/*! 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 32 For each function, find the seco... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

For each function, find the second-order partials a. \(f_{x x}\), b. \(f_{x y}\), c. \(f_{y x}\), and \(\mathrm{d} . f_{y y}\). \(f(x, y)=y \ln x+x e^{y}\)

Short Answer

Expert verified
\( f_{xx} = -\frac{y}{x^2} \), \( f_{xy} = \frac{1}{x} \), \( f_{yx} = \frac{1}{x} \), \( f_{yy} = x e^{y} \).

Step by step solution

01

Calculate the First-order Partial Derivative with respect to x

Differentiate the function \( f(x, y) = y \ln x + x e^{y} \) with respect to \( x \) to find \( f_x \).\[ f_x = \frac{\partial}{\partial x}(y \ln x + x e^{y}) = \frac{y}{x} + e^{y} \]
02

Calculate the First-order Partial Derivative with respect to y

Differentiate the function \( f(x, y) = y \ln x + x e^{y} \) with respect to \( y \) to find \( f_y \).\[ f_y = \frac{\partial}{\partial y}(y \ln x + x e^{y}) = \ln x + x e^{y} \]
03

Calculate the Second-order Partial Derivative \( f_{xx} \)

Differentiate \( f_x = \frac{y}{x} + e^{y} \) with respect to \( x \) to find \( f_{xx} \).\[ f_{xx} = \frac{\partial}{\partial x}\left(\frac{y}{x} + e^{y}\right) = -\frac{y}{x^2} \]
04

Calculate the Second-order Cross Partial Derivative \( f_{xy} \)

Differentiate \( f_x = \frac{y}{x} + e^{y} \) with respect to \( y \) to find \( f_{xy} \).\[ f_{xy} = \frac{\partial}{\partial y}\left(\frac{y}{x} + e^{y}\right) = \frac{1}{x} \]
05

Calculate the Second-order Cross Partial Derivative \( f_{yx} \)

Differentiate \( f_y = \ln x + x e^{y} \) with respect to \( x \) to find \( f_{yx} \).\[ f_{yx} = \frac{\partial}{\partial x}\left(\ln x + x e^{y}\right) = \frac{1}{x} \]
06

Calculate the Second-order Partial Derivative \( f_{yy} \)

Differentiate \( f_y = \ln x + x e^{y} \) with respect to \( y \) to find \( f_{yy} \).\[ f_{yy} = \frac{\partial}{\partial y}\left(\ln x + x e^{y}\right) = x e^{y} \]

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.

Second-order partial derivatives
Second-order partial derivatives are an extension of first-order derivatives and play an essential role in examining how a function behaves. When we have a function with more than one variable, such as \(f(x, y)\), we can consider how this function changes as each variable changes. This leads us to first-order partial derivatives, like \(f_x\) and \(f_y\), which measure the rate of change of the function with respect to each variable independently.

Moving further, we calculate second-order partial derivatives, which involve differentiating the first-order partial derivatives again. For example:
  • \(f_{xx}\) is the derivative of \(f_x\) with respect to \(x\) again.
  • \(f_{yy}\) is the derivative of \(f_y\) with respect to \(y\) again.
  • The mixed partial derivatives \(f_{xy}\) and \(f_{yx}\) involve differentiating \(f_x\) with respect to \(y\) and \(f_y\) with respect to \(x\), respectively.
Second-order partial derivatives give us information about the concavity and interaction effects between the variables, making them a vital tool in multivariable calculus.
Differentiation
Differentiation is a fundamental concept in calculus, referring to the process of finding the derivative of a function. The derivative tells us how a function changes at any given point. In the context of functions with multiple variables, partial differentiation is used.

Partial differentiation differs from ordinary differentiation because it focuses on one variable at a time while treating other variables as constants. This is especially useful for analyzing functions like \(f(x, y) = y \ln x + x e^{y}\), where both \(x\) and \(y\) influence the function's behavior. By obtaining partial derivatives such as \(f_x = \frac{y}{x} + e^{y}\) and \(f_y = \ln x + x e^{y}\), we can understand how slight changes in \(x\) or \(y\) impact the function's value.

Once partial differentiation is mastered, you can extend these principles to higher-order derivatives, revealing deeper insights into the geometry of these multivariable functions.
Multivariable calculus
Multivariable calculus is an extension of single-variable calculus to functions of several variables. It opens new dimensions (literally) in mathematical analysis, allowing us to study surfaces and shapes that are influenced by more than one factor.

This branch of calculus introduces partial derivatives, multiple integrals, and concepts like gradient, divergence, and curl. It is integral to understanding how multi-faceted functions behave, especially in physics, engineering, and economics.

In our example function, \(f(x, y) = y \ln x + x e^{y}\), multivariable calculus lets us explore the terrain of the function’s surface. By utilizing tools such as second-order partial derivatives, one can predict and visualize how small changes in \(x\) or \(y\) will alter the outcome. The knowledge of multivariable calculus is crucial for optimizing functions, as it provides strategies for locating maxima and minima in this expanded framework.

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

Solve each using Lagrange multipliers. (The stated extreme values do exist.) For each Cobb-Douglas production function \(P\) and isocost line (budget constraint, in dollars), find the amounts of labor \(L\) and capital \(K\) that maximize production, and also find the maximum production. Then evaluate and give an interpretation for \(|\lambda|\) and use it to answer the question. a. Maximize \(P=280 L^{3 / 4} K^{1 / 4}\) with budget constraint \(525 L+0.28 K=16,800\). b. Evaluate and give an interpretation for \(|\lambda|\). c. Approximate the increase in production if the budget is increased by \(\$ 300\).

To estimate heating and air conditioning costs, it is necessary to know the volume of a building. A conference center has a curved roof whose height is \(f(x, y)=40-0.006 x^{2}+0.003 y^{2} .\) The building sits on a rectangle extending from \(x=-50\) to \(x=50\) and \(y=-100\) to \(y=100\). Use integration to find the volume of the building. (All dimensions are in feet.)

Solve each using Lagrange multipliers. (The stated extreme values do exist.) Maximize \(f(x, y, z)=x+y+z\) subject to \(x^{2}+y^{2}+z^{2}=12\)

For the given function and values, find: a. \(\Delta f \quad\) b. \(d f\) $$ \begin{array}{l} f(x, y, z)=x y+z^{2}, x=3, \quad \Delta x=d x=0.03 \\ y=2, \quad \Delta y=d y=0.02, z=1, \Delta z=d z=0.01 \end{array} $$

GENERAL: Windchill The windchill index announced during the winter by the weather bureau measures how cold it "feels" for a given temperature \(t\) (in degrees Fahrenheit) and wind speed \(w\) (in miles per hour). It is calculated by the formula \(C(t, w)=35.74+0.6215 t-35.75 w^{0.16}+\) \(0.4275 t w^{0.16} .\) If the temperature is 30 degrees and the wind speed is 10 miles per hour, estimate the change in the windchill temperature if the wind speed increases by 4 miles per hour and the temperature drops by 5 degrees.

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.