Chapter 13: Problem 25
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.
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.
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.