If `e^(x +y) = xy " then " (dy)/(dx)=` ?

Options

A. `(x (1 -y))/(y (x -1))`
B. `(y(1 -x))/(x(y -1))`
C. `((x -xy))/(xy -y)`
D. none of these

Explanation

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

### Step 1: Take the natural logarithm of both sides
We start by taking the natural logarithm of both sides of the equation:
\[ \ln(e^{x+y}) = \ln(xy) \]

### Step 2: Simplify using logarithmic properties
Using the property of logarithms, \( \ln(e^a) = a \) and \( \ln(ab) = \ln(a) + \ln(b) \), we can simplify:
\[ x + y = \ln(x) + \ln(y) \]

### Step 3: Differentiate both sides with respect to \( x \)
Now we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(x + y) = \frac{d}{dx}(\ln(x) + \ln(y)) \]
This gives us:
\[ 1 + \frac{dy}{dx} = \frac{1}{x} + \frac{1}{y} \frac{dy}{dx} \]

### Step 4: Rearrange the equation
Next, we rearrange the equation to isolate \( \frac{dy}{dx} \):
\[ 1 + \frac{dy}{dx} - \frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \]
Factoring out \( \frac{dy}{dx} \):
\[ 1 + \frac{dy}{dx} \left(1 - \frac{1}{y}\right) = \frac{1}{x} \]

### Step 5: Solve for \( \frac{dy}{dx} \)
Now we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} \left(1 - \frac{1}{y}\right) = \frac{1}{x} - 1 \]
\[ \frac{dy}{dx} = \frac{\frac{1}{x} - 1}{1 - \frac{1}{y}} \]

### Step 6: Simplify the expression
To simplify further, we can multiply the numerator and denominator by \( xy \):
\[ \frac{dy}{dx} = \frac{y - x}{x(y - 1)} \]

### Final Answer
Thus, the final expression for \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{y - x}{x(y - 1)} \]