Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find all first partial derivatives of each function. \(f(x, y)=\frac{x^{2}-y^{2}}{x y}\)

Step-by-Step Solution

Answer

Partial derivatives: \( \frac{\partial f}{\partial x} = \frac{1}{y} + \frac{y}{x^2} \) and \( \frac{\partial f}{\partial y} = -\frac{x}{y^2} - \frac{1}{x} \).

Step 1: Understand Partial Derivatives

Partial derivatives represent the rate of change of a function concerning one variable while keeping the other variables constant. In this problem, we need to find \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) for the function \( f(x, y) = \frac{x^2 - y^2}{xy} \).

Step 2: Simplify the Function

Before taking the derivative, simplify the function if possible. Divide each term in the numerator by the denominator: \[ f(x, y) = \frac{x^2}{xy} - \frac{y^2}{xy} = \frac{x}{y} - \frac{y}{x} \].

Step 3: Find the Partial Derivative with Respect to x

To find \( \frac{\partial f}{\partial x} \), treat \( y \) as a constant. Differentiate \( \frac{x}{y} - \frac{y}{x} \): The derivative of \( \frac{x}{y} \) with respect to \( x \) is \( \frac{1}{y} \), and the derivative of \( -\frac{y}{x} \) with respect to \( x \) is \( \frac{y}{x^2} \).So, \( \frac{\partial f}{\partial x} = \frac{1}{y} + \frac{y}{x^2} \).

Step 4: Find the Partial Derivative with Respect to y

To find \( \frac{\partial f}{\partial y} \), treat \( x \) as a constant. Differentiate \( \frac{x}{y} - \frac{y}{x} \): The derivative of \( \frac{x}{y} \) with respect to \( y \) is \( -\frac{x}{y^2} \), and the derivative of \(-\frac{y}{x} \) with respect to \( y \) is \( -\frac{1}{x} \).So, \( \frac{\partial f}{\partial y} = -\frac{x}{y^2} - \frac{1}{x} \).

Key Concepts

Rate of Change

The concept of rate of change is fundamental in calculus and helps us understand how a function behaves as its inputs change. In a single-variable function, the rate of change is represented by the derivative, which shows how a function changes with a slight change in the input. This can be visualized as the slope of the tangent line to the function's graph.
However, in a function of multiple variables like the one provided in the exercise, we focus on how the function changes with respect to one variable while keeping the others constant.
This is known as a partial derivative, and it helps us explore the behavior of the function in response to changes in just one dimension at a time.

• The partial derivative with respect to \(x\) indicates how the function changes when \(x\) changes but \(y\) remains constant.
• Conversely, the partial derivative with respect to \(y\) signifies the change in the function when \(y\) changes while \(x\) remains the same.

Multivariable Calculus

Multivariable calculus extends calculus concepts to functions of several variables. In this realm, we often work with functions like \(f(x, y)=\frac{x^2-y^2}{xy}\), which depend on more than one input.
Working in this domain requires techniques that account for the joint influence of different variables on a function's output. Concepts such as gradient, multiple integration, and especially partial derivatives are central.

• Understanding multivariable functions allows us to model real-world scenarios where multiple factors vary simultaneously, like temperature and pressure in physics.
• It considers changes in one variable while holding others constant to evaluate their individual contributions to the overall function. This is crucial for optimization problems and understanding surfaces and curves in higher dimensions.

First Partial Derivatives

The first partial derivatives of a function provide the primary insights into how the function changes as each variable changes independently. In our exercise, we have determined both \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \).
Let’s break down the calculation steps:

• Initially, we simplified the function from its original form \(f(x, y) = \frac{x^2 - y^2}{xy}\) to \(f(x, y) = \frac{x}{y} - \frac{y}{x}\), which makes taking the derivatives more straightforward.
• To find \( \frac{\partial f}{\partial x} \), we treated \(y\) as a constant, yielding \( \frac{1}{y} + \frac{y}{x^2} \). Here, each term was derived as we would with constant coefficients.
• Conversely, for \( \frac{\partial f}{\partial y} \), treating \(x\) as a constant gave us \(-\frac{x}{y^2} - \frac{1}{x}\).

These first partial derivatives allow us to predict how small changes in each variable will affect the function, useful in fields like engineering and economics, where understanding these rates of change can lead to better decision-making and problem-solving.