If `y^x=x^y ,then find (dy)/(dx)dot`

Explanation

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

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

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

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

Using the product rule on the left side:
\[ \frac{d}{dx}(x \log y) = \log y + x \frac{d(\log y)}{dx} \]
And for the right side:
\[ \frac{d}{dx}(y \log x) = \frac{dy}{dx} \log x + y \frac{1}{x} \]

### Step 4: Set the derivatives equal
Now we set the derivatives equal to each other:
\[ \log y + x \frac{dy}{dx} \frac{1}{y} = \frac{dy}{dx} \log x + \frac{y}{x} \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ x \frac{dy}{dx} \frac{1}{y} - \frac{dy}{dx} \log x = \frac{y}{x} - \log y \]

### Step 6: Factor out \( \frac{dy}{dx} \)
Factoring out \( \frac{dy}{dx} \) from the left side:
\[ \frac{dy}{dx} \left( \frac{x}{y} - \log x \right) = \frac{y}{x} - \log y \]

### Step 7: Solve for \( \frac{dy}{dx} \)
Now we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{\frac{y}{x} - \log y}{\frac{x}{y} - \log x} \]

### Step 8: Simplify the expression
We can simplify this expression further:
\[ \frac{dy}{dx} = \frac{y - x \log y}{x - y \log x} \]

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

---