Question:medium

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

Show Hint

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

The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
This is an infinite series function. Since the pattern repeats infinitely, we can replace the inner part of the square root with \(y\) itself to create a finite algebraic equation.
Step 2: Key Formula or Approach:
1. Recursive substitution: \(y = \sqrt{x + y}\).
2. Implicit differentiation: Differentiate both sides with respect to \(x\).
Step 3: Detailed Explanation:
Given \( y = \sqrt{x + y} \). Square both sides: \[ y^2 = x + y \] Differentiate with respect to \(x\) using the chain rule: \[ 2y \frac{dy}{dx} = 1 + \frac{dy}{dx} \] Rearrange to group \(\frac{dy}{dx}\) terms: \[ 2y \frac{dy}{dx} - \frac{dy}{dx} = 1 \] \[ \frac{dy}{dx} (2y - 1) = 1 \] \[ \frac{dy}{dx} = \frac{1}{2y - 1} \] Note: \(\frac{1}{2y-1}\) is mathematically identical to \(\frac{-1}{1-2y}\).
Step 4: Final Answer:
The derivative is \( \frac{1}{2y - 1} \).
Was this answer helpful?
0