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 find \(\frac{dy}{dx}\) given 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 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: Solve for \(\frac{dy}{dx}\)
Rearranging 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: Substitute \(y\) in terms of \(x\)
From our earlier rearrangement \(y = \frac{x}{\ln x + 1}\). We can substitute this back into our expression for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{1 - \frac{\frac{x}{\ln x + 1}}{x}}{\ln x + 1} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{1 - \frac{1}{\ln x + 1}}{\ln x + 1} \]
Combining the terms gives:
\[ \frac{dy}{dx} = \frac{\ln x}{(\ln x + 1)^2} \]

### Final Answer
Thus, we find that:
\[ \frac{dy}{dx} = \frac{\ln x}{(\ln x + 1)^2} \]

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