Problem 13 - Calculus for Biology and Medicine

Question

In Problems 1-16, find \(\partial f / \partial x\) and \(\partial f / \partial y\) for the given functions. $$ f(x, y)=\ln (2 x+y) $$

Step-by-Step Solution

Answer

\( \partial f / \partial x = \frac{2}{2x + y} \) and \( \partial f / \partial y = \frac{1}{2x + y} \).

Step 1: Identify the function

We are given the function \( f(x, y) = \ln(2x + y) \). We need to find the partial derivatives \( \partial f / \partial x \) and \( \partial f / \partial y \).

Step 2: Calculate the partial derivative with respect to x

To find \( \partial f / \partial x \), treat \( y \) as a constant and differentiate \( f \) with respect to \( x \). The derivative of \( \ln(u) \) is \( 1/u \times \frac{du}{dx} \), where \( u = 2x + y \). Thus, \( \frac{d}{dx}(2x + y) = 2 \). Therefore, \[ \frac{\partial f}{\partial x} = \frac{1}{2x + y} \times 2 = \frac{2}{2x + y}. \]

Step 3: Calculate the partial derivative with respect to y

To find \( \partial f / \partial y \), treat \( x \) as a constant and differentiate \( f \) with respect to \( y \). The derivative of \( \ln(u) \) is \( 1/u \times \frac{du}{dy} \), where \( u = 2x + y \). Thus, \( \frac{d}{dy}(2x + y) = 1 \). Therefore, \[ \frac{\partial f}{\partial y} = \frac{1}{2x + y} \times 1 = \frac{1}{2x + y}. \]

Key Concepts

Understanding Multivariable Calculus

Multivariable calculus is an extension of calculus to functions of more than one variable, like the function you see in this exercise: \( f(x, y) = \ln(2x + y) \). In problems with multiple variables, calculus can be used to understand how changing one variable affects the function while keeping the others constant. Regular functions have derivatives to measure their rate of change, and in multivariable calculus, partial derivatives serve a similar purpose. They represent how the function changes as one specific variable changes, while the others are held constant. This is a crucial tool in optimization problems and helps to analyze mathematical models that involve multiple inputs.

• Function of Two Variables: In our exercise, the function is dependent on two variables: \(x\) and \(y\).
• Goals of Multivariable Calculus: To understand how each individual variable impacts the function on its own.

When working with multivariable calculus, the key is to approach each variable one at a time, treating other variables as constants during differentiation.

The Process of Differentiation

Differentiation is the mathematical process of finding the derivative of a function, which indicates the rate of change. For multivariable functions, partial derivatives allow us to differentiate with respect to one variable at a time. In this exercise, we find the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \). Differentiating involves applying rules for the differentiation of standard functions, like the logarithm:

• Product and Chain Rules: When the function is a combination of other functions, these rules help simplify differentiation. Here, the chain rule is useful for dealing with \( \ln(u) \).
• Treating Variables as Constants: When taking the partial derivative with respect to one variable, consider other variables as constant terms.

For \( \frac{\partial f}{\partial x} \), we focus on how \( f(x, y) \) changes with \( x \). The step-by-step process is: treat \( y \) as a constant and differentiate: first find \( \frac{d}{dx}(2x + y) = 2 \), and use the chain rule.Similarly, for \( \frac{\partial f}{\partial y} \), let \( x \) be constant. The differentiation of \( \ln(2x + y) \) results in \( \frac{1}{2x + y} \times 1 \) because the derivative of \( (2x + y) \) with respect to \( y \) is 1.

The Role of the Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a logarithm to the base \( e \), where \( e \approx 2.71828 \). It's a very common function in calculus due to its unique properties that simplify complex mathematical equations. In derivatives, the natural log function is advantageous because of its straightforward derivative rule.

• Derivative Rule: The derivative of \( \ln(u) \) with respect to \( u \) is \( 1/u \).
• Chain Rule with \( \ln \): When differentiating \( \ln \) of a more complex expression, use the chain rule: \( \frac{1}{u} \times \frac{du}{dx} \).

In our exercise, \( f(x, y) = \ln(2x + y) \), we utilize these rules for both \( \partial f / \partial x \) and \( \partial f / \partial y \). Simplification leads to: For \( \partial f / \partial x \), the derivative is \( \frac{2}{2x + y} \), and for \( \partial f / \partial y \), it's \( \frac{1}{2x + y} \). This reflects how a change in each variable affects the entire logarithmic expression, considering the fundamental properties of natural log.