CHAPTER 59 DIFFERENTIATION OF INVERSE
TRIGONOMETRIC AND HYPERBOLIC FUNCTIONS
EXERCISE 246 Page 667
1. Differentiate with respect to the variable: (a) sin –1 4x
(a) If y = sin −1 4x , then
dy
=
dx
x
dy
(b) If y = sin –1 2 , then
=
dx
4
=
[1 − (4 x)2 ]
1
[ 22 − x 2 ]
=
4
(1 − 16x 2 )
1
( 4 − x2 )
2. Differentiate with respect to the variable: (a) cos –1 3x
(a) If y = cos–1 3x , then
dy
=
dx
−3
[1 − (3x)2 ]
dy 2
2
x
(b) If y = cos −1 , then
= 
dx 3
3
3

=
(a) If y = 3 tan −1 2x , then
1
(b) If y = tan –1
2
(b)
x
2
cos–1 3
3
(b)
1 –1 x
tan
2
−3
(1 − 9x 2 )

 =
( 32 − x 2 )  3
−1
3. Differentiate with respect to the variable:
x
(b) sin –1 2
−2
(9 − x2 )
(a) 3 tan–1 2x


6
dy
2
=
= 3

dx
1 + (2 x) 2  1 + 4x 2
 1 − 12 
x

1
dy 1 2
1
x=
, then
=
=

2
1
d x 2 1 + x  4 x 2 1 + x
( ) 4 x (1 + x)


( )
4. Differentiate with respect to the variable: (a) 2 sec–1 2t
989
3
(b) sec–1 4 x
© 2014, John Bird
(a) If y = 2 sec −1 2t , then

dy
= 2
 2t
dt

3
dy
(b) If y = sec −1 x=
, then
4
dx 3
x
4

 =
2
( 2t ) − 1  t
2
2
( 4t 2 − 1)
3
1
1
4
=
=
=
 3 x  2 
 9x2 
 9 x 2 − 16  x
1
x
x
−




1
−
 

 16

 16 
 4 

x
dy 5
θ
5
= 
(a) If y = cosec −1 , then
d θ 2 θ
2
2

(b) If y = cosec −1 x 2 , then
(a)
θ
5
cosec–1 2
2

 =
(θ 2 − 22 )  θ
−2
− 2x
dy
=
=
d x x 2 ( x 2 )2 − 1
x
(b) If y = cot −1
4
( 9 x 2 − 16 )
(b) cosec–1 x2
−5
(θ 2 − 4 )
−2
( x 4 − 1)
6. Differentiate with respect to the variable: (a) 3 cot–1 2t
(a) If y = 3cot −1 2t , then
( 9 x 2 − 16 )
4
=
5. Differentiate with respect to the variable:
1
2
(b) cot–1 θ − 1
 −2 
dy
−6
= 3
 =
2
dt
1 + ( 2t )  1 + 4t 2
1
1
−
− (θ 2 − 1) 2 2θ
dy
2
− 1 , then
θ 2=
=
dθ 1 +  θ 2 −1 2 
(
) 


x
7. Show that the differential coefficient of tan–1 1 − x 2 is
990
−θ
(θ 2 − 1) [1 + θ 2 − 1]
=
−1
θ
(θ 2 − 1)
1 + x2
1 − x2 + x4
© 2014, John Bird
dy
 x 
If y = tan −1 
=
 then
2
dx
 1− x 
(1 − x 2 )(1) − ( x )( −2 x )
2
x2 )
(1 −=
 x 
1+ 

 1 − x2 
2
1 − x2 + 2 x2
(1 − x 2 )
=
2
(1 − x 2 ) + x 2
2
(1 − x 2 )
2
1 + x2
1 − 2 x2 + x4 + x2
=
8. Differentiate with respect to the variable: (a) 2x sin–1 3x
dy
(a) If y = 2x sin =
3x then
dx
–1
dy
(b) If y = t 2=
sec −1 2t then
dt

3
( 2 x ) 
 1 − ( 3 x )2

(b) t2 sec–1 2t

 + ( sin −1 3 x )( 2 ) =




2
(t 2 ) 
2
 2t ( 2t ) − 1





−2θ
(θ 2 ) 
2
 1 − (θ 2 − 1) 



(1 − 9 x )
( 4t
2
− 1)
+ 2 t sec −1 2t
(b) (1 – x2) tan–1 x


−1
2
 + cos (θ − 1)  ( 2θ )


2θ 3
( 2θ 2 − θ 4 )
2θ 3
= 2θ cos −1 (θ 2 − 1) −
= 2θ cos −1 (θ 2 − 1) −
t
1 − (θ 4 − 2θ 2 + 1) 
= 2θ cos −1 (θ 2 − 1) −
= 2θ cos −1 (θ 2 − 1) −
+ 2sin −1 3 x
2
2θ 3
= 2θ cos −1 (θ 2 − 1) −
(b) If y = (1 – x2) tan–1 x then
6x


−1
 + ( sec 2t )( 2t ) =


9. Differentiate with respect to the variable: (a) θ2 cos–1(θ 2 – 1)
dy
(a) If y = θ 2 cos −1 (θ 2 − 1=
) then
dθ
1 + x2
(1 − x 2 + x 4 )
θ 2 (2 −θ 2 )
2θ 3
θ
(2 −θ 2 )
2θ 2
(2 −θ 2 )
1 − x2
dy
1
=
(1 − x 2 )  2  + ( tan −1 x )( −2 x ) =  2  − 2 x tan −1 x
dx
 1+ x 
 1+ x 
991
© 2014, John Bird
(b) x cosec–1 x
10. Differentiate with respect to the variable: (a) 2 t cot–1t
  1 − 1    −2 t
 −1 
−1 t
+
2 t 
cot
[
]

 2  t 2   =  1 + t 2
 1+ t2 
 
 2
( )
dy
t then
(a) If y = 2 t cot–1=
dt
 −2 t
=
 1+ t2


1 −1
− x 2

dy
2
−1
=
(b) If y = x cosec
x then
( x)
2
dx



 x  x − 1

( )
1

 + [ cot −1 t ]  1

 t2




 1
cot −1 t
 +
t



 + cosec −1 x  (1)

 




−x
 + cosec −1 x 
=

 2 x x ( x − 1)  


= cosec −1 x −
11. Differentiate with respect to the variable: (a)
(a) If y =
sin −1 3x
x2
1
2 ( x − 1)
sin −1 3x
x2
(b)
cos −1 x
1 − x2




3
−1
( x2 ) 
 − ( sin 3 x )( 2 x )
2
 1 − ( 3 x )  
dy

 
then
=
2
dx
( x2 )


3x 2

2
 1 − ( 3 x ) 


= 


−1
 − 2 x ( sin 3 x )


= x 
x4
x4 

1
=
x3
cos −1 x
(b) If y =
1 − x2
dy
then
=
dx
(
1− x
2
)







3x
(1 − 9 x 2 )
3x
(1 − 9 x 2 )

− 2sin −1 3 x 



− 2sin 3 x 

−1

1
1
 − ( cos −1 x )  (1 − x 2 )− 2 ( −2 x ) 
2

[1 − x 2 ] 
−1
(
992
1 − x2
)
2
© 2014, John Bird
−1 +
=
x
1
2 2
(1 − x )
1 − x2
( cos −1 x )
−1 +
=
993
x
( cos −1 x )
1 − x2
1 − x2
© 2014, John Bird
EXERCISE 247 Page 668
1. Use logarithmic equivalents of inverse hyperbolic functions to evaluate correct to 4 decimal
1
places: (a) sinh 2
–1
sinh −1
(b) sinh–1 4
 x + a 2 + x 2
x
= ln 
a
a

(a) sinh −1
(c) sinh–1 0.9



1 + 22 + 12 
 1+ 5 
1
= ln  =

 ln1.618034 = 0.4812
 ln=
2
2
2




(b) sinh −1 4 = sinh −1
 4 + 12 + 42 
4
= ln 
 =ln 4 + 17 =ln 8.123106 = 2.0947
1
1


(
 0.9 + 12 + 0.92
(c) sinh −1 0.9 = ln 
1

)

 = ln 0.9 + 1.81 = ln 2.245362 = 0.8089

(
)
Each answer may be checked using a calculator
2. Use logarithmic equivalents of inverse hyperbolic functions to evaluate correct to 4 decimal
5
places: (a) cosh–1 4
cosh −1
(b) cosh–1 3
(c) cosh–1 4.3
 x + x 2 − a 2 
x
= ln 

a
a


(a) cosh −1
 5 + 52 − 42 
5
 5+3
= ln 
=
 ln=

 ln 2 = 0.6931
4
4
 4 


 3 + 32 − 12
(b) cosh −1 3 = ln 
1


 = ln 3 + 8 = ln 5.828427 = 1.7627

 4.3 + 4.32 − 12
(c) cosh −1 4.3 = ln 
1

(
)

ln 4.3 + 17.49 =
ln 8.482105 = 2.1380
=

(
)
Each answer may be checked using a calculator.
994
© 2014, John Bird
3. Use logarithmic equivalents of inverse hyperbolic functions to evaluate correct to 4 decimal
1
places: (a) tanh 4
–1
(b) tanh–1
5
8
(c) tanh–1 0.7
x 1 a+x
= ln 

a 2 a−x
1  4 +1  1 5
1
(a) tanh −1 = ln 
 = ln = 0.2554
2  4 −1  2 3
4
tanh −1
(b) tanh −1
5 1  8 + 5  1 13
= ln 
= 0.7332
 = ln
8 2  8−5 2 3
(c) tanh −1 0.7 =
1  1 + 0.7  1 1.7
ln 
= 0.8673
 = ln
2  1 − 0.7  2 0.3
Each answer may be checked using a calculator
995
© 2014, John Bird
EXERCISE 248 Page 671
x
1. Differentiate with respect to the variable: (a) sinh 3
–1
x
(a) If y = sinh 3
–1
then
(b) If y = sinh −1 4x then
dy
=
dx
dy
=
dx
1
[ x 2 + 32 ]
1
=
4
( 4 x ) + 1


(b) sinh–1 4x
( x2 + 9)
=
2
4
(16 x 2 + 1)
t
2. Differentiate with respect to the variable: (a) 2 cosh–1 3
(a) If y = 2 cosh −1
t
3
then
dy
 1

= 2
 =
dt
 t 2 − 32 

dy 1
1
(b) If y = cosh–1 2θ then
=
2
dθ 2 

2x
–1
(a) If y = tanh 5
1
cosh–1 2θ
2
2
(t 2 − 9)

 =
2
( 2θ ) − 1 
3. Differentiate with respect to the variable:
(b)
2
1
( 4θ 2 − 1)
2x
(a) tanh–1 5
(b) 3 tanh–1 3x
2
2
2
2 25
×
10
dy
5
5
5
5
1 =
then=
=
=
=
2
2
2
2
4x
25 − 4 x
25 − 4x 2
dx
25 − 4 x
 2x 
1 −   1 − 25
25
 5 
(b) If y = 3 tanh −1 3x then


dy
3
9
= 3
 =
2
dx
1 − ( 3 x )  1 − 9x 2
4. Differentiate with respect to the variable:
3x
(a) sech–1 4
996
1
(b) − sech–1 2x
2
© 2014, John Bird
3x
dy
(a) If y = sech −1 =
then
4
d x 3x
4
3
−
−1
−1
4
=
=
=
2
2
  3x  
 9x 
 16 − 9 x 2  x
1
x
x
−




1 −   
16 

 16  4
  4  
=
(b) If y = −

dy 1
−2
1
sech–1 2x then
=

2
d x 2 2 x 1 − ( 2 x )2

5. Differentiate with respect to the variable:
x
(a) If y = cosech–1 4
then
(16 − 9 x 2 )
−4
x (16 − 9 x 2 )

−1
 =
 2 x (1 − 4 x 2 )

x
(a) cosech–1 4
−4
dy
=
=
d x x x 2 + 42
x
−1
(b)
1
cosech–1 4x
2
−4
( x 2 + 16 )




−1
1
−
y
d
1
4
(b) If y = cosech −1 4 x then
=
= 

2
d x 2  4 x  4 x 2 + 1  2 x (16 x 2 + 1)
( )


6. Differentiate with respect to the variable:
2x
(a) coth–1 7
(b)
1
coth–1 3t
4
2
2
2
2
(49)
2x
14
dy
7
7
7
7
(a) If y = coth −1
then=
=
=
=
=
2
2
2
2
−
x
x
4
49
4
7
49 − 4x 2
dx
49 − 4 x
 2x 
1 −   1 − 49
49
 7 
1
(b) If y = coth–1 3t
4
then
dy 1 3 
3
= 
 =
2
d t 4 1 − ( 3t )  4 (1 − 9t 2 )


7. Differentiate with respect to the variable:
(a) If y =2 sinh–1
2
(a) 2 sinh–1 ( x − 1)
1
 1

−
2 − 1 2 (2 x )
x


(
)
dy
(=
x 2 − 1)
 2

then
2=
2
dx
 


2
  x − 1 + 1 


(
997
)
(b)
1
cosh–1
2
( x 2 + 1)
2x
( x 2 − 1)
1
2
x2 −1 + 1
© 2014, John Bird
2x
=
x2 −1 x2 x
=
2x
( x 2 − 1)
1
 1

−
x 2 + 1) 2 (2 x) 

(
dy 1 2
1

=
(b) If y = cosh −1 =
( x 2 + 1) then
dx 2 
2

2

2
2
  ( x + 1) − 1 


x
=
2
8. Differentiate with respect to the variable:
( x 2 + 1) ( x 2 )
2
=
( x 2 − 1)
x
( x 2 + 1) ( x 2 + 1 − 1)
x
=
( x 2 + 1)
2x
(a) sech–1(x – 1)
1
=
2
( x 2 + 1)
(b) tanh–1(tanh x)
dy
−1
−1
1 ( x − 1)
(a) If y = sech −=
then
=
d x ( x − 1) [1 − ( x − 1) 2 ] ( x − 1) 1 − ( x 2 − 2 x + 1)
=
−1
−1
=
2
( x − 1) (2 x − x )
( x − 1) [ x(2 − x) ]
dy
sech 2 x
sech 2 x
(b) If y = tanh–1(tanh x) =
then
=1
=
2
d x 1 − ( tanh x ) 1 − tanh 2 x
since 1 − tanh 2 x =
sech 2 x from Chapter 24
9. Differentiate with respect to the variable:
 t 
(a) cosh  t − 1 
–1 
(b) coth–1(cos x)
(t − 1)(1) − (t )(1)
 t 
dy

(a) If y = cosh  t −=
1  then d t
–1 
( t − 1)
=
2
2
 t 

 −1
 t −1 
( t − 1)
−1
=
( t − 1)
2
t 2 − ( t − 1)
( t − 1)
2
=
2
=
998
t −1 − t
t2
2
( t − 1)
( t − 1)
2
2
−1
−1
t − t 2 + 2t − 1
(t − 1)
2
−1
( t − 1) 2t − 1
© 2014, John Bird
dy
− sin x
− sin x
− sin x
−1
(b) If y = coth −1 (cos x) then =
=
=
= = = −cosec x
2
2
2
d x 1 − ( cos x ) 1 − cos x sin x sin x
10. Differentiate with respect to the variable: (a) θ sinh–1 θ

dy
then
(a) If y = θ sinh −1 θ=
θ

dθ

(b) If y =

 + ( sinh −1 θ )(1) =
2

(θ + 1) 
1
( x)
dy
then
x cosh–1 x=
dx
=
2sech −1 t
t2
1 −1 
+ ( cosh −1 x )  x 2 
x2 −1
2

x
cosh −1 x
+
x2 −1
2 x
(a)
2sech −1 t
t2
(b)
tanh −1 x
(1 − x 2 )
  1 −1  
 2 − 2 t 2  
  − 2sech −1 t
2
(t )  
 t 1− t 2 


dy


=
2
2
dt
(t )
( )
then
θ
+ sinh −1 θ
2
(θ + 1)
1
11. Differentiate with respect to the variable:
(a) If y =
x cosh–1 x
(b)


2
(t ) 



=
 1  
−  
t 

− 2sech −1 t
t 1− t 


t4
(
−1
 t 1− t

(t 2 ) 
=
(
(

−1 t
 − 4t sech

t4
(
 −1 
−1 t

 − 4sech
−
1
t

= 
t3
999
)
)
=−
) ( 2t )
) ( 2t )
 −t

1− t
= 
(

−1 t
 − 4t sech

t4
)
1 1

+ 4sech −1 t 

t3  1− t

© 2014, John Bird
tanh −1 x
(1 − x 2 )
(b) If y =
12. Show that
dy
=
dθ
then
(1 − x 2 ) 
1 
 − ( tanh x )( −2 x )
 1 − x2 
(1 − x 2 )
2
=
1 + 2 x tanh −1 x
(1 − x 2 )
2
d
[x cosh–1(cosh x)] = 2x
dx
d
=
[ x cosh −1 (cosh x)]
dx


sinh x

 + ( cosh −1 (cosh x) ) (1)
x
( )
 ( cosh 2 x − 1) 


x sinh x
=
since cosh −1 (cosh x) = x
+x
2
( sinh x )
=
x sinh x
+x
sinh x
= x + x = 2x
13. Determine: (a)
(a)
∫
1
( x 2 + 9)
∫
dx=∫
1
( x 2 + 9)
dx
(b)
∫
3
dx
(4 x 2 + 25)
x
1
d x = sinh −1 + c
3
x 2 + 32
3
1
1
3
1
(b)
dx
=
∫ ( 4 x 2 + 25) d x 3=
∫  2 2  d x 3∫ =
∫
5
 2 x  2 
  2 x 2  
2x) + 5
(
2


5    + 1 
  + 1
   5 
 5 

 
2
3
35
5
=  ∫
dx= ∫
2
52
 2 x  2 
+
1
 

 5 

=
14. Determine: (a)
(a)
∫
∫
1
dx
( x − 16)
2
(b)
∫
2
5
 2 x  2 
  + 1
 5 

dx
3
2x
sinh −1
+c
2
5
1
dt
(t − 5)
2
x
1
1
dx=∫
d x = cosh −1 + c
2
2
4
( x − 16)
x −4
2
1000
© 2014, John Bird
1
(b)
=
∫ ( t 2 − 5) d t
1
1
1
dt
=
∫
5
 t  2 

 − 1
 5 

dt
∫=
 t

2
5  5 − 1  

 
=
15. Determine: (a)
dθ
(a) =
∫ (36 + θ 2 )
(b)
3
∫
dθ
(36 + θ 2 )
1
dθ
∫=
6 +θ
2
2
3
(b)
2
 t  2 

 − 1
 5 

1  5
dt


5  1  ∫  t 2 

 − 1
 5 

d t = cosh −1
t
+c
5
3
∫ (16 − 2 x ) d x
2
1
θ
1
6
d θ = tan −1 + c
∫
2
2
6
6
6
6 +θ
d x ∫=
dx
∫ (=
16 − 2 x )
2 (8 − x )
2
∫
1
5
1
5
3
1
3 1
=
dx  
2
∫
∫
2
2 8
8 − x2
( )
=
1001
3
2 8
( 8)
tanh −1
8
2
−x
dx
2
x
+c
8
© 2014, John Bird