Question

If

\[ \sqrt{y-\sqrt{y-\sqrt{y-\ldots \ldots \ldots \infty}}}=\sqrt{x+\sqrt{x+\sqrt{x+\ldots \ldots \ldots \infty}}} \]

then \(\frac{\mathrm{d} y}{\mathrm{~d} x}=\)

• A. \(\frac{y+x+1}{y-x+1}\)
• B. \(\frac{y-x-1}{y-x+1}\)
• C. \(\frac{y-x+1}{y-x-1}\)
• D. 1

Hint:

Treat each infinite radical as a repeating value and square to remove the radical.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Let the right side be \(L=\sqrt{x+L}\), so \(L^2=x+L\). Let the left side be \(M=\sqrt{y-M}\), so \(M^2=y-M\).

Step 2: Use M = L
Since the two sides are equal, \(y=L^2+L=(x+L)+L=x+2L\), so \(2L=y-x\).

Step 3: Differentiate
From \(L^2=x+L\): \((2L-1)\dfrac{dL}{dx}=1\). Also \(y=L^2+L\) gives \(\dfrac{dy}{dx}=(2L+1)\dfrac{dL}{dx}=\dfrac{2L+1}{2L-1}\).

Step 4: Calculation
Using \(2L=y-x\): \(\dfrac{dy}{dx}=\dfrac{y-x+1}{y-x-1}\). This matches option (C).

Final Answer:
The derivative is (y - x + 1)/(y - x - 1). \[ \boxed{\text{(C) }\dfrac{y-x+1}{y-x-1}} \]