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Use the method of Example 6 to find \(\partial w / \partial x\) and \(\partial w / \partial y\) as functions of \(x\) and \(y\). $$ w=u^{2}+v^{2}+x^{2}+y^{2} ; u=x-y, v=x+y $$

Short Answer

Expert verified
\( \frac{\partial w}{\partial x} = 6x \) and \( \frac{\partial w}{\partial y} = 6y \).

Step by step solution

01

Express w in terms of u and v

First, substitute the given expressions for \(u\) and \(v\) into the expression for \(w\). We have \(w = u^2 + v^2 + x^2 + y^2\), where \(u = x - y\) and \(v = x + y\).
02

Substitute the expressions for u and v

Substitute \(u = x-y\) and \(v = x+y\) into the formula for \(w\):\[ w = (x - y)^2 + (x + y)^2 + x^2 + y^2 \].
03

Simplify the expression for w

Expand and simplify the expression obtained from substitution:\[ w = (x^2 - 2xy + y^2) + (x^2 + 2xy + y^2) + x^2 + y^2 \].Combine like terms to get: \[ w = 3x^2 + 3y^2 \].
04

Differentiate w with respect to x

Differentiate the simplified expression for \(w\) with respect to \(x\):\( \frac{\partial w}{\partial x} = \frac{d}{dx}(3x^2) = 6x \).
05

Differentiate w with respect to y

Differentiate the simplified expression for \(w\) with respect to \(y\):\( \frac{\partial w}{\partial y} = \frac{d}{dy}(3y^2) = 6y \).

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Key Concepts

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

Multivariate Calculus
Multivariate calculus is a branch of mathematics that deals with functions of more than one variable. In our problem, we're dealing with a function \( w \) of four variables: \( u, v, x, \) and \( y \). To fully grasp multivariate calculus, it's important to understand that:
  • We can have functions dependent on multiple independent variables.
  • The function's value can change in many directions or 'dimensions'.
  • It allows us to analyze and describe the change in functions when multiple inputs are varied simultaneously.
For instance, in our exercise, the function \( w = u^2 + v^2 + x^2 + y^2 \) incorporates multiple variables to determine the output. By substituting \( u = x - y \) and \( v = x + y \), we revise \( w \) in terms of \( x \) and \( y \) to make the function easier to differentiate with respect to these variables. Revisiting how these variables interact showcases the heart of multivariate calculus.
Differentiation
Differentiation is a fundamental concept in calculus that describes how a function's value changes as its inputs are varied. When dealing with partial derivatives, which are a form of differentiation in multivariate calculus, we focus on how a function changes with respect to one input variable while holding the others constant. This process involves the following steps:
  • Identify the variable with respect to which you want to differentiate.
  • Treat all other variables as constants during differentiation.
  • Apply basic rules of differentiation, such as the power rule.
In our solution, we took the expression \( w = 3x^2 + 3y^2 \), already simplified from our substitutions, and differentiated it with respect to \( x \) and \( y \). This gave us the partial derivatives \( \frac{\partial w}{\partial x} = 6x \) and \( \frac{\partial w}{\partial y} = 6y \). Differentiation here provides insights into how the change in \( x \) or \( y \) individually influences the function \( w \).
Chain Rule
The chain rule is a key tool in calculus that allows us to differentiate composite functions. It's especially handy in multivariate contexts where a function depends on several interrelated variables. While our particular exercise involved straightforward substitutions without explicitly using the chain rule, understanding its potential application can be very useful:
  • The chain rule enables the differentiation of functions of functions.
  • It involves taking derivatives of each layer of the function separately and then combining the results.
  • Helps to handle more complex cases where variables are interdependent.
Imagine if instead of directly substituting \( u = x - y \) and \( v = x + y \), we had a function \( w \) that was more tightly interwoven with \( x \) and \( y \). The chain rule would help us take partial derivatives of each inner function and then link them together. This approach clarifies changes in nested relationships and how alterations ripple through a system, which is essential in more advanced problems.

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