Simplifying Partial Derivatives in Multivariable Calculus

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Homework Statement



Simplify the following two expressions:

y\frac{\partial}{\partial z}z\frac{\partial}{\partial x}

z\frac{\partial}{\partial y}x\frac{\partial}{\partial z}


The Attempt at a Solution



for the first one: y\frac{\partial}{\partial z}z\frac{\partial}{\partial x}

\frac{\partial}{\partial z}z = 1

so therefore y\frac{\partial}{\partial z}z\frac{\partial}{\partial x}= y(1)\frac{\partial}{\partial x}= y\frac{\partial}{\partial x}

How come this is incorrect?




For the second one: z\frac{\partial}{\partial y}x\frac{\partial}{\partial z}

I cannot write that z\frac{\partial}{\partial y}x\frac{\partial}{\partial z}= x\frac{\partial}{\partial z}z\frac{\partial}{\partial y}

so therefore z\frac{\partial}{\partial y}x\frac{\partial}{\partial z}= zx\frac{\partial^2}{\partial y\partial z}

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
<br /> \left( y\frac{\partial}{\partial z}z\frac{\partial}{\partial x} \right) f<br />
do you mean
<br /> y\frac{\partial}{\partial z}\left( z \frac{\partial f}{\partial x} \right)<br />
or, as you imply in your first post,
<br /> y\left( \frac{\partial}{\partial z}z\right)\frac{\partial f}{\partial x}<br />
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 [Lx,Ly]

where:

Lx= (y\frac{\partial}{\partial z} - z\frac{\partial}{\partial y})

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