Problem 26
Question
For Problems calculate all four second-order partial derivatives and confirm that the mixed partials are equal. $$f=e^{x y}$$
Step-by-Step Solution
Verified Answer
All four second-order partial derivatives are calculated, and the mixed partials \( f_{xy} \) and \( f_{yx} \) are equal.
1Step 1: Find the first-order partial derivatives
First, we need to find the partial derivatives of the function \( f(x, y) = e^{xy} \) with respect to \( x \) and \( y \).\1. **Partial derivative with respect to \( x \):**\\[ f_x = \frac{\partial}{\partial x}(e^{xy}) = y \, e^{xy} \] (using the chain rule)\2. **Partial derivative with respect to \( y \):**\\[ f_y = \frac{\partial}{\partial y}(e^{xy}) = x \, e^{xy} \] (also using the chain rule).
2Step 2: Calculate second-order partial derivatives with respect to x
Now, compute the second partial derivative of \( f \) with respect to \( x \). Start from \( f_x = y \, e^{xy} \).\1. **Second partial derivative with respect to \( x \):**\\[ f_{xx} = \frac{\partial}{\partial x}(y \, e^{xy}) = y^2 \, e^{xy} \] (applying the product rule).
3Step 3: Calculate second-order partial derivatives with respect to y
Compute the second partial derivative of \( f \) with respect to \( y \). Start from \( f_y = x \, e^{xy} \).\1. **Second partial derivative with respect to \( y \):**\\[ f_{yy} = \frac{\partial}{\partial y}(x \, e^{xy}) = x^2 \, e^{xy} \] (using the product rule).
4Step 4: Calculate mixed second-order partial derivative \( f_{xy} \)
Find the mixed partial derivative \( f_{xy} \) by differentiating \( f_x = y \, e^{xy} \) with respect to \( y \).\1. **Mixed partial derivative \( f_{xy} \):**\\[ f_{xy} = \frac{\partial}{\partial y}(y \, e^{xy}) = e^{xy} + xy \, e^{xy} = (1 + x y) \, e^{xy} \] (using the product rule).
5Step 5: Calculate mixed second-order partial derivative \( f_{yx} \)
Find the mixed partial derivative \( f_{yx} \) by differentiating \( f_y = x \, e^{xy} \) with respect to \( x \).\1. **Mixed partial derivative \( f_{yx} \):**\\[ f_{yx} = \frac{\partial}{\partial x}(x \, e^{xy}) = e^{xy} + xy \, e^{xy} = (1 + x y) \, e^{xy} \] (again using the product rule).
6Step 6: Verify equality of mixed partials
Finally, verify if the mixed partials \( f_{xy} \) and \( f_{yx} \) are equal:\1. **\( f_{xy} = (1 + x y) \, e^{xy} \)**\2. **\( f_{yx} = (1 + x y) \, e^{xy} \)**\Since both mixed second-order partial derivatives are equal, we have confirmed that \( f_{xy} = f_{yx} \).
Key Concepts
second-order partial derivativesmixed partial derivativeschain ruleproduct rule
second-order partial derivatives
When you're working with functions of multiple variables, calculating partial derivatives helps you understand how the function changes with respect to each variable. Second-order partial derivatives take this concept a step further. They show how the rate of change itself changes. When you calculate a second-order partial derivative of a function, it means you're differentiating twice. The first differentiation gives you the rate of change, and the second tells you how this rate of change shifts.
For example, if we start with the function \( f(x, y) = e^{xy} \), the first-order partial derivatives are \( f_x \) and \( f_y \). From here, second-order derivatives can be calculated:
For example, if we start with the function \( f(x, y) = e^{xy} \), the first-order partial derivatives are \( f_x \) and \( f_y \). From here, second-order derivatives can be calculated:
- \( f_{xx} \) is the derivative of \( f_x \) with respect to \( x \) again.
- \( f_{yy} \) is the derivative of \( f_y \) with respect to \( y \) again.
mixed partial derivatives
Mixed partial derivatives look at the influence of each variable on the other. If you first take the derivative of \( f(x, y) \) with respect to \( x \), and then with respect to \( y \), you get a mixed partial derivative, denoted as \( f_{xy} \). Similarly, \( f_{yx} \) is found by differentiating first with respect to \( y \), then \( x \).
In many practical scenarios, especially under continuous conditions often encountered in real-world problems, mixed partial derivatives satisfy what's called Clairaut's theorem, meaning \( f_{xy} = f_{yx} \).
If you calculate mixed partial derivatives for \( f(x, y) = e^{xy} \), you'll find that both \( f_{xy} \) and \( f_{yx} \) result in \( (1 + xy)e^{xy} \). This symmetry indicates something deep about the potential applications, such as optimizing processes where these derivatives come into play.
In many practical scenarios, especially under continuous conditions often encountered in real-world problems, mixed partial derivatives satisfy what's called Clairaut's theorem, meaning \( f_{xy} = f_{yx} \).
If you calculate mixed partial derivatives for \( f(x, y) = e^{xy} \), you'll find that both \( f_{xy} \) and \( f_{yx} \) result in \( (1 + xy)e^{xy} \). This symmetry indicates something deep about the potential applications, such as optimizing processes where these derivatives come into play.
chain rule
The chain rule is extremely useful for differentiating composite functions. When it comes to partial derivatives, it helps when you're dealing with more layers or levels of functions. It allows you to break down derivatives into simpler parts, making them easier to calculate.
The general idea is if you have a function of a function, such as \( u(g(x, y)) \), to find the derivative, you take the derivative of the outer function \( u \) with respect to \( g \), then multiply it by the derivative of \( g \) with respect to each of your chosen variables.
In the exercise, when calculating the first-order partial derivatives of \( f(x, y) = e^{xy} \), the chain rule helps because the exponent \( xy \) acts as a composite function. You first find the derivative of \( e^{xy} \) with respect to the exponent, \( e^{xy} \), and then multiply by the derivative of \( xy \) with respect to your chosen variable. The chain rule makes navigating and solving each step of the differentiation process manageable.
The general idea is if you have a function of a function, such as \( u(g(x, y)) \), to find the derivative, you take the derivative of the outer function \( u \) with respect to \( g \), then multiply it by the derivative of \( g \) with respect to each of your chosen variables.
In the exercise, when calculating the first-order partial derivatives of \( f(x, y) = e^{xy} \), the chain rule helps because the exponent \( xy \) acts as a composite function. You first find the derivative of \( e^{xy} \) with respect to the exponent, \( e^{xy} \), and then multiply by the derivative of \( xy \) with respect to your chosen variable. The chain rule makes navigating and solving each step of the differentiation process manageable.
product rule
Whenever you encounter a product of two functions, the product rule is your ally. It helps you differentiate each function correctly and then combine their derivatives according to specific rules. This technique is especially crucial when the function involves products in some terms.
The product rule states that the derivative of a product of two functions \( u(x, y) \) and \( v(x, y) \) is given by
\[(uv)' = u'v + uv'\]
When applying to functions of several variables like those in our exercise, the rule remains largely the same, used to differentiate one variable at a time.
Take for instance when differentiating \( y \cdot e^{xy} \) in our function. You apply the product rule to differentiate with respect to \( x \) by finding the derivative of \( y \) alongside \( e^{xy} \) and also \( y \) times the derivative of \( e^{xy} \).
The product rule enables us to be precise with differentiating terms that involve multiple factors, as seen in each step of the exercise.
The product rule states that the derivative of a product of two functions \( u(x, y) \) and \( v(x, y) \) is given by
\[(uv)' = u'v + uv'\]
When applying to functions of several variables like those in our exercise, the rule remains largely the same, used to differentiate one variable at a time.
Take for instance when differentiating \( y \cdot e^{xy} \) in our function. You apply the product rule to differentiate with respect to \( x \) by finding the derivative of \( y \) alongside \( e^{xy} \) and also \( y \) times the derivative of \( e^{xy} \).
The product rule enables us to be precise with differentiating terms that involve multiple factors, as seen in each step of the exercise.
Other exercises in this chapter
Problem 25
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