/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 Find the four second partial der... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 31

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

Short Answer

Expert verified
Question: Find the four second partial derivatives of the function $$f(x, y) = x^2y^3$$. Answer: The four second partial derivatives of the function $$f(x, y) = x^2y^3$$ are: 1. $$\frac{\partial^2 f}{\partial x^2} = 2y^3$$ 2. $$\frac{\partial^2 f}{\partial y^2} = 6x^2y$$ 3. $$\frac{\partial^2 f}{\partial x\partial y} = 6xy^2$$ 4. $$\frac{\partial^2 f}{\partial y\partial x} = 6xy^2$$

Step by step solution

01

Find the first partial derivatives

To find the second partial derivatives, first find the first partial derivatives for the given function. Using the power rule for differentiation, we can differentiate the function with respect to x and y: $$\frac{\partial f}{\partial x} = 2xy^3$$ $$\frac{\partial f}{\partial y} = 3x^2y^2$$
02

Find the second partial derivatives

Now, we can calculate the second partial derivatives by differentiating the first partial derivatives in step 1: 1. Second partial derivative with respect to x twice: $$\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}\left(2xy^3\right) = 2y^3$$ 2. Second partial derivative with respect to y twice: $$\frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y}\left(3x^2y^2\right) = 6x^2y$$ 3. Second partial derivative with respect to x, then y: $$\frac{\partial^2 f}{\partial x\partial y} = \frac{\partial}{\partial x}\left(3x^2y^2\right) = 6xy^2$$ 4. Second partial derivative with respect to y, then x: $$\frac{\partial^2 f}{\partial y\partial x} = \frac{\partial}{\partial y}\left(2xy^3\right) = 6xy^2$$ In conclusion, the four second partial derivatives of the function $$f(x, y) = x^2y^3$$ are: 1. $$\frac{\partial^2 f}{\partial x^2} = 2y^3$$ 2. $$\frac{\partial^2 f}{\partial y^2} = 6x^2y$$ 3. $$\frac{\partial^2 f}{\partial x\partial y} = 6xy^2$$ 4. $$\frac{\partial^2 f}{\partial y\partial x} = 6xy^2$$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Partial Differentiation
Partial differentiation is a fundamental technique used in multivariable calculus to find the derivative of a function with respect to one of its variables, while holding the other variables constant. It is like taking the derivative in single-variable calculus, but here, since we have more than one variable, we differentiate with respect to one variable at a time. For example, if we have a function like \( f(x, y) = x^2y^3 \), to take the partial derivative with respect to \( x \), we treat \( y \) as a constant and only differentiate with respect to \( x \), resulting in \( \frac{\partial f}{\partial x} = 2xy^3 \). Similarly, taking the partial derivative with respect to \( y \) while considering \( x \) as a constant gives us \( \frac{\partial f}{\partial y} = 3x^2y^2 \).
Multivariable Calculus
Multivariable calculus extends the concepts of single variable calculus to functions with more than one variable. It involves operations such as differentiation and integration on these multivariable functions. In our example, \( f(x, y) = x^2y^3 \), the function is defined by two variables \( x \) and \( y \). The goal is to understand how the function changes in response to changes in both \( x \) and \( y \). Multivariable calculus helps in finding the direction and rate of fastest increase or decrease of such functions, optimizing them under constraints, and much more, which is crucial in fields such as engineering, physics, and economics.
Power Rule for Differentiation
The power rule is a basic rule in differentiation that can be used to find the derivative of a function that is the product of a variable raised to a power. In simpler terms, for a single-variable function \( f(x) = x^n \), the power rule states that the derivative of \( f \) with respect to \( x \) is \( f'(x) = nx^{n-1} \). This rule also applies to partial differentiation. When applying the power rule to partial differentiation, as done in the exercise, you differentiate term by term, treating any variable not being differentiated with respect to as a constant.
Mixed Partial Derivatives
Mixed partial derivatives refer to the partial derivatives that are taken with respect to more than one variable in sequence. For instance, if we have a function \( f(x, y) \), the mixed partial derivative first with respect to \( x \) and then with respect to \( y \) is notated as \( \frac{\partial^2 f}{\partial x\partial y} \), while first with respect to \( y \) and then \( x \) is \( \frac{\partial^2 f}{\partial y\partial x} \). The Clairaut's theorem in multivariable calculus usually ensures that if the function and its partial derivatives are continuous, the order of differentiation does not matter, so \( \frac{\partial^2 f}{\partial x\partial y} \) will equal \( \frac{\partial^2 f}{\partial y\partial x} \) most of the time, as seen in our example where both mixed partial derivatives yield the same result, \( 6xy^2 \).

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose you follow the spiral path \(C: x=\cos t, y\) \(=\sin t, z=\) \(t,\) for \(t \geq 0,\) through the domain of the function \(w=f(x, y, z)=x y z /\left(z^{2}+1\right)\) a. Find \(w^{\prime}(t)\) along \(C\) b. Estimate the point \((x, y, z)\) on \(C\) at which \(w\) has its maximum value.

Let \(R\) be a closed bounded region in \(\mathbb{R}^{2}\) and let \(f(x, y)=a x+b y+c,\) where \(a, b\) and \(c\) are real numbers, with \(a\) and \(b\) not both zero. Give a geometrical argument explaining why the absolute maximum and minimum values of \(f\) over \(R\) occur on the boundaries of \(R\).

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 \(k=n\).

Use the definition of differentiability to prove that the following functions are differentiable at \((0,0) .\) You must produce functions \(\varepsilon_{1}\) and \(\varepsilon_{2}\) with the required properties. $$f(x, y)=x+y$$

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

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.