Problem 34 - APEX Calculus

Question

Find \(f_{x}, f_{y}, f_{z}, f_{y z}\) and \(f_{z y}\). $$ f(x, y, z)=\ln (x y z) $$

Step-by-Step Solution

Answer

The derivatives are: \(f_{x} = \frac{1}{x}\), \(f_{y} = \frac{1}{y}\), \(f_{z} = \frac{1}{z}\), \(f_{yz} = 0\), and \(f_{zy} = 0\).

Step 1: Understand the Problem

We need to find five partial derivatives of the function \( f(x, y, z) = \ln(xyz) \). These include the first-order partial derivatives with respect to each variable \(x\), \(y\), and \(z\), and two second mixed partial derivatives \(f_{yz}\) and \(f_{zy}\).

Step 2: Find the First-Order Partial Derivative with respect to x

The partial derivative with respect to \(x\) is found by differentiating \(f(x, y, z) = \ln(xyz)\) with respect to \(x\), treating \(y\) and \(z\) as constants. This yields: \[ f_{x} = \frac{1}{xyz} \cdot yz = \frac{1}{x} \]

Step 3: Find the First-Order Partial Derivative with respect to y

Differentiate \(f(x, y, z) = \ln(xyz)\) with respect to \(y\), considering \(x\) and \(z\) as constants. We get: \[ f_{y} = \frac{1}{xyz} \cdot xz = \frac{1}{y} \]

Step 4: Find the First-Order Partial Derivative with respect to z

Differentiate \(f(x, y, z) = \ln(xyz)\) with respect to \(z\), treating \(x\) and \(y\) as constants. This gives: \[ f_{z} = \frac{1}{xyz} \cdot xy = \frac{1}{z} \]

Step 5: Find the Second Mixed Partial Derivative f_{yz}

First, use \(f_{y} = \frac{1}{y}\) and differentiate it with respect to \(z\). Since \(f_{y}\) doesn't include \(z\), its derivative is: \[ f_{yz} = 0 \]

Step 6: Find the Second Mixed Partial Derivative f_{zy}

Start with \(f_{z} = \frac{1}{z}\) and differentiate it with respect to \(y\). Since \(f_{z}\) doesn't involve \(y\), its derivative is: \[ f_{zy} = 0 \]

Step 7: Verify Consistency of Mixed Derivatives

Verify that \(f_{yz}\) and \(f_{zy}\) are equal as expected for continuous functions. Both are found to be 0, confirming accuracy.

Key Concepts

First-Order Partial Derivative

In multivariable calculus, when we deal with functions that have more than one variable, finding the partial derivative is an essential skill. Let's start by exploring the first-order partial derivative. A first-order partial derivative gives us the rate of change of a function concerning one variable while keeping the others constant.
For instance, consider the function \( f(x, y, z) = \ln(xyz) \). By differentiating with respect to \( x \), treating \( y \) and \( z \) as constants, we get \( f_x = \frac{1}{xyz} \cdot yz = \frac{1}{x} \). Similarly, differentiating with respect to \( y \) and \( z \) yields \( f_y = \frac{1}{y} \) and \( f_z = \frac{1}{z} \). Each of these derivatives shows the effect on the function value as each variable increases separately.

Second Mixed Partial Derivative

Second mixed partial derivatives are a bit of a step up in complexity. They tell us how the rate of change of the first partial derivative of a function changes with respect to another variable. In simple terms, it's like turning the knob an extra time:

• Begin with one first-order derivative like \( f_y = \frac{1}{y} \), then take its derivative with respect to another variable, such as \( z \).
• In this case, as \( f_y \) does not involve \( z \), its derivative, \( f_{yz} \), is \( 0 \).
• The same logic applies for \( f_{zy} \).

This zero result happens because the variables do not interplay in this specific function beyond the logarithm expression itself, leading to constant derivatives.

Differentiation

Differentiation is a fundamental concept used frequently to analyze and understand how functions behave. It's all about figuring out the instantaneous rate of change or slope of a function at a particular point.
When you're dealing with functions of multiple variables, differentiation becomes layer-based. You start with one variable and treat others as constants, applying similar rules learned for single-variable calculus.
For the function \( f(x, y, z) = \ln(xyz) \), applying differentiation, \(f_x\), \(f_y\), and \(f_z\) reveal how changes in \( x \), \( y \), or \( z \) affect the outcome independently. Hence, it's all about selection and exclusion.

Multivariable Calculus

Multivariable calculus extends the ideas of single-variable calculus to functions with multiple variables. This mathematical field allows us to explore complex systems and functions where multiple inputs affect the outcome.
In problems where you have a function like \( f(x, y, z) = \ln(xyz) \), you're navigating a 3-dimensional landscape. Each variable opens up another dimension:

• Interactions among variables are explored using partial derivatives.
• This leads to nuanced understanding through first-order and mixed derivatives.

Through multivariable calculus, you learn about gradients, as they give you direction in multi-dimensional space, illuminating how systems respond to varying factors.