# Derivative with respect to a vector

1. Sep 22, 2005

### 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:

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

Last edited: Sep 22, 2005
2. Sep 23, 2005

### dextercioby

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

Daniel.

3. Sep 25, 2005

### 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)

4. Sep 26, 2005

### robphy

As dextercioby suggests, $$\frac{d \vec{x}^TA}{d \vec{x}}$$ can be written as (assuming that A is independent of x)
$$\frac{\partial (x_a A^a{}_b)}{\partial x^c}=\frac{\partial (x_a )}{\partial x^c}A^a{}_b+x_a \frac{\partial (A^a{}_b)}{\partial x^c} =\delta_{ac}A^a{}_b+0 =A_{cb}$$
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]"?