If `y=sqrt(x+sqrt(y+sqrt(x+sqrt(y+...oo))))`, then `(dy)/(dx)` is equal to

Options

A. `(y+x)/(y^(2)-2)`
B. `(y^(3)-x)/(2y^(2)-2xy-1)`
C. `(y^(3)+x)/(2y^(2)-x)`
D. none of these

Answer

To solve the equation \( y = \sqrt{x + \sqrt{y + \sqrt{x + \sqrt{y + \ldots}}}} \), we will first simplify the expression.

### Step 1: Simplify the equation
We can rewrite the equation as:
\[ y = \sqrt{x + y} \]

### Step 2: Square both sides
Next, we square both sides to eliminate the square root:
\[ y^2 = x + y \]

### Step 3: Rearrange the equation
Now, we rearrange the equation to isolate \( x \):
\[ y^2 - y - x = 0 \]
This can be viewed as a quadratic equation in terms of \( x \).

### Step 4: Differentiate implicitly
Now, we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(y^2 - y - x) = 0 \]
Using the product and chain rules, we get:
\[ 2y \frac{dy}{dx} - \frac{dy}{dx} - 1 = 0 \]

### Step 5: Factor out \( \frac{dy}{dx} \)
Now, we can factor out \( \frac{dy}{dx} \):
\[ (2y - 1) \frac{dy}{dx} - 1 = 0 \]

### Step 6: Solve for \( \frac{dy}{dx} \)
Rearranging gives us:
\[ (2y - 1) \frac{dy}{dx} = 1 \]
Thus, we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{1}{2y - 1} \]

### Final Answer
Therefore, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{1}{2y - 1} \]

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