Problem 7 - College Algebra and Calculus: An Applied Approach

Question

Find the first partial derivatives with respect to \(x\) and with respect to \(y\). $$ f(x, y)=\sqrt{x^{2}+y^{2}} $$

Step-by-Step Solution

Answer

The first partial derivative of \(f(x, y)=\sqrt{x^{2}+y^{2}}\) with respect to \(x\) is given by \(\frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^{2}+y^{2}}}\) and with respect to \(y\) is given by \(\frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^{2}+y^{2}}}\).

Step 1: Understand the Given Function

The function given is \(f(x, y) = \sqrt{x^{2} + y^{2}}\). This is a function of two variables \(x\) and \(y\). Our task is to find its partial derivatives with respect to each of these variables.

Step 2: Calculating the Partial Derivative with respect to \(x\)

The partial derivative of a function with respect to variable \(x\), denoted \(\partial f / \partial x\) or \(f_{x}\), involves taking the derivative of the function with respect to \(x\) while treating all other variables as constants. Using the chain rule, we get \(\frac{\partial f}{\partial x} = \frac{1}{2\sqrt{x^{2}+y^{2}}}(2x) = \frac{x}{\sqrt{x^{2}+y^{2}}}\).

Step 3: Calculating the Partial Derivative with respect to \(y\)

Similarly, the partial derivative of a function with respect to variable \(y\), denoted \(\partial f / \partial y\) or \(f_{y}\), involves taking the derivative of the function with respect to \(y\) while treating all other variables as constants. Calculating in the same way as in the previous step, we get \(\frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^{2}+y^{2}}}\).

Key Concepts

Understanding Partial Derivative with Respect to x

In multivariable calculus, the concept of a partial derivative allows us to see how a function changes as one particular variable changes, with all other variables held constant. When we find the partial derivative with respect to \(x\), we express it as \(\frac{\partial f}{\partial x}\). This is a way to focus on the change of function \(f(x, y)\) in the \(x\)-direction.

Consider the function \(f(x, y) = \sqrt{x^2 + y^2}\). To find \(\frac{\partial f}{\partial x}\), we treat \(y\) as a constant and differentiate \(f(x, y)\) with respect to \(x\). We begin by noticing the function inside the square root, \(x^2+y^2\), resembles the format where the chain rule is useful.

• Apply the chain rule: differentiate \(x^2+y^2\) with respect to \(x\), obtaining \(2x\).
• Respect the outer function, which is a square root: \(\frac{1}{2\sqrt{x^2+y^2}}\).

Thus, applying the chain rule, we find:\[\frac{\partial f}{\partial x} = \frac{1}{2\sqrt{x^2+y^2}} \times 2x = \frac{x}{\sqrt{x^2+y^2}}.\]This expression tells us how \(f\) changes when we shift slightly in the \(x\)-direction.

Understanding Partial Derivative with Respect to y

Partial derivatives allow us to explore the sensitivity of a function with respect to one variable at a time. For the function \(f(x, y) = \sqrt{x^2 + y^2}\), the partial derivative with respect to \(y\) is denoted as \(\frac{\partial f}{\partial y}\). This derivative helps us understand changes in \(f\) as \(y\) varies and \(x\) remains constant.

To calculate \(\frac{\partial f}{\partial y}\):

• Treat \(x\) as a constant, and only focus on how \(y\) changes \(f(x, y)\).
• Consider \(x^2+y^2\) under the square root. Differentiate this with respect to \(y\).
• By the chain rule and differentiating \(x^2+y^2\) with respect to \(y\), we get \(2y\).
• The outer derivative remains \(\frac{1}{2\sqrt{x^2+y^2}}\).

Putting these together:\[\frac{\partial f}{\partial y} = \frac{1}{2\sqrt{x^2+y^2}} \times 2y = \frac{y}{\sqrt{x^2+y^2}}.\]This expression informs us how sensitive the function is to changes in \(y\) while keeping \(x\) fixed.

Exploring the Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate compositions of functions. It allows us to take the derivative of complex, layered functions by breaking them down into simpler parts. In particular, for a function of two variables, like \(f(x, y) = \sqrt{x^2 + y^2}\), the chain rule plays a crucial role in finding partial derivatives.

The chain rule essentially says to:

• First, identify the inner function and the outer function. Here, the inner part is \(x^2 + y^2\), while the outer is the square root function.
• Differentiate the outer function with respect to its argument. For \(\sqrt{u}\), we have \(\frac{1}{2\sqrt{u}}\).
• Then, differentiate the inner function with respect to the variable of interest. For example, with respect to \(x\), give \(2x\), and with respect to \(y\), give \(2y\).
• Multiply these derivatives together to get the partial derivative. This step combines the chain of derivative stages into one expression.

The chain rule is powerful because it streamlines the process of differentiation, especially when functions have multiple layers. It is applicable in both single-variable and multivariable calculus, making it a versatile tool for any student tackling complex derivatives.