Question

Use the definition of differentiability to prove that the following functions are differentiable at $(0,0).$ You must produce functions ${\epsilon }_1$ and ${\epsilon }_2$ with the required properties. $$f(x,y)=x+y$$

Short answer

In the given solution, we proved that the function $f(x,y)=x+y$ is differentiable at $(0,0)$ by finding the partial derivatives with respect to $x$ and $y$, and then showing the existence of limit for the functions ${\epsilon }_1(x,y)$ and ${\epsilon }_2(x,y)$ as $(x,y)\to (0,0)$. So, the function is differentiable at $(0,0)$.

Step-by-step solution

Find the partial derivatives of $f(x,y)$ with respect to $x$ and $y$.

To find the partial derivatives, we will use the following formulas: $$\frac{\partial f}{\partial x}=\underset{h\to 0}{lim}\frac{f(x+h,y)-f(x,y)}{h}$$ $$\frac{\partial f}{\partial y}=\underset{k\to 0}{lim}\frac{f(x,y+k)-f(x,y)}{k}$$ Using the function $f(x,y)=x+y$, let's compute the partial derivatives. $$\frac{\partial f}{\partial x}=\underset{h\to 0}{lim}\frac{(x+h)+y-(x+y)}{h}=\underset{h\to 0}{lim}\frac{h}{h}=1$$ $$\frac{\partial f}{\partial y}=\underset{k\to 0}{lim}\frac{x+(y+k)-(x+y)}{k}=\underset{k\to 0}{lim}\frac{k}{k}=1$$ So, the partial derivatives are: $$\frac{\partial f}{\partial x}=1$$ $$\frac{\partial f}{\partial y}=1$$

Define functions ${\epsilon }_1(x,y)$ and ${\epsilon }_2(x,y)$.

Since we want the functions ${\epsilon }_1(x,y)$ and ${\epsilon }_2(x,y)$ to have the required properties for differentiability, we can define them as the difference between each partial derivative and its corresponding constant. So, we have: $${\epsilon }_1(x,y)=\frac{\partial f}{\partial x}-1$$ $${\epsilon }_2(x,y)=\frac{\partial f}{\partial y}-1$$ By substituting the partial derivatives we found earlier: $${\epsilon }_1(x,y)=1-1=0$$ $${\epsilon }_2(x,y)=1-1=0$$

Prove that the limit of the functions ${\epsilon }_1(x,y)$ and ${\epsilon }_2(x,y)$ exists as $(x,y)\to (0,0)$.

Since both functions ${\epsilon }_1(x,y)$ and ${\epsilon }_2(x,y)$ are equal to 0, it is clear that their limit as $(x,y)\to (0,0)$ exists and is equal to 0. So the limits are: $$\underset{(x,y)\to (0,0)}{lim}{\epsilon }_1(x,y)=0$$ $$\underset{(x,y)\to (0,0)}{lim}{\epsilon }_2(x,y)=0$$ Since both limits exist, we have proved that the function $f(x,y)=x+y$ is differentiable at $(0,0)$ with functions ${\epsilon }_1(x,y)$ and ${\epsilon }_2(x,y)$ having the required properties.

Key concepts

Partial Derivatives

When studying functions in multivariable calculus, partial derivatives are essential for understanding how functions change in different directions. The partial derivative of a function of several variables is its derivative with respect to one of those variables, with all other variables held constant.

For instance, in the function f(x, y) = x + y, the partial derivative with respect to x can be found by treating y as a constant and differentiating with respect to x. The resulting value tells us how the function f changes as x alone changes, which for our simple function is always a constant rate of 1. This concept becomes more interesting and complex with more intricate functions.

Understanding partial derivatives helps students analyze functions in various dimensions and grasps the fundamental notions of rate of change and slope in multivariable functions.

Multivariable Calculus

In multivariable calculus, functions involve more than one variable and can describe more complex phenomena than single-variable functions. The concept of differentiability in multivariable calculus extends the idea of a tangent line to higher dimensions, describing a plane or hyperplane which 'touches' the surface at a point.

The function's differentiability at a point, like (0,0) in our exercise with f(x, y) = x + y, indicates that the function behaves similarly to a plane at that point, which implies smoothness and the possibility to do linear approximations. In practical terms, this concept is widely applied in maximizing profits in business, aerospace engineering, and even in machine learning algorithms.

Limit of a Function

The concept of the limit of a function is foundational in calculus. For single-variable functions, the limit describes the behavior of the function as the variable approaches a certain value. In multivariable calculus, the limit takes on new complexity, dealing with how the function behaves as multiple variables approach a particular point.

In our given function f(x, y), analyzing the limit involves examining the values of f(x, y) as (x, y) gets arbitrarily close to (0,0). The fact that the limit of the functions ${\epsilon }_1(x,y)$ and ${\epsilon }_2(x,y)$ is zero as $x,y$ approaches $0,0$ establishes the differentiability at that point. Thus, understanding limits is crucial for studying continuous behavior and the existence of derivatives in higher dimensions.