If `e^x+e^y=e^(x+y)`, then dy/dx is:

Options

A. `e^(y-x)`
B. `e^(x+y)`
C. `-e^(y-x)`
D. `2e^(x-y)`

Answer

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

### Step 1: Rewrite the equation
We start with the equation:
\[ e^x + e^y = e^{x+y} \]
We can rewrite the right-hand side using the property of exponents:
\[ e^{x+y} = e^x \cdot e^y \]
Thus, we have:
\[ e^x + e^y = e^x \cdot e^y \]

### Step 2: Rearrange the equation
Rearranging gives us:
\[ e^x + e^y - e^x \cdot e^y = 0 \]
This can be factored as:
\[ e^x(1 - e^y) + e^y = 0 \]

### Step 3: Divide by \( e^x e^y \)
To simplify, we can divide the entire equation by \( e^x e^y \) (assuming \( e^x \) and \( e^y \) are not zero):
\[ \frac{e^x}{e^x e^y} + \frac{e^y}{e^x e^y} = \frac{e^{x+y}}{e^x e^y} \]
This simplifies to:
\[ \frac{1}{e^y} + \frac{1}{e^x} = 1 \]
or
\[ e^{-y} + e^{-x} = 1 \]

### Step 4: Differentiate both sides
Now we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(e^{-y}) + \frac{d}{dx}(e^{-x}) = \frac{d}{dx}(1) \]
Using the chain rule, we have:
\[ -e^{-y} \frac{dy}{dx} - e^{-x} = 0 \]

### Step 5: Solve for \( \frac{dy}{dx} \)
Rearranging gives:
\[ -e^{-y} \frac{dy}{dx} = e^{-x} \]
Now, isolating \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -\frac{e^{-x}}{e^{-y}} = -e^{y-x} \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = -e^{y-x} \]