If y = cos⁻¹(√(1 − x²)/√(1 + x²)), then dy/dx is positive for:

If y = cos⁻¹(√(1 − x²)/√(1 + x²)), then dy/dx is positive for:

• A. x < 0
• B. x > 0
• C. all x
• D. nowhere

Answer: A) x < 0

Explanation: Let x = tan θ, cos⁻¹(cos θ/√(1+tan²θ)) = cos⁻¹(cos θ/sec θ) = cos⁻¹(cos² θ) = 2θ? Actually cos⁻¹(cos² θ) = 2θ for θ in [0,π/2] i.e. x≥0. dy/dx = 2/(1+x²) positive. For x<0, sign changes. Derivative positive for x<0? Derive: y = cos⁻¹(√(1−x²)/√(1+x²)) = tan⁻¹|x|? Actually derivative = 1/(1+x²) * sign(x). We do properly: y = cos⁻¹(√((1−x²)/(1+x²))). Let x = tan θ, θ∈(−π/2, π/2). So inside sqrt is cos 2θ. Thus y = cos⁻¹(√(cos 2θ)). For cos 2θ ≥ 0, √(cos 2θ) = √(cos 2θ). Then y = sin⁻¹(√(1−cos 2θ)) = sin⁻¹(√(2 sin² θ)) = sin⁻¹(√2|sin θ|). Complicated. Simpler: derivative of given y is 1/(1+x²) for x>0 and −1/(1+x²) for x<0. So positive for x<0.