Transcript inverse hyperbolic functions

3
DIFFERENTIATION RULES
DIFFERENTIATION RULES
Certain even and odd combinations of
the exponential functions ex and e-x arise so
frequently in mathematics and its applications
that they deserve to be given special names.
DIFFERENTIATION RULES
In many ways, they are analogous to
the trigonometric functions, and they have
the same relationship to the hyperbola that
the trigonometric functions have to the circle.
 For this reason, they are collectively called
hyperbolic functions and individually called
hyperbolic sine, hyperbolic cosine, and so on.
DIFFERENTIATION RULES
3.11
Hyperbolic Functions
In this section, we will learn about:
Hyperbolic functions and their derivatives.
DEFINITION
e e
sinh x 
2
x
x
e x  e x
cosh x 
2
sinh x
tanh x 
cosh x
1
csc h x 
sinh x
1
sec h x 
cosh x
cosh x
coth x 
sinh x
HYPERBOLIC FUNCTIONS
The graphs of hyperbolic sine and cosine
can be sketched using graphical addition,
as in these figures.
HYPERBOLIC FUNCTIONS
Note that sinh has domain and
range , whereas cosh has domain
and range [1, ) .
HYPERBOLIC FUNCTIONS
The graph of tanh is shown.
 It has the horizontal asymptotes y = ±1.
APPLICATIONS
Some mathematical uses of hyperbolic
functions will be seen in Chapter 7.
Applications to science and engineering
occur whenever an entity such as light,
velocity, or electricity is gradually absorbed
or extinguished.
 The decay can be represented by hyperbolic functions.
APPLICATIONS
The most famous application is
the use of hyperbolic cosine to describe
the shape of a hanging wire.
APPLICATIONS
It can be proved that, if a heavy flexible cable
is suspended between two points at the same
height, it takes the shape of a curve with
equation y = c + a cosh(x/a) called a catenary.
 The Latin word
catena means
‘chain.’
APPLICATIONS
Another application occurs in the
description of ocean waves.
 The velocity of a water wave with length L moving
across a body of water with depth d is modeled by
the function
gL
 2 d 
v
tanh 

2
L


where g is the acceleration due to gravity.
HYPERBOLIC IDENTITIES
The hyperbolic functions satisfy
a number of identities that are similar to
well-known trigonometric identities.
HYPERBOLIC IDENTITIES
We list some identities here.
sinh( x)   sinh x
cosh( x)  cosh x
cosh x  sinh x  1
2
2
1  tanh x  sech x
2
2
sinh( x  y )  sinh x cosh y  cosh x sinh y
cosh( x  y )  cosh x cosh y  sinh x sinh y
HYPERBOLIC FUNCTIONS
Prove:
a. cosh2x – sinh2x = 1
b. 1 – tanh2 x = sech2x
Example 1
HYPERBOLIC FUNCTIONS
Example 1 a
2
2
e e  e e 
cosh x  sinh x  
 

2  
2 

2x
2 x
2x
2 x
e 2e
e 2e


4
4
4
 1
4
x
2
2
x
x
x
HYPERBOLIC FUNCTIONS
Example 1 b
We start with the identity proved in (a):
cosh2x – sinh2x = 1
 If we divide both sides by cosh2x, we get:
sinh 2 x
1
1

2
cosh x cosh 2 x
or 1  tanh 2 x  sec h 2 x
HYPERBOLIC FUNCTIONS
The identity proved in Example 1 a
gives a clue to the reason for the name
‘hyperbolic’ functions, as follows.
HYPERBOLIC FUNCTIONS
If t is any real number, then the point
P(cos t, sin t) lies on the unit circle x2 + y2 = 1
because cos2 t + sin2 t = 1.
 In fact, t can be
interpreted as the radian
measure of POQ
in the figure.
HYPERBOLIC FUNCTIONS
For this reason, the trigonometric
functions are sometimes called
circular functions.
HYPERBOLIC FUNCTIONS
Likewise, if t is any real number, then
the point P(cosh t, sinh t) lies on the right
branch of the hyperbola x2 - y2 = 1 because
cosh2 t - sin2 t = 1 and cosh t ≥ 1.
 This time, t does not represent the measure
of an angle.
HYPERBOLIC FUNCTIONS
However, it turns out that t represents twice
the area of the shaded hyperbolic sector in
the first figure.
 This is just as in the trigonometric case t represents
twice the area of the shaded circular sector in the
second figure.
DERIVATIVES OF HYPERBOLIC FUNCTIONS
The derivatives of the hyperbolic
functions are easily computed.
 For example,
d
d  e x  e x  e x  e x
(sinh x)  
 cosh x

dx
dx  2 
2
DERIVATIVES
Table 1
We list the differentiation formulas for
the hyperbolic functions here.
d
(sinh x)  cosh x
dx
d
(cosh x)  sinh x
dx
d
2
(tanh x)  sec h x
dx
d
(csc h x)   csc h x coth x
dx
d
(sec h x)   sec h x tanh x
dx
d
2
(coth x)  csc h x
dx
DERIVATIVES
Note the analogy with the differentiation
formulas for trigonometric functions.
 However, beware that the signs are different
in some cases.
d
(sinh x)  cosh x
dx
d
(cosh x)  sinh x
dx
d
2
(tanh x)  sec h x
dx
d
(csc h x)   csc h x coth x
dx
d
(sec h x)   sec h x tanh x
dx
d
2
(coth x)  csc h x
dx
DERIVATIVES
Example 2
Any of these differentiation rules can
be combined with the Chain Rule.
 For instance,
d
d
sinh x
(cosh x )  sinh x 
x
dx
dx
2 x
INVERSE HYPERBOLIC FUNCTIONS
You can see from the figures that sinh
and tanh are one-to-one functions.
 So, they have inverse functions denoted by
sinh-1 and tanh-1.
INVERSE FUNCTIONS
This figure shows that cosh is not
one-to-one.
However, when restricted to the domain
[0, ∞], it becomes one-to-one.
INVERSE FUNCTIONS
The inverse hyperbolic cosine
function is defined as the inverse
of this restricted function.
INVERSE FUNCTIONS
Definition 2
1

sinh y  x
1

cosh y  x and y  0
1

tanh y  x
y  sinh x
y  cosh x
y  tanh x
 The remaining inverse hyperbolic functions
are defined similarly.
INVERSE FUNCTIONS
By using these figures,
we can sketch the graphs
of sinh-1, cosh-1, and
tanh-1.
INVERSE FUNCTIONS
The graphs of sinh-1,
cosh-1, and tanh-1 are
displayed.
INVERSE FUNCTIONS
Since the hyperbolic functions are defined
in terms of exponential functions, it’s not
surprising to learn that the inverse hyperbolic
functions can be expressed in terms of
logarithms.
INVERSE FUNCTIONS
Defns. 3, 4, and 5
In particular, we have:
1

x  ln  x 

 1
sinh x  ln x  x  1
cosh
1
2
x
2
 1 x 
tanh x  ln 

 1 x 
1
1
2
x
x 1
1 x  1
INVERSE FUNCTIONS
1

Example 3

Show that sinh x  ln x  x  1 .
2
y
y
e

e
 Let y = sinh-1 x. Then, x  sinh y 
2
 So, ey – 2x – e-y = 0
 Or, multiplying by ey, e2y – 2xey – 1 = 0
 This is really a quadratic equation in ey:
(ey)2 – 2x(ey) – 1 = 0
INVERSE FUNCTIONS
Example 3
Solving by the quadratic formula,
we get: y 2 x  4 x 2  4
e 
 x  x 1
2
 Note that ey > 0, but x 
(because x  x2  1 ).
2
x2  1  0
 So, the minus sign is inadmissible and we have:

y
2
y

ln(
e
)

ln
x

x
1
 Thus,

e y  x  x2  1
DERIVATIVES
Table 6
d
1
1
(sinh x) 
dx
1  x2
d
(csc h 1 x)  
dx
x
d
1
cosh x  

dx
d
1
1
(sec h x) 
dx
x 1  x2
d
1
1
(coth x) 
dx
1  x2
1
x2  1
d
1
1
(tanh x) 
dx
1  x2
1
x2  1
DERIVATIVES
The inverse hyperbolic functions are
all differentiable because the hyperbolic
functions are differentiable.
 The formulas in Table 6 can be proved either by
the method for inverse functions or by differentiating
Formulas 3, 4, and 5.
DERIVATIVES
E. g. 4—Solution 1
d
1
1
Prove that
.
(sinh x) 
2
dx
1 x
 Let y = sinh-1 x. Then, sinh y = x.
 If we differentiate this equation implicitly
with respect to x, we get: cosh y dy  1
dx
 As cosh2 y - sin2 y = 1 and cosh y ≥ 0, we have:
cosh y  1  sinh 2 y
1
1
1
 So, dy



2
dx cosh y
1  sinh y
1  x2
DERIVATIVES
E. g. 4—Solution 2
From Equation 3, we have:


d
d
1
2
sinh
x

ln
x

x
1


dx
dx
1
d

x  x2  1
x  x 2  1 dx



x 

1


2
2
x  x 1 
x 1 
1

x 
x2  1  x
x 1
2

x 1
2

1
x2  1
DERIVATIVES
Example 5
d
1
 tanh (sin x)  .
Find
dx
 Using Table 6 and the Chain Rule,
we have: d
1
d
 tanh (sin x)  
(sin x)
2
dx
1  (sin x) dx
1

cos x
2
1  sin x
cos x

 sec x
2
cos x
1