Derivative of distributions inner product The dot product of two distributions $u(s)$ and $v(s)$, for $s$ the parametric coordinates, is written as $u(s)^{\intercal} \, v(s)$. What would a closed form for the derivative $\frac{\partial}{\partial s} \left( u(s)^{\intercal} \, v(s)\right)$ be? My confusion constitutes on the following, for i from 1 to the dimension of $s$:
$\frac{\partial}{\partial s_i} \left( u(s)^{\intercal} \, v(s)\right) = \sum\limits_i \left[\left(\frac{\partial}{\partial s_i} u_i\right) \, v_i + u_i \, \left(\frac{\partial}{\partial s_i} v_i\right) \right]$
Example: $\frac{\partial}{\partial [x \, y]^{\intercal}} \begin{bmatrix} x & y \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2x \\ 2y \end{bmatrix}$
 A: The derivative $\frac{\partial f}{\partial s}$ of $f(s) = u(s)^\top v(s)$ can be presented either as a row vector or a column vector. If it is presented as a row vector, then we have
$$
\frac{\partial f}{\partial s} = \pmatrix{\left[\frac{\partial u}{\partial s_1}\right]^Tv + u^T\left[\frac{\partial v}{\partial s_1}\right] & 
 \cdots &
\left[\frac{\partial u}{\partial s_d}\right]^Tv + u^T\left[\frac{\partial v}{\partial s_d}\right]}.
$$
We note that
$$
\left[\frac{\partial u}{\partial s_k}\right]^Tv + v^T\left[\frac{\partial u}{\partial s_k}\right] = \sum_{i=1}^d \left(\frac{\partial u_i}{\partial s_k}v_i + u_i\frac{\partial vI}{\partial s_k} \right).
$$
This can also be written as
$$
\frac{\partial f}{\partial s} = \left[\frac{\partial u}{\partial s}\right]^Tv + u^T\left[\frac{\partial v}{\partial s}\right],
$$
where $\frac{\partial u}{\partial s}$ denotes the matrix
$$
\frac{\partial u}{\partial s} = \pmatrix{\frac{\partial u}{\partial s_1} & \cdots & \frac{\partial u}{\partial s_d}}.
$$
