KCET · Maths · Differentiation
If \( y=\sqrt{x+\sqrt{x+\sqrt{x+\ldots \infty}}} \), then \( \frac{d y}{d x}= \)
- A \( \frac{1}{y^{2}-1} \)
- B \( \frac{1}{2 y+1} \)
- C \( \frac{2 y}{y^{2}-1} \)
- D \( \frac{1}{2 y-1} \)
Answer & Solution
Correct Answer
(D) \( \frac{1}{2 y-1} \)
Step-by-step Solution
Detailed explanation
Given that, \( y=\sqrt{x+\sqrt{x+\sqrt{x+\ldots \infty}}} \)
\( \Rightarrow y=\sqrt{x+y} \)
\( \Rightarrow y^{2}=x+y \)
\( \Rightarrow 2 y \frac{d y}{d x}=1+\frac{d y}{d x} \)
\( \Rightarrow \frac{d y}{d x}=\frac{1}{2 y-1} \)
\( \Rightarrow y=\sqrt{x+y} \)
\( \Rightarrow y^{2}=x+y \)
\( \Rightarrow 2 y \frac{d y}{d x}=1+\frac{d y}{d x} \)
\( \Rightarrow \frac{d y}{d x}=\frac{1}{2 y-1} \)
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