Inverse hyperbolic functions

The hyperbolic sine function y = sinh x is strictly monotone increasing because d
dx sinh x =
cosh x = (ex + e?x)/2 > 0 always. So it has an inverse function. For x > 0 we have
sinh x = (ex ? e?x)/2 > (ex ? 1)/2 (since e?x < e0 = 1 for x > 0, or ?x < 0). This shows
that limx!1 sinh x = 1. As sinh(?x) = ?sinh x, we also have limx!?1 sinh x = ?1.
The hyperbolic cosine function y = cosh x is always positive. In fact cosh x  cos 0 = 1.
We have d
dx cosh x = sinh x. For x > 0, we have sinh x > sinh 0 = 0 and so cosh x is strictly
monotone increasing for x > 0. On the other hand, for x < 0, sinh x < sinh 0 and so cosh x
is strictly monotone decreasing for x < 0. Also we have
cosh x >
1
2
max(ex, e?x) =
1
2
e|x|
so that limx!1 cosh x = 1 and also limx!?1 cosh x = 1. In fact cosh x grows very
rapidly, comparably fast to the exponential.
Here are graphs of y = sinh x and y = cosh x.
The function y = sinh x has an inverse function sinh?1 : R ! R. We can say then that
y = sinh?1 x means exactly the same as sinh y = x
and the graph of y = sinh?1 x is the reflection of the graph of sinh in the line y = x. We
can find dy/dx for y = sinh?1 x by the theorem on derivatives of inverse functions:
dy
dx
=
 1
dx
dy
 =
1
cosh y
and we can expresse that in terms of x using cosh2 y ?sinh2 y = 1, cosh2 y = 1+sinh2 y =
1 + x2, to get
dy
dx
=
d
dx
sinh?1 x =
1
p
1 + x2
(and it is right to have the square root because cosh y > 0 always).
Proposition 8.6.1. We can express sinh?1 x in terms of the natural logarithm as
sinh?1 x = ln(x +
p
x2 + 1)
MA1131 — Lecture 7 (21/10/2010) 34
Proof. We can solve sinh y = x for y in terms of x as folows.
ey ? e?y
2
= x
ey ? e?y ? 2x = 0
(ey)2 ? 2xey ? 1 = 0
This is a quadradic equation for ey. We get
ey =
2x ±
p
4x2 + 4
2
= x ±
p
x2 + 1
Since
p
x2 + 1 >
p
x2 = |x|, the minus sign would certainly make ey negative — which is
not possible. So we must have ey = x +
p
x2 + 1, and that means y = ln(x +
p
x2 + 1) as
we required.
For cosh?1, we have to interpret what we mean by the inverse because horizontal lines
can cross the graph of cosh more than once.
Definition 8.6.2. By cosh?1 we mean the function cosh?1 : [1,1) ! [0,1) given by the
rule
y = cosh?1 x means cosh y = x and y  0.
We can find the derivative (as long as we don’t go to the end point x = 1) in a similar
way to the way we did above for sinh?1 x. It is
d
dx
cosh?1 x =
1
p
x2 ? 1
(x > 1).
We can also express cosh?1 via the natural logarithm as
cosh?1 x = ln(x +
p
x2 ? 1) (x  1).
Remark 8.6.3. There is an inverse for y = tanh x = sinh x
cosh x = ex?e?x
ex+e?x . It is given by
tanh?1 x =
1
2
ln
1 + x
1 ? x
(?1 < x < 1),
and it has derivative
d
dx
tanh?1 x =
1
1 ? x2 (?1 < x < 1).
Remark 8.6.4. It may be of interest to know that the graph of y = cosh x has the shape of
a ‘catenary’, meaning the shape of a hanging chain (undisturbed by wind). By a ‘chain’ is
meant something like a cable, but one that bends without reistance, yet still has mass.
MA1131 (R. Timoney) October 20, 2010