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

Options

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

Answer

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

### Step 1: Take the natural logarithm of both sides
We start 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: Rearranging the equation
Rearranging the equation gives us:

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

This can be factored as:

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

### Step 4: Solve for \( y \)
Now, we can express \( y \) in terms of \( x \):

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

### Step 5: Differentiate both sides with respect to \( x \)
Now we differentiate both sides with respect to \( x \):

Using the quotient rule:

\[ \frac{dy}{dx} = \frac{(\ln(x) + 1)(1) - x \cdot \frac{1}{x} \cdot 1}{(\ln(x) + 1)^2} \]

This simplifies to:

\[ \frac{dy}{dx} = \frac{\ln(x) + 1 - 1}{(\ln(x) + 1)^2} \]

### Step 6: Final simplification
Thus, we have:

\[ \frac{dy}{dx} = \frac{\ln(x)}{(\ln(x) + 1)^2} \]

### Final Answer
The final answer is:

\[ \frac{dy}{dx} = \frac{\ln(x)}{(1 + \ln(x))^2} \]

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