338 Higher Engineering Mathematics
Problem 6. Differentiate y =
cot
−1 2x
1 + 4x 2
Using the quotient rule:
dy
dx
=
(1 + 4x 2 )
−2
1 + (2x) 2
− (cot −1 2x)(8x)
(1 + 4x 2 ) 2
from Table 33.1(vi)
=
−2(1 +4x cot −1 2x)
(1 +4x 2 )
2
Problem 7. Differentiate y = x cosec
−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
sin t
cos t − 1
then
dy
dt
=
1
2
If
f (t ) =
sin t
cos t − 1
then f (t ) =
(cos t − 1)(cos t ) − (sin t )(−sin t )
(cos t − 1) 2
=
cos 2 t − cos t + sin 2 t
(cos t − 1) 2
=
1 − cos t
(cos t − 1) 2
since sin 2 t + cos 2 t = 1
=
−(cos t − 1)
(cos t − 1) 2 =
−1
cos t − 1
Using Table 33.1(iii), when
y = tan −1
sin t
cos t − 1
then
dy
dt
=
−1
cos t − 1
1 +
sin t
cos t − 1
2
=
−1
cos t − 1
(cos t − 1) 2 + (sin t ) 2
(cos t − 1) 2
=
−1
cos t − 1
(cos t − 1) 2
cos 2 t − 2 cos t + 1 + sin 2 t
=
−(cos t − 1)
2 − 2 cos t
=
1 − cos t
2(1 − cos t )
=
1
2
Now try the following exercise
Exercise 135 Further problems on
differentiating inverse trigonometric
functions
In Problems 1 to 6, differentiate with respect to the
variable.
1. (a) sin −1 4x (b) sin −1 x
2
(a)
4
√
1 − 16x 2
(b)
1
√
4 − x 2
2. (a) cos −1 3x (b)
2
3
cos −1 x
3
(a)
−3
√
1 − 9x 2
(b)
−2
3
√
9 − x 2
3. (a) 3 tan −1 2x (b)
1
2
tan −1 √
x
(a)
6
1 + 4x 2 (b)
1
4
√
x (1 + x)
4. (a) 2 sec −1 2t (b) sec −1 3
4
x
(a)
2
t
√
4t 2 − 1
(b)
4
x
√
9x 2 − 16
5. (a)
5
2
cosec −1 θ
2
(b) cosec −1 x 2
(a)
−5
θ
√
θ 2 − 4
(b)
−2
x
√
x 4 − 1
6. (a) 3 cot −1 2t (b) cot −1 √
θ 2 − 1
(a)
−6
1 + 4t 2 (b)
−1
θ
√
θ 2 − 1