If `y = sqrt (f (x) + sqrt (f (x) + sqrt (f (x) +.....(oo))))` then `(dy)/(dx)` equals

Options

A. `(f (x))/( 2y -1)`
B. `(1)/( f (x) (2 y -1))`
C. `(f '(x))/( 2y -1)`
D. None of these

Answer

To solve the problem, we start with the equation given:

\[ y = \sqrt{f(x) + \sqrt{f(x) + \sqrt{f(x) + \ldots}}} \]

### Step 1: Set up the equation
Since the expression under the square root continues infinitely, we can express it as:

\[ y = \sqrt{f(x) + y} \]

### Step 2: Square both sides
To eliminate the square root, we square both sides of the equation:

\[ y^2 = f(x) + y \]

### Step 3: Rearrange the equation
Rearranging the equation gives us:

\[ y^2 - y - f(x) = 0 \]

### Step 4: Differentiate both sides
Now, we differentiate both sides with respect to \( x \):

Using implicit differentiation, we have:

\[ \frac{d}{dx}(y^2) - \frac{d}{dx}(y) - \frac{d}{dx}(f(x)) = 0 \]

This gives us:

\[ 2y \frac{dy}{dx} - \frac{dy}{dx} - f'(x) = 0 \]

### Step 5: Factor out \( \frac{dy}{dx} \)
Now we can factor out \( \frac{dy}{dx} \):

\[ (2y - 1) \frac{dy}{dx} = f'(x) \]

### Step 6: Solve for \( \frac{dy}{dx} \)
Finally, we can solve for \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{f'(x)}{2y - 1} \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = \frac{f'(x)}{2y - 1} \]

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