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

Options

A. `(1)/( 1-y)`
B. `(y)/(1-y)`
C. `(2y)/(1-y)`
D. None

Answer

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

### Step 1: Define the equation
We start with the equation:
\[ y = e^{x + y} \]
This is because the expression inside the exponent is the same as \( y \) itself.

### Step 2: Take the natural logarithm of both sides
Taking the natural logarithm of both sides gives:
\[ \ln y = x + y \]

### Step 3: Differentiate both sides with respect to \( x \)
Now we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(\ln y) = \frac{d}{dx}(x + y) \]
Using the chain rule on the left side, we have:
\[ \frac{1}{y} \frac{dy}{dx} = 1 + \frac{dy}{dx} \]

### Step 4: Rearrange the equation
Now, we can rearrange the equation to isolate \( \frac{dy}{dx} \):
\[ \frac{1}{y} \frac{dy}{dx} - \frac{dy}{dx} = 1 \]
Factoring out \( \frac{dy}{dx} \):
\[ \left(\frac{1}{y} - 1\right) \frac{dy}{dx} = 1 \]

### Step 5: Solve for \( \frac{dy}{dx} \)
Now, we solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{1}{\frac{1}{y} - 1} \]
Simplifying the right side:
\[ \frac{dy}{dx} = \frac{y}{1 - y} \]

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