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

Options

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

Answer

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

### Step 1: Set up the equation
Since the expression under the square root continues infinitely, we can express it as:
\[ y = \sqrt{x + y} \]

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

### Step 3: Rearrange the equation
Next, we rearrange the equation to isolate terms:
\[ y^2 - y - x = 0 \]

### Step 4: Differentiate both sides
Now, we differentiate both sides with respect to \( x \). Using implicit differentiation:
\[ \frac{d}{dx}(y^2) - \frac{d}{dx}(y) - \frac{d}{dx}(x) = 0 \]
This gives us:
\[ 2y \frac{dy}{dx} - \frac{dy}{dx} - 1 = 0 \]

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

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

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

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