3.

11 Hyperbolic functions

Definition (Hyperbolic functions)

ex − e−x
sinh x = (hyperbolic sine)
2
domain : R range : R

ex + e−x
cosh x = (hyperbolic cosine)
2
domain : R range : [1, ∞)

1
 sinh x
tanh x = (hyperbolic tangent)
cosh x
domain : R range : (−1, 1)

y = ±1 are horizontal asymptotes

1 1 cosh x
cschx = , sechx = , cothx =
sinh x cosh x sinh x

2
Theorem (Hyperbolic identities)
(1) sinh(−x) = − sinh x
(2) cosh(−x) = cosh x
(3) cosh2 x − sinh2 x = 1
(4) 1 − tanh2 x = sech2 x
(5) sinh(x + y) = sinh x cosh y + cosh x sinh y
(6) cosh(x + y) = cosh x cosh y + sinh x sinh y

Proof.
(1) e−x − ex ex − e−x
sinh(−x) = =− = − sinh x
2 2
(3) 2 2
e + e−x ex − e−x
 x  
2 2
cosh x − sinh x = −
2 2
1 2x 1
= (e + 2 + e−2x ) − (e2x − 2 + e−2x )
4 4
=1
(4) cosh2 x − sinh2 x = 1
cosh2 x sinh2 x 1
⇒ 2 − =
cosh x cosh2 x cosh2 x
⇒ 1 − tanh2 x = sech2 x
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4
Theorem (Derivatives of hyperbolic functions)

d d
(1) (sinh x) = cosh x (2) (cschx) = −cschx cothx
dx dx
d d
(3) (cosh x) = sinh x (4) (sechx) = −sechx tanhx
dx dx
d d
(5) (tanh x) = sech2 x (6) (coth x) = −csch2 x
dx dx

Proof. (1)

ex − e−x ex + e−x
 
d d
(sinh x) = = = cosh x
dx dx 2 2

(5)
 
d d sinh x cosh x · cosh x − sinh x · sinh x
(tanh x) = =
dx dx cosh x cosh2 x
cosh2 x − sinh2 x 1
= = = sech2 x
cosh2 x cosh2 x

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Example
√
Differentiate y = cosh x.
Solution. √
d √ √ d √ sinh x
(cosh x) = sinh x · x= √
dx dx 2 x

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Definition (Inverse hyperbolic functions)

y = sinh−1 x ⇔ x = sinh y

y = sinh−1 x
domain = R, range = R

y = cosh−1 x ⇔ x = cosh y and y ≥ 0

y = cosh−1 x
domain = [1, ∞), range = [0, ∞)

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 y = tanh−1 x ⇔ x = tanh y

y = tanh−1 x
domain = (−1, 1), range = R
x = ±1 are vertical asymptote.

sinh and tanh are one-to-one functions and so they have inverse functions
denoted by sinh−1 and tanh−1 .

cosh is not one-to-one, but when restricted to the domain [0, ∞], it becomes
one-to-one. The inverse hyperbolic cosine function is defined as the inverse of
this restricted function.

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Theorem
p
(1) sinh−1 x = ln(x + x2 + 1), x ∈ R
p
(2) cosh−1 x = ln(x + x2 − 1), x ≥ 1
 
−1 1 1+x
(3) tanh x = ln , −1 < x < 1
2 1−x
ey − e−y
Proof. (1) Let y = sinh−1 x. Then x = sinh y = . So
2
ey − 2x − e−y = 0
⇔ e2y − 2x · ey − 1 = 0
⇔ (ey )2 − 2x(ey ) − 1 = 0
Solving by the quadratic formula, we get
√
2x ± 4x2 + 4 p
ey = = x ± x2 + 1
2
√
Note that ey > 0, but x − x2 + 1 < 0. Thus the minus sign is inadmissible
and we have p
ey = x + x2 + 1
Therefore  p 
y = ln(ey ) = ln x + x2 + 1
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Theorem (Derivatives of inverse hyperbolic functions)

d 1 d 1
(1) (sinh−1 x) = √ , x∈R (2) (csch−1 x) = − √ , x ̸= 0
dx 1 + x2 dx |x| x2 + 1
d 1 d 1
(3) (cosh−1 x) = √ , x>1 (4) (sech−1 x) = − √ , 0<x<1
dx x2 − 1 dx x 1 − x2
d 1 d 1
(5) (tanh−1 x) = , |x| < 1 (6) (coth−1 x) = , |x| > 1
dx 1 − x2 dx 1 − x2

Proof.
(1)
(i) Let y = sinh−1 x (x ∈ R). Then x = sinh y. Differentiating x = sinh y
implicitly with respect to x, we get
dy
cosh y · =1
dx
Since cosh2 y − sinh2 y = 1 and cosh y ≥ 0,
q
cosh y = 1 + sinh2 y
So
dy 1 1 1
= = p = √
dx cosh y 1 + sinh2 y 1 + x2
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(ii)
d d  p 
sinh−1 x =


ln x + x2 + 1
dx dx
1 d  p 
= √ x + x2 + 1
x + x + 1 dx
2
 
1 x
= √ 1+ √
x + x2 + 1 x2 + 1
√
2
x +1+x 1
= √
√ = √
x + x2 + 1 x2 + 1 x2 + 1

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Example
d
Find [tanh−1 (sin x)].
dx
Solution.
d  1 d
tanh−1 (sin x) =

(sin x)
dx 1 − (sin x)2 dx
1
= cos x
1 − sin2 x
cos x
= = sec x
cos2 x

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