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

Explanation

To find \(\frac{dy}{dx}\) given the equation \(xy = e^{(x - y)}\), we will follow these steps:

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

### Step 2: Simplify using logarithmic properties
Using the properties of logarithms, we can simplify:
\[ \ln(x) + \ln(y) = (x - y) \cdot \ln(e) \]
Since \(\ln(e) = 1\), this simplifies to:
\[ \ln(x) + \ln(y) = x - y \]

### Step 3: Differentiate both sides with respect to \(x\)
Now we differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(\ln(x) + \ln(y)) = \frac{d}{dx}(x - y) \]
Using the chain rule on the left side:
\[ \frac{1}{x} + \frac{1}{y} \cdot \frac{dy}{dx} = 1 - \frac{dy}{dx} \]

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

### Step 5: Solve for \(\frac{dy}{dx}\)
Now we can solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{1 - \frac{1}{x}}{\frac{1}{y} + 1} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{(x - 1)}{x} \cdot \frac{y}{(y + 1)} \]

### Final Result
Thus, we have:
\[ \frac{dy}{dx} = \frac{(x - 1)y}{x(y + 1)} \]

---