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If \(f(u, v, w)\) is differentiable and \(u=x-y, v=y-z,\) and \(w=z-x,\) show that \(\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z}=0\)

Short Answer

Expert verified
The sum of the partial derivatives is zero: \(\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} + \frac{\partial f}{\partial z} = 0\).

Step by step solution

01

Apply the Chain Rule for Multivariable Functions

First, note that the function \(f(u, v, w)\) is given in terms of the variables \(u, v,\) and \(w\), which are themselves functions of \(x, y,\) and \(z\). The chain rule in the context of multivariable differentiation tells us that \(\frac{\partial f}{\partial x}\), \(\frac{\partial f}{\partial y}\), and \(\frac{\partial f}{\partial z}\) can be calculated using:\[\frac{\partial f}{\partial x} = \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x} + \frac{\partial f}{\partial w} \cdot \frac{\partial w}{\partial x}\]Repeat similarly for \(\frac{\partial f}{\partial y}\) and \(\frac{\partial f}{\partial z}\).
02

Determine Partial Derivatives of Intermediate Variables

Determine the partial derivatives of \(u, v, w\) with respect to \(x, y, z\):\[\begin{align*}\frac{\partial u}{\partial x} &= 1, & \frac{\partial u}{\partial y} &= -1, & \frac{\partial u}{\partial z} &= 0, \\frac{\partial v}{\partial x} &= 0, & \frac{\partial v}{\partial y} &= 1, & \frac{\partial v}{\partial z} &= -1, \\frac{\partial w}{\partial x} &= -1, & \frac{\partial w}{\partial y} &= 0, & \frac{\partial w}{\partial z} &= 1.\end{align*}\]
03

Substitute and Simplify

Using the results from Step 2, substitute into the chain rule expressions:\[\begin{align*}\frac{\partial f}{\partial x} &= \frac{\partial f}{\partial u} \cdot 1 + \frac{\partial f}{\partial v} \cdot 0 + \frac{\partial f}{\partial w} \cdot (-1), \\frac{\partial f}{\partial y} &= \frac{\partial f}{\partial u} \cdot (-1) + \frac{\partial f}{\partial v} \cdot 1 + \frac{\partial f}{\partial w} \cdot 0, \\frac{\partial f}{\partial z} &= \frac{\partial f}{\partial u} \cdot 0 + \frac{\partial f}{\partial v} \cdot (-1) + \frac{\partial f}{\partial w} \cdot 1.\end{align*}\]
04

Add the Partial Derivatives

Add the three partial derivatives:\[\begin{align*}\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} + \frac{\partial f}{\partial z} = \left(\frac{\partial f}{\partial u} - \frac{\partial f}{\partial w}\right) + \left(-\frac{\partial f}{\partial u} + \frac{\partial f}{\partial v}\right) + \left(-\frac{\partial f}{\partial v} + \frac{\partial f}{\partial w}\right).\end{align*}\]Simplifying, observe that all terms cancel out:\[= 0.\]
05

Conclusion: Verify the Result

By the calculations, \(\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} + \frac{\partial f}{\partial z} = 0\), confirming the initial claim that these add up to zero.

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Key Concepts

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

Chain Rule
The chain rule is a fundamental tool in calculus, particularly for dealing with composite functions. It extends the concept of differentiation to functions where variables depend on one another. When we talk about the chain rule in the context of multivariable functions, we're considering a situation where a function like \(f(u, v, w)\) depends on several intermediate variables that in turn depend on other variables, such as \(x, y,\) and \(z\). This requires applying the chain rule across multiple paths of dependency.

When applying the chain rule for such multivariable functions, we express the derivative of the function with respect to one of the initial variables in terms of derivatives with respect to the intermediate variables. Specifically, if \(u, v,\) and \(w\) are functions of \(x, y,\) and \(z\), the chain rule allows us to find \(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y},\) and \(\frac{\partial f}{\partial z}\) in a structured way:
  • Each partial derivative is a sum of products, where each product involves a derivative of \(f\) with respect to one of \(u, v,\) or \(w\), multiplied by the corresponding derivative of that variable with respect to \(x, y,\) or \(z\).

  • This structured approach lets you multiply immediate changes in \(f\) with how the intermediate variables change with respect to your initial variables.
Following through these product sums systematically ensures that all possible impacts on \(f\) through \(u, v,\) and \(w\) are accounted for in the derivatives relative to \(x, y,\) and \(z\).
Multivariable Functions
Multivariable functions take more than one argument or input, meaning they depend on several variables. In our context, \(f(u, v, w)\) is a multivariable function because it is defined in terms of \(u, v,\) and \(w\), which themselves are functions of other variables \(x, y,\) and \(z\). Each of these variables provides a different dimension of input to the function.

Understanding multivariable functions involves recognizing how changes in one or more inputs can affect the output or value of the function. This dynamic is what makes studying partial derivatives key in multivariable calculus, as it allows you to isolate and understand the relationship between specific input variables and the resultant changes in the function.

Key aspects of dealing with multivariable functions include:
  • Domain and Range: The set of possible inputs (domain) and potential outputs (range) of the function needs to be understood.
  • Partial Derivatives: These help us explore how the function changes with respect to just one of its inputs at a time, holding others constant.
  • Interconnections: Often, such functions involve dependencies or interactions between different variables, which the chain rule helps us handle methodically.
These perspectives and tools let us harness the power of multivariable functions to model and interpret complex systems.
Differentiation
Differentiation is the process of finding the rate at which a function changes at any given point. In single-variable calculus, this is straightforward: you find the derivative concerning its variable. However, in the context of multivariable functions, differentiation extends into partial derivatives.

For functions depending on several variables, partial derivatives tell you how a function changes when you alter one variable while holding the others constant. This approach preserves the complexity of interdependent systems that multivariable calculus handles.

Consider how this applies to our original exercise. We derived how \(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y},\) and \(\frac{\partial f}{\partial z}\) were all calculated individually. By collecting all the impacts through partial derivatives and summing them, we were able to observe a significant result, where the derivatives sum to zero:
  • Insight into Change: This summed result provides a global insight into how the function is balanced across its defining variables.
  • Variable Interplay: The zero sum helps highlight how dependencies among \(x, y,\) and \(z\) create a nuanced balance within the function.
This process captures both the raw mechanics and deeper insights that differentiation in multivariable contexts can provide.

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Most popular questions from this chapter

Use a CAS to perform the following steps for each of the functions in Exercises \(69-72 .\) $$ \begin{array}{l}{\text { a. Plot the surface over the given rectangle. }} \\\ {\text { b. Plot several level curves in the rectangle. }} \\ {\text { c. Plot the level curve of } f \text { through the given point. }}\end{array} $$ $$ \begin{array}{l}{f(x, y)=x \sin \frac{y}{2}+y \sin 2 x, \quad 0 \leq x \leq 5 \pi, \quad 0 \leq y \leq 5 \pi} \\ {P(3 \pi, 3 \pi)}\end{array} $$

Three variables Let \(w=f(x, y, z)\) be a function of three independent variables and write the formal definition of the partial derivative \(\partial f / \partial y\) at \(\left(x_{0}, y_{0}, z_{0}\right) .\) Use this definition to find \(\partial f / \partial y\) at \((-1,0,3)\) for \(f(x, y, z)=-2 x y^{2}+y z^{2}\)

Two dependent variables Find \(\partial x / \partial u\) and \(\partial y / \partial u\) if the equations \(u=x^{2}-y^{2}\) and \(v=x^{2}-y\) define \(x\) and \(y\) as functions of the independent variables \(u\) and \(v,\) and the partial derivatives exist. (See the hint in Exercise \(69 . )\) Then let \(s=x^{2}+y^{2}\) and find \(\partial s / \partial u .\)

Display the values of the functions in Exercises \(37-48\) in two ways: (a) by sketching the surface \(z=f(x, y)\) and \((b)\) by drawing an assortment of level curves in the function's domain. Label each level curve with its function value. $$ f(x, y)=1-|y| $$

Suppose that the equation \(g(x, y, z)=0\) determines \(z\) as a differentiable function of the independent variables \(x\) and \(y\) and that \(g_{z} \neq 0 .\) Show that \begin{equation}\left(\frac{\partial z}{\partial y}\right)_{x}=-\frac{\partial g / \partial y}{\partial g / \partial z}.\end{equation}

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