338 Higher Engineering Mathematics
Problem 6. Differentiatey=
cot−^12 x
1 + 4 x^2
Using the quotient rule:
dy
dx
=
( 1 + 4 x^2 )
(
− 2
1 +( 2 x)^2
)
−(cot−^12 x)( 8 x)
( 1 + 4 x^2 )^2
from Table 33.1(vi)
=
−2(1+ 4 xcot−^12 x)
(1+ 4 x^2 )^2
Problem 7. Differentiatey=xcosec−^1 x.
Using the product rule:
dy
dx
=(x)
[
− 1
x
√
x^2 − 1
]
+(cosec−^1 x)( 1 )
from Table 33.1(v)
=
− 1
√
x^2 − 1
+cosec−^1 x
Problem 8. Show that if
y=tan−^1
(
sint
cost− 1
)
then
dy
dt
=
1
2
If f(t)=
(
sint
cost− 1
)
then f′(t)=
(cost− 1 )(cost)−(sint)(−sint)
(cost− 1 )^2
=
cos^2 t−cost+sin^2 t
(cost− 1 )^2
=
1 −cost
(cost− 1 )^2
since sin^2 t+cos^2 t= 1
=
−(cost− 1 )
(cost− 1 )^2
=
− 1
cost− 1
Using Table 33.1(iii), when
y=tan−^1
(
sint
cost− 1
)
then
dy
dt
=
− 1
cost− 1
1 +
(
sint
cost− 1
) 2
=
− 1
cost− 1
(cost− 1 )^2 +(sint)^2
(cost− 1 )^2
=
(
− 1
cost− 1
)(
(cost− 1 )^2
cos^2 t−2cost+ 1 +sin^2 t
)
=
−(cost− 1 )
2 −2cost
=
1 −cost
2 ( 1 −cost)
=
1
2
Now try the following exercise
Exercise 135 Further problems on
differentiating inverse trigonometric
functions
In Problems 1 to 6, differentiatewith respect to the
variable.
- (a) sin−^14 x (b) sin−^1
x
2
[
(a)
4
√
1 − 16 x^2
(b)
1
√
4 −x^2
]
- (a) cos−^13 x(b)
2
3
cos−^1
x
3
[
(a)
− 3
√
1 − 9 x^2
(b)
− 2
3
√
9 −x^2
]
- (a) 3tan−^12 x(b)
1
2
tan−^1
√
x
[
(a)
6
1 + 4 x^2
(b)
1
4
√
x( 1 +x)
]
- (a) 2sec−^12 t(b) sec−^1
3
4
x
[
(a)
2
t
√
4 t^2 − 1
(b)
4
x
√
9 x^2 − 16
]
- (a)
5
2
cosec−^1
θ
2
(b) cosec−^1 x^2
[
(a)
− 5
θ
√
θ^2 − 4
(b)
− 2
x
√
x^4 − 1
]
- (a) 3 cot−^12 t(b) cot−^1
√
θ^2 − 1
[
(a)
− 6
1 + 4 t^2
(b)
− 1
θ
√
θ^2 − 1
]