Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

\(\frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2-y^2}}\), \(\frac{\partial f}{\partial y} = -\frac{y}{\sqrt{x^2-y^2}}\)

Step 1: Understand the Function

We are given the function \(f(x, y) = \sqrt{x^2 - y^2}\). Our objective is to find the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\).

Step 2: Find the Partial Derivative with respect to x

To obtain \(\frac{\partial f}{\partial x}\), we treat \(y\) as a constant and differentiate \(f(x, y)\) with respect to \(x\). The derivative of \(\sqrt{u}\) with respect to \(u\) is \(\frac{1}{2\sqrt{u}}\) multiplied by the derivative of \(u\) with respect to \(x\). Here, \(u = x^2 - y^2\).Thus, \(\frac{\partial f}{\partial x} = \frac{1}{2\sqrt{x^2-y^2}} \cdot \frac{d}{dx}(x^2-y^2) = \frac{1}{2\sqrt{x^2-y^2}} \cdot 2x = \frac{x}{\sqrt{x^2-y^2}}.\)

Step 3: Find the Partial Derivative with respect to y

To find \(\frac{\partial f}{\partial y}\), we treat \(x\) as a constant and differentiate \(f(x, y)\) with respect to \(y\).Using the chain rule again, we have:\(\frac{\partial f}{\partial y} = \frac{1}{2\sqrt{x^2-y^2}} \cdot \frac{d}{dy}(x^2-y^2) = \frac{1}{2\sqrt{x^2-y^2}} \cdot (-2y) = -\frac{y}{\sqrt{x^2-y^2}}.\)

Step 4: Summary of First Partial Derivatives

The first partial derivatives of the given function \(f(x, y) = \sqrt{x^2 - y^2}\) are:1. With respect to \(x\): \(\frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2-y^2}}\).2. With respect to \(y\): \(\frac{\partial f}{\partial y} = -\frac{y}{\sqrt{x^2-y^2}}\).

Key Concepts

Chain Rule

In calculus, the chain rule is a fundamental method used for finding the derivative of a composition of functions. It essentially provides a way to "chain" together the derivatives of nested functions. Think of it as a process similar to peeling an onion, where you differentiate each layer step by step.
For multi-variable functions like our example function \[f(x, y) = \sqrt{x^2 - y^2}\]we often apply the chain rule when finding partial derivatives. In this exercise, we differentiate with respect to one variable while treating the other variable as a constant. This application becomes particularly handy when the function involves nested expressions. For instance, in \[u = x^2 - y^2,\]we use the derivative of the outer function \(\sqrt{u}\), which is \(\frac{1}{2\sqrt{u}}\), and multiply it by the derivative of the inner function \(u\) with respect to \(x\) or \(y\).
Thus, understanding the chain rule is crucial for handling derivatives involving multiple layers of operations.

Multi-Variable Functions

Multi-variable functions are functions with more than one input variable. In the context of the partial derivatives exercise, the function \[f(x, y) = \sqrt{x^2 - y^2}\]is an example of a multi-variable function since it depends on both \(x\) and \(y\). Such functions are quite common in fields like physics, engineering, and economics, where multiple variables describe complex systems.
When working with multi-variable functions, we often calculate partial derivatives. These are derivatives with respect to one of the variables while holding others constant. This gives us insight into how the function changes when we vary just one variable. For \[f(x, y),\]the derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) show how \(f\) changes as we change \(x\) or \(y\) individually, providing a richer analysis than a single-variable function would.

Calculus

Calculus is a vast area of mathematics focused on change and motion, dealing with derivatives and integrals. It provides the foundational tools necessary for analyzing the rates at which quantities change, thus being pivotal in understanding functions like \[f(x, y) = \sqrt{x^2 - y^2}.\]
The branch of calculus that involves derivatives is called "differential calculus." Derivatives measure how a function changes as its inputs change. In this exercise, we computed the first partial derivatives of the function to understand how it evolves as its parameters vary.
These partial derivatives are crucial, especially for multi-variable functions, as they help determine behavior in various directions in a multi-dimensional input space. Consequently, calculus allows us to interpret and predict patterns of movement and change, making it indispensable in both theoretical and practical applications.