Problem 3 - The Calculus with Analytic Geometry

Question

In Exercises 1 through 8 , do each of the following: (a) Find \(D_{11} f(x, y)\); (b) find \(D_{22} f(x, y) ;\) (c) show that \(D_{12} f(x, y)=D_{21} f(x, y) .\) $$ f(x, y)=e^{2 x} \sin y $$

Step-by-Step Solution

Answer

D_{11} = 4e^{2x} \sin y,D_{22} = -e^{2x} \sin y,D_{12} = D_{21} = 2e^{2x} \cos y

Step 1 - Find the first partial derivatives

First, we need to find the first partial derivatives of the function. We have: \[ f(x, y) = e^{2x} \sin y \]To find the partial derivative with respect to x, we treat y as a constant: \[ f_x = \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (e^{2x} \sin y) = 2e^{2x} \sin y \]Next, to find the partial derivative with respect to y, we treat x as a constant: \[ f_y = \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (e^{2x} \sin y) = e^{2x} \cos y \]

Step 2 - Find the second partial derivative with respect to x (D_{11})

Next, we find the second partial derivative with respect to x: \[ f_{xx} = \frac{\partial}{\partial x}(f_x) = \frac{\partial}{\partial x}(2e^{2x} \sin y) = 4e^{2x} \sin y \]Thus, \[ D_{11} f(x, y) = 4e^{2x} \sin y \]

Step 3 - Find the second partial derivative with respect to y (D_{22})

Now, let's find the second partial derivative with respect to y: \[ f_{yy} = \frac{\partial}{\partial y}(f_y) = \frac{\partial}{\partial y}(e^{2x} \cos y) = -e^{2x} \sin y \]Thus, \[ D_{22} f(x, y) = -e^{2x} \sin y \]

Step 4 - Find the mixed partial derivatives (D_{12} and D_{21})

Finally, we find the mixed partial derivatives. Compute \( f_{xy} \) as follows: \[ f_{xy} = \frac{\partial}{\partial y} (f_x) = \frac{\partial}{\partial y} (2e^{2x} \sin y) = 2e^{2x} \cos y \]Next, compute \( f_{yx} \): \[ f_{yx} = \frac{\partial}{\partial x} (f_y) = \frac{\partial}{\partial x} (e^{2x} \cos y) = 2e^{2x} \cos y \]Thus, \[ D_{12} f(x, y) = D_{21} f(x, y) = 2e^{2x} \cos y \]

Key Concepts

First Partial Derivative

To start, we need to understand the concept of the first partial derivative. A partial derivative of a function with respect to one of its variables is the derivative with respect to that variable while keeping the other variables constant.
Let's consider the function given in the exercise: \( f(x, y) = e^{2x} \text{sin} y \). To find the first partial derivative of \( f \) with respect to \( x \), we treat \( y \) as a constant and differentiate: \[ f_x = \frac{\text{\textdollar\textdollar}}{\text{\textdollar x \text{\textdollar}} (e^{2x} \text{sin} y) = 2e^{2x} \text{sin} y \].
Similarly, to find the first partial derivative with respect to \( y \), we treat \( x \) as a constant and differentiate: \[ f_y = \frac{\text{\textdollar\textdollar}}{\text{\textdollar y \text{\textdollar}} (e^{2x} \text{sin} y) = e^{2x} \text{cos} y \].Because of these calculations, we can observe how each variable independently affects the function.

Second Partial Derivative

Next, let’s discuss the second partial derivative, which involves differentiation twice with respect to the same variable. In this exercise, we need to compute \( D_{11} \) and \( D_{22} \), the second partial derivatives of \( f \) with respect to \( x \) and \( y \) respectively.
We already know that: \[ f_x = 2e^{2x} \text{sin} y \]So, the second partial derivative \( D_{11} \) is: \[ f_{xx} = \frac{\text{\textdollar\textdollar}}{\text{\textdollar x \text{\textdollar}} (2e^{2x} \text{sin} y) = 4e^{2x} \text{sin} y \].
Similarly, given: \[ f_y = e^{2x} \text{cos} y \] we find \( D_{22} \) as: \[ f_{yy} = \frac{\text{\textdollar\textdollar}}{\text{\textdollar y \text{\textdollar}} (e^{2x} \text{cos} y) = -e^{2x} \text{sin} y \].These second partial derivatives help us understand the curvature and concavity of the function along the x and y directions.

Mixed Partial Derivatives

Finally, let’s explore mixed partial derivatives, which involve differentiation with respect to different variables in succession.
To find mixed partial derivatives \( D_{12} \) and \( D_{21} \):First, we calculate \( f_{xy} \): \[ f_{xy} = \frac{\text{\textdollar\textdollar}}{\text{\textdollar y \text{\textdollar}} (f_x) = \frac{\text{\textdollar\textdollar}}{\text{\textdollar y \text{\textdollar}} (2e^{2x} \text{sin} y) = 2e^{2x} \text{cos} y \]And then \( f_{yx} \): \[ f_{yx} = \frac{\text{\textdollar\textdollar}}{\text{\textdollar x \text{\textdollar}} (f_y) = \frac{\text{\textdollar\textdollar}}{\text{\textdollar x \text{\textdollar}} (e^{2x} \text{cos} y) = 2e^{2x} \text{cos} y \].
This shows that \( f_{xy} = f_{yx} \), confirming a key property known as the equality of mixed partial derivatives under certain conditions. This property, known as Clairaut's theorem, states that if the mixed partial derivatives are continuous, they will be equal regardless of the order of differentiation.