Problem 13 - Calculus and its applications

Question

Find \(f_{x}\) and \(f_{y}\). $$f(x, y)=x \ln (x-y)$$

Step-by-Step Solution

Answer

\(f_x = \ln(x-y) + \frac{x}{x-y}\), \(f_y = \frac{-x}{x-y}\).

Step 1: Identify the function components

The function given is \(f(x, y) = x \ln(x-y)\). Here, we have a natural logarithm and a subtraction inside the logarithm. This will be useful in applying derivative rules.

Step 2: Determine \(f_x\) using partial differentiation

To find the partial derivative of \(f\) with respect to \(x\), apply the product rule. The product rule states that if \(u(x, y) = x\) and \(v(x, y) = \ln(x-y)\), then \(f_x = u_x v + u v_x\). Compute: \(u_x = 1\) and \(v_x = \frac{1}{x-y}\). Thus, \(f_x = 1 \cdot \ln(x-y) + x \cdot \frac{1}{x-y} = \ln(x-y) + \frac{x}{x-y}\).

Step 3: Simplify \(f_x\)

Combine the terms in \(f_x\):\[ f_x = \ln(x-y) + \frac{x}{x-y}\] The common denominator is already \(x-y\) for the second term, and the first term is already simplified.

Step 4: Determine \(f_y\) using partial differentiation

Find the partial derivative of \(f\) with respect to \(y\) by differentiating \(x \ln(x-y)\). Use the chain rule. Since \(f(x,y) = x \cdot (\ln(x-y))\), \(\frac{d}{dy}[ \ln(x-y)] = \frac{-1}{x-y}\). Hence, \(f_y = x \cdot \frac{-1}{x-y} = \frac{-x}{x-y}\).

Step 5: Simplify \(f_y\)

\(f_y = \frac{-x}{x-y}\) is already in simplest form, with the denominator pointing to a change by \(y\) relative to \(x\).

Key Concepts

Product Rule

The product rule is a fundamental tool in calculus used when differentiating products of two functions. It states: If you have two functions, say \( u \) and \( v \), that are functions of another variable (like \( x \) or \( y \)), the derivative of their product \( uv \) is given by \( u'v + uv' \). Here, \( u' \) is the derivative of \( u \) with respect to the chosen variable, and \( v' \) is the derivative of \( v \).

In our exercise, we're asked to find the partial derivatives of a function \( f(x, y) = x \ln(x-y) \). For this, partial derivatives are needed because \( f \) is a function of two variables, \( x \) and \( y \).

To find \( f_x \), the partial derivative with respect to \( x \), we need the product rule. We think of \( x \) as \( u \) and \( \ln(x-y) \) as \( v \). Applying the rule, we get:

• \( u_x = 1 \)
• \( v_x = \frac{1}{x-y} \)

This results in \( f_x = 1 \cdot \ln(x-y) + x \cdot \frac{1}{x-y} = \ln(x-y) + \frac{x}{x-y} \).

The product rule helps break down the differentiation process, making it easier to handle complex equations.

Chain Rule

The chain rule is a critical concept in calculus, especially useful when dealing with composite functions. Essentially, it helps us differentiate a function that contains another function inside it. In simple terms, for a composite function \( f(g(x)) \), the derivative is \( f'(g(x)) \cdot g'(x) \).

When dealing with partial derivatives, as in our exercise with \( f(x, y) = x \ln(x-y) \), the chain rule assists in finding \( f_y \), the partial derivative with respect to \( y \). The function \( \ln(x-y) \) is composed since it has an inner function \( x-y \).

When differentiating \( \ln(x-y) \) with respect to \( y \), apply the chain rule:

• Derive \( \ln(x-y) \) with respect to its inner function: \( \frac{d}{d(x-y)} \ln(x-y) = \frac{1}{x-y} \).
• The inner function derivative is \( \frac{d}{dy}(x-y) = -1 \).

Combine these using the chain rule to get: \( \frac{d}{dy} \ln(x-y) = \frac{-1}{x-y} \).

This derivative is then multiplied by \( x \) to find \( f_y \), resulting in \( f_y = x \cdot \frac{-1}{x-y} = \frac{-x}{x-y} \). The chain rule simplifies differentiating complex functions.

Logarithmic Differentiation

Logarithmic differentiation is a special technique that involves using logarithms to simplify the process of differentiating functions. It is notably useful for functions involving products, quotients, or powers where direct differentiation could be cumbersome. Here’s how it works:

The steps generally involve taking the natural logarithm of both sides of an equation, differentiating using the properties of logarithms, and then solving for the derivative itself.

• Properties of logarithms, such as logarithm of a product being the sum of logarithms, help break down the expression.
• Its application often intersects with the product rule and chain rule, making it versatile for complex expressions.

In our initial function \( f(x, y) = x \ln(x-y) \), while we did not directly apply logarithmic differentiation, understanding its principles complements our approach to differentiation.

Using natural logs can sometimes make steps clearer and friendlier, especially if dealing with exponential growth models or where derivative rules need reinforcement. While extracting details about \( f(x, y) \) in the exercise didn't necessitate heavy use of this technique, it's imperative for cases with more involved threads of function composition. Logarithmic differentiation serves as a key toolkit extension for differentiation tasks when tackling more intricate functions or transformations.