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The given equation is:

\[
x \ln y + y^5 = 6 \ln x.
\]

We are tasked with finding \(\frac{dy}{dx}\) in terms of \(x\) and \(y\).

### Steps to Solve:

1. **Differentiate both sides implicitly with respect to \(x\):**

   For the left-hand side:
   \[
   \frac{d}{dx}(x \ln y) + \frac{d}{dx}(y^5)
   \]

   Using the product rule for the term \(x \ln y\), we get:
   \[
   \frac{d}{dx}(x \ln y) = \ln y + x \cdot \frac{1}{y} \cdot \frac{dy}{dx}.
   \]

   For the term \(y^5\), we apply the chain rule:
   \[
   \frac{d}{dx}(y^5) = 5y^4 \cdot \frac{dy}{dx}.
   \]

   So the derivative of the left-hand side becomes:
   \[
   \ln y + \frac{x}{y} \cdot \frac{dy}{dx} + 5y^4 \cdot \frac{dy}{dx}.
   \]

2. **Differentiate the right-hand side:**

   The derivative of \(6 \ln x\) is:
   \[
   \frac{d}{dx}(6 \ln x) = \frac{6}{x}.
   \]

3. **Combine the derivatives:**

   Now we have the equation:
   \[
   \ln y + \frac{x}{y} \cdot \frac{dy}{dx} + 5y^4 \cdot \frac{dy}{dx} = \frac{6}{x}.
   \]

4. **Solve for \(\frac{dy}{dx}\):**

   Collect the terms with \(\frac{dy}{dx}\) on one side:
   \[
   \frac{x}{y} \cdot \frac{dy}{dx} + 5y^4 \cdot \frac{dy}{dx} = \frac{6}{x} - \ln y.
   \]

   Factor out \(\frac{dy}{dx}\):
   \[
   \frac{dy}{dx} \left( \frac{x}{y} + 5y^4 \right) = \frac{6}{x} - \ln y.
   \]

   Finally, solve for \(\frac{dy}{dx}\):
   \[
   \frac{dy}{dx} = \frac{\frac{6}{x} - \ln y}{\frac{x}{y} + 5y^4}.
   \]

So, the derivative \(\frac{dy}{dx}\) in terms of \(x\) and \(y\) is:

\[
\frac{dy}{dx} = \frac{\frac{6}{x} - \ln y}{\frac{x}{y} + 5y^4}.
\]

---

Would you like more details or further explanations on any of the steps?

### Related Questions:
1. How does implicit differentiation differ from regular differentiation?
2. What is the product rule, and why is it important in this problem?
3. How would the result change if the equation were \(x \ln y + y^3 = 6 \ln x\) instead?
4. How can we verify the correctness of the derivative we found?
5. Can we solve for \(y\) in terms of \(x\) explicitly from the original equation?

### Tip:
When using implicit differentiation, always treat \(y\) as a function of \(x\), applying the chain rule whenever you differentiate terms involving \(y\).

Find dy/dx in terms of x and y if x ln y + y^5 = 6 ln x.

Learn how to find dy/dx through implicit differentiation from the equation x ln y + y^5 = 6 ln x. This problem applies key concepts like the product rule and chain rule, offering a detailed step-by-step solution. Suitable for high school calculus students.

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