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

Explanation

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

**Step 1:** Start with the equation:
\[ x = e^y + e^{(y + \infty)} \]

**Step 2:** Recognize that the term \( e^{(y + \infty)} \) approaches infinity, and thus the equation simplifies to:
\[ x = e^y + x \]

**Step 3:** Rearranging the equation gives us:
\[ x - x = e^y \]
This implies:
\[ 0 = e^y \]

However, this doesn't seem correct. Let's re-evaluate the infinite series aspect. The correct interpretation of the infinite series is:
\[ x = e^y + e^y + e^y + \ldots \]
This can be interpreted as:
\[ x = e^y + x \]

**Step 4:** Rearranging gives:
\[ x - e^y = x \]
This leads us to:
\[ e^y = x - e^y \]

**Step 5:** Now, we can express \( e^y \) in terms of \( x \):
\[ e^y = x - e^y \]
This means:
\[ 2e^y = x \]
Thus:
\[ e^y = \frac{x}{2} \]

**Step 6:** Taking the natural logarithm of both sides:
\[ y = \ln\left(\frac{x}{2}\right) \]

**Step 7:** Now, differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx} \left( \ln\left(\frac{x}{2}\right) \right) \]

**Step 8:** Using the chain rule, we have:
\[ \frac{dy}{dx} = \frac{1}{\frac{x}{2}} \cdot \frac{d}{dx}\left(\frac{x}{2}\right) \]
\[ \frac{dy}{dx} = \frac{1}{\frac{x}{2}} \cdot \frac{1}{2} \]

**Step 9:** Simplifying gives:
\[ \frac{dy}{dx} = \frac{2}{x} \]

Thus, the final answer is:
\[ \frac{dy}{dx} = \frac{2}{x} \]