Limits, Continuity & Differentiability
Differentiation
Grade None
Question:
<p>If \(f(x) = \sin^{-1}\left(\dfrac{2 \times 3^x}{1+9^x}\right)\), then \(f'\left(-\dfrac{1}{2}\right)\) equals</p>
<p>\(-\sqrt{3}\log_e \sqrt{3}\)</p>
<p>\(\sqrt{3}\log_e \sqrt{3}\)</p>
<p>\(-\sqrt{3}\log_e 3\)</p>
<p>\(\sqrt{3}\log_e 3\)</p>
Step-by-Step Solution
Key Concept: Recognize that the argument of sin⁻¹ has the form of the double angle formula: 2tanθ/(1+tan²θ) = sin(2θ). Substitute 3^x = tanθ to simplify before differentiating.
<p><strong>Step 1:</strong> Recognize the pattern in the argument. Let θ = tan⁻¹(3^x), so 3^x = tanθ.</p><p><strong>Step 2:</strong> Use the identity: 2tanθ/(1+tan²θ) = sin(2θ)</p><p>Therefore: f(x) = sin⁻¹(sin(2tan⁻¹(3^x))) = 2tan⁻¹(3^x)</p><p><strong>Step 3:</strong> Differentiate f(x) = 2tan⁻¹(3^x)</p><p>f'(x) = 2 · 1/(1+(3^x)²) · 3^x·ln(3)</p><p>f'(x) = (2·3^x·ln3)/(1+9^x)</p><p><strong>Step 4:</strong> Evaluate at x = -1/2:</p><p>3^(-1/2) = 1/√3, so 9^(-1/2) = 1/3</p><p>f'(-1/2) = (2·(1/√3)·ln3)/(1+1/3) = (2ln3/√3)/(4/3) = (2ln3/√3)·(3/4) = (3ln3)/(2√3) = (√3·ln3)/2</p><p>∴ Answer: D</p>
Correct Answer: D