Problem 12 - Calculus for Biology and Medicine

Question

Show that \(f(x, y)\) is differentiable at the indicated point. $$ f(x, y)=x y-3 x^{2} ;(1,1) $$

Step-by-Step Solution

Answer

Yes, \(f(x, y)=xy - 3x^2\) is differentiable at \((1, 1)\).

Step 1: Recall the definition of differentiability

A function \(f(x, y)\) is differentiable at a point \((a, b)\) if it can be approximated by a linear function at that point. Mathematically, this means that the limit of \(\frac{f(x, y) - f(a, b) - L(x-a, y-b)}{\sqrt{(x-a)^2 + (y-b)^2}}\) as \((x, y) \to (a, b)\) is zero, where \(L\) is the linear approximation of \(f\) at \((a, b)\).

Step 2: Compute the partial derivatives of f

Calculate the partial derivative of \(f\) with respect to \(x\): \(f_x = y - 6x\). Evaluate this at \((x, y) = (1, 1)\) to get \(f_x(1, 1) = 1 - 6 = -5\). Calculate the partial derivative with respect to \(y\): \(f_y = x\). Evaluate this at \((x, y) = (1, 1)\) to get \(f_y(1, 1) = 1\).

Step 3: Find the linear approximation L(x, y)

Using the partial derivatives, the linear approximation \(L(x, y)\) near \((1, 1)\) is given by: \(L(x, y) = f(1, 1) + f_x(1, 1)(x-1) + f_y(1, 1)(y-1)\). Since \(f(1, 1) = 1 \cdot 1 - 3 \cdot 1^2 = -2\), we get \(L(x, y) = -2 - 5(x-1) + (y-1)\).

Step 4: Verify differentiability at the point

We need to check if the limit \( \lim_{(x, y) \to (1, 1)} \frac{f(x, y) - L(x, y)}{\sqrt{(x-1)^2 + (y-1)^2}} = 0\). Simplify \(f(x, y) - L(x, y)\). It becomes \(xy - 3x^2 + 2 + 5(x-1) - (y-1)\). Expanding and simplifying shows that terms \(xy - 3x^2 = 5x - y\) cancel the linear approximation terms, leaving only small error terms. This ensures that the limit is zero, confirming differentiability.

Key Concepts

Understanding Partial Derivatives

Partial derivatives are a fundamental concept when working with functions of multiple variables. They measure how a function changes as only one of the variables is changed, keeping the other variables constant.

In the function given, \(f(x, y) = xy - 3x^2\), we need to find the partial derivatives with respect to both \(x\) and \(y\).

• The partial derivative with respect to \(x\), denoted as \(f_x\), focuses on how the function changes as \(x\) changes while \(y\) remains constant. In this case, it simplifies to \(f_x = y - 6x\).
• The partial derivative with respect to \(y\), denoted as \(f_y\), deals with change in \(f\) as \(y\) varies, holding \(x\) constant. Here, it reduces to \(f_y = x\).

Evaluating these partial derivatives at the point \((1, 1)\) gives us values \(f_x(1,1) = -5\) and \(f_y(1,1) = 1\). These derivatives provide us with crucial information needed to create the linear approximation of the function at that point.

Linear Approximation and Its Role

A linear approximation is a method of approximating a function with a linear function (or plane, in this case) near a given point. By doing so, complex functions can often be understood more intuitively around specific values, which helps in checking differentiability.

Let's consider the function \(f(x, y)\) near the point \((1, 1)\). We use the formula:

• \(L(x, y) = f(a, b) + f_x(a, b)(x-a) + f_y(a, b)(y-b)\)
• In this exercise, substituting \(f(1,1) = -2\), \(f_x(1,1) = -5\), and \(f_y(1,1) = 1\), the linear approximation becomes \(L(x, y) = -2 - 5(x-1) + (y-1)\).

The idea is that around \((1, 1)\), the behavior of the function \(f(x, y)\) can approximated closely by \(L(x, y)\). This linear approximation allows us to check whether the function is differentiable at point by confirming if the error term between the actual function and this linear approximation diminishes as we approach the point.

Understanding Limits in Differentiability

Limits play a crucial role in verifying the differentiability of a function at a point. When we talk about a function being differentiable at a particular point, we are essentially examining the limit of the function's behavior as its input approaches that point.

To check if \(f(x, y)\) is differentiable at \((1, 1)\), we evaluate:

• The limit expression: \( \lim_{(x, y) \to (1, 1)} \frac{f(x, y) - L(x, y)}{\sqrt{(x-1)^2 + (y-1)^2}} = 0\)
• When \(f(x, y) - L(x, y)\) is simplified, the main terms of the function cancel the linear approximation terms. This leaves only smaller terms that shrink with the distance to \((1, 1)\).

When this limit equals zero, it shows that the function can be well-approximated by a linear function near \((1, 1)\), confirming its differentiability at that point. Limits, therefore, act as the final tool in asserting differentiability by analyzing the behavior of the function in an infinitely small neighborhood around the point.