Question

If $y = \sqrt{x + \sqrt{x + \sqrt{x + ... \infty}$ then $\frac{dy}{dx} = $

• A. $\frac{1}{2y}$
• B. $\frac{1}{1 - 2y}$
• C. $\frac{1}{2(1 - 2y)}$
• D. $\frac{-1}{1 - 2y}$

Hint:

For $y = \sqrt{f(x) + \sqrt{f(x) + ...}}$, the derivative is always $f'(x) / (2y - 1)$.

Answer 1

The Correct Option is D

Step 1: Concept
Use the property of infinite series: $y = \sqrt{x + y}$.

Step 2: Meaning
Square both sides to remove the root: $y^{2} = x + y$.

Step 3: Analysis
Differentiate implicitly: $2y \frac{dy}{dx} = 1 + \frac{dy}{dx} \implies (2y - 1) \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{2y - 1}$.

Step 4: Conclusion
Note that $1/(2y - 1)$ is identical to $-1/(1 - 2y)$, which matches Option (D).
Final Answer: (D)