Problem 33 - Calculus for Biology and Medicine

Question

In Problems \(31-38\), find \(\partial f / \partial x, \partial f / \partial y\), and \(\partial f / \partial z\) for the given functions. $$ f(x, y, z)=x^{3} y^{2} z+\frac{x}{y z} $$

Step-by-Step Solution

Answer

\( \frac{\partial f}{\partial x} = 3x^2 y^2 z + \frac{1}{y z}\), \( \frac{\partial f}{\partial y} = 2x^3 y z - \frac{x}{y^2 z}\), \( \frac{\partial f}{\partial z} = x^3 y^2 - \frac{x}{y z^2} \).

Step 1: Identify the Function Components

The function given is \( f(x, y, z) = x^3 y^2 z + \frac{x}{y z} \). It consists of two distinct terms: \( x^3 y^2 z \) and \( \frac{x}{y z} \). We will differentiate each term separately with respect to each variable \( x, y, \) and \( z \).

Step 2: Compute Partial Derivative with Respect to x

To find \( \frac{\partial f}{\partial x} \), differentiate each term with respect to \( x \). For \( x^3 y^2 z \), the derivative is \( 3x^2 y^2 z \). For \( \frac{x}{y z} \), use the power rule: \( \frac{\partial}{\partial x} \left( \frac{x}{y z} \right) = \frac{1}{y z} \). Therefore, \( \frac{\partial f}{\partial x} = 3x^2 y^2 z + \frac{1}{y z} \).

Step 3: Compute Partial Derivative with Respect to y

To find \( \frac{\partial f}{\partial y} \), differentiate each term with respect to \( y \). For \( x^3 y^2 z \), use the power rule: \( \frac{\partial}{\partial y} (x^3 y^2 z) = 2x^3 y z \). For \( \frac{x}{y z} \), treat \( x \) and \( z \) as constants: \( \frac{\partial}{\partial y} \left( \frac{x}{y z} \right) = -\frac{x}{y^2 z} \). Thus, \( \frac{\partial f}{\partial y} = 2x^3 y z - \frac{x}{y^2 z} \).

Step 4: Compute Partial Derivative with Respect to z

For \( \frac{\partial f}{\partial z} \), differentiate each term with respect to \( z \). For \( x^3 y^2 z \), the derivative is \( x^3 y^2 \). For \( \frac{x}{y z} \), treat \( x \) and \( y \) as constants using the power rule: \( \frac{\partial}{\partial z} \left( \frac{x}{y z} \right) = -\frac{x}{y z^2} \). So, \( \frac{\partial f}{\partial z} = x^3 y^2 - \frac{x}{y z^2} \).

Key Concepts

Multivariable Calculus

Multivariable calculus is a branch of mathematics dealing with functions that depend on multiple variables. It extends the concepts of calculus to higher dimensions. For example, rather than dealing with a function of a single variable like "time," we may study functions of space and time. In our original exercise, the function \(f(x, y, z) = x^{3} y^{2} z + \frac{x}{y z}\) depends on three variables: \(x\), \(y\), and \(z\). This is a typical scenario in multivariable calculus, where we need to study how changes in one of these variables might affect the function's output.

• These functions allow us to model real-world systems where multiple factors interplay.
• Calculations often involve differentiation, helping us identify rates of change concerning these variables.

Understanding how to work with multiple variables is crucial for many fields, including engineering and physics, where systems rarely depend on a single factor alone.

Partial Differentiation

Partial differentiation allows us to find the rate of change of a multivariable function concerning one variable while holding the other variables constant. This technique is fundamental when the function changes with respect to several variables. In the exercise provided, we performed partial differentiation for each variable \(x\), \(y\), and \(z\).
When we calculate \(\frac{\partial f}{\partial x}\), we treat \(y\) and \(z\) as constants. We find how \(f\) changes as \(x\) changes, ignoring changes in \(y\) and \(z\). Similarly, we repeat this process for \(y\) and \(z\).

• Each derivative indicates the sensitivity of the function to each variable.
• Partial derivatives are crucial in optimizing systems with multiple influencing factors.

Thus, partial differentiation helps identify which variables are most influential for a given function, allowing for better analytical understanding and practical application.

Chain Rule

The chain rule in calculus extends to partial differentiation and is incredibly useful in multivariable calculus. It helps us differentiate composite functions depending on multiple variables. In simpler terms, it tells us how to take the derivative of a function that is made up of other functions.
For instance, if we had a scenario where \(z\) is dependent on \(x\) and \(y\), the chain rule would help us find how \(f\) changes indirectly through these relationships:\[\frac{df}{dz} = \frac{df}{dx} \cdot \frac{dx}{dz} + \frac{df}{dy} \cdot \frac{dy}{dz} + \frac{df}{dz} \cdot \frac{dz}{dz}\]

• It's often used to break down complex problems into simpler parts.
• It is a vital tool that ties together direct and indirect rates of change.

Understanding and applying the chain rule helps solve real-world problems where interdependencies and compositional relationships between variables are present.