Consider the equation `x^(y)=e^(x-y)`
What is `(dy)/(dx)` at `x=1` equal to

Options

A. 0
B. 1
C. 2
D. 4

Answer

To find \(\frac{dy}{dx}\) at \(x = 1\) for the equation \(x^y = e^{x - y}\), 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 of both sides gives us:
\[ \ln(x^y) = \ln(e^{x - y}) \]

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

### Step 3: Rearrange the equation
Rearranging the equation gives:
\[ y \ln x + y = x \]
Factoring out \(y\) from the left side:
\[ y(\ln x + 1) = x \]

### Step 4: Solve for \(y\)
Thus, we can express \(y\) as:
\[ y = \frac{x}{\ln x + 1} \]

### Step 5: Differentiate both sides with respect to \(x\)
Now we differentiate \(y\) with respect to \(x\) using the quotient rule:
\[ \frac{dy}{dx} = \frac{(\ln x + 1)(1) - x \left(\frac{1}{x}\right)}{(\ln x + 1)^2} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{\ln x + 1 - 1}{(\ln x + 1)^2} = \frac{\ln x}{(\ln x + 1)^2} \]

### Step 6: Evaluate at \(x = 1\)
Now we substitute \(x = 1\):
\[ \frac{dy}{dx} \bigg|_{x=1} = \frac{\ln(1)}{(\ln(1) + 1)^2} = \frac{0}{(0 + 1)^2} = 0 \]

### Final Answer
Thus, \(\frac{dy}{dx}\) at \(x = 1\) is equal to:
\[ \boxed{0} \]