Question

If y = `tan^-1((2x)/(1 - x^2))`, x ∈ (−1, 1) then `("d"y)/("d"x)` = ______.

Options

• `(-2)/(1 + x^2)`

• 1

• `(2)/(1 + x^2)`

• `1/(1 + x^2)`

Answer 1

If y = `tan^-1((2x)/(1 - x^2))`, x ∈ (−1, 1) then `("d"y)/("d"x)` = `bbunderline((2)/(1 + x^2))`.

Explanation:

Put x = tan θ, `(-pi)/4 < theta < pi/4`

∴ tan⁻¹ x

`therefore y = tan^-1 ((2 tan theta)/(1 - tan^2theta))`

∴ y = tan⁻¹ (tan 2 θ)

∴ y = 20 

∴ y = 2 tan⁻¹ x

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