Problem 23 - Applied Calculus

Question

Calculate all four second-order partial derivatives and confirm that the mixed partials are equal. $$ f(x, y)=\frac{2 x}{y}, \quad y \neq 0 $$

Step-by-Step Solution

Answer

The second-order partial derivatives are: \(f_{xx}=0\), \(f_{yy}=\frac{4x}{y^3}\), \(f_{xy}=f_{yx}=-\frac{2}{y^2}\). Mixed partials are equal.

Step 1: Find the First-Order Partial Derivative with Respect to x

To find the first-order partial derivative of the function \(f(x, y)=\frac{2x}{y}\) with respect to \(x\), treat \(y\) as a constant. The derivative is obtained simply by differentiating the formula: \[ f_x(x, y) = \frac{\partial}{\partial x}\left(\frac{2x}{y}\right) = \frac{2}{y}.\]

Step 2: Find the First-Order Partial Derivative with Respect to y

To find the first-order partial derivative with respect to \(y\), treat \(x\) as a constant and apply the quotient rule: \[f_y(x, y) = \frac{\partial}{\partial y}\left(\frac{2x}{y}\right) = \frac{0 \cdot y - 2x \cdot 1}{y^2} = -\frac{2x}{y^2}.\]

Step 3: Find the Second-Order Partial Derivative with Respect to x, Twice

Differentiate \(f_x(x, y) = \frac{2}{y}\) with respect to \(x\) again. Since \(\frac{2}{y}\) does not depend on \(x\), the derivative is:\[ f_{xx}(x, y) = \frac{\partial}{\partial x}\left(\frac{2}{y}\right) = 0.\]

Step 4: Find the Mixed Partial Derivative f_xy

Differentiate \(f_x(x, y) = \frac{2}{y}\) with respect to \(y\):\[ f_{xy}(x, y) = \frac{\partial}{\partial y}\left(\frac{2}{y}\right) = -\frac{2}{y^2}.\]

Step 5: Find the Mixed Partial Derivative f_yx

Differentiate \(f_y(x, y) = -\frac{2x}{y^2}\) with respect to \(x\):\[ f_{yx}(x, y) = \frac{\partial}{\partial x}\left(-\frac{2x}{y^2}\right) = -\frac{2}{y^2}.\]

Step 6: Check Equality of Mixed Partial Derivatives

Compare the results obtained for \(f_{xy}\) and \(f_{yx}\):Both \(f_{xy}(x, y) = -\frac{2}{y^2}\) and \(f_{yx}(x, y) = -\frac{2}{y^2}\) are equal, confirming the equality of mixed partials.

Step 7: Find the Second-Order Partial Derivative with Respect to y, Twice

Differentiate \(f_y(x, y) = -\frac{2x}{y^2}\) with respect to \(y\) again: Use the power rule on \(-\frac{2x}{y^2}\):\[ f_{yy}(x, y) = \frac{\partial}{\partial y}\left(-\frac{2x}{y^2}\right) = \frac{4x}{y^3}.\]

Key Concepts

Partial Differentiation

Partial differentiation is a technique used to find the derivative of a function with respect to one variable while keeping other variables constant. This is particularly useful in functions of several variables, like the given function, \( f(x, y) = \frac{2x}{y} \). Think of it as slicing through a 3D surface along constant lines of one variable.

• When we differentiate with respect to \( x \), \( y \) is treated as a constant, like it doesn't change.
• This means we focus on how the function changes as \( x \) changes while keeping an eye on \( y \).

For the function \( f(x, y) \), when you compute the first-order partial derivative with respect to \( x \), the result \( f_x(x, y) = \frac{2}{y} \) shows how the function changes with small changes in \( x \), while \( y \) remains fixed.
Similarly, when differentiating with respect to \( y \), we apply rules like the quotient rule. As seen, for \( f_y(x, y) = -\frac{2x}{y^2} \), it helps understand how the function shifts with changes in \( y \) at constant \( x \). Understanding partial differentiation ensures you grasp how each variable individually impacts the behavior of multi-variable functions.

Mixed Partial Derivatives

Mixed partial derivatives involve differentiating a function first with respect to one variable, and then with respect to another. They explore the interplay between different variables.
For instance, we first find \( f_x(x, y) = \frac{2}{y} \) and then differentiate this result with respect to \( y \). This gives the mixed partial \( f_{xy}(x, y) = -\frac{2}{y^2} \). In this step, you see what happens when both variables influence the function.

• You compute \( f_{yx}(x, y) \) by reversing the order; start with \( f_y(x, y) = -\frac{2x}{y^2} \) and differentiate with respect to \( x \). This results in the same \( f_{yx}(x, y) = -\frac{2}{y^2} \).
• A significant property is when the mixed partials \( f_{xy} \) and \( f_{yx} \) are equal, under certain conditions like continuous second derivatives.

The fact that \( f_{xy}(x, y) = f_{yx}(x, y) \) confirms that the function respects the symmetry of partial differentiation. Understanding mixed partial derivatives tells you lots about the dependence between the variables in functions.

Applied Calculus

Applied calculus involves using mathematical principles to solve practical problems in science, engineering, and other fields. Second-order partial derivatives, such as studied in the exercise, are crucial in applications.

• They often describe curvature and behavior of surfaces at given points, akin to how second derivatives inform about the concavity of standard functions.
• In particular, an examination of \( f_{xx}(x, y) \) and \( f_{yy}(x, y) \) offers insights into how a surface might bend in either the \( x \)- or \( y \)-direction.

In the example, \( f_{xx}(x, y) = 0 \) indicates no curvature change purely in the \( x \)-direction, while \( f_{yy}(x, y) = \frac{4x}{y^3} \) showcases how the surface bends significantly in the \( y \)-direction depending on the values of \( x \) and \( y \).
These concepts are applied across different fields such as physics, where you might use them to model wave vibrations, or economics, for optimizing multivariate functions. Understanding these derivatives can solve complex problems by breaking them down into simpler, single-variable components that are easier to manage.