Simplifying Partial Derivatives in Multivariable Calculus

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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]