We are given the equation:

$y^3 + x^2 y^5 = e^{4x}$

To find $\frac{dy}{dx}$, we will differentiate both sides with respect to $x$, applying implicit differentiation. Since $y$ is a function of $x$, we'll need to use the chain rule when differentiating any term involving $y$.

Step-by-Step Differentiation:

1. Differentiate the first term $y^3$ with respect to $x$:

$\frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}$

2. Differentiate the second term $x^2 y^5$ using the product rule. The product rule states that:

$\frac{d}{dx}(u v) = u'v + uv'$

Let $u = x^2$ and $v = y^5$. Then:

$\frac{d}{dx}(x^2 y^5) = \frac{d}{dx}(x^2) \cdot y^5 + x^2 \cdot \frac{d}{dx}(y^5)$

Now, differentiate each part:

$\frac{d}{dx}(x^2) = 2x$

$\frac{d}{dx}(y^5) = 5y^4 \frac{dy}{dx}$

Thus, we get:

$\frac{d}{dx}(x^2 y^5) = 2x y^5 + x^2 \cdot 5y^4 \frac{dy}{dx}$

3. Differentiate the right-hand side $e^{4x}$ with respect to $x$:

$\frac{d}{dx}(e^{4x}) = 4e^{4x}$

Now, putting it all together:

The differentiated equation becomes:

$3y^2 \frac{dy}{dx} + \left(2x y^5 + 5x^2 y^4 \frac{dy}{dx}\right) = 4e^{4x}$

Isolating $\frac{dy}{dx}$:

Move the terms without $\frac{dy}{dx}$ to the right side:

$3y^2 \frac{dy}{dx} + 5x^2 y^4 \frac{dy}{dx} = 4e^{4x} - 2x y^5$

Factor out $\frac{dy}{dx}$ on the left side:

$\frac{dy}{dx}(3y^2 + 5x^2 y^4) = 4e^{4x} - 2x y^5$

Finally, solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{4e^{4x} - 2x y^5}{3y^2 + 5x^2 y^4}$

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

Would you like more details or have any questions?

Related questions:

1. How is implicit differentiation used for higher-order derivatives?
2. What is the geometric interpretation of $\frac{dy}{dx}$ in this equation?
3. Can this formula for $\frac{dy}{dx}$ be simplified further?
4. How would you apply implicit differentiation to trigonometric functions?
5. What role does the chain rule play in implicit differentiation?

Tip: Always apply the product rule carefully when differentiating terms involving two functions of $x$.