(c) y ln(y) − x ln(x) = 0. implicite differentaiation

To implicitly differentiate the equation \( y \ln(y) - x \ln(x) = 0 \), we follow these steps:

### Step 1: Differentiate both sides with respect to \( x \).

The given equation is:

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

We differentiate each term using the product rule, as both \( y \ln(y) \) and \( x \ln(x) \) involve two functions (in the case of \( y \ln(y) \), \( y \) is implicitly a function of \( x \)).

#### Differentiating \( y \ln(y) \):

\[
\frac{d}{dx} \left( y \ln(y) \right) = \ln(y) \frac{dy}{dx} + \frac{y}{y} \frac{dy}{dx} = \ln(y) \frac{dy}{dx} + \frac{dy}{dx}
\]

#### Differentiating \( x \ln(x) \):

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

### Step 2: Set up the differentiated equation.

Now we substitute these results into the equation:

\[
\left( \ln(y) + 1 \right) \frac{dy}{dx} - \left( \ln(x) + 1 \right) = 0
\]

### Step 3: Solve for \( \frac{dy}{dx} \).

To solve for \( \frac{dy}{dx} \), we rearrange the equation:

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

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

This is the implicit derivative of \( y \) with respect to \( x \).

### Summary:
The implicit derivative is:
\[
\frac{dy}{dx} = \frac{\ln(x) + 1}{\ln(y) + 1}
\]

Do you want further details or have any questions on the steps?  
Here are 5 related questions to further explore:

1. How does the product rule work in implicit differentiation?
2. What is the chain rule, and how does it apply when differentiating \( y \ln(y) \)?
3. Can you explain implicit differentiation and how it differs from explicit differentiation?
4. How would you solve \( y \ln(y) - x \ln(x) = 0 \) for \( y \) explicitly?
5. What happens if \( y = x \)? How would that affect the differentiation?

**Tip**: When using implicit differentiation, remember to apply the chain rule whenever you differentiate a term involving \( y \), as \( y \) is a function of \( x \).

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

Explore the implicit differentiation of the equation y ln(y) - x ln(x) = 0 with a step-by-step solution. Learn how to apply the product rule and chain rule to logarithmic functions. This solution is suitable for advanced high school or introductory calculus students.

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