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Question: Answered & Verified by Expert
If \( y=\sqrt{x+\sqrt{x+\sqrt{x+\ldots \infty}}} \), then \( \frac{d y}{d x}= \)
MathematicsDifferentiationKCETKCET 2018
Options:
  • A \( \frac{1}{y^{2}-1} \)
  • B \( \frac{1}{2 y+1} \)
  • C \( \frac{2 y}{y^{2}-1} \)
  • D \( \frac{1}{2 y-1} \)
Solution:
2930 Upvotes Verified Answer
The correct answer is: \( \frac{1}{2 y-1} \)
Given that, \( y=\sqrt{x+\sqrt{x+\sqrt{x+\ldots \infty}}} \)
\( \Rightarrow y=\sqrt{x+y} \)
\( \Rightarrow y^{2}=x+y \)
\( \Rightarrow 2 y \frac{d y}{d x}=1+\frac{d y}{d x} \)
\( \Rightarrow \frac{d y}{d x}=\frac{1}{2 y-1} \)

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