Arkansas Tech University
MATH 2914: Calculus I
Dr. Marcel B. Finan
3.6
Hyperbolic Functions
The hyperbolic functions possess many parallels with the trigonometric functions, both in naming and in their properties and inter-relationships.
The function ex and its reciprocal e−x can be combined to create the following new functions:
Hyperbolic Cosine
cosh (x) =
ex + e−x
.
2
sinh (x) =
ex − e−x
.
2
Hyperbolic Sine
Hyperbolic Tangent
tanh (x) =
ex − e−x
sinh (x)
=
.
ex + e−x
cosh (x)
Hyperbolic Cosecant
csch(x) =
1
sinh (x)
sech(x) =
1
cosh (x)
Hyperbolic Secant
Hyperbolic Cotangent
coth (x) =
cosh (x)
1
=
.
tanh x
sinh (x)
Example 3.6.1
Find cosh (0) and sinh (0).
Solution.
Substituting 0 for x in the formulas for cosh x and sinh x to obtain
cosh (0) =
and
sinh (0) =
e0 + e−0
1+1
=
=1
2
2
e0 − e−0
1−1
=
=0
2
2
1
Hyperbolic Functions Identities
Hyperbolic functions have similar identities than that of trigonometric functions.
Substituting −x for x in the formula of cosh (x) to obtain
cosh (−x) =
e−x + ex
= cosh (x).
2
This says that cosh (x) is an even function. Since sech(x) = cosh1 (x) , the
function sech(x) is also even.
Similarly, substituting −x for x in the formula of sinh (x) to obtain
sinh (−x) =
e−x − ex
= − sinh (x).
2
That is, sinh (x) is an odd function. One can easily show, that tanh (x), csch(x),
and coth (x) are odd functions.
Next, we derive an identity for the hyperbolic functions similar to the
Pythagorean identity for the trigonometric functions.
x
2 x
2
e + e−x
e − e−x
2
2
cosh (x) − sinh (x) =
−
2
2
2x
−2x
2x
e +2+e
e − 2 + e−2x
=
−
= 1.
4
4
Thus, (cosh (x), sinh (x)) is a point on the hyperbola x2 − y 2 = 1, thus,
the name hyperbolic. (Recall that the functions sin x and cos x are called
circular functions since (cos x, sin x) is a point on the unit circle!)
Simple calculations show the following identities
1 − tanh2 (x) =sech2 (x)
coth2 (x) − 1 =csch2 (x)
sinh (x + y) = sinh (x) cosh (y) + cosh (x) sinh (y)
sinh (x − y) = sinh (x) cosh (y) − cosh (x) sinh (y)
cosh (x + y) = cosh (x) cosh (y) + sinh (x) sinh (y)
cosh (x − y) = cosh (x) cosh (y) − sinh (x) sinh (y).
Derivatives of Hyperbolic Functions
Since (ex )0 = ex and (e−x )0 = −e−x , we have
d
d ex + e−x
ex − e−x
(cosh (x)) =
=
= sinh (x).
dx
dx
2
2
2
dy
Now, if y = cosh (u(x)) then by the chain rule, we have dx
= sinh (u) ·
Similarly,
ex + e−x
d
d ex − e−x
=
(sinh (x)) =
= cosh (x)
dx
dx
2
2
du
dx .
and
d
du
(sinh (u(x))) = cosh (u(x)) .
dx
dx
To find the derivative of tanh (x), we use the quotient rule
1
d sinh (x)
cosh2 (x) − sinh2 (x)
d
=
.
(tanh (x)) =
=
2
dx
dx cosh (x)
cosh (x)
cosh2 (x)
Also,
d
du
(tanh (u(x))) = csch2 (u(x)) ·
.
dx
dx
We leave it to the reader to show the following derivatives
d
(sech(x)) = sech(x) tanh (x)
dx
d
du
(sech(u(x))) = sech(u(x)) tanh (u(x)) ·
dx
dx
d
(csch(x)) = −csch(x) coth (x)
dx
d
du
(csch(u(x))) = −csch(u(x)) coth (u(x)) ·
dx
dx
d
(coth (x) = −csch2 (x).
dx
d
du
(coth (u(x)) = −csch2 (u(x)) ·
.
dx
dx
Graphs of Hyperbolic Functions
Since cosh (x) is even, the graph is symmetric about the y−axis so we restrict
ourselves to x ≥ 0 and then we complete the graph by symmetry. Since
x
−x
cosh (0) = 1, the y−intercept is (0, 1). Since (cosh (x))0 = e −e
and ex >
2
−x
e for x > 0, the derivative is positive and so the function is increasing for
x
−x
x > 0. Since (cosh (x))00 = e +e
> 0, the graph is concave up. The graph
2
of cosh (x) is shown in Figure 3.6.1(a).
Likewise, the function sinh (x) is odd so its graph is symmetric about the
origin so we restrict ourselves to x ≥ 0 and then we complete the graph by
3
symmetry. Since sinh (0) = 0, the y−intercept is (0, 0). Since (sinh (x))0 =
ex +e−x
> 0, the derivative is positive and so the function is increasing for x >
2
x
−x
0. Since (sinh (x))00 = e −e
, and ex > e−x for x > 0, we have (sinh (x))00 >
2
0 so that the graph is concave up. The graph of sinh (x) is shown in Figure
3.6.1(b).
Figure 3.6.1
The function tanh (x) is odd so its graph is symmetric about the origin so
we restrict ourselves to x ≥ 0 and then we complete the graph by symmetry. Since tanh (0) = 0, the y−intercept is (0, 0). Since (tanh (x))0 =
sech2 (x) > 0, the derivative is positive and so the function is increasing.
Since (tanh (x))00 = −2sech2 (x) tanh (x), and tanh (x) > 0 for x > 0, we
have (tanh (x))00 < 0 so that the graph is concave down. From the fact that
ex − e−x
1 − e−2x
=
lim
=1
x→∞ ex + e−x
x→∞ 1 + e−2x
lim
we conclude that y = 1 is a horizontal asymptote. The graph of tanh (x) is
shown in Figure 3.6.2.
4
Figure 3.6.2
Inverse Hyperbolic Functions and Their Derivatives
The function f (x) = cosh (x) is always increasing for x ≥ 0 so that it has
an inverse function denoted by f −1 (x) = cosh−1 (x). Thus, we have the
following properties
y = cosh−1 (x) if and only if cosh (y) = x
Dom( cosh−1 (x)) =Range( cosh (x)) = [1, ∞)
Range( cosh−1 (x)) =Dom( cosh (x)) = [0, ∞)
cosh (cosh−1 (x)) =x, x ≥ 1
cosh−1 (cosh (x)) =x, x ≥ 0.
The graph of f (x) = cosh−1 (x) is given in Figure 3.6.3.
Figure 3.6.3
5
To find the derivative of y = cosh−1 (x), we write the equivalent form
cosh (y) = x and use implicit differentiation to obtain
1
1
.
=√
2
sinh (y)
x −1
q
2
2
Note that from cosh (y)−sinh (y) = 1 we find sinh (y) = ± cosh2 (y) − 1 =
√
√
± x2 − 1 = x2 − 1 since y = cosh−1 (x) ≥ 1 and sinh (y) > 0 for y > 0.
Also,
d
1
du
cosh−1 (u(x)) = √
.
·
2
dx
u − 1 dx
y 0 sinh (y) = 1 =⇒ y 0 =
The function f (x) = sinh (x) is always increasing for x ≥ 0 so that it has
an inverse function denoted by f −1 (x) = sinh−1 (x). Thus, we have the
following properties
y = sinh−1 (x) if and only if sinh (y) = x
Dom( sinh−1 (x)) =Range( sinh (x)) = (−∞, ∞)
Range( sinh−1 (x)) =Dom( sinh (x)) = (−∞, ∞)
sinh (sinh−1 (x)) =x
sinh−1 (sinh (x)) =x.
The graph of f (x) = sinh−1 (x) is given in Figure 3.6.4.
Figure 3.6.4
To find the derivative of y = sinh−1 (x), we write the equivalent form
sinh (y) = x and use implicit differentiation to obtain
y 0 cosh (y) = 1 =⇒ y 0 =
6
1
1
=√
.
2
cosh (y)
x +1
Also,
d
1
du
sinh−1 (u(x)) = √
·
.
dx
u2 + 1 dx
The function f (x) = tanh (x) is always increasing for x ≥ 0 so that it has
an inverse function denoted by f −1 (x) = tanh−1 (x). Thus, we have the
following properties
y = tanh−1 (x) if and only if tanh (y) = x
Dom( tanh−1 (x)) =Range( tanh (x)) = (−1, 1)
Range( tanh−1 (x)) =Dom( tanh (x)) = (−∞, ∞)
tanh (tanh−1 (x)) =x
tanh−1 (tanh (x)) =x.
The graph of f (x) = tanh−1 (x) is given in Figure 3.6.5.
Figure 3.6.5
To find the derivative of y = tanh−1 (x), we write the equivalent form
tanh (y) = x and use implicit differentiation to obtain
y 0 sech2 (y) = 1 =⇒ y 0 =
1
1
1
=
=
.
2
2
1 − x2
sech (y)
1 − tanh (y)
Also,
d
1
du
tanh−1 (u(x)) =
·
.
2
dx
1 − u dx
7