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Show that the function is differentiable by finding values for \(\varepsilon_{1}\) and \(\varepsilon_{2}\) as designated in the definition of differentiability, and verify that both \(\varepsilon_{1}\) and \(\varepsilon_{2} \rightarrow 0\) as \((\Delta x, \Delta y) \rightarrow(0,0)\) $$ f(x, y)=x^{2} y $$

Short Answer

Expert verified
The function \(f(x, y)=x^{2} y\) is differentiable everywhere, since both \(\varepsilon_{1}\) and \(\varepsilon_{2}\) go to 0 as \((\Delta x, \Delta y) \rightarrow (0,0)\).

Step by step solution

01

Define the function

The given function is \(f(x, y)=x^{2} y\)
02

Calculate partial derivatives

The partial derivatives of the function are: \(\frac{\partial f}{\partial x} = 2xy\) and \(\frac{\partial f}{\partial y} = x^2\). These will be used to calculate the increments \(\Delta f\) and \(\varepsilon_{1}\), \(\varepsilon_{2}\).
03

Compute the increment \(\Delta f\)

The increment \(\Delta f\) is obtained by evaluating the function at \((x_0 + \Delta x, y_0 + \Delta y)\) and subtracting the value of the function at \((x_0, y_0)\). This gives : \(\Delta f = f(x_0 + \Delta x, y_0 + \Delta y) - f(x_0, y_0) = (x_0 + \Delta x)^2 (y_0 + \Delta y) - x_0^2 y_0\)
04

Compute \(\varepsilon_{1}\) and \(\varepsilon_{2}\)

\(\varepsilon_{1}\) and \(\varepsilon_{2}\) are defined such that \(\Delta f = \frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y + \varepsilon_{1} |\Delta x| + \varepsilon_{2} |\Delta y| \). These can be obtained by rearranging terms and using expressions from Step 2. We get: \(\varepsilon_{1} = (\Delta f - 2x_0 y_0 \Delta x - x_0^2 \Delta y) / |\Delta x|\) and \(\varepsilon_{2} = (\Delta f - 2x_0 y_0 \Delta x - x_0^2 \Delta y) / |\Delta y|\)
05

Let \((\Delta x, \Delta y) \rightarrow (0,0)\)

To show the differentiability of \(f\), we need to show both \(\varepsilon_{1}\) and \(\varepsilon_{2}\) approach 0 as \((\Delta x, \Delta y) \rightarrow (0,0)\). Since \(\varepsilon_{1}\) and \(\varepsilon_{2}\) only involve terms of \(\Delta x\) and \(\Delta y\), we find that as \(\Delta x, \Delta y \rightarrow 0\), indeed \(\varepsilon_{1}, \varepsilon_{2} \rightarrow 0\). So, the function \(f(x, y)=x^{2} y\) is differentiable at all points.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
When we delve into the realm of multivariable calculus, understanding partial derivatives is crucial. They are, simply put, the derivatives of a multivariable function with respect to one variable at a time, holding all other variables constant. In the given function f(x, y) = x^2y, we calculate the partial derivatives with respect to x and y.

For x, the partial derivative is 2xy, indicating how the function's value changes as x varies while y is kept constant. Conversely, the derivative x^2 with respect to y shows the rate of change along the y-direction. These derivatives are key in analyzing the function's behavior and are used to find the increment \(\Delta f\) which helps us further understand the concept of differentiability.
Increment \(\Delta f\)
The increment \(\Delta f\) represents the change in the function's value f given small changes in its inputs x and y. It's calculated by comparing the value of the function at a point enhanced by a small increment (x + \Delta x, y + \Delta y) against the original point (x, y).

In our example, \(\Delta f = (x_0 + \Delta x)^2 (y_0 + \Delta y) - x_0^2 y_0\) encapsulates this change. If we visualize this on a three-dimensional graph, where the z-axis represents the function's output, \(\Delta f\) is the vertical distance between the surface and a plane parallel to the xy-plane at point (x_0, y_0). Understanding this increment is a stepping stone to grasping the concept of differentiability.
Multivariable Calculus
Multivariable calculus extends calculus to functions of multiple variables. While single-variable calculus asks how a function changes as its single input changes, multivariable calculus explores these changes along multiple directions. This is where concepts like partial derivatives and increments become essential.

The function f(x, y) = x^2y is an example where two variable inputs jointly influence the output, and multivariable calculus allows us to dissect these influences individually and together. By mastering this discipline, students can analyze more complex systems encountered in physics, engineering, and economics, where multiple factors simultaneously affect outcomes.
Limits and Continuity
In calculus, the ideas of limits and continuity lay the groundwork for understanding concepts like derivatives and integrals. These notions are just as integral in multivariable calculus.

A function is continuous if, roughly speaking, you can draw it without lifting your pen off the paper. More technically, a function f(x, y) is continuous at a point if the limit of f as (x, y) approaches that point equals the function’s value at that point. Limits, in this context, show us the behavior of the function as we zoom in infinitesimally close to a point.

For differentiability, we need continuity, and we also need the function to have a well-defined tangent plane at that point, which the concept of using limits to ensure \(\epsilon_{1}\) and \(\epsilon_{2}\) approach zero helps to establish. Thereby illustrating that our function behaves predictably as we move infinitesimally close to any given point, making the partial derivatives valid representatives of the function's behavior around that point.

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Most popular questions from this chapter

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