Limits, Continuity & Differentiability
Implicit Differentiation
Grade 12
Question:
<p>If \(x\log_e(\log_e x) - x^2 + y^2 = 4\), then \(\dfrac{dy}{dx}\) at \(x = e\) is equal to:</p>
<p>\(\dfrac{2e-1}{2\sqrt{4+e^2}}\)</p>
<p>\(\dfrac{e}{2\sqrt{4+e^2}}\)</p>
<p>\(\dfrac{2e+1}{2\sqrt{4+e^2}}\)</p>
<p>\(\dfrac{e-1}{\sqrt{4+e^2}}\)</p>
Step-by-Step Solution
Key Concept: Use implicit differentiation on the constraint equation, then substitute x = e carefully. The logarithmic term's derivative simplifies elegantly when x = e since log_e(e) = 1.
<p><strong>Step 1: Differentiate implicitly with respect to x</strong></p><p>Given: x·log_e(log_e x) - x² + y² = 4</p><p>Differentiating both sides:</p><p>d/dx[x·log_e(log_e x)] - 2x + 2y·(dy/dx) = 0</p><p><strong>Step 2: Apply product and chain rules to the first term</strong></p><p>d/dx[x·log_e(log_e x)] = log_e(log_e x) + x·(1/log_e x)·(1/x)</p><p>= log_e(log_e x) + 1/log_e x</p><p><strong>Step 3: Substitute x = e</strong></p><p>At x = e: log_e(e) = 1, so log_e(log_e e) = log_e(1) = 0</p><p>The derivative term becomes: 0 + 1/1 = 1</p><p><strong>Step 4: Find y at x = e</strong></p><p>e·log_e(1) - e² + y² = 4</p><p>0 - e² + y² = 4</p><p>y² = e² + 4, so y = ±√(e² + 4)</p><p><strong>Step 5: Solve for dy/dx at x = e</strong></p><p>1 - 2e + 2y·(dy/dx) = 0</p><p>2y·(dy/dx) = 2e - 1</p><p>dy/dx = (2e - 1)/(2y) = (2e - 1)/(2√(e² + 4))</p><p>∴ Answer: A</p>
Correct Answer: A