Question:medium

If \(y=\sqrt{\sin x + \sqrt{\sin x + \cdots}}\), then \(\frac{dy}{dx}\) is

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For infinite radicals, replace the repeating part by the variable itself.
Updated On: Apr 18, 2026
  • \(\frac{\sin x}{2y-1}\)
  • \(\frac{\cos x}{1-2y}\)
  • \(\frac{\cos x}{2y-1}\)
  • 0
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The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
Since the sequence of square roots is infinite, we can replace the recurring part of the expression with the variable \( y \) itself.
Step 2: Detailed Explanation:
Given: \( y = \sqrt{\sin x + \sqrt{\sin x + \dots}} \).
We can write this as: \[ y = \sqrt{\sin x + y} \] Squaring both sides: \[ y^{2} = \sin x + y \] Differentiating both sides with respect to \( x \): \[ \frac{d}{dx}(y^{2}) = \frac{d}{dx}(\sin x) + \frac{d}{dx}(y) \] \[ 2y \frac{dy}{dx} = \cos x + \frac{dy}{dx} \] Rearranging terms with \( \frac{dy}{dx} \): \[ 2y \frac{dy}{dx} - \frac{dy}{dx} = \cos x \] \[ \frac{dy}{dx} (2y - 1) = \cos x \] \[ \frac{dy}{dx} = \frac{\cos x}{2y - 1} \].
Step 3: Final Answer:
The derivative is \( \frac{\cos x}{2y - 1} \).
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