Question

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

Answer: Yes, the function \(f(x,y) = x+y\) is differentiable at the point \((0,0)\).

Step-by-step solution

Understanding the Definition of Differentiability

We need to prove that there exist \(\varepsilon_1(x,y)\) and \(\varepsilon_2(x,y)\) with the required properties. According to the definition of differentiability, a function \(f(x,y)\) is differentiable at \((x_0,y_0)\) if these exist: 1. Functions \(\varepsilon_1(x,y)\) and \(\varepsilon_2(x,y)\) such that $$f(x,y) = f(x_0,y_0) + (x-x_0) \frac{\partial f}{\partial x}(x_0,y_0) + (y-y_0) \frac{\partial f}{\partial y}(x_0,y_0) + \varepsilon_1(x,y)(x-x_0) + \varepsilon_2(x,y)(y-y_0)$$ 2. Both \(\varepsilon_1(x,y)\) and \(\varepsilon_2(x,y) \to 0\) as \((x,y) \to (x_0,y_0)\). In our problem, we have \(f(x,y) = x+y\), and we want to prove its differentiability at the point \((0,0)\). Let's find the partial derivatives and the required functions \(\varepsilon_1\) and \(\varepsilon_2\).

Partial Derivatives of f(x,y)

For the given function, we can calculate the partial derivatives with respect to x and y as follows: \(\frac{\partial f}{\partial x}(x,y) = 1\) \(\frac{\partial f}{\partial y}(x,y) = 1\) At the point \((0,0)\), these partial derivatives become: \(\frac{\partial f}{\partial x}(0,0) = 1\) \(\frac{\partial f}{\partial y}(0,0) = 1\) Now, we can plug these values into the definition formula and find the functions \(\varepsilon_1\) and \(\varepsilon_2\).

Finding \(\varepsilon_1\) and \(\varepsilon_2\)

We now have the following equation from the definition of differentiability: $$f(x,y) = f(0,0) + (x-0) \frac{\partial f}{\partial x}(0,0) + (y-0) \frac{\partial f}{\partial y}(0,0) + \varepsilon_1(x,y)(x-0) + \varepsilon_2(x,y)(y-0)$$ Plugging the values of \(f(0,0)\), \(\frac{\partial f}{\partial x}(0,0)\), and \(\frac{\partial f}{\partial y}(0,0)\), we get: $$x+y = 0+1\cdot x + 1 \cdot y + \varepsilon_1(x,y) \cdot x + \varepsilon_2(x,y) \cdot y$$ Now, we can solve for \(\varepsilon_1(x,y)\) and \(\varepsilon_2(x,y)\): $$\varepsilon_1(x,y)=0$$ $$\varepsilon_2(x,y)=0$$

Verifying if \(\varepsilon_1\) and \(\varepsilon_2\) Satisfy the Definition of Differentiability

Since both \(\varepsilon_1(x,y)\) and \(\varepsilon_2(x,y)\) are equal to \(0\), they both go to \(0\) as \((x,y) \to (0,0)\). So, the given function \(f(x,y) = x+y\) is differentiable at the point \((0,0)\) with the functions \(\varepsilon_1(x,y)=0\) and \(\varepsilon_2(x,y)=0\) as required.