Problem 32 - Calculus for Biology and Medicine

Question

Find \(\partial f / 2 x, \partial f / \partial y\), and \(\partial f / \partial z\) for the given functions. \(f(x, y, z)=x y(z+x)\)

Step-by-Step Solution

Answer

\(\frac{\partial f}{\partial x} = yz + 2xy\), \(\frac{\partial f}{\partial y} = xz + x^2\), \(\frac{\partial f}{\partial z} = xy\).

Step 1: Identify the Function and the Variables

The function given is \(f(x, y, z) = x y (z + x)\). The variables in the function are \(x\), \(y\), and \(z\). We need to find the partial derivatives with respect to \(x\), \(y\), and \(z\).

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

To find \(\frac{\partial f}{\partial x}\), treat \(y\) and \(z\) as constants and differentiate the function with respect to \(x\).Write the function in expanded form: \(f(x, y, z) = x y z + x^2 y\).Differentiate: \[\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (xyz + x^2 y) = yz + 2xy \].

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

To find \(\frac{\partial f}{\partial y}\), treat \(x\) and \(z\) as constants and differentiate the function with respect to \(y\).The function is \(f(x, y, z) = x y z + x^2 y\).Differentiate: \[\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (xyz + x^2 y) = xz + x^2 \].

Step 4: Partial Derivative with respect to \(z\)

To find \(\frac{\partial f}{\partial z}\), treat \(x\) and \(y\) as constants and differentiate the function with respect to \(z\).Note that \(f(x, y, z) = x y z + x^2 y\) simplifies in this derivative as only the term involving \(z\) contributes.Differentiate: \[\frac{\partial f}{\partial z} = \frac{\partial}{\partial z} (xyz) = xy \].

Key Concepts

Multivariable Calculus

Multivariable calculus extends the principles of calculus to functions of multiple variables. In the exercise, we deal with a function of three variables: \(x\), \(y\), and \(z\). Unlike single-variable calculus, where you work with functions that depend on only one variable, multivariable calculus helps you to study how a function changes when several variables change all at once.
When facing a multivariable function like \(f(x, y, z) = x y(z + x)\), you must consider how each variable contributes to the overall change in the function.

• This means you treat all variables except the one you're differentiating with respect to as constants.
• This process gives you what is known as a partial derivative.

This concept is essential for understanding behaviors in various fields, ranging from physics to economics, where several factors simultaneously impact the outcome.

Differentiation

Differentiation allows us to measure how a function changes with respect to changes in its variables. In our exercise, we differentiate a multivariable function and find its partial derivatives: \(\frac{\partial f}{\partial x}\), \(\frac{\partial f}{\partial y}\), and \(\frac{\partial f}{\partial z}\).

• Finding \(\frac{\partial f}{\partial x}\) involves considering \(y\) and \(z\) as constants, resulting in \(yz + 2xy\).
• For \(\frac{\partial f}{\partial y}\), treating \(x\) and \(z\) as constants gives us \(xz + x^2\).
• Lastly, to find \(\frac{\partial f}{\partial z}\), \(x\) and \(y\) are constants, which simplifies to \(xy\).

These computations show how the function changes as each variable shifts slightly, which is crucial for predicting outcomes and modeling real-world behavior.

Calculus Problem Solving

Solving calculus problems, particularly those involving partial derivatives, requires a methodical approach. The steps in our provided solution outline a clear path:

• Identify: Recognize the function and its variables, ensuring you understand the relationship between them.
• Differentiation: Systematically find each partial derivative by treating non-targeted variables as constants.
• Simplify: Break down complex expressions into simpler terms before differentiating.

By following these steps, you develop an organized strategy that enhances your problem-solving skills. Understanding how each variable independently affects the function allows you to tackle more complex calculus problems with confidence, providing insights into mathematical applications and theoretical exploration.