- #1
wefoust2
- 2
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Hi, I am sort of hung up with a particular step in a derivation, and this has caused me to ponder a few properties of partial derivatives. As a result, I believe I may be correct for the wrong reasons. For this example, the starting term is
([itex]\frac{\partial}{\partial x}[/itex][itex]\frac{\partial v}{\partial t}[/itex]-[itex]\frac{\partial}{\partial y}[/itex][itex]\frac{\partial u}{\partial t}[/itex])
I want to go from the above term to
[itex]\frac{\partial}{\partial t}[/itex] ([itex]\frac{\partial v}{\partial x}[/itex] - [itex]\frac{\partial u}{\partial y}[/itex])
I am a little confused how this is done. I am not sure if you can "factor" out the [itex]\frac{\partial}{\partial t}[/itex] or not. I thought about simply rearranging the partials, but I don't think I can assume the function is smooth or symmetric. Any help or insight you can provide will be appreciated.
Thanks,
wefoust
([itex]\frac{\partial}{\partial x}[/itex][itex]\frac{\partial v}{\partial t}[/itex]-[itex]\frac{\partial}{\partial y}[/itex][itex]\frac{\partial u}{\partial t}[/itex])
I want to go from the above term to
[itex]\frac{\partial}{\partial t}[/itex] ([itex]\frac{\partial v}{\partial x}[/itex] - [itex]\frac{\partial u}{\partial y}[/itex])
I am a little confused how this is done. I am not sure if you can "factor" out the [itex]\frac{\partial}{\partial t}[/itex] or not. I thought about simply rearranging the partials, but I don't think I can assume the function is smooth or symmetric. Any help or insight you can provide will be appreciated.
Thanks,
wefoust