Differentiation of inverse trigonometric and hyperbolic functions 337
is negative between C and D and thus the differential
coefficient
dy
dx
is negative as shown above.
(b) If y = cos −1 f (x) then by letting u = f (x),
y = cos −1 u
Then
dy
du
= −
1
√
1 − u 2
(from part (a))
and
du
dx
= f (x)
From the function of a function rule,
dy
dx
=
dy
du
·
du
dx
= −
1
√
1 − u 2
f
(x)
=
− f (x)
1 − [ f (x)] 2
Hence, when y = cos
−1
(1 − 2x
2
)
then
dy
dx
=
−(−4x)
1 − [1 − 2x 2 ] 2
=
4x
1 − (1 − 4x 2 + 4x 4 )
=
4x
(4x 2 − 4x 4 )
=
4x
[4x 2 (1 − x 2 )]
=
4x
2x
√
1 − x 2
=
2
√
1 − x 2
Problem 3. Determine the differential coefficient
of y = tan −1 x
a
and show that the differential
coefficient of tan −1 2x
3
is
6
9 + 4x 2
If y = tan −1 x
a
then
x
a
= tan y and x = a tan y
dx
dy
= a sec
2 y = a(1 + tan
2 y) since
sec
2 y = 1 + tan
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
The principal value of y = tan −1 x is defined as
the angle lying between −
π
2
and
π
2
and the gradient
i.e.
dy
dx
between these two values is always positive
(see Fig. 33.1(c)).
Comparing tan −1 2x
3
with tan −1 x
a
shows that a =
3
2
Hence if y = tan −1 2x
3
then
dy
dx
=
3
2
3
2
2
+ x 2
=
3
2
9
4
+ x 2
=
3
2
9 + 4x 2
4
=
3
2
(4)
9 + 4x 2 =
6
9 + 4x 2
Problem 4. Find the differential coefficient of
y = ln (cos
−1 3x).
Let u = cos −1 3x then y = ln u.
By the function of a function rule,
dy
dx
=
dy
du
·
du
dx
=
1
u
×
d
dx
(cos
−1 3x)
=
1
cos −1 3x
−3
1 − (3x) 2
i.e.
d
dx
[ln(cos −1 3x)]=
−3
√
1 − 9x 2 cos −1 3x
Problem 5. If y = tan −1 3
t 2 find
dy
dt
Using the general form from Table 33.1(iii),
f (t ) =
3
t 2 = 3t
−2
,
from which f (t ) =
−6
t 3
Hence
d
dt
tan −1 3
t 2
=
f (t )
1 + [ f (t )] 2
=
−
6
t 3
1 +
3
t 2
2
=
−
6
t 3
t 4 + 9
t 4
=
−
6
t 3
t 4
t 4 + 9
= −
6t
t 4 + 9