Question

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

y = `cos^(-1) ((2x)/(1+x^2))`, −1 < x < 1

Answer 1

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

Let, x = tan θ

⇒ θ = tan⁻¹ x

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

= cos⁻¹ (sin 2 θ)

= `cos^-1 {cos (pi/2 - 2  theta)}`

= `pi/2 - 2  theta`

= `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)`