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?

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 $x$.