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Chapter 12: Problem 88

Use the definition of differentiability to prove that the following functions are differentiable at \((0,0) .\) You must produce functions \(\varepsilon_{1}\) and \(\varepsilon_{2}\) with the required properties. $$f(x, y)=x+y$$

Short Answer

Expert verified
Question: Prove that the function \(f(x, y) = x + y\) is differentiable at \((0, 0)\). Answer: To prove the differentiability of \(f(x, y) = x + y\) at point \((0, 0)\), we calculated the partial derivatives \(f_x = 1\) and \(f_y = 1\). We defined the linear functions \(\varepsilon_{1}(x) = x\) and \(\varepsilon_{2}(y) = y\). Next, we computed the limit \(\lim_{(x, y) \to (0, 0)} \frac{f(x, y) - f(0, 0) - \varepsilon_{1}(x) - \varepsilon_{2}(y)}{\sqrt{x^2 + y^2}}\), which resulted in 0. Since this limit is 0, the function \(f(x, y) = x + y\) is differentiable at the point \((0, 0)\).

Step by step solution

01

Determine the partial derivatives of f(x, y)#

To find the linear functions we need to approximate the function \(f(x, y)\), we will first determine the partial derivatives of the function with respect to both x and y. So let's compute \(f_x\) and \(f_y\). $$ f_x = \frac{\partial f(x, y)}{\partial x} = \frac{\partial (x+y)}{\partial x} = 1 $$ $$ f_y = \frac{\partial f(x, y)}{\partial y} =\frac{\partial (x+y)}{\partial y} = 1 $$
02

Define the linear functions \(\varepsilon_{1}\) and \(\varepsilon_{2}\) #

We can now define the linear functions \(\varepsilon_{1}\) and \(\varepsilon_{2}\) using the partial derivatives obtained in the previous step. Since \(f_x = 1\) and \(f_y = 1\), we have: $$ \varepsilon_{1}(x) = x $$ $$ \varepsilon_{2}(y) = y $$
03

Compute the limit#

We will now compute the limit of the following expression as \((x, y) \to (0, 0)\): $$ \lim_{(x,y) \to (0,0)} \frac{f(x, y) - f(0, 0) - \varepsilon_{1}(x) - \varepsilon_{2}(y)}{\sqrt{x^2 + y^2}} $$ Substitute the function \(f(x, y) = x + y\), \(f(0, 0) = 0\), \(\varepsilon_{1}(x) = x\), and \(\varepsilon_{2}(y) = y\) into the expression: $$ \lim_{(x, y) \to (0, 0)} \frac{(x + y) - 0 - x - y}{\sqrt{x^2 + y^2}} $$ Simplifying the expression: $$ \lim_{(x,y) \to (0,0)} \frac{0}{\sqrt{x^2 + y^2}} $$ Since the numerator is 0, the limit is 0: $$ \lim_{(x,y) \to (0,0)} \frac{0}{\sqrt{x^2 + y^2}} = 0 $$
04

Conclusion#

The limit is 0, which means that the function \(f(x, y) = x + y\) is differentiable at the point \((0, 0)\), and the linear functions \(\varepsilon_{1}(x) = x\) and \(\varepsilon_{2}(y) = y\) provide the required properties of differentiability.

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Most popular questions from this chapter

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