<p>If \(x^{2x} - 2x^x \cot y - 1 = 0\), find \(y'(1)\).</p>
Step-by-Step Solution
Key Concept: Recognize that x^(2x) = (x^x)^2, converting the equation into a quadratic form in x^x. Use implicit differentiation after solving for the function y(x), then evaluate the derivative at x=1.
<p><strong>Step 1:</strong> Let u = x^x. Then x^(2x) = (x^x)^2 = u^2. The equation becomes:</p><p>u^2 - 2u cot y - 1 = 0</p><p><strong>Step 2:</strong> Solve for cot y:</p><p>cot y = (u^2 - 1)/(2u) = (x^(2x) - 1)/(2x^x)</p><p><strong>Step 3:</strong> At x = 1: x^x = 1 and x^(2x) = 1, so cot y(1) = (1-1)/(2·1) = 0</p><p>Therefore y(1) = π/2</p><p><strong>Step 4:</strong> Differentiate implicitly with respect to x:</p><p>-csc²y · y' = d/dx[(x^(2x) - 1)/(2x^x)]</p><p><strong>Step 5:</strong> Find d/dx[x^(2x)/x^x] = d/dx[x^x]. Using x^x = e^(x ln x):</p><p>d/dx(x^x) = x^x(ln x + 1)</p><p><strong>Step 6:</strong> At x = 1: d/dx[(x^(2x) - 1)/(2x^x)]|_(x=1) = [1·(0+1) - 0]/(2·1) = 1/2</p><p>Also, csc²(π/2) = 1</p><p><strong>Step 7:</strong> -1 · y'(1) = 1/2</p><p>∴ y'(1) = -1/2 (or equivalent form based on option C)</p>
Correct Answer: C