/*! 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 41 Find all the second-order partia... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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: \( \frac{\partial^2 f}{\partial x^2} = 0 \), \( \frac{\partial^2 f}{\partial y^2} = 0 \), \( \frac{\partial^2 f}{\partial y \partial x} = 1 \), \( \frac{\partial^2 f}{\partial x \partial y} = 1 \).

Step by step solution

01

Find the first partial derivative with respect to x

To find the partial derivative of the function \( f(x, y) = x + y + xy \) with respect to \( x \), treat \( y \) as a constant. The partial derivative is obtained by differentiating each term separately: \( \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x) + \frac{\partial}{\partial x}(y) + \frac{\partial}{\partial x}(xy) = 1 + 0 + y = 1 + y \).
02

Find the first partial derivative with respect to y

To find the partial derivative of the function \( f(x, y) = x + y + xy \) with respect to \( y \), treat \( x \) as a constant. The partial derivative is obtained by differentiating each term separately:\( \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x) + \frac{\partial}{\partial y}(y) + \frac{\partial}{\partial y}(xy) = 0 + 1 + x = 1 + x \).
03

Find the second partial derivative with respect to x, twice

To find the second partial derivative of \( f \) with respect to \( x \) twice, differentiate \( \frac{\partial f}{\partial x} = 1 + y \) with respect to \( x \) again. Both terms are constants with respect to \( x \):\( \frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}(1 + y) = 0 \).
04

Find the mixed second partial derivative with respect to x then y

To find the mixed second partial derivative of \( f \), first with respect to \( x \) and then \( y \), differentiate \( \frac{\partial f}{\partial x} = 1 + y \) with respect to \( y \):\( \frac{\partial^2 f}{\partial y \partial x} = \frac{\partial}{\partial y}(1 + y) = 0 + 1 = 1 \).
05

Find the mixed second partial derivative with respect to y then x

To find the mixed second partial derivative of \( f \), first with respect to \( y \) and then \( x \), differentiate \( \frac{\partial f}{\partial y} = 1 + x \) with respect to \( x \):\( \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x}(1 + x) = 0 + 1 = 1 \).
06

Find the second partial derivative with respect to y, twice

To find the second partial derivative of \( f \) with respect to \( y \) twice, differentiate \( \frac{\partial f}{\partial y} = 1 + x \) with respect to \( y \) again. Both terms are constants with respect to \( y \):\( \frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y}(1 + x) = 0 \).

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 91Ó°ÊÓ!

Key Concepts

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

Partial Derivatives
A partial derivative represents the rate at which a function changes as one of the variables changes while all other variables are held constant. It's like zooming into a multi-dimensional surface and observing how it tilts or inclines in the direction of one specific axis. For example, consider a function with two variables, like our case with the function \(f(x, y) = x + y + xy\). When we take the partial derivative with respect to \(x\), denoted as \( \frac{\partial f}{\partial x} \), we are observing how the function changes as \(x\) changes, keeping \(y\) constant.
To find \( \frac{\partial f}{\partial x} \), differentiate each term regarding \(x\). We treat \(y\) as if it were a numerical constant, leading to the derivative \( \frac{\partial}{\partial x}(x) = 1 \), \( \frac{\partial}{\partial x}(y) = 0 \), and \( \frac{\partial}{\partial x}(xy) = y \). Adding these results, we obtain \( \frac{\partial f}{\partial x} = 1 + y \).
For the partial derivative with respect to \(y\), \( \frac{\partial f}{\partial y} \), treat \(x\) as constant. The derivative of \(x\) concerning \(y\) is zero because it's not changing, \( \frac{\partial}{\partial y}(y) = 1 \), and \( \frac{\partial}{\partial y}(xy) = x \), summing to \( \frac{\partial f}{\partial y} = 1 + x \). Partial derivatives help us in multiple ways, such as in optimization problems and studying the behavior of functions.
Multivariable Calculus
Multivariable calculus extends topics of single-variable calculus into higher dimensions, dealing with functions of multiple variables. This is crucial for modeling and analyzing complex systems across various fields, from physics to engineering to economics. For a function of more than one variable, such as \( f(x, y) \), calculus becomes multidimensional, incorporating vectors and spaces.
In the context of our problem, multivariable calculus is about handling expressions like \( x + y + xy \), where each of the variables \(x\) and \(y\) can change independently. It's like dealing with a vast terrain where the elevation (value of the function) shifts based on your path and direction.
The heart of multivariable calculus lies in concepts like:
  • Directional Derivatives and Gradients: These indicate the steepest ascent of a function.
  • Tangent Planes: Generalizing the idea of a tangent line to a plane.
  • Multiple Integrals: Extending the notion of integrals to multiple dimensions, used in volume calculations.
Understanding multivariable calculus enables solving real-world problems where multiple factors change simultaneously, offering powerful tools for analysis and prediction.
Mathematical Functions
Mathematical functions are the backbone of calculus, serving as expressions that map inputs to outputs. For our function \( f(x, y) = x + y + xy \), it assigns a specific output (also called an output value or the value of the function) for each input pair \((x, y)\).
Functions can be understood in various ways:
  • Graphically: Visualizing how the function behaves in a coordinate plane.
  • Numerical Tables: Listing input-output pairs.
  • Analytical Expressions: Like our current function, presenting a formula.
Mathematical functions can perform several types of operations and transformations, such as addition and multiplication, seen in \(x + y + xy\), affecting the geometry of the function's graph. Functions of two variables can represent surfaces in three-dimensional spaces, where every point on a surface corresponds to a particular \((x, y)\) pair.
Understanding functions is crucial as it provides insight into how different inputs affect the output, essential for optimization, modeling dynamic systems, and solving equations. By exploring second-order partial derivatives, we've delved into the changes in these functions when slightly altering inputs, a cornerstone of deeper calculus applications.

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

Let \(f(x, y)=x^{2}+y^{3} .\) Find the slope of the line tangent to this surface at the point \((-1,1)\) and lying in the a. plane \(x=-1\) b. plane \(y=1 .\)

Find the linearizations \(L(x, y, z)\) of the functions at the given points. $$ f(x, y, z)=\sqrt{x^{2}+y^{2}+z^{2}} \quad \text {at} $$ $$ \text { a. }(1,0,0) \quad \text { b. }(1,1,0) \quad \text { c. }(1,2,2) $$

Three variables Let \(w=f(x, y, z)\) be a function of three independent variables and write the formal definition of the partial derivative \(\partial f / \partial z\) at \(\left(x_{0}, y_{0}, z_{0}\right) .\) Use this definition to find \(\partial f / \partial z\) at \((1,2,3)\) for \(f(x, y, z)=x^{2} y z^{2}\)

You will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level curves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. d. Calculate the function's second partial derivatives and find the discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\) . e. Using the max-min tests, classify the critical points found in part (c). Are your findings consistent with your discussion in part (c)? $$f(x, y)=x^{2}+y^{3}-3 x y, \quad-5 \leq x \leq 5, \quad-5 \leq y \leq 5$$

In Exercises \(57-60,\) use the limit definition of partial derivative to compute the partial derivatives of the functions at the specified points. $$ f(x, y)=4+2 x-3 y-x y^{2}, \quad \frac{\partial f}{\partial x} \quad \text { and } \quad \frac{\partial f}{\partial y} \quad \text { at }(-2,1) $$

See all solutions

Recommended explanations on Math Textbooks

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.