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Chapter 12: Problem 89

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
Answer: Yes, the function \(f(x,y) = xy\) is differentiable at \((0,0)\), because there exist continuous linear functions \(\varepsilon_1\) and \(\varepsilon_2\) such that the definition of differentiability is satisfied.

Step by step solution

01

Compute the difference between f(x, y) and f(0, 0)

First, we need to compute the difference \(f(x, y) - f(0, 0)\). Since our function is \(f(x, y) = xy\), we have: $$f(x, y) - f(0, 0) = xy - 0\cdot 0 = xy$$
02

Find functions 蔚1 and 蔚2

Now we need to express the difference \(f(x, y) - f(0, 0)\) as a sum \(x \varepsilon_1(x, y) + y \varepsilon_2(x, y)\). We can do this by choosing: $$\varepsilon_1(x, y) = y$$ $$\varepsilon_2(x, y) = x$$ With these functions, our difference is expressed as: $$f(x, y) - f(0, 0) = x \varepsilon_1(x, y) + y \varepsilon_2(x, y) = x\cdot y + y\cdot x = 2xy$$
03

Evaluate the limits of 蔚1 and 蔚2 as (x, y) approaches (0, 0)

Finally, we need to evaluate the limits of these functions \(\varepsilon_1\) and \(\varepsilon_2\) as \((x,y) \to (0,0)\). If these limits are 0, then our function is differentiable at (0, 0). We have: $$\lim_{(x,y) \to (0,0)} \varepsilon_1(x, y) = \lim_{(x,y) \to (0,0)} y = 0$$ $$\lim_{(x,y) \to (0,0)} \varepsilon_2(x, y) = \lim_{(x,y) \to (0,0)} x = 0$$ Since both limits are 0, the function \(f(x, y) = xy\) is differentiable at \((0, 0)\).

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Most popular questions from this chapter

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