If `y=sqrt(x+sqrt(y+sqrt(x+......,)))"then"(dy)/(dx)=`

Options

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

Answer

To solve the problem where \( y = \sqrt{x + \sqrt{y + \sqrt{x + \ldots}}} \), we will follow these steps:

### Step 1: Rewrite the equation
Since the expression inside the square root is repeating, we can express it as:
\[ y = \sqrt{x + \sqrt{y}} \]
This allows us to simplify our equation.

### Step 2: Square both sides
To eliminate the square root, we square both sides:
\[ y^2 = x + \sqrt{y} \]

### Step 3: Rearrange the equation
Rearranging the equation gives us:
\[ y^2 - x = \sqrt{y} \]

### Step 4: Square both sides again
Now, we square both sides again to eliminate the square root:
\[ (y^2 - x)^2 = y \]

### Step 5: Expand and rearrange
Expanding the left side:
\[ y^4 - 2xy^2 + x^2 = y \]
Rearranging gives us:
\[ y^4 - 2xy^2 + x^2 - y = 0 \]

### Step 6: Differentiate both sides with respect to \( x \)
Now, we differentiate the entire equation with respect to \( x \):
\[ \frac{d}{dx}(y^4) - \frac{d}{dx}(2xy^2) + \frac{d}{dx}(x^2) - \frac{d}{dx}(y) = 0 \]

Using the chain rule and product rule, we get:
\[ 4y^3 \frac{dy}{dx} - (2y^2 \frac{d}{dx}(x) + 2x \frac{d}{dx}(y^2)) + 2x - \frac{dy}{dx} = 0 \]
This simplifies to:
\[ 4y^3 \frac{dy}{dx} - (2y^2 + 4xy \frac{dy}{dx}) + 2x - \frac{dy}{dx} = 0 \]

### Step 7: Collect terms involving \( \frac{dy}{dx} \)
Rearranging gives:
\[ (4y^3 - 4xy - 1) \frac{dy}{dx} = -2x + 2y^2 \]

### Step 8: Solve for \( \frac{dy}{dx} \)
Finally, we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{-2x + 2y^2}{4y^3 - 4xy - 1} \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{2y^2 - 2x}{4y^3 - 4xy - 1} \]