<p>If \(x^{2x} - 2x^x \cot y - 1 = 0\), then \(\dfrac{dy}{dx}\) at \(\left(1, \dfrac{\pi}{2}\right)\) is:</p>
Step-by-Step Solution
Key Concept: Use implicit differentiation on the equation, recognizing that x^x requires the formula d/dx(x^x) = x^x(ln x + 1), and that at the point (1, π/2), we have x^x = 1 and cot(π/2) = 0.
<p><strong>Step 1:</strong> Given: x^(2x) - 2x^x cot y - 1 = 0</p><p><strong>Step 2:</strong> Differentiate implicitly with respect to x using the chain rule and product rule:</p><p>d/dx(x^(2x)) = x^(2x)·(2ln x + 2)</p><p>d/dx(2x^x cot y) = 2[x^x(ln x + 1)cot y + x^x·(-csc²y)·dy/dx]</p><p><strong>Step 3:</strong> The differentiated equation becomes:</p><p>x^(2x)·(2ln x + 2) - 2x^x(ln x + 1)cot y + 2x^x·csc²y·dy/dx = 0</p><p><strong>Step 4:</strong> At point (1, π/2):</p><p>• x^(2x) = 1^2 = 1</p><p>• x^x = 1^1 = 1</p><p>• ln(1) = 0</p><p>• cot(π/2) = 0</p><p>• csc(π/2) = 1</p><p><strong>Step 5:</strong> Substituting these values:</p><p>1·(2·0 + 2) - 2·1·(0 + 1)·0 + 2·1·1·dy/dx = 0</p><p>2 + 2·dy/dx = 0</p><p><strong>Step 6:</strong> Solving for dy/dx:</p><p>dy/dx = -1</p><p>∴ Answer: B</p>
Correct Answer: B