To find the derivative of $y = y(x)$ given the equation $x^3 \cos y + \sin(2x - 5y) = 4$, we will use implicit differentiation.

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

$\frac{d}{dx}(x^3 \cos y) + \frac{d}{dx}(\sin(2x - 5y)) = 0$

2. Applying the product rule to $x^3 \cos y$:

$\frac{d}{dx}(x^3 \cos y) = 3x^2 \cos y + x^3 (-\sin y \frac{dy}{dx})$

3. For $\sin(2x - 5y)$, we apply the chain rule:

$\frac{d}{dx}(\sin(2x - 5y)) = \cos(2x - 5y) \left(2 - 5\frac{dy}{dx}\right)$

4. Setting the differentiated equation to zero:

$3x^2 \cos y - x^3 \sin y \frac{dy}{dx} + \cos(2x - 5y)(2 - 5\frac{dy}{dx}) = 0$

5. Rearranging the equation to isolate $\frac{dy}{dx}$:

$-x^3 \sin y \frac{dy}{dx} - 5 \cos(2x - 5y) \frac{dy}{dx} = -3x^2 \cos y - 2 \cos(2x - 5y)$

6. Factoring out $\frac{dy}{dx}$:

$\left(-x^3 \sin y - 5 \cos(2x - 5y)\right) \frac{dy}{dx} = -3x^2 \cos y - 2 \cos(2x - 5y)$

7. Finally, solving for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{-3x^2 \cos y - 2 \cos(2x - 5y)}{-x^3 \sin y - 5 \cos(2x - 5y)}$

This simplifies to:

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

Would you like any further details on this process or any specific parts clarified? Here are some follow-up questions you might consider:

1. What does implicit differentiation mean in this context?
2. Can you explain the product and chain rules used in the differentiation?
3. How do we interpret the derivative in terms of $y$ and $x$?
4. What would happen if we set specific values for $x$ or $y$?
5. Are there applications of this type of derivative in real-world scenarios?

Tip: Always keep track of each term when applying the chain and product rules to avoid errors.