Limits, Continuity & Differentiability
Implicit Differentiation
Grade 12
Question:
<p><strong>30.</strong> 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>
<p>(a) \(-\ln 2\)</p>
<p>(b) \(\ln 2\)</p>
<p>(c) \(-1\)</p>
<p>(d) \(1\)</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), then substitute x = 1 to find y'(1).
<p><strong>Step 1:</strong> Start with the implicit equation: x<sup>2x</sup> - 2x<sup>x</sup>·cot(y) - 1 = 0</p><p><strong>Step 2:</strong> Find y(1) by substituting x = 1: 1<sup>2</sup> - 2(1)·cot(y) - 1 = 0 → 1 - 2cot(y) - 1 = 0 → cot(y) = 0 → y(1) = π/4</p><p><strong>Step 3:</strong> Differentiate implicitly with respect to x. For x<sup>2x</sup>: d/dx[x<sup>2x</sup>] = x<sup>2x</sup>·(2ln(x) + 2). For x<sup>x</sup>: d/dx[x<sup>x</sup>] = x<sup>x</sup>·(ln(x) + 1)</p><p><strong>Step 4:</strong> Apply product rule and chain rule to -2x<sup>x</sup>·cot(y): d/dx[-2x<sup>x</sup>·cot(y)] = -2[x<sup>x</sup>(ln(x) + 1)·cot(y) + x<sup>x</sup>·(-csc²(y))·y']</p><p><strong>Step 5:</strong> The equation becomes: x<sup>2x</sup>(2ln(x) + 2) - 2x<sup>x</sup>(ln(x) + 1)·cot(y) + 2x<sup>x</sup>·csc²(y)·y' = 0</p><p><strong>Step 6:</strong> Substitute x = 1 (where x<sup>2x</sup> = 1, x<sup>x</sup> = 1, ln(1) = 0, cot(π/4) = 1, csc²(π/4) = 2): 1(2·0 + 2) - 2·1·(0 + 1)·1 + 2·1·2·y'(1) = 0</p><p><strong>Step 7:</strong> Simplify: 2 - 2 + 4y'(1) = 0 → 4y'(1) = 0 → y'(1) = 0</p><p>∴ Answer: D (0)</p>
Correct Answer: D