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Partial derivative problem

 
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Aug7-12, 11:27 AM   #1
 

Partial derivative problem


Hi!
Here is my function:



My task is to find:



I think I know how to find ∂u/∂x, but I have no idea how to find ∂/∂z(∂u/∂x). Here is how I found ∂u/∂x:

http://oi48.tinypic.com/prsly.jpg

Does someone know how to find ∂/∂z(∂u/∂x)?
I appreciate any help :)
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Aug7-12, 11:58 AM   #2
 
Before the second partial derivative, you should fix the error in your calculation of ∂u/∂x, specifically ∂([itex]\frac{xy}{z}[/itex])/∂x.
Aug7-12, 12:30 PM   #3
 
What's wrong with ∂(xy/z)/∂x? I checked it and it seems correct to me...
Aug8-12, 07:25 AM   #4
 

Partial derivative problem


It's very important so all suggestions are welcome :)
Aug8-12, 01:31 PM   #5
 
Quote by geekba View Post
What's wrong with ∂(xy/z)/∂x? I checked it and it seems correct to me...
Never mind. I hadn't scrolled all the way down, it is correct.

I believe you are having trouble calculating [itex]\frac{∂}{∂z}[/itex]([itex]∂\rho/∂s[/itex]) and [itex]\frac{∂}{∂z}[/itex]([itex]∂\rho/∂t[/itex]) (Let me know if this is not the case).
To simplify this, get rid of s and t by writing [itex]∂\rho/∂s[/itex] and [itex]∂\rho/∂t[/itex] as partial derivatives of [itex]\rho[/itex] w.r.t. x, y and z, using the chain rule. Since you know how s and t depend on x, y and z, this can be done.

Once you have done this, calculating [itex]\frac{∂}{∂z}[/itex]([itex]∂\rho/∂s[/itex]) and [itex]\frac{∂}{∂z}[/itex]([itex]∂\rho/∂t[/itex]) would be straightforward.
Aug8-12, 01:51 PM   #6
 
I got it finally Thaks a lot!
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