Problem 18 - Calculus and its applications

Question

Find \(f_{x}\) and \(f_{y}\). $$f(x, y)=\frac{x}{y}+\frac{y}{5 x}$$

Step-by-Step Solution

Answer

\( f_x = \frac{1}{y} - \frac{y}{5x^2} \) and \( f_y = -\frac{x}{y^2} + \frac{1}{5x} \)."

Step 1: Identify Partial Derivative with respect to x

We want to find the partial derivative of the function \( f(x, y) = \frac{x}{y} + \frac{y}{5x} \) with respect to \( x \). This means we treat \( y \) as a constant and differentiate only with respect to \( x \).

Step 2: Differentiate First Term with respect to x

Differentiate \( \frac{x}{y} \) with respect to \( x \). Treat \( y \) as a constant, so the derivative is \( \frac{1}{y} \) because \( y \) is in the denominator.

Step 3: Differentiate Second Term with respect to x

Next, we differentiate \( \frac{y}{5x} \) with respect to \( x \). Treat \( y \) as constant. This is the same as differentiating \( y \cdot \frac{1}{5x} \). By using the chain rule, the derivative is \( -\frac{y}{5x^2} \).

Step 4: Combine Partial Derivatives to find f_x

Putting together the results from Steps 2 and 3, we find the partial derivative of \( f \) with respect to \( x \): \[ f_x = \frac{1}{y} - \frac{y}{5x^2} \].

Step 5: Identify Partial Derivative with respect to y

We now find the partial derivative of the function \( f(x, y) \) with respect to \( y \). This means we treat \( x \) as a constant and differentiate only with respect to \( y \).

Step 6: Differentiate First Term with respect to y

Differentiate \( \frac{x}{y} \) with respect to \( y \). Treat \( x \) as a constant, so the derivative is \( -\frac{x}{y^2} \) because \( y \) appears in the denominator.

Step 7: Differentiate Second Term with respect to y

Differentiate \( \frac{y}{5x} \) with respect to \( y \). Treat \( x \) as constant. The derivative is \( \frac{1}{5x} \) since \( \frac{y}{5x} \) can be seen as a linear function in \( y \).

Step 8: Combine Partial Derivatives to find f_y

Combine the results from Steps 6 and 7 to find the partial derivative of \( f \) with respect to \( y \): \[ f_y = -\frac{x}{y^2} + \frac{1}{5x} \].

Key Concepts

Partial Derivatives

Partial derivatives are an essential concept in multivariable calculus, especially when dealing with functions involving more than one variable. They help us understand how a function changes as one of its input variables changes, while keeping the others constant. This is crucial when analyzing functions of multiple variables, like those that define surfaces or fields in physics and engineering.

To find a partial derivative, you differentiate the function with respect to one variable, treating all other variables as constants. In the case of the function \( f(x, y) = \frac{x}{y} + \frac{y}{5x} \), we compute the partial derivatives with respect to \( x \) and \( y \) separately to understand how changes in each of these variables independently affect the function.

• For \( f_x \), you treat \( y \) as a constant and differentiate with respect to \( x \).
• Conversely, for \( f_y \), you treat \( x \) as a constant and differentiate with respect to \( y \).

Partial derivatives play a foundational role in varieties of mathematical analyses such as optimization, finding tangents, or engineers' use in heat conduction and stress analysis.

Differentiation

Differentiation is a core concept in calculus, and it's the process used to find the derivative of a function. The derivative represents the rate at which a function is changing at any given point. In multivariable functions, differentiation involves a pattern of continuously accounting for changes in more than one dimension.

When finding the partial derivatives of a function with multiple variables, differentiation allows you to understand how the function behaves as each individual variable changes. For example, differentiating \( \frac{x}{y} \) with respect to \( x \) gives \( \frac{1}{y} \) since you're treating \( y \) as a constant.

• The derivative of \( \frac{y}{5x} \) with respect to \( x \), involves applying the chain rule, resulting in \( -\frac{y}{5x^2} \).
• Similarly, for \( f_y \), differentiating \( \frac{y}{5x} \) yields \( \frac{1}{5x} \) since it's considered as a linear term in \( y \).

Differentiation in multiple variables is critical in fields such as optimization and economic modeling, where it is necessary to understand how changes in one variable can influence the entire system.

Functions of Two Variables

Functions of two variables, typically written as \( f(x, y) \), can represent real-world surfaces or data systems. Each function's value is influenced by two distinct inputs, \( x \) and \( y \). This type of function allows us to create topographic maps of data ranging from elevation across geographic landscapes to profit surfaces in economics.

Understanding how to work with these functions is fundamental in fields like economics, engineering, and the natural sciences. In our exercise, the function \( f(x, y) = \frac{x}{y} + \frac{y}{5x} \) is influenced by the behavior and interaction of \( x \) and \( y \).

• Each variable contributes differently to the function's output, and their changes lead to varying rates of change, highlighted by partial derivatives.
• Working with functions of two variables involves analyzing contour plots or three-dimensional graphs that help visualize these complex relationships.

Developing an intuition for these functions requires understanding concepts like levels curves, gradients, and how these interactions impact the function as a whole. Overall, functions of two variables significantly enhance our ability to model the real-world scenarios effectively.