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In Exercises \(83-86,\) 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 \((\boldsymbol{\Delta x}, \boldsymbol{\Delta} \boldsymbol{y}) \rightarrow(\mathbf{0}, \mathbf{0})\) \(f(x, y)=x^{2}-2 x+y\)

Short Answer

Expert verified
The function \(f(x, y) = x^2 - 2x + y\) does not meet the criteria for differentiability at all points because as \((\Delta x, \Delta y)\) approach (0,0), \(\varepsilon_{1}\) approaches -1 and \(\varepsilon_{2}\) is always 0, hence they do not approach 0.

Step by step solution

01

Calculate Partial Derivatives

Differentiate \(f(x, y) = x^2 - 2x + y\) with respect to both variables \(x\) and \(y\) to obtain the first order partial derivatives. The first order derivatives are \(f_{x}(x, y) = 2x - 2\) and \(f_{y}(x, y) = 1\).
02

Calculate the Change in Function

Calculate the change in the function \(\Delta f(x, y) = f(x + \Delta x, y + \Delta y) - f(x, y)\). This simplifies to \(\Delta f = (x + \Delta x)^2 - 2(x + \Delta x) + (y + \Delta y) - (x^2 - 2x + y) = 2x\Delta x - \Delta x^2 - 2\Delta x + \Delta y\)
03

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

In the equation \(\Delta f = f_{x}(x,y)\Delta x + f_{y}(x,y)\Delta y + \varepsilon_{1} \Delta x + \varepsilon_{2} \Delta y\), coefficients of \(\Delta x\) and \(\Delta y\) must be equal on both sides, hence \(2x\Delta x + \Delta y = (2x - 2)\Delta x + \Delta y + \varepsilon_{1} \Delta x + \varepsilon_{2} \Delta y\). Isolating \(\varepsilon_{1}\) and \(\varepsilon_{2}\) gives us \(\varepsilon_{1} = -\Delta x / \Delta x\) and \(\varepsilon_{2} = 0\) . Note that \(\varepsilon_{1}\) and \(\varepsilon_{2}\) are functions of \(\Delta x\) and \(\Delta y\).
04

Show that \(\varepsilon_{1}\) and \(\varepsilon_{2}\) approach 0

It can be seen that as \(\Delta x, \Delta y\) approach (0,0) then \(\varepsilon_{1}\) approaches -1 and \(\varepsilon_{2}\) is always 0, therefore \(\varepsilon_{1}\) and \(\varepsilon_{2}\) do not approach 0 as \(\Delta x, \Delta y\) approach (0,0). Hence, by the definition of differentiability in several variables, \(f(x, y)\) does not meet the criteria for differentiability 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
Partial derivatives play a crucial role in multivariable calculus, especially in understanding how functions change with respect to each variable independently. Consider a function of two variables, like in our example, f(x, y) = x^2 - 2x + y. The partial derivative of this function with respect to x, denoted as f_x(x, y), is the derivative of f considered as a function of x alone while keeping y constant. Similarly, f_y(x, y) is obtained by differentiating f with respect to y while keeping x constant. These derivatives tell us the rate at which the function changes as we move along the x-axis and the y-axis. In the solution provided, f_x(x, y) was found to be 2x - 2, and f_y(x, y) was found to be 1.

To truly grasp the concept of partial derivatives, one must get comfortable with the notion of holding one variable steady while allowing the other to vary. Imagine walking along a mountain trail鈥攜our altitude changes with each step forward (in the direction of the x-axis), but not necessarily as you step side to side (along the y-axis). This nuance is what partial derivatives encapsulate. As such, the calculation of partial derivatives is a foundational step in analyzing the behavior of multivariable functions.
Multivariable Calculus
Moving beyond single-variable calculus, multivariable calculus explores functions of more than one variable, like our example function f(x, y). It's a vast field that extends concepts such as derivatives, integrals, and limits to higher dimensions. The differentiability of a multivariable function is determined by whether the function is smooth and has a tangent plane at a point, which intuitively means that the function has no sharp corners or edges at that point.

In our step-by-step solution, differentiability was assessed by looking at the behavior of the function as the point (螖x, 螖y) approached the origin. If partial derivatives exist and certain conditions are met鈥攕pecifically, if additional error terms 蔚鈧 and 蔚鈧 in the linear approximation of 螖蹿 go to zero as (螖x, 螖y) approaches the origin鈥攖he function is said to be differentiable. However, the exercise reveals a crucial point: 蔚鈧 does not approach zero, indicating that our function is not differentiable at all points. Thus, the solution showcases a pivotal aspect of multivariable calculus: understanding and applying the criteria for differentiability.
Epsilon-Delta Definition of Limit
The epsilon-delta definition of limit, foundational to calculus, formalizes the concept of approaching a limit. It defines a limit at a point for a function in terms of two values: 蔚 (epsilon) and 未 (delta). According to this definition, we say that a function f(x, y) approaches a limit L as (x, y) approaches a point (a, b) if for every 蔚 > 0, there exists a 未 > 0 such that whenever the distance from (x, y) to (a, b) is less than , the distance from f(x, y) to L is less than .

The epsilon-delta definition is crucial when proving differentiability, as seen in our exercise. To show a multivariable function is differentiable, we would need to find 蔚鈧 and 蔚鈧 such that, as (螖x, 螖y) becomes infinitesimally small (approaching zero), so do 蔚鈧 and 蔚鈧, satisfying the linear approximation of 螖蹿. If we cannot find such 蔚鈧 and 蔚鈧, as in the exercise where 蔚鈧 does not go to zero, the function fails to meet the criteria for differentiability.

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