If ` x=e^(y+e^(y+e^(y+ . . . ""oo)`, x`gt ` 0 then ` (dy)/(dx)`=

Options

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

Answer

To solve the problem, we start with the equation given in the question:

\[ x = e^{y + e^{y + e^{y + \ldots}}} \]

Since the expression inside the exponent repeats infinitely, we can denote it as \( x \):

\[ x = e^{y + x} \]

### Step 1: Rewrite the equation
We can rewrite the equation as:

\[ x = e^y \cdot e^x \]

### Step 2: Take the natural logarithm of both sides
Taking the natural logarithm (ln) of both sides gives us:

\[ \ln(x) = y + x \]

### Step 3: Rearranging the equation
Rearranging the equation to isolate \( y \):

\[ y = \ln(x) - x \]

### Step 4: Differentiate both sides with respect to \( x \)
Now we differentiate both sides with respect to \( x \):

\[ \frac{dy}{dx} = \frac{d}{dx}(\ln(x) - x) \]

Using the derivative rules:

- The derivative of \( \ln(x) \) is \( \frac{1}{x} \).
- The derivative of \( x \) is \( 1 \).

Thus, we have:

\[ \frac{dy}{dx} = \frac{1}{x} - 1 \]

### Step 5: Simplify the expression
Now, we can simplify the expression:

\[ \frac{dy}{dx} = \frac{1 - x}{x} \]

### Final Result
So, the final result for \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = \frac{1 - x}{x} \]

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