Limits, Continuity & Differentiability
Higher order derivatives / inverse trigonometric differentiation
Grade 12
Question:
<p><strong>200.</strong> If \(y = 2\tan^{-1}\!\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)\), then the value of \(\dfrac{d^2y}{dx^2}\) at \(x = 2\) is:</p>
<p>(a) \(\dfrac{2}{36}\)</p>
<p>(b) \(\dfrac{-4}{25}\)</p>
<p>(c) \(\dfrac{-4}{5}\)</p>
<p>(d) \(\dfrac{4}{25}\)</p>
Step-by-Step Solution
Key Concept: Simplify the inverse tangent argument using the substitution x = tan(θ), which converts √(1+x²)-1 to a trigonometric expression, then differentiate twice to find the second derivative.
<p><strong>Step 1:</strong> Simplify the inverse tangent argument. Let x = tan(θ), then √(1+x²) = sec(θ).</p><p>So (√(1+x²)-1)/x = (sec(θ)-1)/tan(θ) = (1-cos(θ))/(sin(θ)) = tan(θ/2) [using half-angle formula]</p><p><strong>Step 2:</strong> Therefore y = 2tan⁻¹(tan(θ/2)) = 2·(θ/2) = θ = tan⁻¹(x)</p><p><strong>Step 3:</strong> Find dy/dx: dy/dx = 1/(1+x²)</p><p><strong>Step 4:</strong> Find d²y/dx²: d²y/dx² = d/dx[1/(1+x²)] = -2x/(1+x²)²</p><p><strong>Step 5:</strong> Evaluate at x = 2: d²y/dx²|ₓ₌₂ = -2(2)/(1+4)² = -4/25</p><p>∴ Answer: B</p>
Correct Answer: B