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Partial derivative problem |
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| Aug7-12, 11:27 AM | #1 |
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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 :) |
| Aug7-12, 11:58 AM | #2 |
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Before the second partial derivative, you should fix the error in your calculation of ∂u/∂x, specifically ∂([itex]\frac{xy}{z}[/itex])/∂x.
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| Aug7-12, 12:30 PM | #3 |
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What's wrong with ∂(xy/z)/∂x? I checked it and it seems correct to me...
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| Aug8-12, 07:25 AM | #4 |
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Partial derivative problem
It's very important so all suggestions are welcome :)
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| Aug8-12, 01:31 PM | #5 |
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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 |
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I got it finally
Thaks a lot!
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