Higher Engineering Mathematics, Sixth Edition

342 Higher Engineering Mathematics

Problem 13. Determine

d

dx

[

cosh−^1

√

(x^2 + 1 )

]

Ify=cosh−^1 f(x),

dy

dx

=

f′(x)

√

[f(x)]^2 − 1

Ify=cosh−^1

√

(x^2 + 1 ),thenf(x)=

√

(x^2 + 1 )and

f′(x)=

1

2

(x+ 1 )−^1 /^2 ( 2 x)=

x

√

(x^2 + 1 )

Hence,

d

dx

[

cosh−^1

√

(x^2 + 1 )

]

=

x

√

(x^2 + 1 )

√[

(√

(x^2 + 1 )

) 2

− 1

]=

x

√

(x^2 + 1 )

√

(x^2 + 1 − 1 )

=

x

√

(x^2 + 1 )

x

=

1

√

(x^2 +1)

Problem 14. Show that

d

dx

[

tanh−^1

x

a

]

=

a

a^2 −x^2

and hence determine the differential

coefficient of tanh−^1

4 x

3

Ify=tanh−^1

x

a

then

x

a

=tanhyandx=atanhy

dx

dy

=asech^2 y=a( 1 −tanh^2 y),since

1 −sech^2 y=tanh^2 y

=a

[

1 −

(x

a

) 2 ]

=a

(

a^2 −x^2

a^2

)

=

a^2 −x^2

a

Hence

dy

dx

=

1

dx

dy

=

a

a^2 −x^2

Comparing tanh−^1

4 x

3

with tanh−^1

x

a

shows thata=

3

4

Hence

d

dx

[

tanh−^1

4 x

3

]

=

3

(^4

3

4

) 2

−x^2

=

3

4

9

16

−x^2

=

3

4

9 − 16 x^2

16

=

3

4

·

16

( 9 − 16 x^2 )

=

12

9 − 16 x^2

Problem 15. Differentiate cosech−^1 (sinhθ).

From Table 33.2(v),

d

dx

[cosech−^1 f(x)]=

−f′(x)

f(x)

√

[f(x)]^2 + 1

Hence

d

dθ

[cosech−^1 (sinhθ)]

=

−coshθ

sinhθ

√

[sinh^2 θ+1]

=

−coshθ

sinhθ

√

cosh^2 θ

sincecosh^2 θ−sinh^2 θ= 1

=

−coshθ

sinhθcoshθ

=

− 1

sinhθ

=−cosechθ

Problem 16. Find the differential coefficient of

y=sech−^1 ( 2 x− 1 ).

From Table 33.2(iv),

d

dx

[sech−^1 f(x)]=

−f′(x)

f(x)

√

1 −[f(x)]^2

Hence,

d

dx

[sech−^1 ( 2 x− 1 )]

=

− 2

( 2 x− 1 )

√

[1−( 2 x− 1 )^2 ]

=

− 2

( 2 x− 1 )

√

[1−( 4 x^2 − 4 x+ 1 )]

=

− 2

( 2 x− 1 )

√

( 4 x− 4 x^2 )

=

− 2

( 2 x− 1 )

√

[4x( 1 −x)]

=

− 2

( 2 x− 1 ) 2

√

[x( 1 −x)]

=

− 1

( 2 x− 1 )

√

[x(1−x)]

Problem 17. Show that

d

dx

[coth−^1 (sinx)]=secx.

From Table 33.2(vi),

d

dx

[coth−^1 f(x)]=

f′(x)

1 −[f(x)]^2