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

Explanation

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

### Step 1: Take the logarithm of both sides
We start by taking the natural logarithm of both sides of the equation:

\[ \ln(y^x) = \ln(x^y) \]

### Step 2: Use the properties of logarithms
Using the property of logarithms that states \( \ln(a^b) = b \ln(a) \), we can rewrite the equation:

\[ x \ln(y) = y \ln(x) \]

### Step 3: Differentiate both sides with respect to \( x \)
Now we differentiate both sides of the equation with respect to \( x \). We will use the product rule on both sides.

For the left side:

\[ \frac{d}{dx}(x \ln(y)) = \frac{d}{dx}(x) \cdot \ln(y) + x \cdot \frac{d}{dx}(\ln(y)) \]

Using the chain rule for \( \ln(y) \):

\[ \frac{d}{dx}(\ln(y)) = \frac{1}{y} \cdot \frac{dy}{dx} \]

So the left side becomes:

\[ \ln(y) + x \cdot \frac{1}{y} \cdot \frac{dy}{dx} \]

For the right side:

\[ \frac{d}{dx}(y \ln(x)) = \frac{d}{dx}(y) \cdot \ln(x) + y \cdot \frac{d}{dx}(\ln(x)) \]

Again using the chain rule for \( \ln(x) \):

\[ \frac{d}{dx}(\ln(x)) = \frac{1}{x} \]

So the right side becomes:

\[ \frac{dy}{dx} \cdot \ln(x) + y \cdot \frac{1}{x} \]

### Step 4: Set the derivatives equal to each other
Now we have:

\[ \ln(y) + x \cdot \frac{1}{y} \cdot \frac{dy}{dx} = \frac{dy}{dx} \cdot \ln(x) + y \cdot \frac{1}{x} \]

### Step 5: Rearrange the equation to isolate \( \frac{dy}{dx} \)
Rearranging gives:

\[ x \cdot \frac{1}{y} \cdot \frac{dy}{dx} - \frac{dy}{dx} \cdot \ln(x) = \frac{1}{x}y - \ln(y) \]

Factoring out \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} \left( \frac{x}{y} - \ln(x) \right) = \frac{y}{x} - \ln(y) \]

### Step 6: Solve for \( \frac{dy}{dx} \)
Now, we can solve for \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{\frac{y}{x} - \ln(y)}{\frac{x}{y} - \ln(x)} \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = \frac{y - x \ln(y)}{x - y \ln(x)} \]

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