We are given the equation $4x^2 y^3 = y e^{2x}$, and we need to find an expression for $\frac{dy}{dx}$, the derivative of $y$ with respect to $x$. This involves implicit differentiation since $y$ is an implicit function of $x$.

Step 1: Differentiate both sides with respect to $x$

Left-hand side:

$\frac{d}{dx}(4x^2 y^3) = 4 \cdot \frac{d}{dx}(x^2 y^3)$ Using the product rule $\frac{d}{dx}(uv) = u'v + uv'$, where $u = x^2$ and $v = y^3$, we get: $= 4 \left( 2x \cdot y^3 + x^2 \cdot 3y^2 \frac{dy}{dx} \right)$ Simplifying this: $= 8x y^3 + 12x^2 y^2 \frac{dy}{dx}$

Right-hand side:

$\frac{d}{dx}(y e^{2x}) = \frac{d}{dx}(y) \cdot e^{2x} + y \cdot \frac{d}{dx}(e^{2x})$ Using the chain rule for $e^{2x}$: $= \frac{dy}{dx} e^{2x} + y \cdot 2e^{2x}$

Step 2: Combine and solve for $\frac{dy}{dx}$

Now, equate the differentiated sides: $8x y^3 + 12x^2 y^2 \frac{dy}{dx} = \frac{dy}{dx} e^{2x} + 2y e^{2x}$

Rearrange the terms to isolate $\frac{dy}{dx}$: $12x^2 y^2 \frac{dy}{dx} - \frac{dy}{dx} e^{2x} = 2y e^{2x} - 8x y^3$ Factor out $\frac{dy}{dx}$ on the left-hand side: $\frac{dy}{dx} (12x^2 y^2 - e^{2x}) = 2y e^{2x} - 8x y^3$ Finally, solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{2y e^{2x} - 8x y^3}{12x^2 y^2 - e^{2x}}$

This is the expression for $\frac{dy}{dx}$.

Would you like further details on this derivation or specific steps?

Here are 5 related questions to expand on this:

1. How do you apply the product rule in differentiation?
2. What are some key cases where implicit differentiation is essential?
3. How does the chain rule work in this context?
4. How does implicit differentiation differ from explicit differentiation?
5. Why do we factor out $\frac{dy}{dx}$ when isolating it?

Tip: Implicit differentiation is particularly useful when dealing with equations where $y$ cannot be easily solved for $x$, as it allows you to differentiate both sides of the equation with respect to $x$ directly.