 y ln(y) − x ln(x) = 0. implicit differentiation

Let's implicitly differentiate the equation \(y \ln(y) - x \ln(x) = 0\) with respect to \(x\).

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### Step-by-Step Solution:

Given:
\[
y \ln(y) - x \ln(x) = 0
\]

1. **Differentiate both sides with respect to \(x\)**:
   - Use the **product rule**: \( \frac{d}{dx}(u \cdot v) = u' v + u v' \).
   - Recall: When differentiating \(y\), treat it as a function of \(x\), so the **chain rule** applies.

   \[
   \frac{d}{dx} \left( y \ln(y) \right) - \frac{d}{dx} \left( x \ln(x) \right) = \frac{d}{dx}(0)
   \]

2. **Differentiate the terms**:
   - **First term** \(y \ln(y)\):
     \[
     \frac{d}{dx}(y \ln(y)) = \ln(y) \frac{dy}{dx} + y \cdot \frac{1}{y} \frac{dy}{dx} = \ln(y) \frac{dy}{dx} + \frac{dy}{dx}
     \]
     (Here we used both the product rule and the chain rule.)

   - **Second term** \(x \ln(x)\):
     \[
     \frac{d}{dx}(x \ln(x)) = \ln(x) + 1
     \]
     (This is a direct application of the product rule.)

3. **Substitute the derivatives into the equation**:
   \[
   \left( \ln(y) + 1 \right) \frac{dy}{dx} - \left( \ln(x) + 1 \right) = 0
   \]

4. **Solve for \(\frac{dy}{dx}\)**:
   \[
   \left( \ln(y) + 1 \right) \frac{dy}{dx} = \ln(x) + 1
   \]
   \[
   \frac{dy}{dx} = \frac{\ln(x) + 1}{\ln(y) + 1}
   \]

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### Final Answer:
\[
\boxed{\frac{dy}{dx} = \frac{\ln(x) + 1}{\ln(y) + 1}}
\]

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Do you have any questions or want further details?  

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Here are five related questions to explore this topic further:  
1. How does the chain rule work in implicit differentiation?  
2. Can you solve for \(y\) in terms of \(x\) directly from \(y \ln(y) - x \ln(x) = 0\)?  
3. What is the product rule and when do you apply it?  
4. What happens if we replace the natural logarithms with logarithms of a different base?  
5. How would the answer change if the equation were \(y^2 \ln(y) - x^2 \ln(x) = 0\)?  

**Tip**: Always apply the chain rule when differentiating a function of \(y\) with respect to \(x\).

y ln(y) − x ln(x) = 0. implicit differentiation

Explore the implicit differentiation of the equation y ln(y) − x ln(x) = 0. This detailed solution covers key concepts such as the product rule, chain rule, and logarithmic differentiation. Suitable for advanced high school and college students.

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