Use the definition of differentiability to prove that the following function is differentiable at​ (0,0). You must produce functions epsilon 1 and epsilon 2 with the required properties.
​f(x,y)equalsxplusy

We are asked to use the **definition of differentiability** to prove that the function \( f(x,y) = x + y \) is differentiable at \( (0,0) \), and to produce functions \( \epsilon_1 \) and \( \epsilon_2 \) with the required properties. Let’s go step by step.

### Definition of Differentiability at a Point

The function \( f(x,y) \) is differentiable at \( (0,0) \) if we can write:
\[
f(x,y) - f(0,0) = A \cdot x + B \cdot y + \epsilon_1(x,y) \cdot x + \epsilon_2(x,y) \cdot y
\]
where:
- \( A \) and \( B \) are constants,
- \( \epsilon_1(x,y) \to 0 \) as \( (x,y) \to (0,0) \),
- \( \epsilon_2(x,y) \to 0 \) as \( (x,y) \to (0,0) \).

### Step 1: Compute \( f(x,y) - f(0,0) \)
For the given function \( f(x,y) = x + y \), we first calculate:
\[
f(0,0) = 0 + 0 = 0.
\]
Thus:
\[
f(x,y) - f(0,0) = (x + y) - 0 = x + y.
\]

### Step 2: Identify the Linear Part

We need to express \( f(x,y) - f(0,0) \) in the form \( A \cdot x + B \cdot y \), plus error terms. Notice that:
\[
x + y = 1 \cdot x + 1 \cdot y.
\]
So, we can choose \( A = 1 \) and \( B = 1 \), meaning the linear part is \( 1 \cdot x + 1 \cdot y \).

### Step 3: Error Terms

Now, we must write the error terms in the form \( \epsilon_1(x,y) \cdot x \) and \( \epsilon_2(x,y) \cdot y \), where \( \epsilon_1(x,y) \to 0 \) and \( \epsilon_2(x,y) \to 0 \) as \( (x,y) \to (0,0) \).

But observe that:
\[
f(x,y) - f(0,0) = x + y = 1 \cdot x + 1 \cdot y.
\]
There are no additional error terms beyond the linear part. So, we can define:
\[
\epsilon_1(x,y) = 0 \quad \text{and} \quad \epsilon_2(x,y) = 0.
\]

Clearly, both \( \epsilon_1(x,y) \to 0 \) and \( \epsilon_2(x,y) \to 0 \) as \( (x,y) \to (0,0) \), which satisfies the condition for differentiability.

### Conclusion

We have shown that \( f(x,y) = x + y \) is differentiable at \( (0,0) \) with the linear part given by \( 1 \cdot x + 1 \cdot y \), and the error terms \( \epsilon_1(x,y) = 0 \) and \( \epsilon_2(x,y) = 0 \).

### Would you like further details or clarification?

#### Follow-up questions:
1. What is the geometric interpretation of differentiability at a point?
2. How does the definition of differentiability relate to partial derivatives?
3. Can we apply the same method to more complicated functions?
4. How does differentiability differ from continuity?
5. How would the proof change if \( f(x,y) \) included higher-order terms (e.g., \( x^2 + y^2 \))?

#### Tip:
When proving differentiability, always start by identifying the linear part of the function and then check how the error terms behave as the input approaches the point of interest.

Use the definition of differentiability to prove that the following function is differentiable at (0,0). You must produce functions epsilon 1 and epsilon 2 with the required properties. f(x,y) = x + y

Explore how to prove that the function f(x, y) = x + y is differentiable at the point (0,0) using the definition of differentiability. The solution includes step-by-step instructions, involving epsilon functions, and is suitable for college-level calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation