If y = tan⁻¹(√(1+x²) + √(1-x²))/(√(1+x²) - √(1-x²)); x² ≤ 1, then find dy/dx.

Given y = tan⁻¹(√(1+x²) + √(1-x²))/(√(1+x²) - √(1-x²))
Put x² = cos2θ ⇒ θ = (1/2)cos⁻¹x² — (1)
= tan⁻¹(√(1+cos2θ) + √(1-cos2θ))/(√(1+cos2θ) - √(1-cos2θ))
⇒ tan⁻¹(√2cosθ + √2sinθ)/(√2cosθ - √2sinθ) [ Since, 1+cos2θ = 2cos²θ and 1-cos2θ = 2sin²θ ]
⇒ tan⁻¹(1 + tanθ)/(1 - tanθ)
⇒ tan⁻¹(tan(π/4 + θ)) = π/4 + θ
⇒ y = π/4 + (1/2)cos⁻¹x²
dy/dx = 0 - 1/(√(1-x⁴)) * (2x)/2 = -x/√(1-x⁴)