Problem 14 - Calculus for Biology and Medicine

Question

Show that \(f(x, y)\) is differentiable at the indicated point. $$ f(x, y)=e^{x-y} ;(0,0) $$

Step-by-Step Solution

Answer

The function \( f(x, y) = e^{x-y} \) is differentiable at \((0, 0)\).

Step 1: Verify Continuity at the Point

To be differentiable at a point, the function must first be continuous at that point. The function given is \( f(x, y) = e^{x-y} \). At the point \((0, 0)\), \( f(0, 0) = e^{0-0} = e^0 = 1\). Check if the limit as \((x, y)\) approaches \((0, 0)\) is 1.\[ \lim_{(x, y) \to (0, 0)} e^{x-y} = e^{0-0} = 1 \] Since the limit equals the function value at \((0, 0)\), \(f(x, y)\) is continuous at \((0, 0)\).

Step 2: Find the Partial Derivatives

Compute the partial derivatives of \(f(x, y)\) with respect to \(x\) and \(y\) at the point \((0, 0)\).The partial derivative of \(f(x, y)\) with respect to \(x\) is:\[ f_x(x, y) = \frac{\partial}{\partial x} e^{x-y} = e^{x-y} \]The partial derivative with respect to \(y\) is:\[ f_y(x, y) = \frac{\partial}{\partial y} e^{x-y} = -e^{x-y} \]

Step 3: Evaluate the Partial Derivatives at the Point

Substitute \((x, y) = (0, 0)\) in the partial derivatives:\[ f_x(0, 0) = e^{0-0} = 1 \]\[ f_y(0, 0) = -e^{0-0} = -1 \]

Step 4: Use the Differentiability Definition

Check the definition of differentiability. A function \(f\) is differentiable at a point \((a, b)\) if:\[ f(x, y) - f(a, b) = f_x(a, b)(x-a) + f_y(a, b)(y-b) + \epsilon \sqrt{(x-a)^2 + (y-b)^2} \]where \( \lim_{(x, y) \to (a, b)} \epsilon = 0 \).For \((0,0)\), try expressing:\[ f(x, y) - 1 = 1 \cdot x - 1 \cdot y + \epsilon \sqrt{x^2 + y^2} \]Simplified, this becomes:\[ e^{x-y} - 1 = x - y + \epsilon \sqrt{x^2 + y^2} \]As \((x, y) \to (0, 0), \epsilon \to 0\) which holds true since the error term diminishes faster than any linear term.

Key Concepts

Partial Derivatives

When working with functions of several variables, like our given function \( f(x, y) = e^{x-y} \), understanding partial derivatives is crucial. Partial derivatives allow us to see how the function changes as we vary one variable while keeping the other constant.
To find the partial derivative of \( f(x, y) \) with respect to \( x \), we treat \( y \) as a constant and differentiate with respect to \( x \). Similarly, to find the partial derivative with respect to \( y \), we treat \( x \) as a constant.

• The partial derivative of \( f \) with respect to \( x \), denoted by \( f_x \), is given by \[f_x(x, y) = \frac{\partial}{\partial x} e^{x-y} = e^{x-y}\]
• The partial derivative of \( f \) with respect to \( y \), denoted by \( f_y \), is \[f_y(x, y) = \frac{\partial}{\partial y} e^{x-y} = -e^{x-y}\]

Each partial derivative describes how \( f \) changes as each individual variable changes.

Continuity

A function is continuous at a point if you can draw its graph without lifting your pen. This means that the limit of the function as it approaches the point is equal to the function's value at that point.
For the function \( f(x, y) = e^{x-y} \), we need to ensure that it is continuous at the point \((0, 0)\).

• The value of the function at \( (0, 0) \) is \[f(0, 0) = e^{0-0} = 1\]
• We check the limit as \( (x, y) \) approaches \( (0, 0) \): \[\lim_{(x, y) \to (0, 0)} e^{x-y} = 1\]

Since the limit equals the function value at \((0, 0)\), \( f(x, y) \) is continuous at this point. This is an essential precursor to proving differentiability.

Limit

The concept of a limit in calculus is what allows us to define continuity and differentiability formally. A limit describes the behavior of a function as the input approaches a particular point.
For multidimensional functions, examining limits involves approaching the point from all directions, ensuring the function behaves consistently.
For the function \( f(x, y) = e^{x-y} \), we determined the limit at \( (0, 0) \):

• As \( (x, y) \to (0, 0) \), \[\lim_{(x, y) \to (0, 0)} e^{x-y} = 1\]

This confirms the function is continuous at the point, as the limit and the function’s value coincide. Testing the limit is a crucial step in confirming both continuity and differentiability.

Differentiability Definition

Differentiability extends the concept of derivatives to functions of multiple variables. For a function to be differentiable at a point, it must first be continuous there. Differentiability implies that the function can be approximated by a linear function near that point.
The formal definition states a function \( f \) is differentiable at a point \( (a, b) \) if the following condition is satisfied:\[ f(x, y) - f(a, b) = f_x(a, b)(x-a) + f_y(a, b)(y-b) + \epsilon \sqrt{(x-a)^2 + (y-b)^2}\]where \( \lim_{(x, y) \to (a, b)} \epsilon = 0 \).
For our function at \( (0, 0) \):

• The error term \( \epsilon \) diminishes faster than linear terms \[e^{x-y} - 1 = x - y + \epsilon \sqrt{x^2 + y^2}\]

This shows that \( f(x, y) \) is differentiable at \((0, 0)\), indicating its behavior near the point can be captured by linear approximations.