Differentiation of inverse trigonometric and hyperbolic functions 339
- 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.
- (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
⎤
⎥
⎥
⎦
- (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
⎤
⎥
⎥
⎥
⎦
- (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 )
⎤
⎥
⎥
⎦
- (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