Derivative with respect to a vector

[songCalculus]

Hi, I have a question about doing derivative with respect to a vector, can someone help please.

Problem:
Suppose A is a (nxn) dimensional symmetric matrix, \vec{x} is a (nx1) column vector.
We know that

\frac{d A\vec{x}}{d \vec{x}}=A

and

\frac{d \vec{x}^TA\vec{x}}{d \vec{x}}=2A\vec{x} ( A is symmetric)

question:

\frac{d \vec{x}^TA}{d \vec{x}}=?

many thanks in advance!

 

[dextercioby]

Do you know how to work with tensor formalism? Using subscripts and superscripts.

Daniel.

 

[jaap de vries]

this should be A too. When A is not a function of x then take it out of the derivative and realize that dxi/dxj is simply (d)ij where (d) is the isentropic replacement tensor.
(Sorry I didn't feel like latex today)

 

[robphy]

As dextercioby suggests, \frac{d \vec{x}^TA}{d \vec{x}} can be written as (assuming that A is independent of x)
\frac{\partial (x_a A^a{}_b)}{\partial x^c}=\frac{\partial (x_a )}{\partial x^c}A^a{}_b+x_a \frac{\partial (A^a{}_b)}{\partial x^c}<br /> =\delta_{ac}A^a{}_b+0<br /> =A_{cb}<br />
which is not exactly the object A^a{}_b that we started with.

But, now, what is the interpretation of a "derivative with respect to a vector [or a tensor]"?