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Magazine Chapter 14: Problem 41
Find all the second-order partial derivatives of the functions in Exercises \(41-50 .\) $$f(x, y)=x+y+x y$$
Short Answer
Expert verified
The second-order partial derivatives are \( f_{xx} = 0 \), \( f_{yy} = 0 \), \( f_{xy} = 1 \), and \( f_{yx} = 1 \).
Step by step solution
01
Identify the Function
We start by identifying the function given in the problem. The function is \( f(x, y) = x + y + xy \).
02
Find the First-order Partial Derivative with Respect to x
To find the first-order partial derivative of \( f(x, y) \) with respect to \( x \), denoted as \( f_x \), we differentiate each term of the function that contains \( x \) with respect to \( x \):- The derivative of \( x \) with respect to \( x \) is \( 1 \).- The derivative of \( y \) with respect to \( x \) is \( 0 \) since \( y \) is treated as a constant.- The derivative of \( xy \) with respect to \( x \) is \( y \).So, \( f_x = 1 + y \).
03
Find the First-order Partial Derivative with Respect to y
To find the first-order partial derivative of \( f(x, y) \) with respect to \( y \), denoted as \( f_y \), we differentiate each term of the function that contains \( y \) with respect to \( y \):- The derivative of \( x \) with respect to \( y \) is \( 0 \) since \( x \) is treated as a constant.- The derivative of \( y \) with respect to \( y \) is \( 1 \).- The derivative of \( xy \) with respect to \( y \) is \( x \).So, \( f_y = 1 + x \).
04
Find the Second-order Partial Derivative with Respect to x
Now, we find the second-order partial derivative by differentiating \( f_x \) with respect to \( x \). The function \( f_x = 1 + y \) contains no \( x \) in terms other than \( 1 \), so the derivative is \( 0 \). Thus, \( f_{xx} = 0 \).
05
Find the Second-order Partial Derivative with Respect to y
Next, we differentiate \( f_y \) with respect to \( y \) to find \( f_{yy} \). The function \( f_y = 1 + x \) contains no \( y \) in terms other than \( 1 \), so the derivative is \( 0 \). Thus, \( f_{yy} = 0 \).
06
Find the Mixed Second-order Partial Derivative
We need to find the mixed second-order partial derivatives \( f_{xy} \) and \( f_{yx} \). First, differentiate \( f_x = 1 + y \) with respect to \( y \):- The derivative of \( 1 \) is \( 0 \).- The derivative of \( y \) with respect to \( y \) is \( 1 \).So, \( f_{xy} = 1 \).Then, differentiate \( f_y = 1 + x \) with respect to \( x \):- The derivative of \( 1 \) is \( 0 \).- The derivative of \( x \) with respect to \( x \) is \( 1 \).So, \( f_{yx} = 1 \).
07
Conclusion
The second-order partial derivatives are \( f_{xx} = 0 \), \( f_{yy} = 0 \), \( f_{xy} = 1 \), and \( f_{yx} = 1 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-Order Partial Derivatives
When working with functions of multiple variables, we often need to find derivatives with respect to one variable at a time. This is where first-order partial derivatives come in handy. Let's consider the function given: \( f(x, y) = x + y + xy \). We need to find partial derivatives \( f_x \) and \( f_y \).
- For \( f_x \), or the derivative of \( f \) with respect to \( x \), we focus on how \( f \) changes as \( x \) changes, treating \( y \) as a constant. Here, the derivative \( 1 \) comes from the term \( x \), and \( y \) is considered a constant multiplier in \( xy \), giving a derivative of \( y \). So \( f_x = 1 + y \).
- For \( f_y \), or the derivative of \( f \) with respect to \( y \), we focus on changes in \( f \) as \( y \) changes. Treat \( x \) as a constant. Therefore, the derivative of \( y \) is \( 1 \), and since \( xy \) behaves like \( x \) times a variable, the derivative is \( x \). Thus, \( f_y = 1 + x \).
Mixed Partial Derivatives
Mixed partial derivatives show how a function changes when two different variables are considered, one after the other. For the function \( f(x, y) \), these are represented as \( f_{xy} \) and \( f_{yx} \).
- To find \( f_{xy} \), differentiate the partial derivative \( f_x = 1 + y \) with respect to \( y \). The constant term disappears, and the derivative of \( y \) is \( 1 \), so \( f_{xy} = 1 \).
- Similarly, to find \( f_{yx} \), differentiate \( f_y = 1 + x \) with respect to \( x \). Here, the derivative of \( x \) is also \( 1 \), leading to \( f_{yx} = 1 \).
Differentiation
Differentiation is the process of finding derivatives, which measure how a function changes as its inputs change. In calculus, we often deal with functions of more than one variable and look at partial derivatives.
- With Respect to a Single Variable: Differentiation with respect to a single variable involves treating all other variables as constants. In this exercise, we found partial derivatives of \( f(x, y) \) first with respect to \( x \), then with respect to \( y \).
- Second-Order Derivatives: These are derivatives of the first-order partial derivatives. We had \( f_{xx} \) and \( f_{yy} \), both equals to \( 0 \) because differentials of constants vanish. This means no additional slope change in those pure variable directions.
