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

Options

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

Answer

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

### Step 1: Rewrite the equation
Since the expression inside the exponent is the same as \( y \), we can rewrite the equation as:
\[ y = e^{x + y} \]

### Step 2: Differentiate both sides
Now we will differentiate both sides with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(e^{x + y}) \]

### Step 3: Apply the chain rule
Using the chain rule on the right-hand side, we have:
\[ \frac{dy}{dx} = e^{x + y} \left( \frac{d}{dx}(x + y) \right) \]
This gives:
\[ \frac{dy}{dx} = e^{x + y} \left( 1 + \frac{dy}{dx} \right) \]

### Step 4: Substitute \( e^{x + y} \)
Since \( e^{x + y} = y \), we can substitute this into our equation:
\[ \frac{dy}{dx} = y \left( 1 + \frac{dy}{dx} \right) \]

### Step 5: Expand and rearrange
Expanding the right-hand side:
\[ \frac{dy}{dx} = y + y \frac{dy}{dx} \]
Now, we can rearrange this to isolate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} - y \frac{dy}{dx} = y \]
Factoring out \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} (1 - y) = y \]

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

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

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