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The problem asks to find \(\frac{dy}{dx}\) using implicit differentiation for the equation:

\[
5x^4 + y^3 + e^{xy} = 0
\]

### Step-by-Step Solution:

1. **Differentiate each term with respect to \(x\)**, remembering that \(y\) is a function of \(x\), so when differentiating \(y\), we apply the chain rule and multiply by \(\frac{dy}{dx}\).

\[
\frac{d}{dx}\left(5x^4\right) + \frac{d}{dx}\left(y^3\right) + \frac{d}{dx}\left(e^{xy}\right) = 0
\]

2. **Differentiate each term**:
   - \( \frac{d}{dx}(5x^4) = 20x^3 \)
   - \( \frac{d}{dx}(y^3) = 3y^2 \cdot \frac{dy}{dx} \)
   - For \( e^{xy} \), we use the product rule:
     \[
     \frac{d}{dx}(e^{xy}) = e^{xy} \cdot \left(\frac{d}{dx}(xy)\right)
     \]
     Using the product rule on \(xy\):
     \[
     \frac{d}{dx}(xy) = y + x \cdot \frac{dy}{dx}
     \]
     So,
     \[
     \frac{d}{dx}(e^{xy}) = e^{xy} \cdot \left(y + x \cdot \frac{dy}{dx}\right)
     \]

3. **Write the full differentiated equation**:
   \[
   20x^3 + 3y^2 \cdot \frac{dy}{dx} + e^{xy} \cdot \left(y + x \cdot \frac{dy}{dx}\right) = 0
   \]

4. **Isolate \(\frac{dy}{dx}\)**:
   Collect the terms involving \(\frac{dy}{dx}\) on one side:
   \[
   3y^2 \cdot \frac{dy}{dx} + e^{xy} \cdot x \cdot \frac{dy}{dx} = -20x^3 - e^{xy} \cdot y
   \]

5. **Factor out \(\frac{dy}{dx}\)**:
   \[
   \frac{dy}{dx} \left( 3y^2 + e^{xy} \cdot x \right) = -20x^3 - e^{xy} \cdot y
   \]

6. **Solve for \(\frac{dy}{dx}\)**:
   \[
   \frac{dy}{dx} = \frac{-20x^3 - e^{xy} \cdot y}{3y^2 + e^{xy} \cdot x}
   \]

Thus, the solution for \(\frac{dy}{dx}\) is:

\[
\frac{dy}{dx} = \frac{-20x^3 - e^{xy} \cdot y}{3y^2 + e^{xy} \cdot x}
\]

Would you like more details or have any questions?

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Here are 5 related questions for further exploration:
1. What is implicit differentiation, and how does it differ from explicit differentiation?
2. How would you apply implicit differentiation to an equation involving a trigonometric function, like \(\sin(xy)\)?
3. Can you apply implicit differentiation to parametric equations? How?
4. How do higher-order derivatives work with implicit differentiation?
5. How does the chain rule apply to implicit differentiation when dealing with multiple variables?

**Tip:** When differentiating implicitly, always apply the chain rule to the terms involving \(y\), treating \(y\) as a function of \(x\).

Use implicit differentiation to find dy/dx, where 5x^4 + y^3 + e^(xy) = 0.

Learn how to find dy/dx using implicit differentiation for the equation 5x^4 + y^3 + e^(xy) = 0. This solution involves the product rule and chain rule, making it suitable for students in calculus or higher-level mathematics courses.

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