If ` x^(y) = e^(2(x-y)), "then" (dy)/(dx) ` is equal to

Options

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

Explanation

To solve the problem \( x^y = e^{2(x-y)} \) and find \( \frac{dy}{dx} \), we will follow these steps:

### Step 1: Take the natural logarithm of both sides
We start with the equation:
\[ x^y = e^{2(x-y)} \]
Taking the natural logarithm on both sides gives us:
\[ \ln(x^y) = \ln(e^{2(x-y)}) \]

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

### Step 3: Rearranging the equation
Expanding the right side:
\[ y \ln x = 2x - 2y \]
Now, we can rearrange this to isolate terms involving \( y \):
\[ y \ln x + 2y = 2x \]
Factoring out \( y \) from the left side:
\[ y(\ln x + 2) = 2x \]

### Step 4: Solve for \( y \)
Now, we can solve for \( y \):
\[ y = \frac{2x}{\ln x + 2} \]

### Step 5: Differentiate both sides with respect to \( x \)
Now we differentiate \( y \) with respect to \( x \) using the quotient rule:
Let \( u = 2x \) and \( v = \ln x + 2 \). Then:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Calculating \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
- \( \frac{du}{dx} = 2 \)
- \( \frac{dv}{dx} = \frac{1}{x} \)

Substituting these into the quotient rule:
\[ \frac{dy}{dx} = \frac{(\ln x + 2)(2) - (2x)(\frac{1}{x})}{(\ln x + 2)^2} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{2(\ln x + 2) - 2}{(\ln x + 2)^2} \]
\[ = \frac{2\ln x + 4 - 2}{(\ln x + 2)^2} \]
\[ = \frac{2\ln x + 2}{(\ln x + 2)^2} \]
\[ = \frac{2(\ln x + 1)}{(\ln x + 2)^2} \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{2(\ln x + 1)}{(\ln x + 2)^2} \]