Limits, Continuity & Differentiability
Differentiation
Grade 12
Question:
<p>If \( x = e^{y + e^{y + \cdots}} \), \( x > 0 \), then \( \dfrac{dy}{dx} \) is:</p>
<p>\(\dfrac{x}{1+x}\)</p>
<p>\(\dfrac{1}{x}\)</p>
<p>\(\dfrac{1-x}{x}\)</p>
<p>\(\dfrac{1+x}{x}\)</p>
Step-by-Step Solution
Key Concept: Recognize that the infinite tower e^(y + e^(y + ...)) converges to x itself, creating a self-referential equation x = e^(y+x). Differentiate implicitly using the substitution insight rather than trying to expand the infinite expression.
<p><strong>Step 1:</strong> Recognize the self-referential structure. Since x = e^(y + e^(y + ...)) and the infinite tower e^(y + e^(y + ...)) = x, we have:</p><p>x = e^(y + x)</p><p><strong>Step 2:</strong> Take natural logarithm of both sides:</p><p>ln(x) = y + x</p><p>Therefore: y = ln(x) - x</p><p><strong>Step 3:</strong> Differentiate both sides with respect to x:</p><p>dy/dx = d/dx[ln(x) - x]</p><p>dy/dx = 1/x - 1</p><p>dy/dx = (1 - x)/x</p><p><strong>Step 4:</strong> Simplify to standard form:</p><p>dy/dx = 1/x - 1, or equivalently dy/dx = (1-x)/x</p><p>∴ Answer: C</p>
Correct Answer: C