Higher Engineering Mathematics

340 DIFFERENTIAL CALCULUS

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 that

a=

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

since cosh^2 θ−sinh^2 θ= 1

=

−coshθ

sinhθcoshθ

=

− 1

sinhθ

=−cosechθ

Problem 16. Find the differential coefficient of

y=sech−^1 (2x−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 (2x−1)]

=

− 2

(2x−1)

√

[1−(2x−1)^2 ]

=

− 2

(2x−1)

√

[1−(4x^2 − 4 x+1)]

=

− 2

(2x−1)

√

(4x− 4 x^2 )

=

− 2

(2x−1)

√

[4x(1−x)]

=

− 2

(2x−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

Hence

d

dx

[coth−^1 (sinx)]=

cosx

[1−(sinx)^2 ]

=

cosx

cos^2 x

since cos^2 x+sin^2 x= 1

=

1

cosx

=secx

Problem 18. Differentiate

y=(x^2 −1) tanh−^1 x.

Using the product rule,

dy

dx

=(x^2 −1)

(

1

1 −x^2

)

+( tanh−^1 x)(2x)

=

−(1−x^2 )

(1−x^2 )

+ 2 xtanh−^1 x= 2 xtanh−^1 x− 1