  sinh x sinh y

e x  e x e y  e  y e x  e x e y  e  y
   
2 2 2 2
e x e y  e x e  y  e x e y  e x e  y e x e y  e x e  y  e x e y  e x e  y
 
4 4
2e  
 2e  
x y x y

4
e e   
x y  x y

2
 cosh( x  y )

P19
Exercise 16:
Prove the identity cosh 2x  cosh x  sinh x
2 2
•

• Solution  x

2
 x

2
e x
 e e x
 e
cosh 2 x  sinh 2 x     
 2   2 
 2x 1 2 x   2x 1 2 x 
 e  2 e x
x
 e   e  2e x
x
 e 
 e e
  
 4   4 
   
 e 2 x  2  e 2 x   e 2 x  2  e 2 x 
   
 4   4 
 e 2 x  2  e 2 x  e 2 x  2  e 2 x 
 
 4 
 2e 2 x  2e 2 x 
 
 4 
 e 2 x  e 2 x 
 
 2 
P21
 cosh 2x
Recall:  Power Rule 
d n 1 d
f  x    n f  x   f x 
n
dx dx

Exercise 32
• Find the derivative of y  sin 2 

• Solution
y  sin 2 
dy d
  sin   2
dx dx
d
 2  sin    sin  
dx
 2sin  cos
P22
Exercise 34:
• Find the derivative of h  x   ln  cosh x 

• Solution

h  x   ln  cosh x 
1 d
h x  
'
 cosh x 
cosh x dx
1
 sinh x
cosh x
sinh x

cosh x
 h '  x   tanh x

P22
Recall:
d
dx
ef  (x )
 e f (x ) d
dx
f (x )

Exercise 37:
• Find the derivative of y  e cosh(3x )

• Solution
y  e cosh(3x )
dy

dx dx

d cosh(3x )
e 
cosh(3x ) d
e  cosh(3x ) 
dx
d
 e cosh(3x ) sinh(3x )  3x 
dx
 3e cosh(3x ) sinh(3x )

P24
Exercise 40
• Find the derivative of y  sinh  cosh x 

• Solution

y  sinh  cosh x 
dy d
  sinh(cosh x ) 
dx dx
d
 cosh(cosh x )  cosh x 
dx
 cosh(cosh x )sinh x

P25