Problem 21 - The Calculus with Analytic Geometry

Question

In Exercises 13 through 24 , find the indicated partial derivatives by holding all but one of the variables constant and applying theorems for ordinary differentiation. $$ f(x, y, z)=4 x y z+\ln (2 x y z) ; f_{3}(x, y, z) $$

Step-by-Step Solution

Answer

\( \frac{\partial f}{\partial z} = 4xy + \frac{1}{z} \)

Step 1: Identify the Function and Desired Partial Derivative

Given the function: \[ f(x, y, z) = 4xyz + \ln(2xyz) \]The exercise asks to find the partial derivative with respect to the third variable, which is typically noted as \(f_3(x, y, z)\) or \(\frac{\partial f}{\partial z}\).

Step 2: Hold Other Variables Constant

When finding \( \frac{\partial f}{\partial z} \), treat \(x\) and \(y\) as constants.

Step 3: Differentiate the First Term

For the term \( 4xyz \), differentiating with respect to \(z\) yields:\[ \frac{\partial}{\partial z}(4xyz) = 4xy \]

Step 4: Differentiate the Second Term

For the term \( \ln(2xyz) \), use the chain rule:Let \( u = 2xyz \), then \( \frac{\partial}{\partial z}(\ln(u)) = \frac{1}{u} \frac{\partial u}{\partial z} \). Since \( u = 2xyz \), we find \( \frac{\partial u}{\partial z} = 2xy \). Therefore:\[ \frac{\partial}{\partial z}(\ln(2xyz)) = \frac{1}{2xyz} \cdot 2xy = \frac{2xy}{2xyz} = \frac{1}{z} \]

Step 5: Combine Results

Add the derivatives from the two terms to get the final partial derivative:\[ \frac{\partial f}{\partial z} = 4xy + \frac{1}{z} \]

Key Concepts

multivariable calculus

Multivariable calculus involves functions with more than one variable. Unlike single-variable calculus, where we only deal with one independent variable, here we examine how changes in several variables affect a function. For instance, if we have a function \(f(x, y, z) = 4xyz + \ln(2xyz)\), changes in any of \(x, y,\) or \(z\) can influence the outcome of \(f\). This field is essential for understanding real-world problems involving multiple changing conditions, such as physical systems, economics, and engineering.

chain rule

The chain rule is a fundamental principle in calculus used for differentiating compositions of functions. In multivariable calculus, it has a special application. When differentiating a function like \(\ln(2xyz)\) with respect to one variable, say \(z\), we treat other variables \(x\) and \(y\) as constants.

Here’s how it works: Let \(u = 2xyz\). The chain rule tells us that:

• First, differentiate \(\ln(u)\) with respect to \(u\): \( \frac{d}{du}[\ln(u)] = \frac{1}{u} \).
• Then, differentiate \(u = 2xyz\) with respect to \(z\): \( \frac{\partial u}{\partial z} = 2xy \).

Combining these, the partial derivative \( \frac{\partial}{\partial z}[\ln(2xyz)] \) becomes \( \frac{1}{2xyz} \cdot 2xy = \frac{1}{z} \).
This method simplifies complex differentiations significantly.

ordinary differentiation

Ordinary differentiation focuses on finding the derivative of functions with a single variable. In the context of our exercise, we apply the same principles but adapt them for partial derivatives.

Consider the term \(4xyz\) in our function \(f(x, y, z)\). To find the partial derivative with respect to \(z\), we treat \(x\) and \(y\) as constants and differentiate as usual:

• For \(4xyz\), \( \frac{\partial}{\partial z}[4xyz] = 4xy \).

This straightforward approach allows us to break down complex multivariable functions into simpler pieces.
By mastering ordinary differentiation, students build a foundation for understanding more advanced topics like partial derivatives and the chain rule in multivariable calculus.