Question

Find `bb(dy/dx)` in the following:

y = `sin^(-1) ((1-x^2)/(1+x^2))`, 0 < x < 1

Answer 1

y = `sin^-1 ((1 - x^2)/(1 + x^2))`

Putting x = tan θ

⇒ θ = tan⁻¹ x

∴ y = `sin^-1 ((1 - tan^2 theta)/(1 + tan^2 theta))`

= sin⁻¹ (cos 2 θ)

= `sin^-1 [sin (pi/2 - 2 theta)]`

= `pi/2 - 2theta`

= `pi/2 - 2 tan^-1 x`

On differentiating with respect to x,

`dy/dx = pi/2 d/dx (1) - 2 d/dx tan^-1 x`

`dy/dx = 0 - 2 xx 1/(1 + x^2)`

`dy/dx = -2/(1 + x^2)`