<p>Let \( y \) be an implicit function of \( x \) defined by \( x^{2x} - 2x^x \cot y - 1 = 0 \). Then \( y'(1) \) equals</p>
Step-by-Step Solution
Key Concept: At x=1, substitute to find y, then differentiate the implicit equation with respect to x and evaluate at the point (1,y). The trick is recognizing that x^(2x) = x^x · x^x and carefully handling the derivative of x^x using d/dx(x^x) = x^x(ln x + 1).
<p><strong>Step 1:</strong> Find y at x = 1. Substitute x = 1 into the equation:</p><p>1^2 - 2(1)^1 cot y - 1 = 0</p><p>1 - 2 cot y - 1 = 0</p><p>cot y = 0 ⟹ y = π/2</p><p><strong>Step 2:</strong> Differentiate the equation x^(2x) - 2x^x cot y - 1 = 0 with respect to x:</p><p>d/dx(x^(2x)) = x^(2x)·d/dx(2x ln x) = x^(2x)·(2 ln x + 2)</p><p>d/dx(2x^x cot y) = 2[x^x(ln x + 1) cot y + x^x · (-csc²y) · y']</p><p><strong>Step 3:</strong> The differentiated equation becomes:</p><p>x^(2x)(2 ln x + 2) - 2x^x(ln x + 1) cot y + 2x^x csc²y · y' = 0</p><p><strong>Step 4:</strong> Evaluate at x = 1, y = π/2 (where ln 1 = 0, cot(π/2) = 0, csc(π/2) = 1):</p><p>1^2(2·0 + 2) - 2·1^1(0 + 1)·0 + 2·1^1·1·y' = 0</p><p>2 + 2y' = 0</p><p>y'(1) = -1</p><p>∴ Answer: A</p>
Correct Answer: A