Question

If f(x) = sin^-1`(sqrt((1 - x)/2))`, then f'(x) = ?

पर्याय

• `1/(2sqrt(1 + x^2))`

• `(-1)/(2sqrt(1 + x^2))`

• `1/(sqrt(1 - x^2))`

• `(-1)/(2sqrt(1 - x^2))`

Answer 1

`(-1)/(2sqrt(1 - x^2))`

Explanation:

We have,

f(x) = `sin^-1 (sqrt((1 - x)/2))`

Put, x = cos 2θ ⇒ θ = `1/2 cos^-1 x`

∴ f(x) = `sin^-1sqrt((1 - cos 2theta)/2)`

`=> "f"(x) = sin^-1 sqrt((2 sin^2 theta)/2) = sin^-1 (sin theta)`

⇒ f(x) = θ = `1/2 cos^-1 x`

⇒ f'(x) = `1/2 ((- 1)/sqrt(1 - x^2)) = (- 1)/(2 sqrt(1 - x^2))`