Higher Engineering Mathematics

336 DIFFERENTIAL CALCULUS

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−2 cost+ 1 +sin^2 t

)

=

−(cost−1)

2 −2 cost

=

1 −cost

2(1−cost)

=

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−^14 x(b) sin−^1

x

2

[

(a)

4

√

1 − 16 x^2

(b)

1

√

4 −x^2

]

2. (a) cos−^13 x (b)

2

3

cos−^1

x

3

[

(a)

− 3

√

1 − 9 x^2

(b)

− 2

3

√

9 −x^2

]

3. (a) 3 tan−^12 x (b)

1

2

tan−^1

√

x

[

(a)

6

1 + 4 x^2

(b)

1

4

√

x(1+x)

]

4. (a) 2 sec−^12 t (b) sec−^1

3

4

x

[

(a)

2

t

√

4 t^2 − 1

(b)

4

x

√

9 x^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−^12 t (b) cot−^1

√

θ^2 − 1

[

(a)

− 6

1 + 4 t^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) 2xsin−^13 x(b)t^2 sec−^12 t
   ⎡

⎢

⎣

(a)

6 x

√

1 − 9 x^2

+2 sin−^13 x

(b)

t

√

4 t^2 − 1

+ 2 tsec−^12 t

⎤

⎥

⎦