If `e^(x)+e^(y)=e^(x+y)" then "y_(1)=`

Options

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

Answer

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

### Step-by-Step Solution:

1. **Differentiate Both Sides**:
We start with the equation:
\[ e^x + e^y = e^{x+y} \]
Now, we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(e^x) + \frac{d}{dx}(e^y) = \frac{d}{dx}(e^{x+y}) \]

2. **Apply the Chain Rule**:
The derivative of \( e^x \) is \( e^x \), and for \( e^y \), we apply the chain rule:
\[ e^x + e^y \frac{dy}{dx} = e^{x+y} \left( \frac{d}{dx}(x+y) \right) \]
The derivative of \( x+y \) is \( 1 + \frac{dy}{dx} \):
\[ e^x + e^y \frac{dy}{dx} = e^{x+y} (1 + \frac{dy}{dx}) \]

3. **Expand the Right Side**:
Expanding the right side gives:
\[ e^x + e^y \frac{dy}{dx} = e^{x+y} + e^{x+y} \frac{dy}{dx} \]

4. **Rearranging the Equation**:
We can rearrange the equation to isolate terms involving \( \frac{dy}{dx} \):
\[ e^y \frac{dy}{dx} - e^{x+y} \frac{dy}{dx} = e^{x+y} - e^x \]
Factoring out \( \frac{dy}{dx} \) on the left side:
\[ \left( e^y - e^{x+y} \right) \frac{dy}{dx} = e^{x+y} - e^x \]

5. **Solve for \( \frac{dy}{dx} \)**:
Now we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{e^{x+y} - e^x}{e^y - e^{x+y}} \]

6. **Simplifying the Expression**:
We can simplify the expression further:
\[ \frac{dy}{dx} = \frac{e^x(e^y - 1)}{e^y - e^{x+y}} = \frac{e^x(e^y - 1)}{e^y(1 - e^x)} \]

### Final Result:
Thus, the derivative \( y' \) is given by:
\[ \frac{dy}{dx} = \frac{e^x(e^y - 1)}{e^y(1 - e^x)} \]