Partial Derivative of a outer product in Vector Calculus I am trying to compute the partial derivative of certain vector products for calculating the stiffness matrix. So we already know that 
For any vector $\textbf{x}$, we have 
1) The derivative of the vector magnitude with itself as 
$$\frac{\partial{|\textbf{x}|}}{\partial{\textbf{x}}} = \hat{\textbf{x}}^T$$
2) The derivative of a normalized vector $\hat{\textbf{x}}$ with respect to itself is 
$$\frac{\partial{\hat{\textbf{x}}}}{\partial{\textbf{x}}} = \frac{I - \hat{\textbf{x}} \hat{\textbf{x}}^T}{|\textbf{x}|} $$
My question is : What is the derivative of the outer product with respect to itself ? 
So far I have got (by applying the product rule) : 
$$\frac{\partial \left({\hat{\textbf{x}} \hat{\textbf{x}}}^T\right)}{\partial{\textbf{x}}} =   
\hat{\textbf{x}}^T \left(\frac{I - \hat{\textbf{x}} \hat{\textbf{x}}^T}{|\textbf{x}|}\right) + \hat{\textbf{x}} \frac{\partial{\hat{\textbf{x}}^T}}{\partial{\textbf{x}}} $$
I am confused about how to evaluate $$\frac{\partial{\hat{\textbf{x}}^T}}{\partial{\textbf{x}}} ?$$ I know it will be a Rank 3 Tensor Matrix. Any help is appreciated. 
 A: The question, (in Gibbs/dyadic notation) is to evaluate the third-order tensor
$$
  T = \frac{\partial}{\partial x}\bigg(\frac{xx}{x\cdot x}\bigg)
$$ 
The only derivative that we need to know is
$$\,\,\frac{\partial x}{\partial x}=I$$
which allows us to expand the RHS using the product rule
$$\eqalign{
  T &= \frac{Ix+xI}{x\cdot x} - \frac{xx\,\,(I\cdot x+x\cdot I)}{(x\cdot x)^2} \cr
    &= \frac{Ix+xI}{x\cdot x} - \frac{2\,xxx}{(x\cdot x)^2} \cr
  &= \frac{Iu+uI-2\,uuu}{\|x\|} \cr
}$$
where the normalized vector has been written as 
$$u=\frac{x}{\|x\|}$$
Please note that $(Iu)$ does not denote the usual matrix-vector product, but the dyadic (aka tensor) product, and thus represents a third-order tensor with components 
$$(Iu)_{ijk} = \delta_{ij} u_k$$ 
The normal matrix-vector product is denoted by an explicit dot product, e.g. $(I\cdot x)$, in the above derivation.
A: Sorry, I realize now that I misunderstood your question! I've left my old answer below though.
In regards to your actual question, isn't it the case that:
$$
\frac{\partial}{\partial \mathbf{x}}\hat{\mathbf{x}}^T = \nabla \times \hat{\mathbf{x}}?
$$
In general for the product of a scalar and vector function we get:
$$\nabla \times \left(f(\mathbf{x})\mathbf{v}(\mathbf{x})\right)=\nabla f(\mathbf{x})\times \mathbf{v}(\mathbf{x})+f(\mathbf{x})\left(\nabla\times \mathbf{v}(\mathbf{x})\right).$$
See the second curl identity here.
For your example $f(\mathbf{x})=\frac{1}{|\mathbf{x}|}$, and $\mathbf{v}(\mathbf{x})=\mathbf{x}$.
I got 
$$\nabla \times \mathbf{x}=\mathbf{I}$$ 
and 
$$\nabla \frac{1}{|\mathbf{x}|}=-\frac{\mathbf{I}\cdot\hat{\mathbf{x}}^T}{|\mathbf{x}|^2}.$$
Putting the two together gives
$$
\begin{aligned}
\frac{\partial}{\partial \mathbf{x}}\hat{\mathbf{x}}^T &= \nabla \times \hat{\mathbf{x}}\\
&=\mathbf{I}-\frac{\mathbf{I}\cdot\hat{\mathbf{x}}^T}{|\mathbf{x}|^2}\\
&=\mathbf{I}\left(\mathbf{1}^T-\frac{\hat{\mathbf{x}}^T}{|\mathbf{x}|^2}\right).
\end{aligned}
$$
Where $\mathbf{1}$ is the vector of ones.

Original off-topic answer:
I got zero: $\hat{\mathbf{x}}$ is a unit vector, therefor its inner product with itself is $1$, a constant.
For the derivative of the inner product:
$$
\begin{aligned}
\frac{d}{d\mathbf{x}}\left(\hat{\mathbf{x}}^T\hat{\mathbf{x}}\right)=
\frac{d}{d\mathbf{x}}\left\langle\frac{\mathbf{x}}{|\mathbf{x}|},\frac{\mathbf{x}}{|\mathbf{x}|}\right\rangle
&=2\left\langle\frac{d}{d\mathbf{x}}\left(\frac{\mathbf{x}}{|\mathbf{x}|}\right),\frac{\mathbf{x}}{|\mathbf{x}|}\right\rangle\\
&=2\left\langle
\frac{|\mathbf{x}|-\mathbf{x}\cdot\frac{\mathbf{x}}{|\mathbf{x}|}}{|\mathbf{x}|^2}
,\frac{\mathbf{x}}{|\mathbf{x}|}
\right\rangle \\
&=2\left\langle 0 ,\frac{\mathbf{x}}{|\mathbf{x}|}\right\rangle=0.
\end{aligned}
$$
In the first equality, I used the product rule and commutivity of inner products. Then I used a quotient rule. In the second to last equality, I used the fact that $\mathbf{x}\cdot\mathbf{x}=|\mathbf{x}|^2$.
Notice that I'm using notation $\mathbf{a}\cdot\mathbf{b}=\left\langle\mathbf{a},\mathbf{b}\right\rangle=\mathbf{a}^T\mathbf{b}$ for the inner product.
