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CHAPTER 2
DIFFERENTIATION
2.1
Differentiation of hyperbolic functions
2.2
Differentiation of inverse trigonometric
functions
2.3
Differentiation of inverse hyperbolic
functions
*Recall: Methods of differentiation
- Chain rule
- Product differentiation
- Quotient differentiation
- Implicit differentiation
1
2.1
Differentiation of Hyperbolic Functions
Recall: Definition:
e x  e-x
cosh x 
2
e x  e-x
sinh x 
2
sinh x
e x  e-x
tanh x 
 x
cosh x
e  e-x
1
cosh x
coth x 

tanh x sinh x
cosech x =
sech x 
1
sinh x
1
cosh x
2
Derivatives of hyperbolic functions
Example 2.1: Find the derivatives of
(a) sinh x
(b) cosh x
(c) tanh x
Solution:
d  e x  e x
(a)

sinh x 
dx
dx 
2
d

1
2




( e x  e  x )  cosh x
 e x  e x
d
d
(b)

cosh x 
dx
dx 
2

1
2




( e x  e  x )  sinh x
3
 e x  e x
d
d
(c)

tanh x 
dx
dx  e x  e  x




Using quotient diff:

e x  e  x e x  e  x  e x  e  x  e x  e  x 

x
x 2
e  e 


e 2 x  e 2 x  2  e 2 x  e 2 x  2
e
x
e
4
e
x
e
x

2

x
2

2


x
x
e

e



 
 cosh x
1








2
2
 sech 2 x
Using the same methods, we can obtain the derivatives
of the other hyperbolic functions and these gives us the
standard derivatives.
4
Standard Derivatives
y  f ( x)
dy
 f ( x)
dx
cosh x
sinh x
sinh x
cosh x
tanh x
sech 2 x
sech x
cosech x
coth x
sech x tanh x
cosech x coth x
cosech 2 x
5
Example 2.2:
1.Find the derivatives of the following functions:
a)
b)
c)
2. Find the derivatives of the following functions:
(a) y  cosh  3x 
(b) r  sinh(2t 2  1)
(c) g ( x)  ( x  1)3 sech 2 x
(d) y  tanh(ln x)
3.(Implicit differentiation)
Find
dy
from the following expressions:
dx
(a) x  y 2 sinh 4 x  cosh y
(b) y  tanh ( x  y )
6
2.2 Differentiation Involving Inverse Trigonometric
Functions
Recall: Definition of inverse trigonometric functions
Function
Domain
sin 1 x
1  x  1
cos1 x
1  x  1
tan 1 x
  x  
sec1 x
cot 1 x
cosec1 x
x 1
Range

 y
2

2
0  y 


 y
2
0 y

2


2

2
 y 
0  y 
  x  
x 1



2
 y  00  y 

2
7
Derivatives of Inverse Trigonometric Functions
Standard Derivatives:
d
1
1
1.
(sin x) 
dx
1 x2
d
1
1
(cos x) 
2.
dx
1 x2
d
3.
dx
d
4.
dx
d
5.
dx
d
6.
dx
(tan
(cot
(sec
1
1
1
x)
x)
1
1 x 2
1
1 x 2
1
x)
x
(csc
1
x 2 1
1
x)
x
x 2 1
8
2.2.1
Derivatives of y  sin 1 x . (proof)
Recall: y  sin 1 x  x  sin y
for x  [1, 1] and y  [  2 ,  2] .
Because the sine function is differentiable on
[  2 ,  2] , the inverse function is also differentiable.
To find its derivative we proceed implicitly:
Given sin y  x . Differentiating w.r.t. x:
d
d
(sin y )  ( x)
dx
dx
cos y

Since 

2
 y

2
dy
1
dx
dy
1

dx cos y
, cos y  0 , so
dy
1
1
1



dx cos y
1  sin 2 y
1  x2
9
Example 2.3:
1. Differentiate each of the following functions.
(a) f ( x)  tan1 x
(b) g (t )  sin 1(1  t )
(c) h( x)  sec1 e2 x
2. Find the derivative of:
(a) y  (tan1 x 2 )4
(b) f ( x)  ln(sin1 4 x)
3. Find the derivative of
4. Find the derivative
y  tan 1 (tan(3t 2  1)) .
dy
if
dx
(a) x tan1 y  x 2  y
(b) sin 1 ( xy) 

2
 cos1 y
10
Summary
If u is a differentiable function of x, then
d
1
du
(sin 1 u ) 

1. dx
1  u 2 dx
d
1
du
(cos1 u )  

2. dx
1  u 2 dx
d
1
du
1
3. dx (tan u )  1  u 2  dx
d
1
du
1
(cot
u
)



4. dx
1  u 2 dx
d
1
du
1
(sec
u
)


5. dx
u u 2  1 dx
d
1
du
1
(csc
u
)



6. dx
u u 2  1 dx
11
2.3 Derivatives of Inverse hyperbolic Functions
Recall: Inverse Hyperbolic Functions
Function
Domain
Range
y  sinh 1 x
(  ,  )
(  ,  )
y  cosh 1x
[ 1,  )
[ 0 , )
y  tanh 1 x
(1, 1)
(  ,  )
y  coth 1 x
(  , 1)  (1,  )
(  , 0 )  ( 0 ,  )
y = sech1 x
(0, 1]
[ 0 , )
y = cosech1x
(  , 0 )  ( 0 ,  )
(  , 0 )  ( 0 ,  )
Function
Logarithmic form
y  sinh 1 x
ln x  x 2  1
y  cosh 1x
y  tanh 1 x

ln  x 

x 1
2
1  1 x 
ln 
; x  1
2  1 x 
12
2.3.1 Proof: (
d
1
)
(sinh 1 x) 
2
dx
1 x
Recall: y  sinh 1 x  x  sinh y
To find its derivative we proceed implicitly:
 Given x  sinh y . Differentiating w.r.t. x:
d
d
( x)  (sinh y )
dx
dx
dy
1  cosh y
dx
dy
1


dx cosh y
 Since    y   , cosh y  0 , so using the identity
cosh 2 y  sinh 2 y  1:
dy
1
1
1



dx cosh y
1  sinh2 y
1 x 2

d
1
(sinh 1 x) 
dx
1  x2
13
 Other ways to obtain the derivatives are:
(a)
y  sinh 1 x  x  sinh y then
e y  e y
dy
x
. Hence, find
.
2
dx


(b) y  sinh 1 x = ln x  x 2  1 .
Hence, find dy .
dx
14
Standard Derivatives
Function, y
Derivatives,
sinh 1 x
dy
dx
1
x2  1
cosh 1 x
1
x 1
2
; x 1
tanh 1 x
1
; x 1
2
1 x
coth 1 x
1
; x 1
2
1 x
sech 1 x
cosech1x

1
x 1 x
2
; 0  x 1
1
x 1 x
2
; x0
15
Generalised Form
y  f (u );
u  g ( x)
sinh 1 u
dy dy du
 
dx du dx
1
du
u 2  1 dx
cosh 1u
du
; u 1
2
u  1 dx
tanh 1 u
1 du
; u 1
2
1  u dx
coth 1u
1 du
; u 1
2
1  u dx
sech 1u
cosech1u
1

1
du
; 0  u 1
2 dx
u 1 u
1
du
; u0
2 dx
u 1 u
16
Example 2.4: Find the derivatives of
(a) y  sinh 1 (1  3x)
1
(b) y  cosh 1 
 x
(c) y  e xsech 1x
(d) y  sinh 1 (tan 3x)
1 2
tanh
t
(e) f (t ) 
1  sec t
(f) y  coth 1 u
(g) y  cos 4 x cosh 1 4 x
(h) y3  sinh 1 xy  0
17