Question

7. sin-1 (2x2 - 1), 0<x<1​

Answer 1

Step-by-step explanation:

Let x = sinA. Then 1–2x^2=1–2sin^2 A

=cos 2A = sin(π/2 - 2A)

d{sin^-1 (1–2x^2)}/dx

=d[sin^-1{sin(π/2 - 2A)}]/dx

=d(π/2 - 2A)/dx

=d {π/2 - 2 sin^-1 x}/dx

=d(π/2)/dx -2 d(sin^-1 x)/dx

= 0 -2×{1/√(1-x^2)} =-2/√(1-x^2).