If `x=e^(y+e^(y+e^(y+...oo))),xgt0`, then `(dy)/(dx)` is equal to

Options

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

Answer

To solve the problem where \( x = e^{y + e^{y + e^{y + \ldots}}} \) and \( x > 0 \), we need to find \( \frac{dy}{dx} \).

### Step-by-step Solution:

1. **Understanding the Infinite Exponential Expression**:
We start with the equation:
\[ x = e^{y + e^{y + e^{y + \ldots}}} \]
Since the expression inside the exponent is the same as \( x \), we can rewrite it as:
\[ x = e^{y + x} \]

2. **Taking the Natural Logarithm**:
To simplify this equation, we take the natural logarithm of both sides:
\[ \ln(x) = y + x \]

3. **Rearranging the Equation**:
Rearranging the equation gives us:
\[ y = \ln(x) - x \]

4. **Differentiating Both Sides**:
Now, we differentiate both sides with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(\ln(x) - x) \]
Using the derivative rules:
\[ \frac{dy}{dx} = \frac{1}{x} - 1 \]

5. **Simplifying the Derivative**:
We can simplify this expression:
\[ \frac{dy}{dx} = \frac{1 - x}{x} \]

### Final Result:
Thus, we have:
\[ \frac{dy}{dx} = \frac{1 - x}{x} \]