Hence, the derivative of d/dx (x*x') is (d/dx)x*x'+x*x

[Fernando Revilla]

I quote a question from Yahoo! Answers

What is the derivative of d/dx (x*x') where x is a vector and x' denotes x transpose (note that x*x' is a matrix, and not the norm of x!)

I have given a link to the topic there so the OP can see my response.

 

[Fernando Revilla]

In general, if $A=[a_{ij}(x)],\;B=[b_{ij}(x)]$ are matrices $n\times n$ with $a_{ij}(x)$ and $b_{ij}(x)$ differentiable real funtions defined on the open interval $(\alpha,\beta)$, its derivative matrices are defined by $$\frac{d}{dx}A=\left[\frac{d}{dx}a_{ij}(x)\right],\;\frac{d}{dx}B=\left[\frac{d}{dx}b_{ij}(x)\right]$$ Using the product formula: $$A(x)B(x)=C(x)=[c_{ij}(x)] \quad \left(c_{ij}(x)=\sum_{k=1}^na_{ik}(x)b_{kj}(x) \right)$$
it is easy to prove the relation on $(\alpha,\beta)$ $$\frac{d}{dx}(AB)=\left(\frac{d}{dx}A\right)B+A \left(\frac{d}{dx}B\right) $$
as a consequence $$\frac{d}{dx}(AA^T)=\left(\frac{d}{dx}A\right)A^T+A\left(\frac{d}{dx}A^T\right)=\left(\frac{d}{dx}A\right)A^T+A\left(\frac{d}{dx}A\right)^T$$