Question

The derivative of `sin^-1 (2"x" sqrt(1 - "x"^2))` w.r.t sin⁻¹ x, `-1/sqrt2 < "x" < 1/sqrt2`, is:

पर्याय

• 2

• `π/2 - 2`

• `π/2`

• −2

Answer 1

2

Explanation:

Let u = `sin^-1 (2"x" sqrt(1 - "x"^2))`

and v = sin⁻¹ x, `-1/sqrt2 < "x" < 1/sqrt2`

sin v = x .........(1)

Using (1), we get:

sin⁻¹ (2 sin v cos v) = sin⁻¹ x (sin 2v)

u = 2v, `-π/2 < 2"v" < π/2`

Differentiating u with respect to v, we get:

`("du")/"dv"` = 2