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I came across this derivative in a paper:

$\frac{\partial{\|{A}\|}}{\partial A} = \frac{A}{\|A\|}$

where A is a rank 2 tensor. And I wonder what is the definition of this norm. I believe it is not a general expression.

One more thing, what would be the second derivative if this expression is correct:

$\frac{\partial^2{\|{A}\|}}{\partial A^2} = ?$

Thank you all.

Vahid
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    ...which paper? – Carlo Beenakker Jan 26 '14 at 22:29
  • The norm is not going to be differentiable at $A=0$; also, your notation is mixed-up; e.g., if you had vectors, then $|x|_2$ is subdifferentiable at $x=0$, but of course not differentiable. Also, I think this question might fit math.SE well (after you've clarified the notation, however). – suvrit Jan 26 '14 at 22:35
  • @carlo: The paper is about recrystallization in polycrystals and A is there the rate of plastic strain. – Vahid Jan 26 '14 at 23:45
  • @survit: that is my question. There was no subscript so I could know the type of the norm. – Vahid Jan 26 '14 at 23:47
  • I second Carlo's request: which paper? As $A$ is a rank 2 tensor, it is possible that the notation $\dots/A$ means $\dots A^{-1}$ where $A$ is interpreted as a matrix (or an element of an algebra). But then again, it can also mean something entirely different. You really need to give more information. – Willie Wong Jan 27 '14 at 09:14
  • Really sorry guys, I had a mistake in the question which I corrected ($\frac{\partial{|{A}|}}{\partial A} = \frac{A}{|A|}$). The paper you can find here: http://onlinelibrary.wiley.com/doi/10.1002/jgrb.50125/abstract (See equations 46d, 55 and 60). As I mentioned $A$ (a rank 2 tensor) is there the rate of the plastic strain ($\dot{\epsilon_p}$). Again, $A$ is a second order tensor not anything else. If you need more information please let me know. Thank you very much again. – Vahid Jan 27 '14 at 10:20

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