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Calculate all four second-order partial derivatives and check that \(f_{x y}=f_{y x} .\) Assume the variables are restricted to a domain on which the function is defined. $$f(x, y)=(x+y)^{3}$$

Short Answer

Expert verified
All second-order partial derivatives are 6(x+y), and f_{xy} = f_{yx}.

Step by step solution

01

Find the First-Order Partial Derivatives

First, we need to find the partial derivatives of the given function with respect to both variables. The function is \(f(x, y) = (x+y)^3\).- **Differentiate with respect to \(x\):** \[ f_x = \frac{d}{dx}((x+y)^3) = 3(x+y)^2 \]- **Differentiate with respect to \(y\):** \[ f_y = \frac{d}{dy}((x+y)^3) = 3(x+y)^2 \]
02

Find the Second-Order Partial Derivatives

Now, we will differentiate the first-order derivatives with respect to both \(x\) and \(y\) again.- **Differentiate \(f_x\) with respect to \(x\):** \[ f_{xx} = \frac{d}{dx}(3(x+y)^2) = 6(x+y) \]- **Differentiate \(f_x\) with respect to \(y\):** \[ f_{xy} = \frac{d}{dy}(3(x+y)^2) = 6(x+y) \]- **Differentiate \(f_y\) with respect to \(x\):** \[ f_{yx} = \frac{d}{dx}(3(x+y)^2) = 6(x+y) \]- **Differentiate \(f_y\) with respect to \(y\):** \[ f_{yy} = \frac{d}{dy}(3(x+y)^2) = 6(x+y) \]
03

Verify the Equality of Mixed Partial Derivatives

To check if the mixed partial derivatives are equal, we compare \(f_{xy}\) and \(f_{yx}\).- We have calculated \(f_{xy} = 6(x+y)\) and \(f_{yx} = 6(x+y)\).- Since \(f_{xy} = f_{yx}\), the equality holds.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First-order partial derivatives
To understand a function's behavior in a multivariable context, we begin with first-order partial derivatives. These derivatives help us examine how the function changes as each variable varies, while keeping the other variables constant. In our exercise, we have a function \(f(x, y) = (x+y)^3\). Here, we need to find how \(f(x, y)\) changes with respect to \(x\) and \(y\) separately.

- **With respect to \(x\):** When calculating the partial derivative with respect to \(x\), we differentiate \(f\) while treating \(y\) as a constant. Thus, the derivative \(f_x = \frac{\partial f}{\partial x} = 3(x+y)^2\).
- **With respect to \(y\):** Similarly, to find the derivative with respect to \(y\), treat \(x\) as a constant. Then, the derivative \(f_y = \frac{\partial f}{\partial y} = 3(x+y)^2\).

These computations tell us that the rate of change of \(f\) with respect to \(x\) and \(y\) is the same in form, reflecting the symmetry in \((x+y)^3\).
Mixed partial derivatives
Mixed partial derivatives involve taking derivatives with respect to different variables. In simpler terms, to find a mixed derivative, we first differentiate with respect to one variable, then take a derivative of that result with respect to the other variable. This process helps us examine how the variables jointly affect the function's rate of change.

In the given exercise, the function is \(f(x, y) = (x+y)^3\). Let's see how mixed partial derivatives are calculated:
  • For \(f_{xy}\), we differentiate \(f_x = 3(x+y)^2\) with respect to \(y\), resulting in \(f_{xy} = 6(x+y)\).
  • For \(f_{yx}\), we differentiate \(f_y = 3(x+y)^2\) with respect to \(x\), resulting in \(f_{yx} = 6(x+y)\).
We notice that \(f_{xy}\) and \(f_{yx}\) are equal, and this equality is often expected for "nice" functions—those that are continuous and have continuous second partial derivatives—which fits our scenario.
Partial differentiation
Partial differentiation is a powerful method used to study functions of several variables. Unlike ordinary differentiation, partial differentiation focuses on changing one variable at a time, holding others constant. This technique is essential in multivariable calculus, especially for functions where variables interact in complex ways.

Let's break it down using our function \(f(x, y) = (x+y)^3\):
  • By partially differentiating with respect to \(x\) and \(y\), we gain insight into how each individual variable influences the function, as seen with \(f_x = 3(x+y)^2\) and \(f_y = 3(x+y)^2\).
  • Second-order partial derivatives like \(f_{xx}\), \(f_{yy}\), \(f_{xy}\), and \(f_{yx}\) provide more detailed information on the curvature and direction of the function's change.
Partial differentiation is critical not only in mathematics but also in real-world applications such as physics and engineering, where it helps describe systems with multiple interacting factors. Understanding how to compute and interpret these derivatives allows for deeper insight into the behavior of functions in multidimensional spaces.

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Most popular questions from this chapter

(a) Find the equation of the plane tangent to the graph of \(f(x, y)=x^{2} e^{x y}\) at (1,0) (b) Find the linear approximation of \(f(x, y)\) for \((x, y)\) near (1,0) (c) Find the differential of \(f\) at the point (1,0)

Find the gradient of the function. Assume the variables are restricted to a domain on which the function s defined. $$z=x e^{y}$$

The tastiness, \(T\), of a soup depends on the volume, \(V\) of the soup in the pot and the quantity, \(S,\) of salt in the soup. If you have more soup, you need more salt to make it taste good. Match the three stories (a)-(c) to the three statements (I)-(III) about partial derivatives. (a) I started adding salt to the soup in the pot. At first the taste improved, but eventually the soup became too salty and continuing to add more salt made it worse. (b) The soup was too salty, so I started adding unsalted soup. This improved the taste at first, but eventually there was too much soup for the salt, and continuing to add unsalted soup just made it worse. (c) The soup was too salty, so adding more salt would have made it taste worse. I added a quart of unsalted soup instead. Now it is not salty enough, but I can improve the taste by adding salt. (I) \(\partial^{2} T / \partial V^{2} < 0\) (II) \(\partial^{2} T / \partial S^{2} < 0\) (III) \(\partial^{2} T / \partial V \partial S > 0\)

Let \(f(x, y)=x^{2} y^{3} .\) At the point \((-1,2),\) find a vector (a) In the direction of maximum rate of change. (b) In the direction of minimum rate of change. (c) In a direction in which the rate of change is zero.

A student was asked to find the directional derivative of \(f(x, y)=x^{2} e^{y}\) at the point (1,0) in the direction of \(\vec{v}=4 \vec{i}+3 \vec{j} .\) The student's answer was $$ f_{i i}(1,0)=\operatorname{grad} f(1,0) \cdot \vec{u}=\frac{8}{5} \vec{i}+\frac{3}{5} \vec{j} $$ (a) At a glance, how do you know this is wrong? (b) What is the correct answer?

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