Problem 15 - Calculus for Biology and Medicine

Question

In Problems 1-16, find \(\partial f / \partial x\) and \(\partial f / \partial y\) for the given functions. $$ f(x, y)=\log _{3}\left(y^{2}-x^{2}\right) $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Function

We have the function \( f(x, y) = \log_3(y^2 - x^2) \). Our goal is to find the partial derivatives with respect to \( x \) and \( y \).

Step 2: Convert to Natural Logarithm

To simplify differentiation, convert the base 3 logarithm to a natural logarithm: \( f(x, y) = \frac{\ln(y^2 - x^2)}{\ln(3)} \).

Step 3: Differentiate with Respect to x

Using the chain rule, the partial derivative of \( f \) with respect to \( x \) is \( \frac{\partial f}{\partial x} = \frac{1}{\ln(3)} \cdot \frac{1}{y^2-x^2} \cdot (-2x) = \frac{-2x}{\ln(3) (y^2 - x^2)} \).

Step 4: Differentiate with Respect to y

Again using the chain rule, the partial derivative of \( f \) with respect to \( y \) is \( \frac{\partial f}{\partial y} = \frac{1}{\ln(3)} \cdot \frac{1}{y^2-x^2} \cdot (2y) = \frac{2y}{\ln(3) (y^2 - x^2)} \).

Step 5: Compile Results

The partial derivatives are: \( \frac{\partial f}{\partial x} = \frac{-2x}{\ln(3) (y^2 - x^2)} \) and \( \frac{\partial f}{\partial y} = \frac{2y}{\ln(3) (y^2 - x^2)} \).

Key Concepts

Chain Rule

The chain rule is a crucial concept in calculus that helps us find derivatives of composite functions. In simpler terms, it's used when we have a function inside another function. For example, in our exercise, the function we are dealing with is a composite function:

• The outer function is the logarithm, specifically the natural logarithm after conversion.
• The inner function is the expression inside the log, which is \( y^2 - x^2 \).

When differentiating such a function with respect to \( x \) or \( y \), the chain rule tells us to multiply the derivative of the outer function evaluated at the inner function by the derivative of the inner function itself. In our specific exercise:

• For \( x \), the derivative of the outer function \( \ln(u) \) is \( \frac{1}{u} \), where \( u = y^2 - x^2 \), and the derivative of the inner function is \( -2x \).
• For \( y \), similarly, the derivative of the outer function is \( \frac{1}{u} \), and the derivative of the inner function is \( 2y \).

Thus, the partial derivatives are calculated using these rules, showcasing the power and utility of the chain rule in calculus.

Natural Logarithm

The natural logarithm, represented as \( \ln \), is logarithm to the base \( e \), where \( e \) is approximately equal to 2.71828. It plays a vital role in calculus due to its unique properties, especially because the derivative of \( \ln(x) \) is simply \( 1/x \). When dealing with logarithms of any other base, such as base 3 in our exercise, it is often beneficial to convert them into natural logarithms. This is because it simplifies differentiation and integration.

• To convert a logarithm to a different base, we use the formula: \[\log_b(a) = \frac{\ln(a)}{\ln(b)}\] In the given function, \( \log_3(y^2 - x^2) \) becomes \( \frac{\ln(y^2 - x^2)}{\ln(3)} \).

This transformation allows us to apply standard differentiation rules seamlessly, simplifying our task. Understanding natural logarithms and their properties helps make complex calculus operations more manageable.

Partial Derivative

Partial derivatives are an extension of the derivative concept for functions of multiple variables. Instead of looking at the rate of change of the function when the input changes, partial derivatives focus on how the function changes as one variable changes, keeping the others constant.

• In the function \( f(x, y) = \log_3(y^2 - x^2) \), finding the partial derivative with respect to \( x \) means we view \( y \) as a constant.
• Similarly, when finding the partial derivative with respect to \( y \), we keep \( x \) constant.

In the context of our exercise:

• For partial derivative with respect to \( x \), the rule tells us to focus on \( x \) while treating \( y \) like a constant factor, resulting in \( \frac{-2x}{\ln(3) (y^2 - x^2)} \).
• For partial derivative with respect to \( y \), the process is reversed, leading to \( \frac{2y}{\ln(3) (y^2 - x^2)} \).

Utilizing partial derivatives allows mathematicians and engineers to understand how complex systems behave when some of their components are changed, while others remain unchanged.