Limits, Continuity & Differentiability
Implicit Differentiation
Grade 12
Question:
<p>If \( x\log_e(\log_e x) - x^2 + y^2 = 4 \) \((y > 0)\), then \( \dfrac{dy}{dx} \) at \( x = e \) is equal to:</p>
<p>\(\dfrac{e}{\sqrt{4+e^2}}\)</p>
<p>\(\dfrac{(2e-1)}{2\sqrt{4+e^2}}\)</p>
<p>\(\dfrac{(1+2e)}{\sqrt{4+e^2}}\)</p>
<p>\(\dfrac{(1+2e)}{2\sqrt{4+e^2}}\)</p>
Step-by-Step Solution
Key Concept: Differentiate the implicit equation term-by-term using the chain rule and product rule, then substitute x = e to find dy/dx. The logarithmic term simplifies nicely at x = e since log_e(e) = 1.
<p><strong>Step 1:</strong> Differentiate the equation x·log_e(log_e x) - x² + y² = 4 implicitly with respect to x.</p><p><strong>Step 2:</strong> For the first term, use the product rule: d/dx[x·log_e(log_e x)] = log_e(log_e x) + x · d/dx[log_e(log_e x)]</p><p>Using chain rule on log_e(log_e x): d/dx[log_e(log_e x)] = 1/(log_e x) · 1/x = 1/(x·log_e x)</p><p>So: d/dx[x·log_e(log_e x)] = log_e(log_e x) + x · 1/(x·log_e x) = log_e(log_e x) + 1/(log_e x)</p><p><strong>Step 3:</strong> The full differentiation gives: log_e(log_e x) + 1/(log_e x) - 2x + 2y·dy/dx = 0</p><p><strong>Step 4:</strong> At x = e: log_e(log_e e) = log_e(1) = 0, and 1/(log_e e) = 1/1 = 1</p><p>Substituting x = e: 0 + 1 - 2e + 2y·dy/dx = 0</p><p><strong>Step 5:</strong> From the original equation at x = e: e·log_e(1) - e² + y² = 4 → 0 - e² + y² = 4 → y² = e² + 4 → y = √(e² + 4)</p><p><strong>Step 6:</strong> Solving for dy/dx: 2y·dy/dx = 2e - 1 → dy/dx = (2e - 1)/(2√(e² + 4))</p><p>∴ Answer: D</p>
Correct Answer: D