Limits, Continuity & Differentiability
Higher order derivatives and 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 half-angle substitution: if tan(θ/2) = t, then tan⁻¹(t) = θ/2. Here, (√(1+x²)-1)/x = tan(θ/2) where θ relates to x, allowing y to be expressed as a simple function before differentiation.
<p><strong>Step 1: Recognize the half-angle form</strong></p><p>Let x = tan(α). Then √(1+x²) = sec(α), so:</p><p>(√(1+x²)-1)/x = (sec(α)-1)/tan(α) = tan(α/2)</p><p>This is a standard half-angle identity.</p><p><strong>Step 2: Simplify y</strong></p><p>y = 2tan⁻¹(tan(α/2)) = 2·(α/2) = α = tan⁻¹(x)</p><p><strong>Step 3: Find first derivative</strong></p><p>dy/dx = d/dx[tan⁻¹(x)] = 1/(1+x²)</p><p><strong>Step 4: Find second derivative</strong></p><p>d²y/dx² = d/dx[1/(1+x²)] = -2x/(1+x²)²</p><p><strong>Step 5: Evaluate at x = 2</strong></p><p>d²y/dx²|ₓ₌₂ = -2(2)/(1+4)² = -4/25</p><p>∴ Answer: <strong>-4/25</strong> (Option B)</p>
Correct Answer: B