Higher Engineering Mathematics

338 DIFFERENTIAL CALCULUS

Thus,

x

a

=

e^2 y− 1

e^2 y+ 1

from which, x(e^2 y+1)=a(e^2 y−1)

Hencex+a=ae^2 y−xe^2 y=e^2 y(a−x)

from which e^2 y=

(

a+x

a−x

)

Taking Napierian logarithms of both sides gives:

2 y=ln

(

a+x

a−x

)

and y=

1

2

ln

(

a+x

a−x

)

Hence,tanh−^1

x

a

=

1

2

ln

(

a+x

a−x

)

Substitutingx=3 anda=5 gives:

tanh−^1

3

5

=

1

2

ln

(

5 + 3

5 − 3

)

=

1

2

ln 4

=0.6931, correct to 4 decimal places

Problem 11. Prove that

cosh−^1

x

a

=ln

{

x+

√

x^2 −a^2

a

}

and hence evaluate cosh−^1 1.4 correct to

4 decimal places.

Ify=cosh−^1

x

a

then

x

a

=cosy

ey=coshy+sinhy=coshy±

√

cosh^2 y− 1

=

x

a

±

√[

(x

a

) 2

− 1

]

=

x

a

±

√

x^2 −a^2

a

=

x±

√

x^2 −a^2

a

Taking Napierian logarithms of both sides gives:

y=ln

{

x±

√

x^2 −a^2

a

}

Thus, assuming the principal value,

cosh−^1

x

a

=ln

{

x+

√

x^2 −a^2

a

}

cosh−^11. 4 =cosh−^1

14

10

=cosh−^1

7

5

In the equation for cosh−^1

x

a

, letx=7 anda= 5

Then cosh−^1

7

5

=ln

{

7 +

√

72 − 52

5

}

=ln 2. 3798 =0.8670,

correct to 4 decimal places

Now try the following exercise.

Exercise 137 Further problems on logarith-

mic forms of the inverse hyperbolic functions

In Problems 1 to 3 use logarithmic equivalents of

inverse hyperbolic functions to evaluate correct

to 4 decimal places.

1. (a) sinh−^1

1

2

(b) sinh−^1 4 (c) sinh−^1 0.9

[(a) 0.4812 (b) 2.0947 (c) 0.8089]

2. (a) cosh−^1

5

4

(b) cosh−^1 3 (c) cosh−^1 4.3

[(a) 0.6931 (b) 1.7627 (c) 2.1380]

3. (a) tanh−^1

1

4

(b) tanh−^1

5

8

(c) tanh−^1 0.7

[(a) 0.2554 (b) 0.7332 (c) 0.8673]

33.4 Differentiation of inverse

hyperbolic functions

Ify=sinh−^1

x

a

then

x

a

=sinhyandx=asinhy

dx

dy

=acoshy(from Chapter 32).

Also cosh^2 y−sinh^2 y=1, from which,

coshy=

√

1 +sinh^2 y=

√[

1 +

(x

a

) 2 ]

=

√

a^2 +x^2

a

Hence

dx

dy

=acoshy=

a

√

a^2 +x^2

a

=

√

a^2 +x^2

Then

dy

dx

=

1

dx

dy

=

1

√

a^2 +x^2