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Chapter 12: Problem 4

Find the four second partial derivatives of \(f(x, y)=3 x^{2} y+x y^{3}\).

Short Answer

Expert verified
Answer: The four second partial derivatives of the function \(f(x,y) = 3x^2y + xy^3\) are: \(\frac{\partial^2 f}{\partial x^2} = 6y\), \(\frac{\partial^2 f}{\partial y^2} = 6x\), \(\frac{\partial^2 f}{\partial x \partial y} = 6y+6y^2\), and \(\frac{\partial^2 f}{\partial y \partial x} = 6x+6x^2\).

Step by step solution

01

Find the first partial derivatives of the function with respect to x and y

Differentiate the given function with respect to x and y separately: \(\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (3x^2y+xy^3)\) \(\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (3x^2y+xy^3)\)
02

Compute the first partial derivatives

After differentiating, we get the first partial derivatives as follows: \(\frac{\partial f}{\partial x} = 6xy+y^3\) \(\frac{\partial f}{\partial y} = 3x^2+3xy^2\)
03

Find the second partial derivatives

Now, differentiate the first partial derivatives again with respect to x and y separately: \(\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x} (6xy+y^3)\) \(\frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y} (3x^2+3xy^2)\) \(\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x} (3x^2+3xy^2)\) \(\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial}{\partial y} (6xy+y^3)\)
04

Compute the second partial derivatives

After differentiating the first partial derivatives, we get the second partial derivatives as follows: \(\frac{\partial^2 f}{\partial x^2} = 6y\) \(\frac{\partial^2 f}{\partial y^2} = 6x\) \(\frac{\partial^2 f}{\partial x \partial y} = 6y+6y^2\) \(\frac{\partial^2 f}{\partial y \partial x} = 6x+6x^2\)
05

Write down the final results

The four second partial derivatives of the function \(f(x, y) = 3x^2y+xy^3\) are: \(\frac{\partial^2 f}{\partial x^2} = 6y\) \(\frac{\partial^2 f}{\partial y^2} = 6x\) \(\frac{\partial^2 f}{\partial x \partial y} = 6y+6y^2\) \(\frac{\partial^2 f}{\partial y \partial x} = 6x+6x^2\)

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Key Concepts

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

Partial Differentiation
Partial differentiation is a technique used to find the derivative of a function with more than one variable. Unlike ordinary differentiation where we find the derivative of a function with respect to one variable, partial differentiation involves taking the derivative with respect to one variable at a time, while keeping the other variables constant.

In the context of our exercise, the function given is \(f(x, y)=3 x^{2} y+x y^{3}\). The partial derivatives with respect to \(x\) and \(y\) tell us how the function changes as we change only one of the variables, holding the other fixed. The notation \(\frac{\partial}{\partial x}\) and \(\frac{\partial}{\partial y}\) are used to represent partial differentiation with respect to \(x\) and \(y\), respectively. This concept plays a fundamental role in multivariable calculus, as it helps to understand the behavior of functions of several variables.
Multivariable Calculus
Multivariable calculus extends the principles of single-variable calculus to functions of several variables. It introduces students to concepts such as partial derivatives, multiple integrals, and vector calculus. Crucial to this field is the understanding of functions with more than one independent variable, just like our example \(f(x, y)\).

In the exercise, we encounter second partial derivatives, which are derived by differentiating the first partial derivatives once more with respect to either of the original variables. These second-order derivatives can reveal additional insights about the curvature and behavior of the function's graph. The mixed second partial derivatives, such as \(\frac{\partial^2 f}{\partial x \partial y}\) and \(\frac{\partial^2 f}{\partial y \partial x}\), help us understand how the function changes in a mixed scenario, changing both variables simultaneously. These concepts are foundational for more advanced studies in mathematics, physics, engineering, and economics.
Calculus Early Transcendentals
The phrase 'Calculus Early Transcendentals' refers to a calculus approach and also to textbooks, like the one from Stewart, that introduce transcendental functions early in the curriculum. Transcendental functions include exponential, logarithmic, and trigonometric functions. While our function \(f(x, y) = 3x^2y+xy^3\) does not include transcendental functions, the principles of taking derivatives remain similar.

These textbooks are designed to lead students through the journey of understanding calculus in a step-by-step method, integrating concepts such as limits, derivatives, and integrals. In the case of our exercise, the step-by-step approach aids students in understanding how to apply the rules of differentiation to find both first and second partial derivatives, which is a vital skill in applying calculus to real-world problems.

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Most popular questions from this chapter

In its many guises, the least squares approximation arises in numerous areas of mathematics and statistics. Suppose you collect data for two variables (for example, height and shoe size) in the form of pairs \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right)\) The data may be plotted as a scatterplot in the \(x y\) -plane, as shown in the figure. The technique known as linear regression asks the question: What is the equation of the line that "best fits" the data? The least squares criterion for best fit requires that the sum of the squares of the vertical distances between the line and the data points is a minimum. Let the equation of the best-fit line be \(y=m x+b,\) where the slope \(m\) and the \(y\) -intercept \(b\) must be determined using the least squares condition. First assume that there are three data points \((1,2),(3,5),\) and \((4,6) .\) Show that the function of \(m\) and \(b\) that gives the sum of the squares of the vertical distances between the line and the three data points is $$ \begin{aligned} E(m, b)=&((m+b)-2)^{2}+((3 m+b)-5)^{2} \\ &+((4 m+b)-6)^{2} \end{aligned}. $$ Find the critical points of \(E\) and find the values of \(m\) and \(b\) that minimize \(E\). Graph the three data points and the best-fit line.

A classical equation of mathematics is Laplace's equation, which arises in both theory and applications. It governs ideal fluid flow, electrostatic potentials, and the steadystate distribution of heat in a conducting medium. In two dimensions, Laplace's equation is $$\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0.$$ Show that the following functions are harmonic; that is, they satisfy Laplace's equation. $$u(x, y)=e^{a x} \cos a y, \text { for any real number } a$$

An important derivative operation in many applications is called the Laplacian; in Cartesian coordinates, for \(z=f(x, y),\) the Laplacian is \(z_{x x}+z_{y y} .\) Determine the Laplacian in polar coordinates using the following steps. a. Begin with \(z=g(r, \theta)\) and write \(z_{x}\) and \(z_{y}\) in terms of polar coordinates (see Exercise 64). b. Use the Chain Rule to find \(z_{x x}=\frac{\partial}{\partial x}\left(z_{x}\right) .\) There should be two major terms, which, when expanded and simplified, result in five terms. c. Use the Chain Rule to find \(z_{y y}=\frac{d}{\partial y}\left(z_{y}\right) .\) There should be two major terms, which, when expanded and simplified, result in five terms. d. Combine parts (b) and (c) to show that $$z_{x x}+z_{y y}=z_{r r}+\frac{1}{r} z_{r}+\frac{1}{r^{2}} z_{\theta \theta}$$

What point on the plane \(x-y+z=2\) is closest to the point (1,1,1)\(?\)

In its many guises, the least squares approximation arises in numerous areas of mathematics and statistics. Suppose you collect data for two variables (for example, height and shoe size) in the form of pairs \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right)\) The data may be plotted as a scatterplot in the \(x y\) -plane, as shown in the figure. The technique known as linear regression asks the question: What is the equation of the line that "best fits" the data? The least squares criterion for best fit requires that the sum of the squares of the vertical distances between the line and the data points is a minimum. Generalize the procedure in Exercise 70 by assuming that \(n\) data points \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right)\) are given. Write the function \(E(m, b)\) (summation notation allows for a more compact calculation). Show that the coefficients of the best-fit line are $$ \begin{aligned} m &=\frac{\left(\sum x_{k}\right)\left(\sum y_{k}\right)-n \sum x_{k} y_{k}}{\left(\sum x_{k}\right)^{2}-n \sum x_{k}^{2}} \text { and } \\ b &=\frac{1}{n}\left(\sum y_{k}-m \Sigma x_{k}\right) \end{aligned}, $$ where all sums run from \(k=1\) to \(k=n\).
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