Problem 15 - Calculus and its applications

Question

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

Step-by-Step Solution

Answer

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

Step 1: Differentiate with Respect to x

To find the partial derivative of the function with respect to \(x\), treat \(y\) as a constant. The function is \(f(x, y) = x \ln(xy)\). Apply the product rule: \((u\cdot v)' = u'v + uv'\). Here, \(u = x\) and \(v = \ln(xy)\). Then, \(u' = 1\) and \(v' = \frac{1}{xy} \cdot y = \frac{y}{xy} = \frac{1}{x}\). Thus, \(f_x = 1 \cdot \ln(xy) + x \cdot \frac{1}{x} = \ln(xy) + 1\).

Step 2: Differentiate with Respect to y

Now find the partial derivative with respect to \(y\), treating \(x\) as a constant. Again use the product rule where \(u = x\) and \(v = \ln(xy)\). Thus \(u' = 0\) and \(v' = \frac{1}{xy} \cdot x = \frac{x}{xy} = \frac{1}{y}\). This gives \(f_y = x \cdot \frac{1}{y} = \frac{x}{y}\).

Key Concepts

Product Rule

The product rule is an essential tool when differentiating functions that are the product of two or more functions. It helps you understand how to take the derivative of such product functions accurately.

When you have two functions multiplied together, say \(u\) and \(v\), the product rule states that the derivative of their product \((u \, v)'\) is given by:

• \((u \, v)' = u' \, v + u \, v'\)

This means you take the derivative of the first function \(u\), multiply it by the second function \(v\), and then add the product of \(u\) and the derivative of \(v\).

In our exercise, we applied this rule to the function \(f(x, y) = x \ln(xy)\). Here, \(u\) is \(x\) and \(v\) is \(\ln(xy)\). By evaluating \(u'\) and \(v'\) separately and substituting into the product rule, the derivatives are calculated.

Differentiate with Respect to a Variable

To differentiate a function with respect to a specific variable, all other variables are held constant. This process is particularly important when dealing with functions of more than one variable, such as \(f(x, y)\).

Partial derivatives represent these scenarios. For partial differentiation with respect to \(x\):

• Treat \(y\) as a constant.

For partial differentiation with respect to \(y\):

• Treat \(x\) as a constant.

In the exercise, we first found \(f_x\), treating \(y\) as constant and then \(f_y\), treating \(x\) as constant. By using the product rule appropriately, this approach provides an accurate way to explore the behavior of functions in multi-variable contexts.

Natural Logarithm Function

The natural logarithm function, denoted \(\ln\), forms the basis for logarithms in calculus and is particularly prominent due to its exceptional properties when differentiated. The function itself is the inverse of the exponential function \(e^x\).

The differentiation rule for the natural logarithm function \(\ln(u)\) is as follows:

• If \(u\) is a differentiable function of \(x\), then \(\frac{d}{dx}\ln(u) = \frac{1}{u} \cdot \frac{du}{dx}\).

In the exercise, the natural logarithm \(\ln(xy)\) was differentiated in partial terms by recognizing \(xy\) as a single entity. This ability to differentiate seamlessly while considering the component parts is a powerful tool in calculus.