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# Derivative of an expression

Tue, 2010-04-06 10:26 - Anton Ishmurzin

Hello iMechanica!

While reading a paper, I've tried to repeat a derivation of a simple tensorial expression given in the paper and my result differs from the result in the paper. Could you please look in to the PDF-File (just 1 page long!) that I have attached to my post and see if I derived everything right? That would be great!

Thanks a lot in advance!

Anton

Attachment | Size |
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sde.pdf | 44.3 KB |

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## Re: Derivative of J_m

Your expresion is

J_m = \eta |I_1|^l \sign(I_1)

The least confusing way of proceeding to find dJ_m/d\sigma is to break up the absolute value function into two parts.

If I_1 >= 0,

J_m = \eta I_1^l => dJ_m/d\sigma = \eta l I_1^{l-1} dI_1/d\sigma

If I_1 < 0

J_m = -\eta (-I_1)^l => dJ_m/d\sigma = - \eta l (-I_1)^{l-1} (-dI_1/d\sigma)

Combine them together to get

dJ_m/d\sigma = \eta l |I_1|^{l-1} dI_1/d\sigma

I can't see the extra sign term in the derivative but that may be just an oversight on my part. Please check for correctness.

-- Biswajit

## Thanks for your anwer,

Thanks for your anwer, Biswajit.

Your way of deriving is definetely less confusing and so far I couldn't spot any oversights in it.

I've also tried to do obtain the derivative using SAGE. The result is the same as in your and mine derivations. I made it available here: http://www.sagenb.org/home/pub/1894/ .

Thanks once again!