# Homework Help: Symplify the expression.

1. Jun 8, 2009

### hoch449

1. The problem statement, all variables and given/known data

Simplify the following two expressions:

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

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

3. 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..

2. Jun 8, 2009

### CompuChip

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
$$\left( y\frac{\partial}{\partial z}z\frac{\partial}{\partial x} \right) f$$
do you mean
$$y\frac{\partial}{\partial z}\left( z \frac{\partial f}{\partial x} \right)$$
or, as you imply in your first post,
$$y\left( \frac{\partial}{\partial z}z\right)\frac{\partial f}{\partial x}$$
or yet something else?

3. Jun 8, 2009

### Dick

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.

4. Jun 8, 2009

### hoch449

I don't believe there are any specific parentheses. The exact question is to find the following commutator $$[Lx,Ly]$$

where:

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

$$Ly= (z\frac{\partial}{\partial x} - x\frac{\partial}{\partial z})$$

5. Jun 8, 2009

### Dick

In that case you should treat it as
$$y\frac{\partial}{\partial z}\left( z \frac{\partial f}{\partial x} \right)$$