Differentiation of inverse trigonometric and hyperbolic functions 343
Hence
d
dx
[coth−^1 (sinx)]=
cosx
[1−(sinx)^2 ]
=
cosx
cos^2 x
since cos^2 x+sin^2 x= 1
=
1
cosx
=secx
Problem 18. Differentiate
y=(x^2 − 1 )tanh−^1 x.
Using the product rule,
dy
dx
=(x^2 − 1 )
(
1
1 −x^2
)
+(tanh−^1 x)( 2 x)
=
−( 1 −x^2 )
( 1 −x^2 )
+ 2 xtanh−^1 x= 2 xtanh−^1 x− 1
Problem 19. Determine
∫
dx
√
(x^2 + 4 )
.
Since
d
dx
(
sinh−^1
x
a
)
=
1
√
(x^2 +a^2 )
then
∫
dx
√
(x^2 +a^2 )
=sinh−^1
x
a
+c
Hence
∫
1
√
(x^2 + 4 )
dx =
∫
1
√
(x^2 + 22 )
dx
=sinh−^1
x
2
+c
Problem 20. Determine
∫
4
√
(x^2 − 3 )
dx.
Since
d
dx
(
cosh−^1
x
a
)
=
1
√
(x^2 −a^2 )
then
∫
1
√
(x^2 −a^2 )
dx=cosh−^1
x
a
+c
Hence
∫
4
√
(x^2 − 3 )
dx= 4
∫
1
√
[x^2 −(
√
3 )^2 ]
dx
=4cosh−^1
x
√
3
+c
Problem 21. Find
∫
2
( 9 − 4 x^2 )
dx.
Since tanh−^1
x
a
=
a
a^2 −x^2
then
∫
a
a^2 −x^2
dx=tanh−^1
x
a
+c
i.e.
∫
1
a^2 −x^2
dx=
1
a
tanh−^1
x
a
+c
Hence
∫
2
( 9 − 4 x^2 )
dx= 2
∫
1
4
( 9
4 −x
2 )dx
=
1
2
∫
1
[(
3
2
) 2
−x^2
]dx
=
1
2
[
1
( 3
2
)tanh−^1
x
( 3
2
)+c
]
i.e.
∫
2
( 9 − 4 x^2 )
dx=
1
3
tanh−^1
2 x
3
+c
Now try the following exercise
Exercise 137 Further problems on
differentiation of inverse hyperbolic
functions
In Problems 1 to 11, differentiate with respect to
the variable.
- (a) sinh−^1
x
3
(b) sinh−^14 x
[
(a)
1
√
(x^2 + 9 )
(b)
4
√
( 16 x^2 + 1 )
]
- (a) 2 cosh−^1
t
3
(b)
1
2
cosh−^12 θ
[
(a)
2
√
(t^2 − 9 )
(b)
1
√
( 4 θ^2 − 1 )
]
- (a) tanh−^1
2 x
5
(b) 3 tanh−^13 x
[
(a)
10
25 − 4 x^2
(b)
9
( 1 − 9 x^2 )
]
- (a) sech−^1
3 x
4
(b)−
1
2
sech−^12 x
[
(a)
− 4
x
√
( 16 − 9 x^2 )
(b)
1
2 x
√
( 1 − 4 x^2 )
]