Question

Verify that \(f_{x y}=f_{y x}\) for the following functions. $$f(x, y)=e^{x+y}$$

Short answer

Based on the given function $$f(x, y) = e^{x + y}$$, we have found the respective second-order mixed partial derivatives and determined that $$f_{xy}(x, y) = e^{x + y}$$ and $$f_{yx}(x, y) = e^{x + y}$$. As both mixed partial derivatives are equal, we can conclude that $$f_{xy}(x, y) = f_{yx}(x, y)$$.

Step-by-step solution

Find the first order partial derivative with respect to x

To find the first order partial derivative of the function $$f(x, y) = e^{x + y}$$ with respect to $$x$$, we will differentiate the function with respect to $$x$$ while treating $$y$$ as a constant. This can be written as: $$f_x(x, y) = \frac{\partial}{\partial x}(e^{x + y})$$ Using the chain rule for differentiation, we get: $$f_x(x, y) = e^{x + y} \cdot \frac{\partial}{\partial x}(x + y)$$ Since $$\frac{\partial}{\partial x}(x + y) = 1$$, we have: $$f_x(x, y) = e^{x + y}$$

Find the first order partial derivative with respect to y

Now, we will find the first order partial derivative of the function $$f(x, y) = e^{x + y}$$ with respect to $$y$$, by differentiating the function with respect to $$y$$ while treating $$x$$ as a constant: $$f_y(x, y) = \frac{\partial}{\partial y}(e^{x + y})$$ Using the chain rule for differentiation, we get: $$f_y(x, y) = e^{x + y} \cdot \frac{\partial}{\partial y}(x + y)$$ Since $$\frac{\partial}{\partial y}(x + y) = 1$$, we have: $$f_y(x, y) = e^{x + y}$$

Find the second order mixed partial derivative with respect to x and y

With the first order partial derivatives found, we can now calculate the second order mixed partial derivative of the function with respect to $$x$$ and $$y$$: $$f_{xy} (x, y) = \frac{\partial}{\partial y} (f_x(x, y))$$ We already have $$f_x(x, y) = e^{x + y}$$, so we just need to differentiate this expression with respect to $$y$$: $$f_{xy}(x, y) = \frac{\partial}{\partial y}(e^{x + y})$$ Using the chain rule for differentiation, we get: $$f_{xy}(x, y) = e^{x + y} \cdot \frac{\partial}{\partial y}(x + y)$$ Since $$\frac{\partial}{\partial y}(x + y) = 1$$, we have: $$f_{xy}(x, y) = e^{x + y}$$

Find the second order mixed partial derivative with respect to y and x

Similarly, we will find the second order mixed partial derivative of the function with respect to $$y$$ and $$x$$: $$f_{yx}(x, y) = \frac{\partial}{\partial x} (f_y(x, y))$$ We already have $$f_y(x, y) = e^{x + y}$$, so we just need to differentiate this expression with respect to $$x$$: $$f_{yx}(x, y) = \frac{\partial}{\partial x}(e^{x + y})$$ Using the chain rule for differentiation, we get: $$f_{yx}(x, y) = e^{x + y} \cdot \frac{\partial}{\partial x}(x + y)$$ Since $$\frac{\partial}{\partial x}(x + y) = 1$$, we have: $$f_{yx}(x, y) = e^{x + y}$$

Compare the mixed partial derivatives

Now that we have found both $$f_{xy}(x, y)$$ and $$f_{yx}(x, y)$$, we can compare them to verify if they are equal. We have: $$f_{xy}(x, y) = e^{x + y}$$ and $$f_{yx}(x, y) = e^{x + y}$$ Since both mixed partial derivatives are equal, we can conclude that $$f_{xy}(x, y) = f_{yx}(x, y)$$.

Key concepts

Partial Derivatives

Partial derivatives are used to examine how a function changes as each of its input variables changes, keeping the other variables constant. When we have a function of two variables, such as \( f(x, y) = e^{x + y} \), we consider each variable separately to find the partial derivatives.

To compute the partial derivative of \( f \) with respect to \( x \), denoted as \( f_x(x, y) \), we treat \( y \) as a constant and differentiate the function with respect to \( x \). Similarly, for the partial derivative with respect to \( y \), denoted as \( f_y(x, y) \), we hold \( x \) constant and derive with respect to \( y \).
- **Example:** For \( f(x, y) = e^{x + y} \), both partial derivatives \( f_x \) and \( f_y \) equal \( e^{x + y} \) because the exponential function is differentiated using the chain rule.

Chain Rule

The chain rule is a fundamental technique in calculus used to differentiate composite functions. In the context of partial derivatives, it helps when we are differentiating a function of two variables with respect to one variable, but the expression involves the other variable.
- **Example:** For the function \( f(x, y) = e^{x + y} \), using the chain rule, when differentiating with respect to \( x \), the derivative of \( x+y \) with respect to \( x \) is 1, leaving us with \( f_x(x, y) = e^{x+y} \cdot 1 = e^{x+y} \).

This rule simplifies the process as it breaks down the differentiation of complex functions into manageable parts. It ensures that we seamlessly differentiate functions like these by efficiently applying the chain rule.

Differentiation

Differentiation is the process of finding the rate at which a function changes with respect to its variables. In the context of functions of multiple variables, it involves taking derivatives with respect to each variable, leading to partial derivatives.
- **Key Steps in Differentiation:** - Identify the variables - Differentiate with respect to one variable while keeping others constant - Apply rules like the chain rule where needed

These steps were followed in differentiating \( f(x, y) = e^{x+y} \), allowing us to find partial derivatives in each direction. Differentiation is at the heart of calculus, enabling us to understand and describe how functions behave as their inputs change. This fundamental skill leads us to more complex operations like finding mixed derivatives.

Function of Two Variables

A function of two variables is an expression where the output depends on two input variables, typically denoted as \( x \) and \( y \). Such functions are often written in the form \( f(x, y) \).
- **Characteristics of Two-Variable Functions:** - They can describe surfaces or planes - Each combination of \( x \) and \( y \) inputs produces a unique output

The function \( f(x, y) = e^{x+y} \) demonstrates how inputs combine additively in the exponent, resulting in exponential growth along both axes. Analyzing how these functions behave includes computing partial derivatives and understanding their geometric implications. By studying these derivatives, we get insights into properties like slope and curvature, critical for applications in physics, economics, and engineering.