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

Options

A. ` e^(x-y) ((1-e^(y))/( 1-e^(x)) ) `
B. ` -e^(x-y) ((1-e^(y))/( 1-e^(x)) ) `
C. ` e^(y-x) ((1-e^(y))/( 1-e^(x)) ) `
D. `- e^(y-x) ((1-e^(y))/( 1-e^(x)) ) `

Answer

To solve the equation \( e^x + e^y = e^{x+y} \) and find \( \frac{dy}{dx} \), we will differentiate both sides of the equation with respect to \( x \).

### Step-by-Step Solution:

1. **Differentiate both sides with respect to \( x \)**:
\[ \frac{d}{dx}(e^x + e^y) = \frac{d}{dx}(e^{x+y}) \]

2. **Apply the differentiation rules**:
- The derivative of \( e^x \) with respect to \( x \) is \( e^x \).
- For \( e^y \), we use the chain rule, giving us \( e^y \cdot \frac{dy}{dx} \).
- The derivative of \( e^{x+y} \) also requires the chain rule, resulting in \( e^{x+y} \cdot (1 + \frac{dy}{dx}) \).

Thus, we have:
\[ e^x + e^y \cdot \frac{dy}{dx} = e^{x+y} \cdot (1 + \frac{dy}{dx}) \]

3. **Rearrange the equation**:
\[ e^x + e^y \cdot \frac{dy}{dx} = e^{x+y} + e^{x+y} \cdot \frac{dy}{dx} \]

4. **Collect all terms involving \( \frac{dy}{dx} \) on one side**:
\[ e^y \cdot \frac{dy}{dx} - e^{x+y} \cdot \frac{dy}{dx} = e^{x+y} - e^x \]

5. **Factor out \( \frac{dy}{dx} \)**:
\[ \left( e^y - e^{x+y} \right) \cdot \frac{dy}{dx} = e^{x+y} - e^x \]

6. **Solve for \( \frac{dy}{dx} \)**:
\[ \frac{dy}{dx} = \frac{e^{x+y} - e^x}{e^y - e^{x+y}} \]

7. **Simplify the expression**:
Notice that \( e^{x+y} = e^x \cdot e^y \), thus:
\[ \frac{dy}{dx} = \frac{e^x(e^y - 1)}{e^y - e^x e^y} \]
\[ = \frac{e^x(e^y - 1)}{e^y(1 - e^x)} \]

8. **Final form**:
We can express this as:
\[ \frac{dy}{dx} = -\frac{e^y}{e^x} \]

### Final Answer:
\[ \frac{dy}{dx} = -\frac{e^y}{e^x} \]