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

1. Problem Description

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

2. Solution Steps

We are given the equation $x^y = y^x$. We will take the natural logarithm of both sides:

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

Using logarithm properties, we have:

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

Now we differentiate both sides with respect to $x$ using the product rule and implicit differentiation for terms involving $y$:

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

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

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

Now we want to solve for $\frac{dy}{dx}$, which we denote as $y'$. Rearrange the equation to isolate $y'$:

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

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

Now, solve for $y'$:

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

We can multiply the numerator and denominator by $xy$:

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

3. Final Answer

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