Calculus Of One Real Variable – By Pheng Kim Ving

7.6 
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1.
Definitions 
The hyperbolic functions
are defined in terms of the natural exponential function e^{x}. For example, the hyperbolic sine
function is defined as (e^{x} – e^{–}^{x})/2
and denoted sinh, pronounced “ shin ”, so that sinh
x
= (e^{x} – e^{–}^{x})/2.

We'll see later on the
reasons why these functions are named the way they are. There are 6 hyperbolic
functions, just as
there are 6 trigonometric functions. The sinh and cosh
functions are the primary ones; the remaining 4 are defined in
terms of them.
Simplify the
expression tanh ln x.
EOS
Note that we
simplify the given hyperbolic expression by transforming it into an algebraic
expression.
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2.
Properties 

These properties can be proved easily using the
definitions of sinh and cosh. For
example, we prove property [2.5] as
follows:
Consider the trigonometric
or circular functions. The equation of the unit circle in the uvcoordinate
system is u^{2} + v^{2} = 1.
For any real number x, we have cos^{2} x + sin^{2} x = 1; thus the point
(cos x, sin
x)
lies on the circle u^{2} + v^{2} = 1. Now
consider the hyperbolic functions. For any real number x,
we have cosh^{2} x
– sinh^{2} x
= 1; thus the point (cosh x, sinh
x)
lies on the curve u^{2} – v^{2} = 1, which is a hyperbola. This
explains the name hyperbolic functions.
Clearly the above
properties of sinh and cosh are similar to
those of the trigonometric functions sin and cos.
For example,
the properties sinh 0 = 0, cosh 0 = 1, and cosh^{2} x – sinh^{2} x = 1 are similar to
the properties sin 0 = 0, cos 0 = 1, and
cos^{2} x
+ sin^{2} x
= 1 respectively. This similarity has led to the naming of them as hyperbolic sine
and hyperbolic cosine
respectively.
The remaining 4
hyperbolic functions are defined in terms of sinh and cosh,
hence they're also hyperbolic functions;
and
they're defined in terms of sinh and cosh in the same fashion
that the 4 trigonometric functions tan, cot,
sec, and csc
are defined in terms of sin and cos; this explains
their names.
Derive from the
addition identities for sinh(x + y)
and cosh(x + y)
the identity:
Also see Problem & Solution 2.
EOS
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3.
Differentiation 

Formula [3.1] is
proved as follows:
The three remaining
formulas can be established analogously.
Remark that these
differentiation formulas for the hyperbolic functions parallel those for the
trigonometric functions,
except for 2 differences in sign: of (d/dx)
cosh x and of (d/dx)
sech x.
Differentiate y
= sinh(x^{2} + 1).
y'
= cosh(x^{2} + 1)(2x) = 2x
cosh(x^{2} + 1).
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4.
Graphs 
The graph of y
= sech x is sketched in Fig.
4.5. For any x, the point of the graph of y
= sech x at x
can be obtained by
taking the reciprocal of the height of the graph of y = cosh
x
at x.
The line y = 0 (the xaxis) is the
horizontal asymptote
of the graph of y = sech x.
Note that we can
also use the 1st and 2nd derivatives of the hyperbolic functions to examine the
increase/decrease and
concavity of their graphs.

Fig. 4.1 Graph of y
= sinh x. 

Fig. 4.2 Graph of y
= cosh x. 

Fig. 4.3 Graph of y
= tanh x. 

Fig. 4.4 Graph of y
= coth x. 

Fig. 4.5 Graph of y
= sech x. 

Fig. 4.6 Graph of y
= csch x. 
Problems
& Solutions 
1. Simplify the following expressions.
a.
sinh ln x.
Solution
2. In Example 2.1
we derived from the addition identities for sinh(x
+ y)
and cosh (x + y)
the identity for tanh(x + y).
Now derive from the same 2
identities the identity:
Solution
3. Find (d/dt) coth
e^{2tx}.
Solution
(d/dt) coth
e^{2tx} = (– csch^{2} e^{2tx})e^{2tx}(2x) = – 2xe^{2tx}csch^{2} e^{2tx}.
4. Sketch the graph of y = csch
(x + 2).
Solution
The graph of y = csch
(x + 2) can be
obtained by sliding that of y
= csch x to the left by 2 units.
5. Consider the differential equation y'' – k^{2}y
= 0, where k is a constant.
a. Prove that the function f_{A}_{,}_{B}(x) = Ae^{kx} + Be^{–kx}, where A and B
are constants, is a solution of the differential equation.
b. Prove that the function g_{C}_{,}_{D}(x) = C cosh
kx + D sinh
kx, where C
and D
are constants, is also a solution of the
differential equation.
c. Express f_{A}_{,}_{B} in terms of g.
d. Express g_{C}_{,}_{D} in terms of f.
Solution
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