Problem 30 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Evaluate the first partial derivatives of the function at the given point. \(f(x, y)=\frac{x+y}{x-y} ;(1,-2)\)

Step-by-Step Solution

Answer

The first partial derivatives of the function at the given point (1, -2) are: \(\left.\frac{\partial f}{\partial{x}}\right|_{(1,-2)} = \frac{4}{9},\) and \(\left.\frac{\partial f}{\partial{y}}\right|_{(1,-2)} = \frac{2}{9}.\)

Step 1: Find the partial derivative with respect to x

First, we will find the partial derivative of the function with respect to x. To do that, differentiate the function while treating y as a constant: \[ \begin{aligned} \frac{\partial}{\partial{x}}\left(\frac{x+y}{x-y}\right) \end{aligned} \] Since the given function is a quotient of two functions, we can use the quotient rule: \[ \begin{aligned} \frac{\partial}{\partial{x}}\left(\frac{u}{v}\right) = \frac{vu' - uv'}{v^2} \end{aligned} \] Here, \(u = x+y\) and \(v = x-y\). Differentiate both u and v with respect to x: \[ \begin{aligned} u' &= \frac{\partial}{\partial{x}}(x+y) = 1 \\ v' &= \frac{\partial}{\partial{x}}(x-y) = 1 \\ \end{aligned} \] Now apply the quotient rule: \[ \begin{aligned} \frac{\partial}{\partial{x}}\left(\frac{x+y}{x-y}\right) = \frac{(x-y)(1) - (x+y)(1)}{(x-y)^2} = \frac{-2y}{(x-y)^2} \end{aligned} \]

Step 2: Find the partial derivative with respect to y

Next, we will find the partial derivative of the function with respect to y. To do that, differentiate the function while treating x as a constant: \[ \begin{aligned} \frac{\partial}{\partial{y}}\left(\frac{x+y}{x-y}\right) \end{aligned} \] As before, we can use the quotient rule: \[ \begin{aligned} \frac{\partial}{\partial{y}}\left(\frac{u}{v}\right) = \frac{vu' - uv'}{v^2} \end{aligned} \] We have the same u and v as before, but now we'll differentiate both u and v with respect to y: \[ \begin{aligned} u' &= \frac{\partial}{\partial{y}}(x+y) = 1 \\ v' &= \frac{\partial}{\partial{y}}(x-y) = -1 \\ \end{aligned} \] Now apply the quotient rule: \[ \begin{aligned} \frac{\partial}{\partial{y}}\left(\frac{x+y}{x-y}\right)=\frac{(x-y)(1)-(x+y)(-1)}{(x-y)^2}=\frac{2x}{(x-y)^2} \end{aligned} \]

Step 3: Evaluate the partial derivatives at the given point

Now that we have the expressions for the partial derivatives with respect to x and y, we can evaluate them at the given point (1, -2): \[ \begin{aligned} \left.\frac{\partial f}{\partial{x}}\right|_{(1,-2)} &= \frac{-2(-2)}{(1-(-2))^2} \\ &= \frac{4}{9} \\ \end{aligned} \] \[ \begin{aligned} \left.\frac{\partial f}{\partial{y}}\right|_{(1,-2)} &= \frac{2(1)}{(1-(-2))^2} \\ &= \frac{2}{9} \\ \end{aligned} \]

Step 4: Final answer

The first partial derivatives of the function at the given point (1, -2) are: \[ \begin{aligned} \left.\frac{\partial f}{\partial{x}}\right|_{(1,-2)} = \frac{4}{9} \\ \left.\frac{\partial f}{\partial{y}}\right|_{(1,-2)} = \frac{2}{9} \end{aligned} \]

Key Concepts

Quotient Rule

Understanding the quotient rule is essential when working with the division of two functions, which frequently arises in differential calculus. In essence, the quotient rule provides us with a systematic way to find the derivative of a quotient without having to perform the division directly.

Let's consider a function that is defined as the quotient of two differentiable functions, denoted as \( f(x) = \frac{u(x)}{v(x)} \), where both \( u(x) \) and \( v(x) \) are differentiable, and \( v(x) \) is not equal to zero. The quotient rule states that:

\[\begin{equation}\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{vu' - uv'}{v^2}\end{equation}\]
where \( u' \) represents the derivative of \( u \) with respect to \( x \) and \( v' \) is the derivative of \( v \) with respect to \( x \). In the given exercise, applying the quotient rule allows us to find the partial derivatives efficiently and correctly without the complicated algebra that can come from manual manipulation.

Differential Calculus

Differential calculus is a subfield of calculus concerned with the concept of change and rates of change. The fundamental operation in differential calculus is differentiation, which is essentially about finding rates at which variables change with respect to one another. When we differentiate a function, we're finding its derivative, which provides the slope of the function at any given point. The derivative at a point gives us the rate of change or the slope of the tangent line at that specific point on the graph of the function.

In our exercise involving the function \( f(x, y) = \frac{x + y}{x - y} \), differential calculus allows us to compute the partial derivatives of the function, which measure how \( f \) changes as each variable \( x\) and \( y \) is varied while the other variable is held constant. This provides us with critical information about the function’s behavior in the multidimensional space.

Functions of Several Variables

Functions of several variables, such as the one presented in our exercise, \( f(x, y) \), are characterized by having more than one independent variable. These multivariable functions extend the concept of single-variable functions, necessitating new techniques and methods such as partial differentiation.

Partial derivatives are a critical tool in understanding these multivariable functions. They allow us to explore the function’s behavior with respect to each variable independently. The first partial derivative with respect to \( x \) tells us how the function changes as \( x \) increases, with \( y \) held steady, and vice versa for the partial derivative with respect to \( y \).

In practical terms, these derivatives help us identify slopes along \( x \) and \( y \) directions and are foundational in fields like economics, engineering, and the physical sciences, where systems depend on several variables.