Question

If \(y=\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y+\cdots}}}}\), then \(\dfrac{dy}{dx}\) is equal to

• A. \(\dfrac{y+x}{y^2-2x}\)
• B. \(\dfrac{y^3-x}{2y^2-2xy-1}\)
• C. \(\dfrac{y^3+x}{2y^2-x}\)
• D. None of these

Hint:

For infinite nested radicals, use repetition property: set the whole expression equal to \(y\), express inner radical as \(y\), then solve and differentiate.

Answer 1

The Correct Option is D

Step 1: Use repeating nature of expression.
Given:
\[ y=\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y+\cdots}}}} \] The expression after first \(\sqrt{}\) repeats itself, so:
\[ y=\sqrt{x+\sqrt{y+y}} \] But more accurately:
Let inner part after first root be \(y\) itself:
\[ y=\sqrt{x+y} \] Step 2: Square both sides.
\[ y^2=x+y \] \[ y^2-y-x=0 \] Step 3: Differentiate implicitly.
Differentiate w.r.t. \(x\):
\[ 2y\frac{dy}{dx}-\frac{dy}{dx}-1=0 \] \[ (2y-1)\frac{dy}{dx}=1 \] \[ \frac{dy}{dx}=\frac{1}{2y-1} \] Step 4: Compare with given options.
None of the options matches \(\frac{1}{2y-1}\).
Final Answer: \[ \boxed{\text{None of these}} \]