342 Higher Engineering Mathematics
Problem 13. Determine
d
dx
cosh −1
(x 2 + 1)
If y = cosh −1 f (x),
dy
dx
=
f (x)
[ f (x)] 2 − 1
If y = cosh −1
(x 2 + 1), then f (x) =
(x 2 + 1) and
f (x) =
1
2
(x + 1) −1/2 (2x) =
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 4x
3
If y = tanh −1 x
a
then
x
a
= tanh y and x = a tanh y
dx
dy
= a sech 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 4x
3
with tanh −1 x
a
shows that a =
3
4
Hence
d
dx
tanh
−1 4x
3
=
3
4
3
4
2
− x 2
=
3
4
9
16
− x 2
=
3
4
9 − 16x 2
16
=
3
4
·
16
(9 − 16x 2 )
=
12
9 −16x 2
Problem 15. Differentiate cosech −1 (sinh θ).
From Table 33.2(v),
d
dx
[cosech
−1 f (x)] =
− f
(x)
f (x)
[ f (x)] 2 + 1
Hence
d
dθ
[cosech −1 (sinh θ)]
=
−cosh θ
sinh θ
[sinh
2
θ + 1]
=
−cosh θ
sinh θ
√
cosh 2 θ
since cosh
2
θ − sinh
2
θ = 1
=
−cosh θ
sinh θ cosh θ
=
−1
sinh θ
= −cosech θ
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 − 4x + 1)]
=
−2
(2x − 1)
(4x − 4x 2 )
=
−2
(2x −1)
√
[4x(1−x)]
=
−2
(2x − 1)2
√
[x(1 − x)]
=
−1
(2x − 1)
√
[x(1 −x)]
Problem 17. Show that
d
dx
[coth −1 (sin x)] = sec x.
From Table 33.2(vi),
d
dx
[coth
−1 f (x)] =
f (x)
1 − [ f (x)] 2