Derivative involving Summation Notation

1. Mar 20, 2012

Lucid Dreamer

Hello, I am looking at a derivation that involves (note x is a column vector)
$$\frac {d(\vec{x}^T\vec{x})} {d\vec{x}} = \vec{x}^{T}$$

So I convert to summation notation and evaluate as follows
$$\sum_{i,j} \frac {d(x_{i}x^{i})} {dx^{j}}$$
$$\sum_{i,j} \frac {dx_{i}} {dx^{j}} x^{i} + \sum_{i,j} x_{i} \frac {dx^{i}} {dx^{j}}$$
$$\sum_{i,j} \frac {dx_{i}} {dx^{j}} x^{i} + \sum_{i,j} x_{i} \delta_{ij}$$
$$\sum_{i,j} \frac {dx_{i}} {dx^{j}} x^{i} + \sum_{i} x_{i}$$

So I do recover the answer in the second term, but I am not sure what the derivative in the first term means. This derivative is of a row vector with respect to a column vector.

Last edited: Mar 20, 2012
2. Mar 21, 2012

mathman

I am not familiar with the notation here, but I suspect dxi/dxj = 0 for i ≠ j.

3. Mar 21, 2012

slider142

I have only encountered this notation in metric spaces.
You will need to write the covector field Fi(x) = xi as a function of the vector field Gi(x) = xi. This depends on the inner product of the vector space you are working in. If the induced metric of your space is M, then xi = Mijxj. Thus, (d/dxj)(xi) = (d/dxj)(Mikxk). If the components of the metric Mik are constants, then we have (d/dxj)(Mikxk) = Mik(d/dxj)(xk) = Mikδkj = Mij.
If your metric is the standard Euclidean metric, then Mij = δij, so
$$\sum_{i, j} \frac{dx_i}{dx^j}x^i = \sum_{i, j} \delta_{ij} x^i = \sum_j x_j$$
If your metric is not standard Euclidean, you replace it as above.

Last edited: Mar 21, 2012