If `x^(y)=e^(x-y)`, then `dy//dx` is equal to which one of the following ?

Options

A. `((x-y))/((1+logx)^(2))`
B. `(y)/((1+logx))`
C. `((x+y))/((1+logx))`
D. `((logx)/(1+logx)^(2))`

Explanation

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

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

### 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:
\[ y \ln x + y = x \]
or
\[ y (\ln x + 1) = x \]

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

### Step 5: Rearrange to isolate \( \frac{dy}{dx} \)
Now, we rearrange the equation to isolate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx}(\ln x + 1) = 1 - \frac{y}{x} \]
Thus,
\[ \frac{dy}{dx} = \frac{1 - \frac{y}{x}}{\ln x + 1} \]

### Step 6: Simplify the expression
To simplify further, we can express \( 1 - \frac{y}{x} \) as:
\[ 1 - \frac{y}{x} = \frac{x - y}{x} \]
So we have:
\[ \frac{dy}{dx} = \frac{x - y}{x(\ln x + 1)} \]

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