Problem 139 - Calculus 3

Question

Given \(f(x, y, z)=x y z, \quad\) find \(\quad f_{x y y}, f_{y x y}, \quad\) and \(f_{y y x}\)

Step-by-Step Solution

Answer

\(f_{xyy} = 0, f_{yxy} = 0, f_{yyx} = z\)

Step 1: First Partial Derivative with respect to x

To find the mixed partial derivatives, we first need to calculate each partial derivative step by step. Let's find the partial derivative of \(f(x, y, z) = x y z\) with respect to \(x\). The partial derivative of \(f\) with respect to \(x\) is:\[ f_x = y z \]

Step 2: Partial Derivative of f_x with respect to y

Next, we take the partial derivative of \(f_x = y z\) with respect to \(y\).\[ f_{xy} = z \]

Step 3: Second Partial Derivative with respect to y

Now, we find the second partial derivative of \(f_{xy} = z\) with respect to \(y\).Since \(f_{xy} = z\) is independent of \(y\), its derivative with respect to \(y\) is:\[ f_{xyy} = 0 \]

Step 4: First Partial Derivative with respect to y

Now, let's find the partial derivative of \(f(x, y, z) = x y z\) with respect to \(y\).The partial derivative of \(f\) with respect to \(y\) is:\[ f_y = x z \]

Step 5: Partial Derivative of f_y with respect to x

Next, take the partial derivative of \(f_y = x z\) with respect to \(x\).\[ f_{yx} = z \]

Step 6: Second Partial Derivative with respect to y on f_{yx}

Now, take the partial derivative of \(f_{yx} = z\) with respect to \(y\). Since it is independent of \(y\), we have:\[ f_{yxy} = 0 \]

Step 7: Second Partial Derivative with respect to y

Now, let's find the second partial derivative of \(f(x, y, z)\) with respect to \(y\).The second partial derivative of \(f\) with respect to \(y\) gives us:\[ f_{yy} = x z \]

Step 8: Partial Derivative with respect to x

Finally, we take the partial derivative of \(f_{yy} = x z\) with respect to \(x\).\[ f_{yyx} = z \]

Key Concepts

Partial Derivatives

When working with functions of several variables, such as in multivariable calculus, we often need to determine how the function changes with respect to each variable independently. This is done using partial derivatives. A partial derivative assesses how a function changes as one variable changes while keeping the other variables constant. For example, if you have a function \( f(x, y, z) = xy z \), and you take the partial derivative with respect to \( x \), treating \( y \) and \( z \) as constants, you end up with \( f_x = yz \).
To find mixed partial derivatives, you take the derivative of a partial derivative with respect to another variable. As seen in the original exercise, first \( f_x \) was found, and then \( f_{xy} \) was determined by differentiating \( f_x \) with respect to \( y \). Continuing with \( f_{xy} \) and differentiating with respect to \( y \) again yields \( f_{xyy} = 0 \) because \( z \) does not depend on \( y \).
A good understanding of partial derivatives is essential for analyzing functions in multivariable calculus, as they provide insight into the function's behavior along different directions.

Multivariable Calculus

Multivariable calculus extends the techniques of calculus to functions that depend on more than one variable. In the context of this exercise, we are working with the function \( f(x, y, z) = xyz \), a function of three variables. Each variable can represent a different dimension in space, and the interactions between these dimensions are explored using calculus.

A key aspect of multivariable calculus is understanding how these variables influence the function's output together and separately. Calculating partial derivatives is a fundamental method for examining individual variable impacts. Various derivatives, like the mixed partial derivatives shown in the solution, help illustrate these interactions.

• For example, \( f_{xyy} \) shows how the output changes when \( x \) changes first, followed by two changes in \( y \).
• On the other hand, \( f_{yxy} \) and \( f_{yyx} \) evaluate different sequences of changes, which in some cases, like this exercise, result in the same value.

Multivariable calculus uses these concepts to solve complex problems in physics, engineering, and economics where multiple factors influence outcomes.

Differentiation Rules

To effectively take partial derivatives, it is crucial to understand the differentiation rules that apply to them. In single-variable calculus, you are introduced to rules such as the power rule, product rule, and chain rule. These rules also have versions for multivariable functions.

In the exercise, the product rule was implicitly applied when taking the derivative of \( f(x, y, z) = xyz \) with respect to each variable separately. However, when the function is treated with respect to a single variable, the other variables are seen as constants, simplifying the application of these rules. For example, the derivative of \( xyz \) with respect to \( x \) becomes simply \( yz \), as \( y \) and \( z \) are treated as constant coefficients.

Understanding these rules ensures that when you take a partial derivative, you do so correctly and efficiently. Other rules include the chain rule for composite functions involving multiple variables. Utilizing these rules aids in solving complicated derivatives, as illustrated in finding \( f_{xyy}, f_{yxy}, \) and \( f_{yyx} \) effectively.