<p>Let \(y\) be an implicit function of \(x\) defined by \(x^{2x} - 2x^x \cot y - 1 = 0\). The value of \(y'(1)\), where \(y'\) denotes the first derivative of \(y\), is:</p>
Step-by-Step Solution
Key Concept: Use implicit differentiation on the equation x^(2x) - 2x^x·cot(y) - 1 = 0, recognizing that d/dx[x^(2x)] = x^(2x)·(2ln(x) + 2) and d/dx[x^x] = x^x·(ln(x) + 1). At x=1, these derivatives simplify because ln(1)=0 and x^x=1, making the algebra tractable.
<p><strong>Step 1:</strong> From the given equation x^(2x) - 2x^x·cot(y) - 1 = 0, find y(1).</p><p>At x=1: 1^2 - 2(1)·cot(y) - 1 = 0 → 1 - 2cot(y) - 1 = 0 → cot(y) = 0 → y(1) = π/2</p><p><strong>Step 2:</strong> Differentiate implicitly with respect to x.</p><p>d/dx[x^(2x)] = x^(2x)·(2ln(x) + 2)</p><p>d/dx[x^x] = x^x·(ln(x) + 1)</p><p>So: x^(2x)·(2ln(x) + 2) - 2[x^x·(ln(x) + 1)·cot(y) + x^x·(-csc²(y))·y'] = 0</p><p><strong>Step 3:</strong> Evaluate at x=1 where ln(1)=0, x^(2x)=1, x^x=1, and y=π/2 (so cot(π/2)=0, csc(π/2)=1).</p><p>1·(0 + 2) - 2[1·(0+1)·0 + 1·(-1)·y'] = 0</p><p>2 - 2(-y') = 0</p><p>2 + 2y' = 0</p><p>y'(1) = -1</p><p>∴ Answer: D</p>
Correct Answer: D