Question

If y = `tan^-1  (sqrt(1 + x^2) - sqrt(1 - x^2))/(sqrt(1 + x^2) + sqrt(1 - x^2))`, then `dy/dx` is equal to ______.

Options

• `x^2/sqrt(1 - x^4)`

• `x^2/sqrt(1 + x^4)`

• `x/sqrt(1 + x^4)`

• `x/sqrt(1 - x^4)`

Answer 1

If y = `tan^-1  (sqrt(1 + x^2) - sqrt(1 - x^2))/(sqrt(1 + x^2) + sqrt(1 - x^2))`, then `dy/dx` is equal to `underlinebb(x/sqrt(1 - x^4))`.

Explanation:

Given, y = `tan^-1  (sqrt(1 + x^2) - sqrt(1 - x^2))/(sqrt(1 + x^2) + sqrt(1 - x^2))`

Put x² = cos 2θ in the given equation,

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

= `tan^-1  (cos θ - sin θ)/(cos θ + sin θ)`

= `tan^-1  (cosθ/cosθ - sinθ/cosθ)/(cosθ/cosθ + sinθ/cosθ)`

= `tan^-1  ((1 - tanθ))/((1 + tanθ))`

= `tan^-1 {tan(π/4 - θ)}`

`\implies` y = `π/4 - θ = π/4 - 1/2 cos^-1x^2`

`\implies dy/dx = 0 - 1/2((-2x)/sqrt(1 - x^4)) = x/sqrt(1 - x^4)`