The problem asks us to use implicit differentiation to find $y'$, given the equation $x^y = y^x$.

1. Problem Description

The problem asks us to use implicit differentiation to find $y'$, given the equation $x^y = y^x$.

2. Solution Steps

We are given $x^y = y^x$. Taking the natural logarithm of both sides, we get

$$\ln(x^y) = \ln(y^x)$$

Using the logarithm property $\ln(a^b) = b \ln(a)$, we have

$$y \ln(x) = x \ln(y)$$

Now we differentiate both sides with respect to $x$ using the product rule.

$$\frac{d}{dx}(y \ln(x)) = \frac{d}{dx}(x \ln(y))$$

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

$$y' \ln(x) + \frac{y}{x} = \ln(y) + \frac{x}{y} y'$$

Now we want to isolate $y'$.

$$y' \ln(x) - \frac{x}{y} y' = \ln(y) - \frac{y}{x}$$

$$y' \left( \ln(x) - \frac{x}{y} \right) = \ln(y) - \frac{y}{x}$$

$$y' = \frac{\ln(y) - \frac{y}{x}}{\ln(x) - \frac{x}{y}}$$

Multiply both the numerator and the denominator by $xy$:

$$y' = \frac{xy \ln(y) - y^2}{xy \ln(x) - x^2}$$

3. Final Answer

$$y' = \frac{xy \ln(y) - y^2}{xy \ln(x) - x^2}$$