Limits, Continuity & Differentiability
Differentiation of inverse trigonometric functions
Grade 12
Question:
<p>Given: \(f(x) = \sin^{-1}\left(\dfrac{2 \cdot 3^x}{1+(3^x)^2}\right)\). Find \(f'\!\left(-\dfrac{1}{2}\right)\).</p>
<p>\(\sqrt{3}\ln\sqrt{3}\)</p>
<p>\(\sqrt{3}\log_e\sqrt{3}\)</p>
<p>\(\sqrt{3}\ln 3\)</p>
<p>\(\sqrt{3}\log_e 3\)</p>
Step-by-Step Solution
Key Concept: Recognize that the argument of sin⁻¹ matches the derivative formula for tan⁻¹: d/dx[tan⁻¹(u)] = u'/(1+u²). Substitute u = 3ˣ to transform the composition and simplify the derivative.
<p><strong>Step 1: Recognize the substitution pattern.</strong> Let u = 3ˣ. Then:</p><p>f(x) = sin⁻¹(2u/(1+u²))</p><p><strong>Step 2: Use the inverse tangent identity.</strong> Recall that 2tan⁻¹(u) has the derivative form: d/dx[2tan⁻¹(u)] = 2u'/(1+u²).</p><p>Notice that sin⁻¹(2u/(1+u²)) = 2tan⁻¹(u) for the appropriate domain.</p><p><strong>Step 3: Apply differentiation.</strong> With u = 3ˣ, we have du/dx = 3ˣ·ln(3):</p><p>f'(x) = 2·(du/dx)/(1+(3ˣ)²) = 2·3ˣ·ln(3)/(1+3²ˣ)</p><p><strong>Step 4: Evaluate at x = -1/2.</strong></p><p>• 3⁻¹/² = 1/√3</p><p>• 3⁻¹ = 1/3</p><p>• 1 + 3⁻¹ = 1 + 1/3 = 4/3</p><p>f'(-1/2) = (2·(1/√3)·ln(3))/(4/3) = (2ln(3)/√3)·(3/4) = (3ln(3))/(2√3) = (√3·ln(3))/2</p><p>∴ Answer: B</p>
Correct Answer: B