differentiation

336 DIFFERENTIAL CALCULUS
Problem 7. Differentiate y = x cosec −1 x.
Using the product rule:
[ ]
dy
dx = (x) −1
x √ + (cosec −1 x) (1)
x 2 − 1
from Table 33.1(v)
= √ −1
x 2 − 1 + cosec−1 x
Problem 8. Show that if
( ) sin t
y = tan −1 then dy
cos t − 1 dt = 1 2
If
( ) sin t
f (t) =
cos t − 1
then f ′ (cos t − 1)(cos t) − (sin t)(−sin t)
(t) =
(cos t − 1) 2
= cos2 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
( ) sin t
y = tan −1 cos t − 1
−1
then dy
dt = cos t − 1
( ) sin t 2
1 +
cos t − 1
−1
=
cos t − 1
(cos t − 1) 2 + (sin t) 2
(cos t − 1) 2
( )( −1
(cos t − 1) 2 )
=
cos t − 1 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 136 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
[
]
4
(a) √ (b) 1
√
1 − 16x 2 4 − x 2
2. (a) cos −1 3x (b) 2 x 3 cos−1 3
[
]
−3
(a) √ (b) −2
1 − 9x 2 3 √ 9 − x 2
3. (a) 3 tan −1 2x (b) 1 √ 2 tan−1 x
[
]
6
(a)
1 + 4x 2 (b) 1
4 √ x(1 + x)
4. (a) 2 sec −1 2t (b) sec −1 3 4 x
[
]
2
(a)
t √ 4t 2 − 1 (b) 4
x √ 9x 2 − 16
5. (a) 5 θ 2 cosec−1 2 (b) cosec−1 x 2
[
]
−5
(a)
θ √ θ 2 − 4 (b) −2
x √ x 4 − 1
6. (a) 3 cot −1 2t (b) cot −1 √ θ 2 − 1
[
]
−6
(a)
1 + 4t 2 (b) −1
θ √ θ 2 − 1
7. Show that the differential coefficient of
tan −1 x
1 − x 2 is 1 + x 2
1 − x 2 + x 4
In Problems 8 to 11 differentiate with respect to
the variable.
8. (a) 2x sin −1 3x (b) t 2 sec −1 2t
⎡
6x
⎤
(a) √
⎢
+ 2 ⎣ 1 − 9x 2 sin−1 3x
⎥
t
(b) √
4t 2 − 1 + 2t ⎦
sec−1 2t