Applications of Derivatives
Implicit Differentiation
Grade 12
Question:
<p>Let <span>\(y = \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}}\)</span>, then <span>\(\frac{dy}{dx}\)</span> equals</p>
<p>(a) <span>\(\frac{1}{2y - 1}\)</span></p>
<p>(b) <span>\(\frac{x}{x + 2y}\)</span></p>
<p>(c) <span>\(\frac{1}{1 - 4x}\)</span></p>
<p>(d) <span>\(\frac{2x}{x + y}\)</span></p>
Step-by-Step Solution
Key Concept: Recognize the infinite nested radical as a self-similar expression and use implicit differentiation.
<p>Since <span>$y = \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}}$</span>, we have <span>$y = \sqrt{x + y}$</span></p><p>Squaring: <span>$y^2 = x + y$</span></p><p>Differentiating with respect to <span>$x$</span>: <span>$2y\frac{dy}{dx} = 1 + \frac{dy}{dx}$</span></p><p><span>$\frac{dy}{dx}(2y - 1) = 1$</span></p><p><span>$\frac{dy}{dx} = \frac{1}{2y - 1}$</span></p>
Correct Answer: A