Problem 14 - Calculus for Biology and Medicine

Question

Find \(\partial f / \partial x\) and \(\partial f / \partial y\) for the given functions. \(f(x, y)=\ln \left(3 x^{2}-x y\right)\)

Step-by-Step Solution

Answer

\(\frac{\partial f}{\partial x} = \frac{6x - y}{3x^2 - xy}\) and \(\frac{\partial f}{\partial y} = \frac{-x}{3x^2 - xy}\).

Step 1: Understand the Function

We are given the function \(f(x, y) = \ln (3x^2 - xy)\) and are tasked with finding the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\). This involves differentiating \(f\) with respect to \(x\) while treating \(y\) as a constant, and vice versa.

Step 2: Calculate the Partial Derivative with Respect to x

First, recall that the derivative of \(\ln u\) with respect to \(x\) is \(\frac{1}{u} \cdot \frac{du}{dx}\). Here, \(u = 3x^2 - xy\). So:\[\frac{\partial f}{\partial x} = \frac{1}{3x^2 - xy} \cdot \left( \frac{d}{dx}(3x^2) - \frac{d}{dx}(xy) \right).\]Calculate \(\frac{d}{dx}(3x^2) = 6x\) and \(\frac{d}{dx}(xy) = y\). Therefore:\[\frac{\partial f}{\partial x} = \frac{1}{3x^2 - xy} \cdot (6x - y).\]

Step 3: Calculate the Partial Derivative with Respect to y

Apply the same principle, taking \(y\) as the variable and \(x\) as a constant:\[\frac{\partial f}{\partial y} = \frac{1}{3x^2 - xy} \cdot \left( \frac{d}{dy}(3x^2) - \frac{d}{dy}(xy) \right).\]Calculate \(\frac{d}{dy}(3x^2) = 0\) because it is constant with respect to \(y\), and \(\frac{d}{dy}(xy) = x\). Thus:\[\frac{\partial f}{\partial y} = \frac{1}{3x^2 - xy} \cdot (-x).\]

Step 4: Simplify the Partial Derivatives

For \(\frac{\partial f}{\partial x}\), we derived\[\frac{\partial f}{\partial x} = \frac{6x - y}{3x^2 - xy}.\]For \(\frac{\partial f}{\partial y}\), we derived\[\frac{\partial f}{\partial y} = \frac{-x}{3x^2 - xy}.\] These are the simplified partial derivatives of the function.

Key Concepts

Multivariable Calculus

Multivariable calculus extends the concepts of calculus into more than one dimension. This is essential when dealing with functions that have multiple variables, like the function in our exercise: \( f(x, y) = \ln(3x^2 - xy) \).

• Partial Derivatives: In multivariable calculus, a partial derivative represents how a function changes as one of the variables changes, while keeping the other variables constant.
• Function of Two Variables: With the function \( f(x, y) \), we have two independent variables. To analyze how \( f \) behaves, we calculate partial derivatives with respect to each variable separately.
• Dimensional Analysis: Multivariable functions provide a richer understanding of phenomena that depend on several factors, helping in fields ranging from physics to economics.

Understanding how to navigate functions of multiple variables is crucial for advancing in calculus and applying it to real-world situations.

Natural Logarithm Functions

The natural logarithm function, denoted as \( \ln \), is the inverse of the exponential function with base \( e \), where \( e \approx 2.718 \). The natural logarithm possesses some unique properties that make it particularly useful in calculus:

• Derivative Rule: The derivative of \( \ln(u) \), where \( u \) is a function of \( x \), is \( \frac{1}{u} \cdot \frac{du}{dx} \). This is used frequently when finding derivatives of logarithmic functions.
• Logarithmic Nature: Logarithms convert multiplicative relationships into additive ones, simplifying complex calculations such as those in compound interest or decay.
• Simplification: When handling a function like \( \ln(3x^2 - xy) \), it's crucial to identify the inner function \((3x^2 - xy)\) and use the chain rule for differentiation.

Natural logarithms are prevalent in calculus due to their natural occurrence in continuous growth and decay processes.

Differentiation Techniques

Differentiation techniques are essential tools in calculus. They help us find the rate of change of functions, especially when working with complex expressions like the one in our exercise. Let's delve into some key techniques:

• Chain Rule: This is crucial when differentiating compositions of functions. When you have a function inside a function, like \( \ln(u) \), you need to apply the chain rule: differentiate the outer function and then multiply by the derivative of the inner function.
• Product Rule: Used when differentiating expressions where two functions are multiplied together, differentiating \( xy \) with respect to \( x \) or \( y \) requires the product rule: \( (uv)' = u'v + uv' \).
• Simplification: Always simplify derivatives at the end. Simplifying can reveal further insights into the function's behavior and help identify errors.

Mastering these differentiation techniques ensures you can tackle a wide variety of problems across mathematics and physics with confidence.