Problem 4 - The Calculus with Analytic Geometry

Question

In Exercises 1 through 8 , do each of the following: (a) Find \(D_{11} f(x, y)\); (b) find \(D_{22} f(x, y) ;\) (c) show that \(D_{12} f(x, y)=D_{21} f(x, y) .\) $$ f(x, y)=e^{-x / y}+\ln \frac{y}{x} $$

Step-by-Step Solution

Answer

(a) \( \frac{1}{y^2} e^{-x/y} + \frac{1}{x^2} \). (b) \( -\frac{x^2}{y^3} e^{-x/y} - \frac{1}{y^2} \). (c) \( \frac{x}{y^2} e^{-x/y} = \frac{x}{y^2} e^{-x/y} \).

Step 1 - Find the First Partial Derivatives

First, find the partial derivative of the function with respect to x, \( \frac{\partial f}{\partial x} \), and with respect to y, \( \frac{\partial f}{\partial y} \).

Step 1a - Partial Derivative with respect to x

\( f(x, y) = e^{-x/y} + \ln \frac{y}{x} \). To find \( \frac{\partial f}{\partial x} \), we need to use the chain rule on both terms. \[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} e^{-x/y} + \frac{\partial}{\partial x} \ln \frac{y}{x} \]. The first term is \( -\frac{1}{y} e^{-x/y} \) and the second term is \( -\frac{1}{x} \). So, \[ \frac{\partial f}{\partial x} = -\frac{1}{y} e^{-x/y} - \frac{1}{x} \].

Step 1b - Partial Derivative with respect to y

Using a similar process for \( \frac{\partial f}{\partial y} \): \[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} e^{-x/y} + \frac{\partial}{\partial y} \ln \frac{y}{x} \]. The first term is \( \frac{x}{y^2} e^{-x/y} \) and the second term is \( \frac{1}{y} \). So, \[ \frac{\partial f}{\partial y} = \frac{x}{y^2} e^{-x/y} + \frac{1}{y} \].

Step 2 - Evaluate the Second Partial Derivatives

Now, find the second partial derivatives of the function: \( \frac{\partial^2 f}{\partial x^2} \) and \( \frac{\partial^2 f}{\partial y^2} \).

Step 2a - Second Partial Derivative with respect to x

Differentiate \( \frac{\partial f}{\partial x} \) with respect to x: \[ \frac{\partial}{\partial x} \left( -\frac{1}{y} e^{-x/y} - \frac{1}{x} \right) \]. The first term is \( \frac{1}{y^2} e^{-x/y} \) and the second term is \( \frac{1}{x^2} \). So, \[ \frac{\partial^2 f}{\partial x^2} = \frac{1}{y^2} e^{-x/y} + \frac{1}{x^2} \].

Step 2b - Second Partial Derivative with respect to y

Differentiate \( \frac{\partial f}{\partial y} \) with respect to y: \[ \frac{\partial}{\partial y} \left( \frac{x}{y^2} e^{-x/y} + \frac{1}{y} \right) \]. The first term is \( -\frac{x^2}{y^3} e^{-x/y} \) and the second term is \( -\frac{1}{y^2} \). So, \[ \frac{\partial^2 f}{\partial y^2} = -\frac{x^2}{y^3} e^{-x/y} - \frac{1}{y^2} \].

Step 3 - Show Equality of Mixed Partial Derivatives

Find \( \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right) \) and \( \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial y} \right) \) to show that \( D_{12} f(x, y) = D_{21} f(x, y) \).

Step 3a - Evaluate \( D_{12} f(x, y) \)

\[ \frac{\partial}{\partial y} \left( -\frac{1}{y} e^{-x/y} - \frac{1}{x} \right) \]. The first term is \( \frac{x}{y^2} e^{-x/y} \) and the second term is 0. So, \[ D_{12} f(x, y) = \frac{x}{y^2} e^{-x/y} \].

Step 3b - Evaluate \( D_{21} f(x, y) \)

\[ \frac{\partial}{\partial x} \left( \frac{x}{y^2} e^{-x/y} + \frac{1}{y} \right) \]. The first term is \( \frac{y-x}{y^2} e^{-x/y} \) and the second term is 0. So, \[ D_{21} f(x, y) = \frac{x}{y^2} e^{-x/y} \]. Therefore, \( D_{12} f(x, y) = D_{21} f(x, y) \).

Key Concepts

Multivariable Calculus

Multivariable calculus extends the principles of calculus to functions of two or more variables. This means we deal with functions like \( f(x, y) \), which depends on both \( x \) and \( y \). These types of functions are crucial for describing real-world systems where factors interact in complex ways.

In this context, we perform operations such as differentiation and integration with respect to multiple variables. This will become especially useful for analyzing surfaces and curves in three-dimensional space.

Partial Derivatives

Partial derivatives are a fundamental tool in multivariable calculus. They represent the rate of change of a function with respect to one variable while keeping the others constant.

For a function \( f(x, y) \):

• The partial derivative with respect to \( x \) is denoted as \( \frac{\partial f}{\partial x} \) and shows how \( f \) changes as only \( x \) changes.
• The partial derivative with respect to \( y \) is denoted as \( \frac{\partial f}{\partial y} \) and shows how \( f \) changes as only \( y \) changes.

For example, in the given exercise, partial derivatives are calculated as:
\[ \frac{\partial f}{\partial x} = -\frac{1}{y} e^{-x/y} - \frac{1}{x} \]
\[ \frac{\partial f}{\partial y} = \frac{x}{y^2} e^{-x/y} + \frac{1}{y} \]

Mixed Partial Derivatives

In multivariable calculus, mixed partial derivatives involve taking the partial derivative of a function multiple times, with respect to different variables in succession.
For example, for the function \( f(x, y) \):

• The mixed partial derivative \( \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right) \) is denoted as \( D_{12} f(x, y) \).
• The mixed partial derivative \( \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial y} \right) \) is denoted as \( D_{21} f(x, y) \).

For a well-behaved function, these mixed partial derivatives are equal, which is confirmed by showing:
\[ D_{12} f(x, y) = D_{21} f(x, y) = \frac{x}{y^2} e^{-x/y} \]

Chain Rule

The chain rule is an essential technique for computing the derivative of a composition of functions. In multivariable calculus, it extends to functions of several variables. Consider a function \( f(x, y) \) where both \( x \) and \( y \) themselves are functions of other variables, say \( u \) and \( v \).

The chain rule allows us to differentiate \( f \) with respect to \( u \) or \( v \) by chaining together the derivatives appropriately.

In the given exercise, we apply the chain rule to find the partial derivatives of the function \( f(x, y) = e^{-x/y} + \ln \frac{y}{x} \) with respect to both \( x \) and \( y \).

Functions of Two Variables

Functions of two variables, like \( f(x, y) \), are essential in capturing relationships where outcomes depend on more than one input factor.

Such functions are frequently encountered in fields like physics, economics, and engineering.

The function's graph is a surface in three-dimensional space, enabling us to visually interpret changes with respect to both variables.

In the exercise, we work with the function: \[ f(x, y) = e^{-x/y} + \ln \frac{y}{x} \]
by finding its partial and mixed partial derivatives to understand how \( f \) behaves as \( x \) and \( y \) change.