Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

If \(F(x, y)=\frac{2 x-y}{x y}\), find \(F_{x}(3,-2)\) and \(F_{y}(3,-2) .\)

Step-by-Step Solution

Answer

\( F_x(3, -2) = -\frac{1}{18} \) and \( F_y(3, -2) = \frac{1}{9} \)

Step 1: Understand the Problem

We need to find the partial derivatives of the function \( F(x, y) = \frac{2x - y}{xy} \). Specifically, we want to find \( F_{x}(3, -2) \) and \( F_{y}(3, -2) \). This means we must first compute the partial derivatives \( F_x(x, y) \) and \( F_y(x, y) \) and then evaluate them at the given point \((3, -2)\).

Step 2: Compute the Partial Derivative with Respect to x

Using the quotient rule for partial differentiation with respect to \( x \), we have the numerator \( u = 2x - y \) and the denominator \( v = xy \). The derivative \( \frac{\partial}{\partial x} \frac{u}{v} \) is found using:\[ F_x = \frac{v \frac{\partial u}{\partial x} - u \frac{\partial v}{\partial x}}{v^2} \]where:- \( \frac{\partial u}{\partial x} = 2 \)- \( \frac{\partial v}{\partial x} = y \)Substitute into the formula:\[ F_x = \frac{xy \cdot 2 - (2x - y) \cdot y}{(xy)^2} = \frac{2xy - (2xy - y^2)}{x^2 y^2} = \frac{y^2}{x^2 y^2} \]Simplifying gives:\[ F_x = \frac{1}{x^2 y} \]

Step 3: Evaluate F_x at (3, -2)

Substitute \( x = 3 \) and \( y = -2 \) into \( F_x = \frac{1}{x^2 y} \),\[ F_x(3, -2) = \frac{1}{3^2 \cdot (-2)} = \frac{1}{9 \cdot (-2)} = \frac{1}{-18} = -\frac{1}{18} \]

Step 4: Compute the Partial Derivative with Respect to y

Using the quotient rule for partial differentiation with respect to \( y \), we have the same expressions for \( u \) and \( v \) as before. The derivative \( \frac{\partial}{\partial y} \frac{u}{v} \) is:\[ F_y = \frac{v \frac{\partial u}{\partial y} - u \frac{\partial v}{\partial y}}{v^2} \]where:- \( \frac{\partial u}{\partial y} = -1 \)- \( \frac{\partial v}{\partial y} = x \)Substitute into the formula:\[ F_y = \frac{xy \cdot (-1) - (2x - y) \cdot x}{(xy)^2} = \frac{-xy - (2x^2 - xy)}{x^2 y^2} = \frac{-2x^2}{x^2 y^2} \]Simplifying gives:\[ F_y = \frac{-2}{y \cdot x^2} \]

Step 5: Evaluate F_y at (3, -2)

Substitute \( x = 3 \) and \( y = -2 \) into \( F_y = \frac{-2}{y \cdot x^2} \),\[ F_y(3, -2) = \frac{-2}{(-2) \cdot 3^2} = \frac{-2}{-2 \cdot 9} = \frac{-2}{-18} = \frac{1}{9} \]

Key Concepts

Quotient Rule

The quotient rule is a fundamental concept in calculus, particularly useful in differentiation. This rule helps us find the derivative of a quotient of two functions. If we have a function expressed as a ratio, such as \( \frac{u}{v} \), the quotient rule becomes handy.

• Let \( u \) be the numerator and \( v \) be the denominator.
• The derivative of the quotient \( \frac{u}{v} \) is given by:

\[ \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \]
In this formula:

• \( \frac{du}{dx} \) is the derivative of \( u \).
• \( \frac{dv}{dx} \) is the derivative of \( v \).
• \( v \) and \( v^2 \) are the original and squared denominator respectively.

With partial differentiation, the procedure is similar, but we differentiate with respect to one variable, keeping others constant.

Partial Differentiation

Partial differentiation extends the idea of differentiation to functions of multiple variables. In essence, it allows us to see the effect of changing one variable while keeping others constant. For a function \( F(x, y) \), partial derivatives are denoted as \( F_x \) and \( F_y \), representing the partial derivatives with respect to the variables \( x \) and \( y \) respectively.

• With \( F(x, y) = \frac{2x - y}{xy} \), we apply the quotient rule to find these derivatives.
• Find \( F_x \) by treating \( y \) as constant and differentiating with respect to \( x \).
• Find \( F_y \) by treating \( x \) as constant and differentiating with respect to \( y \).

Each partial derivative shows how the function changes as just one variable changes, proving to be a powerful tool in multivariable calculus.

Function Evaluation

Once the partial derivatives are computed, the next step is to evaluate the derivatives at a specific point. This process is called function evaluation, where we substitute given values for the variables into the function or its derivatives.

• Given \( F(x, y) = \frac{2x - y}{xy} \), we found:
• \( F_x(x, y) = \frac{1}{x^2 y} \)
• \( F_y(x, y) = \frac{-2}{y \cdot x^2} \)
• Now, evaluate \( F_x(3, -2) \) by substituting \( x = 3 \) and \( y = -2 \):
• \( F_x(3, -2) = -\frac{1}{18} \)
• Similarly, for \( F_y(3, -2) \):
• \( F_y(3, -2) = \frac{1}{9} \)

Function evaluation is critical as it gives specific insights into the function's behavior at particular points in its domain.