Hyperbolic functions
Definitions
1 x
e  e x 

2
1
cosh x   e x  e  x 
2
e x  e x
tanh x  x
e  e x
2
e  e x
2
sech x  x
e  e x
e x  e x
coth x  x
e  e x
sinh x 
cosech x 
x
Graphs
It is useful to first consider the graphs of y  e x and y  e x .
y  e x
y  ex
y
y
1
0
1
x
0
x
y  sinh x 
1 x
e  e x 

2
1 0
1
e  e0   1  1  0

2
2
1
1
As x  , sinh x  e x and as x  , sinh x   e  x
2
2
y
sinh 0 
0
x
1
y  cosh x 
1 x
e  e x 

2
1 0
1
e  e 0   1  1  1

2
2
1
1
As x  , cosh x  e x and as x  , cosh x  e  x
2
2
y
cosh 0 
1
0
x
e x  e x
y  tanh x  x
e  e x
e0  e0 0
tanh 0  0
 0
e  e0 2
ex  0
0  e x
As x  , tanhx  x
 1 and as x  , tanhx 
 1
e 0
0  e x
y
1
0
x
1
2
Identities
cosech x 
1
sinh x
sech x 
1
cosh x
coth x 
1
follow directly from the definitions.
tanh x
cosh 2 x  sinh 2 x  1
cos 2 x  sin 2 x  1
1  tanh 2 x  sech 2 x
1  tan 2 x  sec 2 x
coth 2 x  1  cosech 2 x
cot 2 x  1  cosec 2 x
2
2
1 x
1
e  e x    e x  e x 

4
4
1 2x
  e  2  e 2 x  e 2 x  2  e 2 x 
4
1
 cosh 2 x  sinh 2 x  1 ………………..(A)
cosh 2 x  sinh 2 x 
The other two may be done in a similar way, or, once (A) has been done, deviding it by
cosh 2 x leads to 1  tanh 2 x  sech 2 x and deviding it by sinh 2 x leads to coth 2 x  1  cosech 2 x
sinh 2x  2sinh x cosh x
cosh 2 x  cosh 2 x  sinh 2 x
sin 2x  2sin x cos x
cos 2 x  cos 2 x  sin 2 x
 1  2sinh 2 x
 1  2sin 2 x
 2 cosh 2 x  1
2 tanh x
tanh 2 x 
1  tanh 2 x
 2 cos 2 x  1
2 tan x
tan 2 x 
1  tan 2 x
Compare these identities with the trigonometric identities. What do you notice?
Wherever sin 2 x is involved in a trigonometric identity, the corresponding hyperbolic identity
contains  sinh 2 x . This is known as Osborne’s rule.
Derivatives of the hyperbolic functions.
Function
Derivative
sinh x
cosh x
tanh x
coth x
sech x
cosech x
cosh x
sinh x
sech 2 x
cosech 2 x
sech x tanh x
cosech x coth x
3
The derivatives also follow the pattern of the trigonometric functions with the following two
d
d
cosh x  sinh x
cos x   sin x
exceptions:
and
dx
dx
d
d
sech x  sech x tanh x
sec x  sec x tan x
and
dx
dx
The derivatives of the hyperbolic functions are obtained by differentiating their definitions.
d
d 1 x
 1
cosh x 
e  e  x     e x  e  x   sinh x


dx
dx  2
 2
0  2  e x  e x 
e x  e x 

d
d
2
2
sech x 

 x
. x
 sech x tanh x
x
x
x
x 2
dx
dx  e x  e  x 
e

e
e

e




e

e


and so on.
4
Revision of differentiation
Standard derivatives
Function
Derivative
c
xn
ax
ex
0
nx n 1
a x ln a
ex
1
x ln a
1
x
log a x
ln x
cos x
 sin x
sec 2 x
cosec 2 x
sin x
cos x
tan x
cot x
sec x
cosec x
sec x tan x
cosec x cot x
sinh x
cosh x
tanh x
coth x
sech x
cosech x
cosh x
sinh x
sech 2 x
cosech 2 x
sech x tanh x
cosech x coth x
Differentiation Rules
Let u  f  x  and v  g  x  .
Sum and difference:
If y  u  v then y   u   v 
Example: If y  x5  7 x  e x  ln x then
1
y   5 x 4  7 x ln 7  e x 
x
5
dy dy du dv dw




.
dx du dv dw dx
An alternative way to look at it is: If y  f  g  x   then y   f   g  x    g   x 
If y  u  v  and v  w  x  then
Chain Rule:
Example: If y  ln tan x3 then y  ln u, u  tan v where v  x3 .
dy dy du dv 1



  sec2 v  3x 2
Thus
dx du dv dx u
1

 sec2 x3  3x 2
3
tan x
Alternative:
y 
1
 sec 2 x3  3 x 2
3
tan x
[First do the ln function, then the tan function
and finally the x 3 . ]
Constant times function:
Example:
If y  8sin 3x then y   8
If y  e4 x tan 5 x then
y    e 4 x .4  .tan 5 x  e 4 x .  sec 2 5 x.5 
Quotient Rule:
Example:
d
sin 3 x  8cos 3 x  3  24 cos 3 x
dx
If y  u.v then y   u v  uv  .
Product Rule:
Example:
If y  c  f  x  then y   c  f   x  .
If y 
u
u v  uv 
then y  
.
v
v2
sinh x
then
cosh x
cosh x.cosh x  sinh x.sinh x
y 
 1  tanh 2 x  sech 2 x which
2
cosh x
sinh x
is the derivative of tanh x 
.
cosh x
If y 
Exercise.
6
Inverse Trigonometric- and Hyperbolic functions
The method to find the standard derivatives of these functions is as follows:
Example: To find the derivative of sin 1 f  x   arcsin f  x  .
Let y  sin 1 f  x  . Thus sin y  f  x  .
dy
 f  x
dx
f  x
f  x
dy f   x 




2
dx cos y
1  sin 2 y
1   f  x  
f  x
 cos y.
Thus
d
sin 1 f  x  
2
dx
1   f  x  
All the other standard derivatives are obtained in a similar manner.
The table of integrals will be used in order to differentiate the inverse functions.
f  x
f  x
d
sin 1

dx
a
The table gives
a
 f  x 
1 

 a 


2
f  x
a   f  x  
2
Examples.
d
sin 1  4 x 3  
1
dx
f  x
 f  x 
a 1 

 a 
dx  sin 1
2
12 x 2
1   4x

3 2
2
f  x
a

f  x
a 2   f  x 
2
c
[Let f  x   4x3 and a  1 ]
2
d
 tan x 
sinh 1 

dx
 4 
3
5
d
5x4
15 x 4
1 x
[Note: Remember to multiply by 3]
tan
 3
 10
2
dx
3
 x5   32 x  9
We have used
sec 2 x
tan 2 x  16
f  x
  f  x 

2
2
 a
dx 
[Let f  x   tan x and a  4 ]
f  x
1
tan 1
c
a
a
7
Parametric differentiation
If a function is defined by y  f  x  , where x  g  t  and y  h t  and t is a parameter, then
dy
dy dy dt


 dt
dx dt dx dx
dt
and
d
d
2
d y d dy
dy

 dt

2
dx dx dx dx
dx
dt
 dy 
dt  dx 
dx
dt
We use the derivatives as follows:
dy
2
dy
 dt
dx dx
dt
Example; Determine
d
d y

dx 2
 dy 
dt  dx 
dx
dt
dy
d2 y
and
if x  2  4cos  and y  3  4sin 
dx
dx 2
dx
dy
 4sin 
 4 cos 
d
d
dy
dy
4 cos 
 d 
  cot 
dx dx
4sin 
d
2
d y

dx 2
d
 dy 
d  dx  cosec 2
1


dx
4sin 
4sin 3 
d
Exercise: Page 314 Exercise 6(f): 1a-d, 2b. Also find
d2 y
. 8, 9
dx 2
Additional:
1.
dy
d2y
Find
:
and
dx
dx 2
y  cos 2t ; x  sin t (b)
(a)
(c)
2.
y  3sin   sin 3  , x  cos3 
x  a  cos   sin   , y  a sin    cos 
Show that
dy 1
2  3t
3  2t
 if x 
and y 
dx 5
1 t
1 t
8