If `x=e^y+e^((y+ tooo))` , where `x >0,t h e n (dy)/(dx)`

Explanation

To solve the problem where \( x = e^y + e^{(y + \infty)} \) and we need to find \( \frac{dy}{dx} \), we can follow these steps:

### Step 1: Rewrite the equation
Given the equation:
\[ x = e^y + e^{(y + \infty)} \]
Since \( e^{(y + \infty)} \) approaches infinity, we can simplify this to:
\[ x = e^y + \infty \]
This suggests that we can treat the second term as a repeating term, leading us to:
\[ x = e^y + e^y = 2e^y \]
Thus, we can express this as:
\[ x = 2e^y \]

### Step 2: Solve for \( e^y \)
From the equation \( x = 2e^y \), we can isolate \( e^y \):
\[ e^y = \frac{x}{2} \]

### Step 3: Take the natural logarithm
Now, taking the natural logarithm on both sides gives us:
\[ y = \ln\left(\frac{x}{2}\right) \]

### Step 4: Differentiate both sides
Next, we differentiate both sides with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx} \left( \ln\left(\frac{x}{2}\right) \right) \]
Using the chain rule, we have:
\[ \frac{dy}{dx} = \frac{1}{\frac{x}{2}} \cdot \frac{d}{dx}\left(\frac{x}{2}\right) = \frac{1}{\frac{x}{2}} \cdot \frac{1}{2} = \frac{1}{x} \]

### Step 5: Final expression
Thus, we can express \( \frac{dy}{dx} \) as:
\[ \frac{dy}{dx} = \frac{1}{x} \]

### Summary
The final answer is:
\[ \frac{dy}{dx} = \frac{1}{x} \]

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