Problem 14 - Calculus and its applications

Question

Find \(f_{x}\) and \(f_{y}\). $$f(x, y)=y \ln (x+2 y)$$

Step-by-Step Solution

Answer

\(f_x = \frac{y}{x+2y}\); \(f_y = \ln(x+2y) + \frac{2y}{x+2y}.\)

Step 1: Identify Partial Derivative with Respect to x

To find the partial derivative of a function with respect to a variable, we treat any other variables as constants. For the function \(f(x, y) = y \ln(x + 2y)\), we will first find \(f_x\), the partial derivative with respect to \(x\). The derivative of \(\ln(x+2y)\) with respect to \(x\) is \(\frac{1}{x+2y}\). Now multiply by the constant \(y\).

Step 2: Compute f_x

Using the chain rule, we compute:\[ f_x = y \cdot \frac{d}{dx}[ \ln(x + 2y) ] = y \cdot \frac{1}{x + 2y}. \] Therefore, the partial derivative of \(f\) with respect to \(x\) is \(f_x = \frac{y}{x+2y}.\)

Step 3: Identify Partial Derivative with Respect to y

Next, we'll find \(f_y\), the partial derivative with respect to \(y\). We need to consider both the product rule and chain rule, as \(y\) appears both inside the logarithm and as a multiplier.

Step 4: Compute f_y

First, apply the product rule for \(y \ln(x+2y)\):\(f_y = \ln(x+2y) \cdot \frac{d}{dy}[y] + y \cdot \frac{d}{dy}[\ln(x+2y)].\) The derivative \(\frac{d}{dy}[y] = 1\), and \(\frac{d}{dy}[\ln(x + 2y)] = \frac{1}{x+2y} \cdot 2\) by chain rule.Thus, the derivative is:\[ f_y = \ln(x+2y) + \frac{2y}{x+2y}. \]

Key Concepts

Chain Rule

The chain rule is an essential tool in calculus when dealing with composite functions—functions made up of other functions. When you are tasked with finding the derivative of a composite function, the chain rule becomes your best friend. In simple terms, it works by differentiating the outer function and then multiplying it by the derivative of the inner function. This allows us to handle intricate functions that would be tricky to differentiate directly.
For example, in the exercise provided, when calculating the partial derivative of the function \(f(x, y) = y \ln(x + 2y)\) with respect to \(x\), the logarithmic component \(\ln(x + 2y)\) is a composite function. By applying the chain rule, you take the derivative of the outer function, \(\ln(x + 2y)\), which is \(\frac{1}{x+2y}\), and multiply it by the derivative of the inner function \((x+2y)\) with respect to \(x\), which is 1. The chain rule simplifies the process, allowing us to work through potentially daunting derivatives efficiently.

Product Rule

The product rule is another powerful differentiation tool used when you need to distinguish between functions that are multiplied together. The core principle is that the derivative of a product is not simply the product of the derivatives. Instead, it involves a clever twist: you'll differentiate each function separately, then apply a sum.

• Take the derivative of the first function and multiply it by the second one.
• Add the product of the first function and the derivative of the second function.

In our example, \(f(x, y) = y \ln(x+2y)\), finding the partial derivative \(f_y\) involves the product rule. The process starts by treating \(y\) and \(\ln(x+2y)\) as two distinct parts of the function. The derivative of \(y\) is 1, and the derivative of \(\ln(x+2y)\) is \(\frac{1}{x+2y} \cdot 2\) by employing the chain rule. With this, we apply the product rule to get:
\[ f_y = \ln(x + 2y) \cdot 1 + y \cdot \frac{2}{x + 2y} = \ln(x+2y) + \frac{2y}{x+2y}. \] This shows how both the chain and product rules work hand in hand for a complete solution.

Logarithmic Differentiation

Logarithmic differentiation deals primarily with functions involving logarithms. It's particularly useful when differentiating functions that would be otherwise complex or awkward using standard differentiation rules. By taking the logarithm of both sides of an equation, differentiation can sometimes be simplified into more manageable pieces.
In particular, when you have a function in the form of a logarithm like \(y \ln(x+2y)\), you leverage the properties of logarithms (such as \(\ln(ab) = \ln(a) + \ln(b)\)) to separate and simplify the differentiation process. Even though this exercise didn't fully require a direct application of logarithmic differentiation, understanding this concept can be handy. It shows when simplification can transform laborious differentiation into easier steps, especially when dealing with more intricate multiplicative or power functions.
Logarithmic differentiation can be a lifesaver in calculus when faced with complex exponents or products within a function, and its principles underpin many of the methods used in tackling challenging derivatives.