/*! 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 50 Verify that \(w_{x y}=w_{y x}\).... [FREE SOLUTION] | 91Ó°ÊÓ

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Verify that \(w_{x y}=w_{y x}\). \(w=x \sin y+y \sin x+x y\)

Short Answer

Expert verified
Yes, \(w_{xy} = w_{yx}\), so the equality is verified.

Step by step solution

01

Determine First Partial Derivative with Respect to x

To find the partial derivative of \(w\) with respect to \(x\), treat \(y\) as a constant. The partial derivative is obtained by differentiating \(x \sin y + y \sin x + xy\) term by term. - The derivative of \(x \sin y\) with respect to \(x\) is \(\sin y\), as \(\sin y\) is treated as a constant.- The derivative of \(y \sin x\) with respect to \(x\) is \(y \cos x\).- The derivative of \(xy\) with respect to \(x\) is \(y\).So, \(w_x = \sin y + y \cos x + y\).
02

Determine the Partial Derivative \(w_{xy}\)

Now, we'll differentiate the expression for \(w_x = \sin y + y \cos x + y\) with respect to \(y\).- The derivative of \(\sin y\) with respect to \(y\) is \(\cos y\).- The derivative of \(y \cos x\) with respect to \(y\) is \(\cos x\).- The derivative of \(y\) with respect to \(y\) is 1.Thus, \(w_{xy} = \cos y + \cos x + 1\).
03

Determine First Partial Derivative with Respect to y

To find the partial derivative of \(w\) with respect to \(y\), treat \(x\) as a constant. The partial derivative is obtained by differentiating \(x \sin y + y \sin x + xy\) term by term.- The derivative of \(x \sin y\) with respect to \(y\) is \(x \cos y\).- The derivative of \(y \sin x\) with respect to \(y\) is \(\sin x\).- The derivative of \(xy\) with respect to \(y\) is \(x\).So, \(w_y = x \cos y + \sin x + x\).
04

Determine the Partial Derivative \(w_{yx}\)

Now, we'll differentiate the expression for \(w_y = x \cos y + \sin x + x\) with respect to \(x\).- The derivative of \(x \cos y\) with respect to \(x\) is \(\cos y\).- The derivative of \(\sin x\) with respect to \(x\) is \(\cos x\).- The derivative of \(x\) with respect to \(x\) is 1.Thus, \(w_{yx} = \cos y + \cos x + 1\).
05

Verify \(w_{xy} = w_{yx}\)

We have found: - \(w_{xy} = \cos y + \cos x + 1\)- \(w_{yx} = \cos y + \cos x + 1\)Since \(w_{xy} = w_{yx}\), the mixed partial derivatives are equal, verifying the given exercise.

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Key Concepts

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

Mixed Partial Derivatives
When working with multivariable calculus, mixed partial derivatives play an essential role. These arise when we take the partial derivatives of a function more than once, with respect to different variables. In other words, if you have a function that depends on multiple variables, you can differentiate it first with respect to one variable and then with respect to another.
This process gives us what we call mixed partial derivatives.
For example, consider a multivariable function, like the one in the exercise: \[ w = x \sin y + y \sin x + xy \]When calculating mixed partial derivatives such as \( w_{xy} \) and \( w_{yx} \), we say they are mixed because we first take a derivative with one variable and then a derivative with another.
A fundamental property of mixed partial derivatives is that if the function is well-behaved (meaning it is continuous and has continuous derivatives), these mixed partials should be equal, \( w_{xy} = w_{yx} \). This equality is known as the symmetry of second derivatives and is a result of Clairaut's theorem.
Function Differentiation
Function differentiation involves finding the derivative of a function concerning one of its variables. When the function depends on more than one variable, we use partial derivatives, treating all other variables as constants in the differentiation process.
Let's break it down for our given example:- To find the partial derivative of \( w \) with respect to \( x \), consider \( y \) as a constant and differentiate each term:
  • The derivative of \( x \sin y \) with respect to \( x \) is \( \sin y \)
  • The derivative of \( y \sin x \) with respect to \( x \) is \( y \cos x \)
  • The derivative of \( xy \) with respect to \( x \) is \( y \)
Thus, \( w_x = \sin y + y \cos x + y \).
Using similar steps, we differentiate with respect to \( y \) and find that: \( w_y = x \cos y + \sin x + x \).
These basic principles of differentiating functions allows us to handle and work with functions of several variables, opening up insights for deeper analysis.
Mathematical Verification
Mathematical verification is about confirming that a mathematical statement or result is true. In the context of this exercise, it means demonstrating that the mixed partial derivatives are indeed equal, showing \( w_{xy} = w_{yx} \). This verification involves calculating each derivative step-by-step and checking their equality.
After finding \( w_{xy} = \cos y + \cos x + 1 \) and \( w_{yx} = \cos y + \cos x + 1 \), we notice they are identical. This confirms the mathematical principle that, under typical conditions for smooth functions, mixed partial derivatives should be equal. This concept is rooted in the assumptions of differentiability and continuity that often apply in real-world scenario functions.
Verification is not just a repetitive calculation but an essential check to ensure our mathematical expressions and results meet the underlying rules and principles of calculus. Ensuring accuracy in such calculations supports stability and correctness in theoretical work as well as practical applications.

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