91Ó°ÊÓ

Partial derivatives are a way to measure how a multivariable function changes as one of its variables is altered, while the others are held constant. In simpler terms, it describes the rate of change of a function with respect to each individual variable. For the function given, \( f(x, y) = x^2 + 3y \), we find the partial derivatives:

• With respect to \( x \): \( f_x = \frac{\partial f}{\partial x} = 2x \)
• With respect to \( y \): \( f_y = \frac{\partial f}{\partial y} = 3 \)

This tells us that for every unit increase in \( x \), \( f(x, y) \) increases by \( 2x \), and for every unit increase in \( y \), \( f(x, y) \) increases by 3. Identifying partial derivatives is crucial for understanding differentiability and is a fundamental step in solving such problems.

Limits help us handle values approaching a particular point, even if they never actually reach that point. In the context of differentiability, they allow us to explore how a function behaves as changes in the inputs become very small. For this problem, we are interested in \( \lim_{(\Delta x, \Delta y) \to (0,0)} \varepsilon_1 \Delta x + \varepsilon_2 \Delta y \). As the changes \( \Delta x \) and \( \Delta y \) reduce to zero, the terms \( \varepsilon_1 \) and \( \varepsilon_2 \) should also approach zero in a way that ensures continuity and differentiability. This aligns with understanding the continuity and smoothness of the function \( f(x, y) \). The property of limits guarantees that our function satisfies the required conditions for differentiability wherever the limit holds true.

Function expansion allows us to express a function in terms of its variables with slightly altered values. This method helps us comprehend how a function diverges or resembles a simpler form when each variable is incremented by a small amount, \( \Delta x \) and \( \Delta y \). In this exercise: We substitute \( x + \Delta x \) and \( y + \Delta y \) into \( f(x, y) \).

• Compute the squared term: \( (x + \Delta x)^2 = x^2 + 2x\Delta x + (\Delta x)^2 \)
• Add the linear term adjustment: \( 3(y + \Delta y) = 3y + 3\Delta y \)

The expansion yields: \( x^2 + 2x\Delta x + (\Delta x)^2 + 3y + 3\Delta y \). This expansion is critical to breaking down and evaluating how each segment contributes to the function's total change \( \Delta z \). It is particularly useful in crafting expressions that demonstrate differentiability.

The epsilon-delta definition is a formal way of proving that a function is continuous (and thus, differentiable) at a certain point. This approach involves showing that for every small value \( \varepsilon \) there exists another small value \( \delta \) such that changes in the input \( \Delta x, \Delta y \) affect the change in output \( \Delta z \) in a controlled manner. In this problem, we need \( \varepsilon_1 \) and \( \varepsilon_2 \) to approach zero as \( \Delta x \) and \( \Delta y \) tend to zero.

• Consider \( \varepsilon_1 = \Delta x \) and \( \varepsilon_2 = 0 \).
• As \( \Delta x \to 0 \) and \( \Delta y \to 0 \), both \( \varepsilon_1 \Delta x \) and \( \varepsilon_2 \Delta y \) effectively become zero.

By fulfilling these conditions, we demonstrate that the change in function value \( \Delta z \) remains within acceptable bounds for every small \( \varepsilon \), making \( f(x, y) \) differentiable at any point \((x, y)\). This foundational idea is a cornerstone of calculus, ensuring that the behavior of functions is consistent and predictable.