If `x^(y) = y^(x) " then " (dy)/(dx)=` ?

Options

A. `((y - x log y))/((x - y log x))`
B. `(y(y -x log y))/(x(x - y logx))`
C. `(y(y + x log y))/(x(x + y logx))`
D. none of these

Explanation

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

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

### Step 2: Apply the logarithmic identity
Using the property of logarithms that states \( \log(a^b) = b \log(a) \), we can rewrite the equation as:
\[ y \log(x) = x \log(y) \]

### Step 3: Differentiate both sides with respect to \( x \)
Now we differentiate both sides with respect to \( x \). We will use implicit differentiation:
\[ \frac{d}{dx}(y \log(x)) = \frac{d}{dx}(x \log(y)) \]

Using the product rule on both sides:
- For the left side, \( \frac{d}{dx}(y \log(x)) = \frac{dy}{dx} \log(x) + y \cdot \frac{1}{x} \)
- For the right side, \( \frac{d}{dx}(x \log(y)) = \log(y) + x \cdot \frac{1}{y} \frac{dy}{dx} \)

So we have:
\[ \frac{dy}{dx} \log(x) + \frac{y}{x} = \log(y) + x \cdot \frac{1}{y} \frac{dy}{dx} \]

### Step 4: Rearrange the equation
Now we rearrange the equation to isolate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} \log(x) - \frac{x}{y} \frac{dy}{dx} = \log(y) - \frac{y}{x} \]

Factoring out \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} \left( \log(x) - \frac{x}{y} \right) = \log(y) - \frac{y}{x} \]

### Step 5: Solve for \( \frac{dy}{dx} \)
Now, we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{\log(y) - \frac{y}{x}}{\log(x) - \frac{x}{y}} \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is given by:
\[ \frac{dy}{dx} = \frac{\log(y) - \frac{y}{x}}{\log(x) - \frac{x}{y}} \]