Limits, Continuity & Differentiability
Differentiation
Grade 12
Question:
<p><strong>320.</strong> If \(y = \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}}}\), where \(x > 0\), then \(\dfrac{dy}{dx}\) can be:</p>
<p>(a) \(\dfrac{1}{2y-1}\)</p>
<p>(b) \(\dfrac{x}{x+2y}\)</p>
<p>(c) \(\dfrac{1}{\sqrt{1+4x}}\)</p>
<p>(d) \(\dfrac{y}{2x+y}\)</p>
Step-by-Step Solution
Key Concept: Recognize that the nested radical converges to a finite value y, then use the self-referential property y = √(x + y) to create an algebraic equation that can be differentiated implicitly.
<p><strong>Step 1:</strong> Since the nested radical converges (for x > 0), let y = √(x + √(x + √(x + ...)))</p><p><strong>Step 2:</strong> Observe the self-similar structure: the expression inside the first square root is x + √(x + √(x + ...)) = x + y</p><p>Therefore: <strong>y = √(x + y)</strong></p><p><strong>Step 3:</strong> Square both sides: y² = x + y</p><p><strong>Step 4:</strong> Rearrange: y² - y - x = 0</p><p><strong>Step 5:</strong> Differentiate implicitly with respect to x:</p><p>2y(dy/dx) - dy/dx - 1 = 0</p><p><strong>Step 6:</strong> Solve for dy/dx:</p><p>dy/dx(2y - 1) = 1</p><p><strong>dy/dx = 1/(2y - 1)</strong></p><p><strong>Step 7:</strong> From y² - y - x = 0, we get y = (1 + √(1 + 4x))/2 (taking positive root since y > 0)</p><p>Therefore: <strong>2y - 1 = √(1 + 4x)</strong></p><p>∴ <strong>dy/dx = 1/√(1 + 4x)</strong> (Answer: A)</p>
Correct Answer: A