If ` x^y =y^x ,` then ` (dy)/(dx)= (y (xlog y-y))/(x(y log x-x))`

Answer

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

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

### Step 2: Apply the power rule of logarithms
Using the property of logarithms \( \log(a^b) = b \log(a) \), 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, we have:
\[ \frac{dy}{dx} \log x + y \cdot \frac{1}{x} = \log y + x \cdot \frac{dy}{dx} \cdot \frac{1}{y} \]

### Step 4: Rearranging the equation
Now we will rearrange the equation to isolate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} \log x - x \cdot \frac{dy}{dx} \cdot \frac{1}{y} = \log y - \frac{y}{x} \]
Factoring out \( \frac{dy}{dx} \) from the left-hand side gives:
\[ \frac{dy}{dx} \left( \log x - \frac{x}{y} \right) = \log y - \frac{y}{x} \]

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

### Step 6: Simplifying the expression
To express \( \frac{dy}{dx} \) in the required form, we can multiply the numerator and denominator by \( xy \):
\[ \frac{dy}{dx} = \frac{y(\log y - \frac{y}{x})}{x(\log x - \frac{x}{y})} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{y(x \log y - y)}{x(y \log x - x)} \]

Thus, we have shown that:
\[ \frac{dy}{dx} = \frac{y(x \log y - y)}{x(y \log x - x)} \]