/*! 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 10 If \(z=x+f(u),\) where \(u=x y,\... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If \(z=x+f(u),\) where \(u=x y,\) show that $$x \frac{\partial z}{\partial x}-y \frac{\partial z}{\partial y}=x$$

Short Answer

Expert verified
The given expression is true; the terms cancel leaving \(x\).

Step by step solution

01

Define the problem and given expressions

We have the expression \( z = x + f(u) \) where \( u = xy \). We are asked to show that \( x \frac{\partial z}{\partial x} - y \frac{\partial z}{\partial y} = x \).
02

Calculate the partial derivative of z with respect to x

First, differentiate \( z = x + f(u) \) with respect to \( x \). Using the chain rule, we get:\[\frac{\partial z}{\partial x} = 1 + \frac{df}{du} \cdot \frac{\partial u}{\partial x}\]Since \( u = xy \), \( \frac{\partial u}{\partial x} = y \), so:\[\frac{\partial z}{\partial x} = 1 + \frac{df}{du} \cdot y\]
03

Calculate the partial derivative of z with respect to y

Now, differentiate \( z = x + f(u) \) with respect to \( y \). Again, using the chain rule, we have:\[\frac{\partial z}{\partial y} = \frac{df}{du} \cdot \frac{\partial u}{\partial y}\]Since \( u = xy \), \( \frac{\partial u}{\partial y} = x \), thus:\[\frac{\partial z}{\partial y} = \frac{df}{du} \cdot x\]
04

Substitute the expressions into the given equation

Substitute the expressions from Steps 2 and 3 into the given equation:\[x \left( 1 + \frac{df}{du} \cdot y \right) - y \left( \frac{df}{du} \cdot x \right)\]Simplify the expression:\[x + x \cdot y \cdot \frac{df}{du} - y \cdot x \cdot \frac{df}{du} = x\]
05

Simplify the resulting expression

Notice that the terms \( x \cdot y \cdot \frac{df}{du} \) and \( - y \cdot x \cdot \frac{df}{du} \) cancel each other out:\[x + 0 = x\]Thus, we have shown that:\[x \frac{\partial z}{\partial x} - y \frac{\partial z}{\partial y} = x\]

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.

Multivariable Calculus
Multivariable calculus is a branch of mathematics that extends the principles of calculus to functions of multiple variables. Unlike single-variable calculus, which deals with functions governed by a single independent variable, multivariable calculus investigates functions dependent on several inputs. This expansion helps to explore complex systems that occur in areas like physics, engineering, and economics.
Multivariable calculus encompasses several key concepts:
  • Partial Derivatives: Represents how a function changes as each variable is independently varied. For a function of two variables, like our example with \( z = x + f(u) \), partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) describe the rate of change of \( z \) with respect to \( x \) and \( y \), respectively.
  • Gradient: A vector containing all the partial derivatives of a function, which points in the direction of greatest increase of the function.
  • Multiple Integrals: Allow integration over regions in space, generalizing the concept of a single integral over an interval.
In our exercise, calculating partial derivatives helps identify how changing one variable affects the output, proving crucial in solving the given problem.
Chain Rule
The chain rule is a fundamental theorem in calculus used for finding the derivative of a composite function. In the context of multivariable calculus, the chain rule adapts to deal with functions where variables themselves are tied to other variables. This linkage allows for differentiation in more intricate settings, as seen in our problem where \( u = xy \) is a function of \( x \) and \( y \).
Using the chain rule involves several steps:
  • Differentiate the outer function while treating the inner function as a constant. In our case, \( z = x + f(u) \).
  • Multiply this result by the derivative of the inner function with respect to your variable of interest. This means finding \( \frac{df}{du} \cdot \frac{\partial u}{\partial x} \) when differentiating with respect to \( x \) and \( \frac{df}{du} \cdot \frac{\partial u}{\partial y} \) when differentiating with respect to \( y \).
In practice, this rule simplifies the process of finding derivatives for complex functions, enabling the solution of systems reliant on interconnected variables.
Differentiation
Differentiation is the process of finding the derivative of a function, which measures how a function's output changes as its inputs change. In the realm of multivariable calculus, differentiation extends to functions involving more than one input, requiring methods like partial differentiation and employing rules such as the chain rule.
Differentiation processes include:
  • Ordinary Derivatives: Determines the rate of change of a function concerning a single variable.
  • Partial Derivatives: Essential for dealing with functions with multiple inputs. They express the function's change concerning one variable at a time, keeping others constant.
Particularly in our exercise, differentiation demonstrates how alterations in \( x \) and \( y \) independently influence the value of \( z \). By calculating these individual effects using partial derivatives, we established the relationship outlined, meeting the problem's requirements. This analysis highlights differentiation's significance in understanding the sensitivity and behavior of multivariable functions.

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

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)? $$\begin{aligned} &f(x, y)=5 x^{6}+18 x^{5}-30 x^{4}+30 x y^{2}-120 x^{3},\\\ &-4 \leq x \leq 3,-2 \leq y \leq 2 \end{aligned}$$

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)? $$\begin{aligned} &f(x, y)=x^{4}+y^{2}-8 x^{2}-6 y+16, \quad-3 \leq x \leq 3,\\\ &-6 \leq y \leq 6 \end{aligned}$$

(A) find the function's domain, (b) find the function's range, (c) describe the function's level curves, (d) find the boundary of the function's domain, (e) determine if the domain is an open region, a closed region, or neither, and (f) decide if the domain is bounded or unbounded. $$f(x, y)=\ln \left(9-x^{2}-y^{2}\right)$$

Find the limits by rewriting the fractions first. $$\lim _{P \rightarrow(2,-3,6)} \ln \sqrt{x^{2}+y^{2}+z^{2}}$$

Show that each function satisfies a Laplace equation. \(f(x, y, z)=\left(x^{2}+y^{2}+z^{2}\right)^{-1 / 2}\)

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.