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Chapter 22: Problem 5
If \(z=f(x-y, y-x)\) show that \((\partial z / \partial x)+(\partial z / \partialy)=0\).
Short Answer
Expert verified
Using the chain rule for partial derivatives to compute \((\partial z / \partial x)\) and \((\partial z / \partial y)\), and then adding them shows that the terms cancel each other, proving that \((\partial z / \partial x) + (\partial z / \partial y) = 0\).
Step by step solution
01
Understand partial derivatives
Understand that partial derivatives measure how a function changes as its input variables are changed individually, holding the other variables constant.
02
Apply the chain rule
To find \((\partial z / \partial x)\) and \((\partial z / \partial y)\), apply the chain rule for partial derivatives taking into account that \(z\) depends on \(x-y\) and \(y-x\).
03
Compute the partial derivative with respect to x
Compute the partial derivative of \(z\) with respect to \(x\) by taking the derivative of the inside function \(x-y\) which is 1 and the inside function \(y-x\) which is -1. Then, multiply each by their respective derivative of the outside function (\(f\)):\[\frac{\partial z}{\partial x} = \frac{\partial f}{\partial (x-y)} \cdot \frac{\partial (x-y)}{\partial x} + \frac{\partial f}{\partial (y-x)} \cdot \frac{\partial (y-x)}{\partial x} = \frac{\partial f}{\partial (x-y)} \cdot 1 - \frac{\partial f}{\partial (y-x)} \cdot 1\]
04
Compute the partial derivative with respect to y
Similarly, compute the partial derivative of \(z\) with respect to \(y\) by differentiating the inside functions \(x-y\) which is -1 and \(y-x\) which is 1. Then, multiply each by their respective derivative of the outside function:\[\frac{\partial z}{\partial y} = \frac{\partial f}{\partial (x-y)} \cdot \frac{\partial (x-y)}{\partial y} + \frac{\partial f}{\partial (y-x)} \cdot \frac{\partial (y-x)}{\partial y} = -\frac{\partial f}{\partial (x-y)} \cdot 1 + \frac{\partial f}{\partial (y-x)} \cdot 1\]
05
Add the partial derivatives
Add the partial derivatives \((\partial z / \partial x)\) and \((\partial z / \partial y)\):\[\frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = (\frac{\partial f}{\partial (x-y)} - \frac{\partial f}{\partial (y-x)}) + (-\frac{\partial f}{\partial (x-y)} + \frac{\partial f}{\partial (y-x)}) = 0\]The terms cancel each other out, proving that \((\partial z / \partial x)+ (\partial z / \partial y)=0\).
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule
When dealing with functions that are composed of other functions, the chain rule is an essential tool. It's a formula used to compute the derivative of a composite function. In the context of partial derivatives and multivariable calculus, the chain rule lets us understand how a function behaves with respect to changes in each variable independently.
For instance, if you have a function like z = f(u, v), where both u and v are functions of x and y, applying the chain rule lets you find how z changes when you change either x or y. It involves taking the derivatives of u and v with respect to x and y, and then using those to find the partial derivatives of z.
This principle makes it easier to deal with complex scenarios where variables are interdependent and leads to a deeper understanding of the function behavior in multiple dimensions.
Function Behavior
Understanding the behavior of functions in calculus is integral to predicting how changes in input values affect the output. This is particularly true in the realm of multivariable calculus, where functions can be dependent on two or more variables.
In our exercise example, the function z is defined in terms of two other variables that are expressions involving both x and y. A keen understanding of function behavior allows you to anticipate and describe how z is affected as x and y change, either in tandem or independently. Understanding that z remains constant if the change in x is accompanied by an equivalent and opposite change in y is a nuanced insight that calculus helps us deduce.
Computing Partial Derivatives
When you're computing partial derivatives, you're focusing on the rate of change of a function with respect to one variable at a time, while keeping other variables constant.
In our example, the computation is done by considering the composite nature of the function z = f(x-y, y-x) and how changes in x and y propagate through to z. By breaking it down into simpler parts and computing the rates of change step by step, we determine the overall derivative with respect to each variable. This process is not just arithmetic—it reflects the fundamental relationships between the quantities in the function and requires both algebraic manipulation and conceptual understanding.
Multivariable Calculus
Multivariable calculus expands the principles of calculus to functions of more than one variable. Unlike single-variable calculus, where we only consider the change along one axis, multivariable calculus allows us to explore spaces that are higher-dimensional.
This is where partial derivatives come in—they're derivatives of multivariable functions taken with respect to one variable at a time. In this context, the activity we engage in when we calculate partial derivatives or apply the chain rule represents the slicing of the multivariable function along one direction/variable, allowing us to understand and analyze the behavior of these higher-dimensional functions piece by piece. This division into simpler slices is crucial for problem-solving in fields like physics, engineering, and economics, where multivariable functions frequently come into play.
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