Problem 16 - Calculus and its applications

Question

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

Step-by-Step Solution

Answer

\(f_x = \frac{y}{x}\), \(f_y = \ln(xy) + 1\).

Step 1: Determine the Partial Derivative with Respect to x

To find the partial derivative of the function with respect to \(x\), treat \(y\) as a constant. Apply the product rule to differentiate the term \(y \ln(xy)\). This requires differentiating \(\ln(xy)\) with respect to \(x\).The differentiation of \(\ln(xy)\) with respect to \(x\) is: \[\frac{d}{dx}[\ln(xy)] = \frac{1}{xy} \cdot y = \frac{y}{xy} = \frac{1}{x}\]Using the product rule: \[f_{x} = \frac{d}{dx}[y \ln(xy)] = y \cdot \frac{1}{x} = \frac{y}{x}\]

Step 2: Determine the Partial Derivative with Respect to y

Next, find the partial derivative with respect to \(y\), treating \(x\) as a constant. Again apply the product rule to the function \(y \ln(xy)\).First, differentiate \(y\) we simply obtain 1, and the derivative of \(\ln(xy)\) with respect to \(y\) is:\[\frac{d}{dy}[\ln(xy)] = \frac{1}{xy} \cdot x = \frac{x}{xy} = \frac{1}{y}\]Using the product rule:\[f_{y} = \ln(xy) + y \cdot \frac{1}{y} = \ln(xy) + 1\]

Key Concepts

Product Rule

The product rule is a fundamental technique in calculus used to differentiate functions that are products of two or more functions. It can be particularly useful when dealing with multivariable functions, where you need to find partial derivatives. For any two functions, say \( u(x) \) and \( v(x) \), the product rule states: \[ \frac{d}{dx} [u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \] Applying this to partial derivatives involves treating one variable as a constant while differentiating with respect to the other. In our exercise, we used the product rule to find the partial derivatives of the function \( y \ln(xy) \):

• When computing \( f_{x} \), we set \( y \) as a constant, differentiated \( \ln(xy) \) with respect to \( x \), and multiplied it by \( y \).
• Conversely, when computing \( f_{y} \), we treated \( x \) as a constant and differentiated accordingly.

This approach is essential for correctly solving problems involving functions of multiple variables.

Natural Logarithm Differentiation

Differentiating natural logarithms, \( \ln(x) \), is a crucial skill for calculus, especially when working with exponential functions. The derivative of \( \ln(x) \) with respect to \( x \) is:\[ \frac{d}{dx} [\ln(x)] = \frac{1}{x} \]When dealing with multivariable expressions inside the logarithm, like \( \ln(xy) \), the chain rule comes into play. We differentiate \( \ln(xy) \) by:\[ \frac{d}{dx} [\ln(xy)] = \frac{1}{xy} \cdot \text{(derivative of xy with respect to x)} \]This results in \( \frac{y}{x} \) for \( x \) and \( \frac{x}{y} \) for \( y \). The meticulous handling of each variable ensures accuracy in computations and solidifies understanding of how logarithms operate under differentiation.

Multivariable Functions

Multivariable functions involve more than one input variable and often require techniques beyond single variable calculus. Partial derivatives are a key concept in multivariable calculus, allowing us to understand how a function changes with respect to each variable independently.In this exercise, the function \( f(x, y) = y \ln(xy) \) requires us to consider how it varies with each input variable \( x \) and \( y \) using partial derivatives:

• For \( f_{x} \), we examine the change in the function as \( x \) changes, treating \( y \) as constant.
• For \( f_{y} \), we look at how the function changes with \( y \), keeping \( x \) constant.

Multivariable calculus extends the breadth of calculus by exploring these dynamic interactions, providing insights into more complex systems or functions that appear in real-world scenarios.