If `x^(y)=e^(x-y)`, then `(dy)/(dx)` is equal to

Options

A. `(1+logx)^(-1)`
B. `(1+logx)^(-2)`
C. `logx(1+logx)^(-2)`
D. none of these

Answer

To solve the equation \( x^y = e^{x - y} \) for \( \frac{dy}{dx} \), 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:

\[ \ln(x^y) = \ln(e^{x - y}) \]

### Step 2: Simplify using logarithmic properties
Using the properties of logarithms, we can simplify both sides:

\[ y \ln x = x - y \]

### Step 3: Rearrange the equation
Rearranging the equation gives us:

\[ y \ln x + y = x \]

This can be factored as:

\[ y(\ln x + 1) = x \]

### Step 4: Differentiate both sides with respect to \( x \)
Now we differentiate both sides with respect to \( x \):

\[ \frac{d}{dx}[y(\ln x + 1)] = \frac{d}{dx}[x] \]

Using the product rule on the left side, we have:

\[ \frac{dy}{dx}(\ln x + 1) + y \cdot \frac{1}{x} = 1 \]

### Step 5: Solve for \( \frac{dy}{dx} \)
Now we can isolate \( \frac{dy}{dx} \):

\[ \frac{dy}{dx}(\ln x + 1) = 1 - \frac{y}{x} \]

\[ \frac{dy}{dx} = \frac{1 - \frac{y}{x}}{\ln x + 1} \]

### Step 6: Simplify the expression
We can rewrite the expression for \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{x - y}{x(\ln x + 1)} \]

### Final Result
Thus, the final expression for \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = \frac{x - y}{x(1 + \ln x)} \]