Problem 13 - Calculus for Biology and Medicine

Question

Show that \(f(x, y)\) is differentiable at the indicated point. \(f(x, y)=\cos (x+y) ;(0,0)\)

Step-by-Step Solution

Answer

The function is differentiable at (0, 0).

Step 1: Review the Definition of Differentiability

A function is differentiable at a point if it is continuous at that point and has a total derivative at the point. The function is differentiable at \( (x_0, y_0) \) if there exist linear functions \( L_x(h, k) \) and \( L_y(h, k) \) such that the following limit exists: \[\lim_{{(h, k) \to (0, 0)}} \frac{f(x_0 + h, y_0 + k) - f(x_0, y_0) - L_x(h, k) - L_y(h, k)}{\sqrt{h^2 + k^2}} = 0.\]

Step 2: Compute the Partial Derivatives

First, compute the partial derivatives of the function \( f(x, y) = \cos(x + y) \) at the point \( (0,0) \). The partial derivative with respect to \( x \) is: \[f_x(x, y) = -\sin(x + y).\]At \( (0,0) \), \( f_x(0, 0) = -\sin(0 + 0) = 0 \).The partial derivative with respect to \( y \) is: \[f_y(x, y) = -\sin(x + y).\]At \( (0,0) \), \( f_y(0, 0) = -\sin(0 + 0) = 0 \).

Step 3: Construct the Linear Function

Since both partial derivatives at \( (0,0) \) are zero, \( L(h, k) = 0 \). Therefore, \( L_x(h, k) = 0 \) and \( L_y(h, k) = 0 \).

Step 4: Substitute into Differentiability Limit

Substitute the results from Steps 2 and 3 into the differentiability condition: \[\lim_{{(h, k) \to (0, 0)}} \frac{f(h, k) - f(0, 0)}{\sqrt{h^2 + k^2}} = \lim_{{(h, k) \to (0, 0)}} \frac{\cos(h + k) - 1}{\sqrt{h^2 + k^2}}.\]

Step 5: Apply Limit and Squeeze Theorem

As \( (h, k) \to (0, 0) \), \( h + k o 0 \). Therefore, \( \cos(h + k) \to \cos(0) = 1 \) and \( \cos(h+k) - 1 \to 0 \). The numerator approaches zero faster than the denominator, and we apply the squeeze theorem to conclude:\[\lim_{{(h, k) \to (0, 0)}} \frac{\cos(h + k) - 1}{\sqrt{h^2 + k^2}} = 0.\] Thus, \( f(x, y) = \cos(x+y) \) is differentiable at \( (0, 0) \).

Key Concepts

Partial Derivatives

Partial derivatives are a way to explore how a multivariable function changes as each individual variable changes, while keeping others constant. Imagine you're smoothing butter on a slice of bread. As you move horizontally or vertically, you're doing similar to what partial derivatives do in calculus. For the function \(f(x, y) = \cos(x + y)\), you calculate the rate at which the function changes as you only vary \(x\) while keeping \(y\) steady, and vice versa.

• For the partial derivative with respect to \(x\): Take the derivative of the inside function \(x+y\) treating \(y\) as a constant. This gives you \(f_x(x, y) = -\sin(x+y)\).
• For \(y\): Repeat the process, this time treating \(x\) as a constant, resulting in \(f_y(x, y) = -\sin(x+y)\).

At the point \((0,0)\), these derivatives both evaluate to zero because \(-\sin(0+0) = 0\). This means small changes in \(x\) or \(y\) at \((0,0)\) do not influence \(f\). Partial derivatives help determine how to "linearize" a function at a point, leading to finding if it is differentiable.

Continuity

Continuity is the idea that a small change in the input of a function results in a small change in the output. Imagine a smooth road; if you move a step forward, your position changes smoothly without any jumps or gaps. For a function of two variables like \(f(x, y) = \cos(x + y)\), continuity implies that as you inch towards \((0, 0)\) along any direction in the \((x, y)\) plane, the function \(f(x, y)\) approaches \(f(0, 0)\) smoothly. For \(\cos(x+y)\), as \((x, y) \rightarrow (0, 0)\), \(\cos(x+y)\) approaches \(\cos(0) = 1\), showing that \(f(x,y)\) behaves nicely and predictively as you near the origin. A function must be continuous at the point where you wish to prove differentiability because differentiability is a stronger condition; it implies, but is not implied by, continuity. So, verifying continuity is a critical part of checking differentiability.

Squeeze Theorem

The Squeeze Theorem is a technique to find the limit of a function when the limit is tough to calculate directly. The idea is to "squeeze" your function between two other functions whose limits are easier to compute and are the same. Think of a sandwich where the unknown is squeezed between two known slices, and they all shrink down to the same crumb. In the exercise, the limit expression \(\lim_{{(h, k) \to (0, 0)}} \frac{\cos(h + k) - 1}{\sqrt{h^2 + k^2}}\) is tricky because both the numerator and denominator go to zero. However, since we know that \(\cos(h+k) - 1\) approaches zero faster than \(\sqrt{h^2 + k^2}\) when \((h+k)\) is very small, we can use inequalities bounding \(\cos(h+k) - 1\) to fit the Squeeze Theorem mechanics.

• If the numerator approaches zero more quickly, it helps to conclude the overall limit tends to zero.
• By comparing it to simpler bounding functions both becoming zero, we see the limit in the middle must also become zero.

This confirms the differentiability of \(f(x, y)\) at \((0, 0)\), owing much to the clever insight from the Squeeze Theorem.