If f (x) = `sqrt (x + sqrt(x + sqrt (x + sqrt(…., oo))))` , then what is f'(x) equal

Options

A. `(1)/(1 - 2f (x))`
B. `(1)/(2 f (x) - 1)`
C. `(1)/(1 + 2f (x))`
D. `(1)/(2 + f (x))`

Answer

To find the derivative \( f'(x) \) of the function defined as

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

we can follow these steps:

### Step 1: Define the function
Since the expression under the square root is the same as \( f(x) \), we can write:

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

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

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

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

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

### Step 4: Differentiate both sides
Now, we differentiate both sides with respect to \( x \). Using implicit differentiation, we have:

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

Using the chain rule on the first term:

\[ 2f(x)f'(x) - f'(x) - 1 = 0 \]

### Step 5: Factor out \( f'(x) \)
Now, we can factor out \( f'(x) \):

\[ f'(x)(2f(x) - 1) = 1 \]

### Step 6: Solve for \( f'(x) \)
Now, we can solve for \( f'(x) \):

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

### Final Result
Thus, the derivative \( f'(x) \) is:

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