1.4 Hyperbolic Functions
The hyperbolic functions are combinations of exponential functions that have many similarities to the
trigonometric functions. In fact, for almost every formula involving a trigonometric function there is an
analogous formula involving a hyperbolic function. The hyperbolic functions are useful for doing integrals
and they also arise is the solution of problems involving mechanical vibrations and electric circuits where
one frequently encounters combinations of exponentials such as 2e3x – 5e-2x. For certain calculations it is
convenient to express such combinations in terms of the two special combinations
ex - e-x
2
ex + e-x
2
and
These two special combinations are the first two hyperbolic functions. The remaining four are related to
these in the same way as the remaining trigonometric functions are related to the sine and cosine.
(1)
hyperbolic sine of x
= sinh x =
(2)
hyperbolic cosine of x
= cosh x =
(3)
hyperbolic tangent of x
= tanh x =
(4)
hyperbolic cotangent of x = coth x =
(5)
hyperbolic secant of x
= sech x =
(6)
hyperbolic cosecant of x
= csch x =
ex - e-x
2
ex + e-x
2
sinh x
=
cosh x
1
=
tanh x
1
=
cosh x
1
=
sinh x
ex - e-x
1 - e-2x
=
ex + e-x
1 + e-2x
cosh x
ex + e-x
= x -x
sinh x
e -e
2
ex + e-x
2
ex - e-x
Example 1.
sinh 2 =
e2 - e-2
7.38905 - 0.13533

 3.62686
2
2
Here are their graphs. As we go along we will explain why these graphs have the shape they do.
x
x
sinh y
cosh y
x
x
x
tanh y
x
10
10
1.0
8
0.5
5
6
3
2
1
1
2
3
y
5
10
3
4
2
1
1
1
0.5
2
3
2
1
2
1.4.1 - 1
3
y
1.0
2
3
y
We are particularly interested in the derivatives of the hyperbolic functions and integrals involving the
hyperbolic functions. The following table contains the derivatives of the hyperbolic functions. The
formulas are similar to the trigonometric functions, but the sign is opposite in (8) and (11).
d
sinh x
dx
d
cosh x
dx
d
tanh x
dx
d
coth x
dx
d
sech x
dx
d
csch x
dx
(7)
(8)
(9)
(10)
(11)
(12)
= cosh x
= sinh x
= sech2 x
compare
with
= - csch2 x
= - tanh x sech x
= - coth x csch x
d
sin x
dx
d
cos x
dx
d
tan x
dx
d
cot x
dx
d
sec x
dx
d
csc x
dx
= cos x
= - sin x
= sec2 x
= - csc2 x
= tan x sec x
= - cot x csc x
In order to prove these derivative formulas we will need the following formulas which are similar to the
corresponding formulas for the trigonometric functions.
(13)
cosh2x – sinh2x = 1
(14)
sech2x + tanh2x = 1
(15)
coth2x – csch2x = 1
compare
with
cos2x + sin2x = 1
sec2x - tan2x = 1
cot2x – csc2x = 1
Proposition 1. (13) – (15) hold.
Proof. To prove (13) one has
cosh2x – sinh2x =
x
-x 2
x
-x 2
2x
-2x
2x
-2x
e + e  - e - e  = e + 2 + e - e - 2 + e = 1
4
4
 2 
 2 
To prove (14), divide (13) by cosh2x. To prove (15), divide (13) by sinh2x. //
Proposition 2. (7) – (12) hold.
Proof. To prove (7) and (8) one has
d
d ex - e-x
sinh x =
=
dx
dx  2 
d
d ex + e-x
cosh x =
=
dx
dx  2 
x
-x
e + e  = cosh x
 2 
x
-x
e - e  = sinh x
 2 
Formulas (9) – (12) follow from (7), (8), (14) and (15) and the rules for derivatives. //
1.1 - 2
If we combine the formulas for the derivatives of the
hyperbolic functions with the chain rule we get the
formulas at the right which are useful for the situation
where what is inside a hyperbolic is an expression u(x)
(16)
(17)
(18)
involving x instead of just x.
(19)
dy
Example 2. Find if
dx
(20)
a.
y = cosh( ln x )
b.
y = tan-1( sinh x )
c.
y = tanh( 1 + e2x )
d.
y = sinh( sin-1 x )
e.
y = coth( sec-1 x )
(21)
d
sinh u(x)
dx
d
cosh u(x)
dx
d
tanh u(x)
dx
d
coth u(x)
dx
d
sech u(x)
dx
d
csch u(x)
dx
du
dx
du
= sinh u(x)
dx
du
= sech2 u(x)
dx
du
= - csch2 u(x)
dx
= cosh u(x)
du
dx
du
= - coth u(x) csch u(x)
dx
= - tanh u(x) sech u(x)
dy
d
d
sinh( ln x )
=
(cosh( ln x )) = sinh( ln x ) ( ln x ) =
. For b, using the
dx
dx
dx
x
dy
d
1
d
 cosh x 
chain rule one has
=
(tan-1( sinh x )) = 
dx
dx
1 + sinh2x dx ( sinh x ) = 1 + sinh2x. For c, using (18)
dy
d
d
one has
=
(tanh( 1 + e2x )) = sech2( 1 + e2x ) ( 1 + e2x ) =
dx
dx
dx
(22)

 cosh x dx = sinh x
2e2x sech2( 1 + e2x ).
(23)

 sinh x dx = cosh x
Each of the differentiation formulas (7) – (12) has a corresponding
2
(24)

 sech x dx = tanh x
integration formula. These are given in the table at the right.
For a, using (17) one has
cosh( x)
Example 3. Find 
dx

x
Let u =
x = x1/2. Then du = (1/2)x-1/2 dx =
the integral becomes 

cosh( x)
dx =
x
dx
2
x
. So 2 du =
dx
. So
x
(25)
2

 csch x dx = - coth x
(26)
 tanh x sech x dx = - sech x

(27)
 coth x csch x dx = - csch x

 2 cosh u du = 2 sinh u

= 2 sinh( x).
Problem 1. Find
c.
a.
2 2

 s sech (x ) dx

 sinh(1 + 4x) dx
d.
b.

 tanh x dx
cosh x
dx

 (sinh x) e
Now let's consider why the graphs of the hyperbolic functions look the way they do. First we consider
positive x.
1.1 - 3
First, note that sinh x > 0 for x > 0 since ex > e-x. Also cosh x > 0 for x > 0 since ex and e-x are positive for
all x. It follows that the remaining four hyperbolic functions are positive for positive x.
Second, recall that a function with a positive derivative is increasing and a function with a negative
derivative is decreasing. So, it follows from the derivative formulas that sinh x, cosh x and tanh x are
increasing for x > 0 and coth x, sech x and csch x are decreasing for x > 0.
Third, note that as x increases e-x goes to zero. In fact e-4  0.02. So by the time x  4 one has sinh x 
ex
2
ex
and coshh x  . In particular, sinh x and cosh x approach  as x approaches . On the other hand, it
2
1 - e-2x
follows from tanh x =
that tanh x  1 as x  . From the reciprocal formulas it follows that
1 + e-2x
coth x  1 and sech x  0 and csch x  0 as x  .
The graphs for negative x follow from the graphs for positive x and the fact that sinh x, tanh x, coth x and
csch x are odd functions and cosh x and sech x are even functions.
Here are the addition formulas for sinh x and cosh x. Note their similarity to the corresponding formulas for
sin x and cos x.
Proposition 3. For all x and y one has
(28)
sinh(x + y) = sinh x cosh y + cosh x sinh y
(29)
cosh(x + y) = cosh x cosh y + sinh x sinh y
sin(x + y) = sin x cos y + cos x sin y
compare
with
cos(x + y) = cos x cos y - sin x sin y
To prove (28) one has
sinh x cosh y + cosh x sinh y =
=
x
-x
y
-y
x
-x
y
-y
e - e e + e  + e + e e - e 
 2  2   2  2 
ex+y + ex-y - ey-x - e-x-y
ex+y - ex-y + ey-x - e-x-y
ex+y - e-x-y
+
=
= sinh(x + y)
4
4
2
To prove (29) one has
cosh x cosh y + sinh x sinh y =
=
x
-x
y
-y
x
-x
y
-y
e + e e + e  + e - e e - e 
 2  2   2  2 
ex+y + ex-y + ey-x + e-x-y
ex+y - ex-y - ey-x + e-x-y
ex+y + e-x-y
+
=
= cosh(x + y)
4
4
2
The following double angle formulas are a corollary. They will be useful when we do integrals.
Proposition 4. For all x one has
(30)
sin 2x = 2 sin x cos y
sinh 2x = 2 sinh x cosh x
compare
with
1.1 - 4
cos 2x = cos2 x - sin2 x
sin2x =
cos2x =
1 - cos 2x
2
1 + cos 2x
(31)
cosh 2x = cosh2 x + sinh2 x
(32)
sinh2x =
cosh 2x - 1
2
(33)
cosh2x =
cosh 2x + 1
2
Proof. (30) and (31) follow from (28) and (29) by replacing y by x. (32) and (33) follow from (31) and the
identity cosh2x – sinh2x = 1. //
Example 4. A cable hangs from two poles that are 40 feet tall and separated by 100 feet. In the middle the
cable is 38 feet off the ground. Find
a.
A formula for the height of the cable above the ground for a general point between the two cables.
b.
The angle the cable makes with the horizontal where it meets the pole.
Solution. Choose a coordinate system with the x axis lying along the ground, the y axis pointing upward
and half way between the poles. Then be base of the poles are at (- 50, 0) and (50, 0) and the tops of the
poles are at (- 50, 40) and (50, 40). The midpoint of the cable is at (0, 38). See picture at right.
In more advanced books it is shown that the equation of the cable is
x
y = c + a cosh 
a
where c and a are numbers depending of heights of the poles and the distance between them. Plugging in
x = 0 and y = 38 we get
38 = c + a
Plugging in x = 50 and y = 40 we get
50
40 = c + a cosh 
a
Subtracting these two equations we get
50
a cosh  - a = 2
a
If we let u =
50
then this becomes
a
cosh u = 1 +
u
25
1.1 - 5
This is an equation that we can't solve algebraically, so we need to solve it numerically. Using some
50
mathematical software (e.g. www.wolframalpha.com), we get u  0.08000. So a 
 625 and
0.08000
1
c  38 – 625 = -587. Since  0.0016, the equation of the cable is approximately
a
y = 625 cosh(0.0016x) - 587
We need to find
dy
. One has
dx x = 50
|
dy
d
x
x 1
x
=
(c + a cosh ) = a sinh    = sinh 
dx
dx
a
a a
a
dy
50
= sinh  = sinh u = sinh(0.08) = 0.08004
dx x = 50
a
|
dy
 = tan-1 dx |x = 50 = tan-1(0.08004) = 0.0798 radians = 4.58
Inverse Hyperbolic Functions
The inverse hyperbolic functions are formed from the hyperbolic functions in the same way the inverse
trigonometric functions are formed from the regular trigonometric functions. The only difference is the
restrictions one places on the hyperbolic functions to make them one-to-one. To summarize
(34)
y = sinh-1 x = that number y such that sinh y = x
(35)
y = cosh-1 x = that number y such that y  0 and cosh y = x
(36)
y = tanh-1 x = that number y such that tanh y = x
(37)
y = coth-1 x = that number y such that coth y = x
(38)
y = sech-1 x = that number y such that y  0 and sech y = x
(39)
y = csch-1 x = that number y such that csch y = x
As with the inverse trigonometric functions, we get the graphs of the inverse hyperbolic functions by
reflecting the graphs of the hyperbolic functions across the
line y = x. Their graphs are the following.
As opposed to the inverse trigonometric functions, there
(40)
sinh-1 x = ln(x +
x2 + 1)
(41)
cosh-1 x = ln(x +
x2 - 1)
(42)
tanh-1 x =
1 1+x
ln
2  1-x 
(43)
coth-1 x =
1  x +1
ln
2  x - 1
(44)
sech-1 x = ln
(45)
csch-1 x = ln
are formulas for the inverse hyperbolic involving
logarithms. At the right are these formulas.
1.1 - 6
1 + 1 - x2
x


1 + x2 + 1
x


Example 5. Show how to obtain the formula sinh-1 x = ln(x +
x2 + 1).
Solution. We use the fact that y = sinh-1x is that number y such that sinh y = x. Using the definition of
1
1
ey - y
uy
-y
e
u
e -e
1
sinh y this gives x =
=
. Now let u = ey. So x =
and 2x = u - . This implies
2
2
2
u
u2 - 2xu - 1 = 0. Using the quadratic formula, this gives u =
sign since u > 0. So ey = u = x +
2x 
4x2 +4
=x
2
x2 +1. Taking ln of both sides gives y = ln(x +
x2 +1. We choose the +
x2 + 1).
In calculus one of the main applications of hyperbolic functions is to do certain integrals. In order to see
how they are used, we first have to look at the derivative formulas for the inverse hyperbolic functions.
1
1 + x2
1
=
x2 - 1
1
=
1 - x2
1
=
1 - x2
-1
=
x 1 - x2
-1
=
| x| 1 + x2
1
1 - x2
-1
=
1 - x2
1
=
1 + x2
-1
=
1 - x2
1
=
x x2 - 1
-1
=
x x2 - 1
(46)
d
sinh-1 x =
dx
d
sin-1 x =
dx
(47)
d
cosh-1 x
dx
d
cos-1 x
dx
(48)
d
tanh-1 x
dx
(49)
d
coth-1 x
dx
(50)
d
sech-1 x
dx
(51)
d
csch-1 x
dx
Example 6. Show that
d
sinh-1 x =
dx
compare
with
d
tan-1 x
dx
d
cot-1 x
dx
d
sec-1 x
dx
d
csc-1 x
dx
1
1 + x2
Solution. Let y = sinh-1x so that x = sinh y. Use implicit differentiation. Take the derivative of both sides
dy
dy
1
of this last formula with respect to x to get 1 = (cosh y) . Thus =
. Using cosh2x – sinh2x = 1
dx
dx cosh y
dy
1
1
gives =
=
.
2
dx
1 + sinh y
1 + x2
1.1 - 7