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 step by step, we start with the given equation:

### Step 1: Set up the equation
We have:
\[ x = e^{y + e^{y + e^{y + \ldots}}} \]

### Step 2: Recognize the infinite pattern
Notice that the right-hand side has a repeating structure. We can rewrite it as:
\[ x = e^{y + x} \]

### Step 3: Take the natural logarithm of both sides
To simplify the expression, we take the natural logarithm (ln) of both sides:
\[ \ln(x) = \ln(e^{y + x}) \]

### Step 4: Simplify using properties of logarithms
Using the property of logarithms (\(\ln(e^a) = a\)), we get:
\[ \ln(x) = y + x \]

### Step 5: Rearrange the equation
Now, we can rearrange this equation to isolate \(y\):
\[ y = \ln(x) - x \]

### Step 6: Differentiate both sides with respect to \(x\)
Now we differentiate both sides with respect to \(x\):
\[ \frac{dy}{dx} = \frac{d}{dx}(\ln(x) - x) \]

### Step 7: Apply differentiation rules
Using the rules of differentiation:
- 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 8: Simplify the expression
We can combine the terms:
\[ \frac{dy}{dx} = \frac{1 - x}{x} \]

### Final Result
So the final expression for \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{1 - x}{x} \]

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