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SECTION 5.7
HYPERBOLIC
FUNCTIONS
INVERSE FUNCTIONS
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.
5.7 P2
DEFINITION OF THE HYPERBOLIC FUNCTIONS
e x  e x
sinh x 
2
e e
cosh x 
2
x
x
sinh x
tanh x 
cosh x
1
csc h x 
sinh x
1
sec h x 
cosh x
cosh x
coth x 
sinh x
5.7 P3
HYPERBOLIC FUNCTIONS
The graphs of hyperbolic sine and cosine can be
sketched using graphical addition, as in Figures
1 and 2.
5.7 P4
HYPERBOLIC FUNCTIONS
Note that sinh has domain
whereas cosh has domain
and range ,
and range [1, ).
5.7 P5
HYPERBOLIC FUNCTIONS
The graph of tanh is shown in Figure 3.

It has the horizontal asymptotes y = ±1.
5.7 P6
APPLICATIONS
Applications of hyperbolic functions to science
and engineering occur whenever an entity such
as light, velocity, electricity, or radioactivity is
gradually absorbed or extinguished, for the
decay can be represented by hyperbolic
functions.
5.7 P7
APPLICATIONS
The most famous application is the use of
hyperbolic cosine to describe the shape of a
hanging wire.
5.7 P8
APPLICATIONS
It can be proved that, if a heavy flexible cable is
suspended between two points at the same
height, it takes the shape of a curve with
equation y = c + a cosh(x/a) called a catenary.


See Figure 4.
The Latin word catena
means ‘‘chain.’’
5.7 P9
HYPERBOLIC IDENTITIES
The hyperbolic functions satisfy a number of
identities that are similar to well-known
trigonometric identities.
5.7 P10
HYPERBOLIC IDENTITIES
sinh( x)   sinh x
cosh( x)  cosh x
cosh x  sinh x  1
2
2
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
5.7 P11
Example 1
 Prove (a) cosh2x – sinh2x = 1 (b) 1 – tanh2x =
sech2x
 SOLUTION (a)
2
2
e e  e e 
cosh x  sinh x  
 

2
2

 

e 2 x  2  e 2 x e 2 x  2  e 2 x


4
4
4
 1
4
x
2
x
x
x
2
5.7 P12
Example 1(b) SOLUTION
We start with the identity proved in (a) cosh2x –
sinh2x = 1

If we divide both sides by cosh2x, we get:
sinh 2 x
1
1

2
cosh x cosh 2 x
or 1  tanh 2 x  sec h 2 x
5.7 P13
HYPERBOLIC FUNCTIONS
The identity proved in Example 1(a) gives a
clue to the reason for the name ‘hyperbolic’
functions, as follows:
5.7 P14
HYPERBOLIC FUNCTIONS
If t is any real number, then the point P (cos t,
sin t) lies on the unit circle x2 + y2 = 1 because
cos2 t + sin2 t = 1.

In fact, t can be interpreted
as the radian measure of
POQ in Figure 5.
For this reason, the
trigonometric functions
are sometimes called
circular functions.
5.7 P15
HYPERBOLIC FUNCTIONS
Likewise, if t is any real number, then the point
P(cosh t, sinh t) lies on the right branch of the
hyperbola x2 – y2 = 1 because cosh2 t – sin2 t = 1
and cosh t ≥ 1.

This time, t does not represent the measure of an
angle.
5.7 P16
HYPERBOLIC FUNCTIONS
However, it turns out that t represents twice the
area of the shaded hyperbolic sector in Figure 6.

This is just as in the trigonometric case t represents
twice the area of the shaded circular sector in Figure
5.
5.7 P17
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
5.7 P18
DERIVATIVES OF HYPERBOLIC FUNCTIONS
We list the differentiation formulas for the
hyperbolic functions as Table 1.
d
(sinh x)  cosh x
dx
d
(cosh x)  sinh x
dx
d
(tanh x)  sec h 2 x
dx
d
(csc h x)   csc h x coth x
dx
d
(sec h x)   sec h x tanh x
dx
d
(coth x)  csc h 2 x
dx
5.7 P19
DERIVATIVES OF HYPERBOLIC FUNCTIONS
Note the analogy with the differentiation
formulas for trigonometric functions.

However, beware that the signs are different in
some cases.
d
(sinh x)  cosh x
dx
d
(cosh x)  sinh x
dx
d
(tanh x)  sec h 2 x
dx
d
(csc h x)   csc h x coth x
dx
d
(sec h x)   sec h x tanh x
dx
d
(coth x)  csc h 2 x
dx
5.7 P20
Example 2
Any of these differentiation rules can be
combined with the Chain Rule.
SOLUTION

For instance,
d
d
sinh x
(cosh x )  sinh x 
x
dx
dx
2 x
5.7 P21