Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ f(x, y)=x y+\frac{2}{x}+\frac{4}{y} $$

Step-by-Step Solution

Answer

Find the partial derivatives: \( \frac{\partial f}{\partial x} = y - \frac{2}{x^2} \) and \( \frac{\partial f}{\partial y} = x - \frac{4}{y^2} \).

Step 1: Understand the Function

The function given is a two-variable function \( f(x, y) = xy + \frac{2}{x} + \frac{4}{y} \). This function consists of a term that involves the product of \( x \) and \( y \), and two rational terms, \( \frac{2}{x} \) and \( \frac{4}{y} \). The goal is typically to analyze or find specific values, behavior, or derivatives of the function.

Step 2: Identify Objective

Since no specific task was provided with the function, determine a common task such as finding critical points, computing partial derivatives, or evaluating at a specific point. For this example, let's decide to find the partial derivatives to analyze how the function changes with respect to \( x \) and \( y \).

Step 3: Compute Partial Derivative with Respect to x

To find \( \frac{\partial f}{\partial x} \), differentiate the function \( f(x, y) = xy + \frac{2}{x} + \frac{4}{y} \) with respect to \( x \). Using the power rule and the quotient rule, we get:\[\frac{\partial f}{\partial x} = y - \frac{2}{x^2}\]

Step 4: Compute Partial Derivative with Respect to y

To find \( \frac{\partial f}{\partial y} \), differentiate the function \( f(x, y) = xy + \frac{2}{x} + \frac{4}{y} \) with respect to \( y \). This gives:\[\frac{\partial f}{\partial y} = x - \frac{4}{y^2}\]

Step 5: Summary of Partial Derivatives

The partial derivatives of the function are:- With respect to \( x \): \( \frac{\partial f}{\partial x} = y - \frac{2}{x^2} \)- With respect to \( y \): \( \frac{\partial f}{\partial y} = x - \frac{4}{y^2} \)

Key Concepts

Two-variable function

A two-variable function is a function that depends on two independent variables, typically denoted as \(x\) and \(y\). These functions describe a relationship where the output or dependent variable depends on the values of both inputs. An example of such a function is \(f(x, y) = xy + \frac{2}{x} + \frac{4}{y}\). Here, each term in the function depends on either \(x\), \(y\), or both.

• The term \(xy\) shows direct multiplication of the two variables.
• The term \(\frac{2}{x}\) demonstrates a rational expression involving only \(x\).
• The term \(\frac{4}{y}\) features \(y\) in a similar context.

Two-variable functions are represented on a three-dimensional graph, where each point \((x, y, z)\) represents a position in space, with \(z = f(x, y)\). Understanding these functions is crucial for fields like physics, engineering, and economics, where relationships between variables are essential.

Critical points

Critical points in two-variable functions are similar to those in single-variable functions. They indicate where the function's rate of change is zero, which can be useful for finding maxima, minima, or saddle points.
To find critical points:

• Calculate the partial derivatives of the function with respect to each variable.
• Set these partial derivatives equal to zero.
• Solve the system of equations to find points \((x, y)\) where the function's rate of change is zero.

For the function \(f(x, y) = xy + \frac{2}{x} + \frac{4}{y}\), critical points are obtained by solving:- \(\frac{\partial f}{\partial x} = y - \frac{2}{x^2} = 0\)- \(\frac{\partial f}{\partial y} = x - \frac{4}{y^2} = 0\)Finding critical points helps in understanding how the function behaves and where potential peaks or troughs occur.

Power rule

The power rule is a fundamental principle of differentiation used to find the derivative of functions with powers. It states that if \(f(x) = x^n\), then the derivative \(f'(x) = nx^{n-1}\). This rule is simple but incredibly powerful, allowing swift calculation of derivatives.
When applying it to two-variable functions for partial derivatives, it helps differentiate terms like \(x^n\) or \(y^n\) with respect to one variable:

• For \(f(x, y) = xy\), treating \(y\) as a constant yields \(\frac{\partial }{\partial x}(xy) = y\).
• For \(\frac{2}{x} = 2x^{-1}\), using the power rule gives \(-2x^{-2}\).

The power rule simplifies the process, focusing only on the exponents and coefficients associated with the variable being considered.

Quotient rule

The quotient rule is used to differentiate functions where variables are divided by each other. Specifically, if \(f(x) = \frac{u(x)}{v(x)}\), then the derivative is:\[\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2}\]This rule is critical in handling terms like \(\frac{2}{x}\) when finding partial derivatives.
For two-variable functions, apply the quotient rule when dealing with terms that divide by a variable. For example:

• The derivative with respect to \(x\) of \(\frac{2}{x}\) uses \(u = 2\) (which has derivative 0) and \(v = x\) to find \(-\frac{2}{x^2}\).

Applying the quotient rule correctly is vital for accurate calculation of derivatives in multi-variable calculus.