Problem 85 - Calculus Early Transcendentals

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. $$f(x, y)=x y$$

Step-by-Step Solution

Answer

Answer: Yes, the function \(f(x,y)=xy\) is differentiable at the point \((0,0)\) because its partial derivatives exist at this point, and the limiting functions \(\varepsilon_1(x,y)\) and \(\varepsilon_2(x,y)\) both go to \(0\) as \((x,y)\to(0,0)\), which satisfies the definition of differentiability for multivariable functions.

Step 1: Recall the definition of differentiability for multivariable functions

A function \(f(x,y)\) is said to be differentiable at a point \((a,b)\) if its partial derivatives exist at that point and there exist functions \(\varepsilon_1(x,y)\) and \(\varepsilon_2(x,y)\) such that \begin{align*} f(x,y)-f(a,b) &= (\varepsilon_1(x,y))(x-a) + (\varepsilon_2(x,y))(y-b), \end{align*} where \begin{align*} \lim_{(x,y)\to(a,b)}\varepsilon_1(x,y) = 0 \quad\text{and}\quad \lim_{(x,y)\to(a,b)}\varepsilon_2(x,y) = 0. \end{align*} In this exercise, we need to verify if the given function \(f(x,y)=xy\) is differentiable at \((0,0)\) using this definition.

Step 2: Compute the partial derivatives of \(f(x,y)=xy\) at \((0,0)\)

First, we need to compute the partial derivatives of \(f(x,y)\) with respect to \(x\) and \(y\) at the point \((0,0)\). \begin{align*} \frac{\partial f}{\partial x} &= y \\ \frac{\partial f}{\partial y} &= x \end{align*} At \((0,0)\), both partial derivatives exist, and they are equal to \(0\).

Step 3: Define the functions \(\varepsilon_1(x,y)\) and \(\varepsilon_2(x,y)\)

By substituting \(f(x,y)=xy\) into the differentiability equation and plugging in \((a,b)=(0,0)\), we have: \begin{align*} f(x,y)-f(0,0) &= (\varepsilon_1(x,y))(x-0) + (\varepsilon_2(x,y))(y-0) \\ xy - 0 &= (\varepsilon_1(x,y))x + (\varepsilon_2(x,y))y \end{align*} From this equation, we can define the functions \(\varepsilon_1(x,y)\) and \(\varepsilon_2(x,y)\) as \begin{align*} \varepsilon_1(x,y) &= \frac{1}{x}\left(xy-0\right) = y, \\ \varepsilon_2(x,y) &= \frac{1}{y}\left(xy-0\right) = x. \end{align*}

Step 4: Check if \(\lim_{(x,y)\to(0,0)}\varepsilon_1(x,y)=0\) and \(\lim_{(x,y)\to(0,0)}\varepsilon_2(x,y)=0\)

Now, we need to show that the limits of the functions \(\varepsilon_1(x,y)\) and \(\varepsilon_2(x,y)\) go to \(0\) as \((x,y)\to(0,0)\): \begin{align*} \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. \end{align*} Since both limits go to \(0\), the function \(f(x,y)=xy\) is differentiable at the point \((0,0)\) according to the definition of differentiability for multivariable functions.