, x ∈ (– 1, 1) To make this result a bit more attractive, we carry out the following manipulation Recall that for x ∈ (– 1, 1), sin (sin–1 x) = x and hence
cos2 y = 1 – (sin y)2 = 1 – (sin (sin–1 x))2 = 1 – x2
Also, since y ∈ 
,
2 2
π π


−


, cos y is positive and hence cos y = 
2
1
x
−
Thus, for x ∈ (– 1, 1),
2
1
1
cos
1
dy
dx
y
x
=
=
−
2
1
1
x
−
2
1
1
x
−
−
2
1
1
x
+
f(x)                    sin–1 x                      cos-1 x                   tan-1x
Domain off               (-1, 1)                     (-1, 1)                        R
f 1(x)
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
125
EXERCISE 5 3
Find dy
dx  in the following:
1