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Calculus 1 Problems & Solutions  –  Chapter 4  –  Section 4.3.1

 

4.3.1
The Hyperbolic Functions

 

 

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Review

 

1. Definitions

 

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

 

Definition 1.1

 

 

 

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.

 

Example 1.1

 

Simplify the expression tanh ln x.

 

Solution


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:

 

 

Reasons For The Naming

 

Consider the trigonometric or circular functions. The equation of the unit circle in the uv-coordinate system is u2 + v2 = 1.
For any real number x, we have cos2x + sin2x = 1; thus the point (cos x, sin x) lies on the circle u2 + v2 = 1. Now
consider the hyperbolic functions. For any real number x, we have cosh2 x – sinh2 x = 1; thus the point (cosh x, sinh x)
lies on the curve u2 – v2 = 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 cosh2 x – sinh2 x = 1 are similar to the properties sin 0 = 0, cos 0 = 1, and
cos2 x + sin2 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.

 

Example 2.1

 

Derive from the addition identities for sinh(x + y) and cosh(x + y) the identity:

 

 

Also see Problem & Solution 2.

 

Solution

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.

 

Example 3.1

 

Differentiate y = sinh(x2 + 1).

 

Solution

y' = cosh(x2 + 1)(2x) = 2x cosh(x2 + 1).

EOS

 

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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 x-axis) 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.

 

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Problems & Solutions

 

1. Simplify the following expressions.

 

a.  sinh ln x.

 

 

Solution

 

 

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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

 

 

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3. Find (d/dt) coth e2tx.

 

Solution

 

(d/dt) coth e2tx = (– csch2 e2tx)e2tx(2x) = – 2xe2tx csch2 e2tx.

 

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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.

 

 

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5. Consider the differential equation y'' – k2y = 0, where k is a constant.
a. Prove that the function fA,B(x) = Aekx + Be–kx, where A and B are constants, is a solution of the differential equation.
b. Prove that the function gC,D(x) = C cosh kx + D sinh kx, where C and D are constants, is also a solution of the
    differential equation.
c. Express fA,B in terms of g.
d. Express gC,D in terms of f.

 

Solution

 

 

 

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