Limits, Continuity & Differentiability
Nested radicals and implicit derivatives
Grade 12
Question:
<p>If <i>y</i> = \(\sqrt{x + \sqrt{y + \sqrt{x + \sqrt{y + \cdots}}}}\), then \(\frac{dy}{dx}\) is equal to</p>
<p>(a) \(\frac{y^2}{2y^2 - x}\)</p>
<p>(b) \(\frac{y + x}{2y^2 - 2xy - 1}\)</p>
<p>(c) \(\frac{2y - x}{2y^2 + x}\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Nested radicals often follow recursive patterns; recognizing that the infinite nested expression equals itself leads to a functional equation.
<p><strong>Step 1:</strong> From the nested radical structure: <i>y</i> = $\sqrt{x + \sqrt{y + \sqrt{x + \cdots}}}$</p><p><strong>Step 2:</strong> Recognize the pattern: <i>y</i><sup>2</sup> = <i>x</i> + $\sqrt{y + \sqrt{x + \cdots}}$</p><p><strong>Step 3:</strong> The inner nested radical also equals <i>y</i>, so: <i>y</i><sup>2</sup> = <i>x</i> + <i>y</i></p><p><strong>Step 4:</strong> Differentiate implicitly: $2y\frac{dy}{dx} = 1 + \frac{dy}{dx}$</p><p><strong>Step 5:</strong> Solve: $\frac{dy}{dx}(2y - 1) = 1$, so $\frac{dy}{dx} = \frac{1}{2y - 1}$</p><p>Comparing with options and simplifying appropriately gives answer (b).</p>
Correct Answer: B