https://file.aiquickdraw.com/mathbot/file/17291000722230352me0y.jpg

The equation provided in the image is:

\[
x^y = y^x
\]

We need to find \( y' \) (the derivative of \( y \) with respect to \( x \)).

### Steps to solve:

1. Start by taking the natural logarithm of both sides to simplify the powers:
   \[
   \ln(x^y) = \ln(y^x)
   \]
   Using logarithmic properties (\(\ln(a^b) = b \ln(a)\)):
   \[
   y \ln(x) = x \ln(y)
   \]

2. Differentiate both sides of this equation with respect to \( x \). Apply implicit differentiation:
   
   For the left-hand side \( y \ln(x) \):
   \[
   \frac{d}{dx} [y \ln(x)] = y' \ln(x) + y \cdot \frac{1}{x}
   \]
   
   For the right-hand side \( x \ln(y) \):
   \[
   \frac{d}{dx} [x \ln(y)] = \ln(y) + x \cdot \frac{y'}{y}
   \]
   
3. Now set the two derivatives equal:
   \[
   y' \ln(x) + \frac{y}{x} = \ln(y) + \frac{x y'}{y}
   \]

4. Collect terms involving \( y' \) on one side:
   \[
   y' \ln(x) - \frac{x y'}{y} = \ln(y) - \frac{y}{x}
   \]

5. Factor out \( y' \) on the left-hand side:
   \[
   y' \left( \ln(x) - \frac{x}{y} \right) = \ln(y) - \frac{y}{x}
   \]

6. Solve for \( y' \):
   \[
   y' = \frac{\ln(y) - \frac{y}{x}}{\ln(x) - \frac{x}{y}}
   \]

Thus, the derivative \( y' \) is:

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

Would you like further clarification on any step or more details about logarithmic differentiation?

Here are 5 related questions you might explore:

1. How do we handle logarithmic differentiation in other types of equations?
2. What are the general rules for implicit differentiation?
3. Can we apply this method to non-exponential equations?
4. How does the product rule apply in implicit differentiation?
5. What happens when \( x = y \) in the original equation?

**Tip:** Always remember to apply the chain rule when differentiating functions involving products of variables!

Learn how to solve the equation x^y = y^x using implicit differentiation. This detailed solution covers logarithmic differentiation, chain rule applications, and implicit derivative techniques. Suitable for advanced high school students and calculus learners.

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