Simplifying Partial Derivatives in Multivariable Calculus

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Homework Help Overview

The discussion centers around simplifying expressions involving partial derivatives in the context of multivariable calculus. Participants are examining two specific expressions and their simplifications, questioning the notation and the application of derivative rules.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to simplify two expressions involving partial derivatives but questions the correctness of their reasoning. Some participants suggest clarifying notation and the placement of parentheses to understand the expressions better. Others mention the need for the product rule in the simplification process.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the expressions and providing guidance on how to approach the simplifications. There is a focus on ensuring clarity in notation and the implications of derivative operations.

Contextual Notes

Participants note the importance of specifying how expressions are to be interpreted, particularly regarding the placement of parentheses and the use of test functions in the context of partial derivatives.

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Homework Statement



Simplify the following two expressions:

[tex]y\frac{\partial}{\partial z}z\frac{\partial}{\partial x}[/tex]

[tex]z\frac{\partial}{\partial y}x\frac{\partial}{\partial z}[/tex]


The Attempt at a Solution



for the first one: [tex]y\frac{\partial}{\partial z}z\frac{\partial}{\partial x}[/tex]

[tex]\frac{\partial}{\partial z}z = 1[/tex]

so therefore [tex]y\frac{\partial}{\partial z}z\frac{\partial}{\partial x}= y(1)\frac{\partial}{\partial x}= y\frac{\partial}{\partial x}[/tex]

How come this is incorrect?




For the second one: [tex]z\frac{\partial}{\partial y}x\frac{\partial}{\partial z}[/tex]

I cannot write that [tex]z\frac{\partial}{\partial y}x\frac{\partial}{\partial z}= x\frac{\partial}{\partial z}z\frac{\partial}{\partial y}[/tex]

so therefore [tex]z\frac{\partial}{\partial y}x\frac{\partial}{\partial z}= zx\frac{\partial^2}{\partial y\partial z}[/tex]

The second one I believe is correct though..
 
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You need to be a little more clear on your notation. Suppose that f is some arbitrary function that we can let the expressions act on.
Then by
[tex] \left( y\frac{\partial}{\partial z}z\frac{\partial}{\partial x} \right) f[/tex]
do you mean
[tex] y\frac{\partial}{\partial z}\left( z \frac{\partial f}{\partial x} \right)[/tex]
or, as you imply in your first post,
[tex] y\left( \frac{\partial}{\partial z}z\right)\frac{\partial f}{\partial x}[/tex]
or yet something else?
 
Depends on where you put the parentheses. If the first one means y*d/dz(z*d/dx) you need to use a product rule on the z*d/dx product. To make it clearer write it as y*d/dz(z*df/dx) where f is a test function.
 
I don't believe there are any specific parentheses. The exact question is to find the following commutator [tex][Lx,Ly][/tex]

where:

[tex]Lx= (y\frac{\partial}{\partial z} - z\frac{\partial}{\partial y})[/tex]

[tex]Ly= (z\frac{\partial}{\partial x} - x\frac{\partial}{\partial z})[/tex]
 
In that case you should treat it as
[tex] <br /> y\frac{\partial}{\partial z}\left( z \frac{\partial f}{\partial x} \right)<br /> [/tex]
 

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