Higher Engineering Mathematics, Sixth Edition

Differentiation of inverse trigonometric and hyperbolic functions 339

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

+2sin−^13 x

(b)

t

√

4 t^2 − 1

+ 2 tsec−^12 t

⎤

⎥

⎥

⎦

9. (a)θ^2 cos−^1 (θ^2 − 1 )(b)( 1 −x^2 )tan−^1 x
   ⎡
   ⎢
   ⎢
   ⎢
   ⎣

(a) 2θcos−^1 (θ^2 − 1 )−

2 θ^2

√

2 −θ^2

(b)

(

1 −x^2

1 +x^2

)

− 2 xtan−^1 x

⎤

⎥

⎥

⎥

⎦

10. (a) 2

√

tcot−^1 t(b)xcosec−^1

√ x ⎡ ⎢ ⎢ ⎣

(a)

− 2

√

t

1 +t^2

+

1

√

t

cot−^1 t

(b) cosec−^1

√

x−

1

2

√

(x− 1 )

⎤

⎥

⎥

⎦

11. (a)

sin−^13 x

x^2

(b)

cos−^1 x

√

1 −x^2

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

(a)

1

x^3

{

3 x

√

1 − 9 x^2

−2sin−^13 x

}

(b)

− 1 +

x

√

1 −x^2

cos−^1 x

( 1 −x^2 )

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

33.3 Logarithmic forms of inverse

hyperbolic functions

Inverse hyperbolic functions may be evaluated most
conveniently when expressed in a logarithmic
form.
For example, ify=sinh−^1

x

a

then

x

a

=sinhy.

From Chapter 5, ey=coshy+sinhyand
cosh^2 y−sinh^2 y=1, from which,
coshy=

√
1 +sinh^2 ywhich is positive since coshyis
always positive (see Fig. 5.2, page 43).

Hence ey=

√

1 +sinh^2 y+sinhy

=

√[

1 +

(x

a

) 2 ]

+

x

a

=

√(

a^2 +x^2

a^2

)

+

x

a

=

√

a^2 +x^2

a

+

x

a

or

x+

√

a^2 +x^2

a

Taking Napierian logarithms of both sides gives:

y=ln

{

x+

√

a^2 +x^2

a

}

Hence, sinh−^1

x

a

=ln

{

x+

√

a^2 +x^2

a

}

(1)

Thus to evaluate sinh−^1

3

4

,letx=3anda=4in

equation (1).

Then sin h−^1

3

4

=ln

{

3 +

√

42 + 32

4

}

=ln

(

3 + 5

4

)

=ln2= 0. 6931

By similar reasoning to the above it may be shown that:

cosh−^1

x

a

=ln

{

x+

√

x^2 −a^2

a

}

and tanh−^1

x

a

=

1

2

ln

(

a+x

a−x

)

Problem 9. Evaluate, correct to 4 decimal places,

sinh−^1 2.

From above, sinh−^1

x

a

=ln

{

x+

√

a^2 +x^2

a

}

Withx=2anda=1,

sinh−^12 =ln

{

2 +

√

12 + 22

1

}

=ln( 2 +

√

5 )=ln4. 2361

= 1. 4436 ,correct to 4 decimal places

Using a calculator,

(i) press hyp

(ii) press 4 and sinh−^1 ( appears

(iii) type in 2