Derivative of a product of linear combinations let $f_i(e_i) = \sum_j^N a_{ij}x_j$ be a linear combination of N variables $x_1 \dots x_n$.
Let $f(x_1, \cdots, x_n) = \prod_i^N f_i(e_i)$ be a product of linear combinations.
To derive $f$ I used the following steps:
$\prod_i^N f_i(e_i) = exp(log(\prod_i^N f_i(e_i))) = exp(\sum_i^N log(f_i(e_i))) $
$\frac{\partial}{\partial x_1 \cdots \partial x_n} f(x_1, \cdots, x_n) = \left( \frac{\partial}{\partial x_1 \cdots \partial x_n} \sum_i^N log(f_i(e_i))) \right) \cdot exp(\sum_i^N log(f_i(e_i))) \\= \left(\sum_i^N \frac{\frac{\partial}{\partial x_1 \cdots \partial x_n}f_i(e_i)}{f_i(e_i)}\right) \cdot \prod_i^N f_i(e_i) $
As far as I know these steps are all legal and are similary found in derivatives for linear models.
The usecase is, that I have a cumulative desnity functions over linearcombinations of $x_1, \dots, x_n$ and i want to get from the integral form of the product to the density. Sadly, when i put in a very simple linear combination and a uniform distribution I dont get what I expected.
Let $e_1 = x_1 = \mathcal{U}(0,1); e_2 = x_2 = \mathcal{U}(0,1)$, then
$f(x_1, x_2) = \prod_i^2 F_i(e_i)     = \left\{ \begin{array}{lr}
         x_1 \cdot x_2 & \text{if } x_1, x_2 \in [0, 1]\\
         0 & \text{else} 
    \end{array} \right\}\\$
and the derivative
$f'(x_1, x_2) = (\frac{1}{x_1} + \frac{1}{x_2}) \cdot x_1 \cdot x_2$.
But for a 2D uniform distribution over the unit space I would expect a density of 1. Where am I wrong?
 A: You made a mistake at this point :
$$
\frac{\partial^n \ln f_i}{\partial x_1 \ldots \partial x_n} \neq \frac{1}{f_i}\frac{\partial^n f_i}{\partial x_1 \ldots \partial x_n} \verb+  +\mathrm{when}\verb+  + n\ge2
$$
Indeed, let's take a counter-example with $f_i(x_1,x_2) = x_1+x_2$; one has then
$$
\frac{\partial^2 \ln f_i}{\partial x_1 \partial x_2} = \frac{\partial^2}{\partial x_1 \partial x_2}\ln(x_1+x_2) = \frac{\partial}{\partial x_1}\frac{1}{x_1+x_2} = -\frac{1}{(x_1+x_2)^2}
$$
while
$$
\frac{1}{f_i}\frac{\partial^2 f_i}{\partial x_1 \partial x_2} = \frac{1}{x_1+x_2}\frac{\partial^2}{\partial x_1 \partial x_2}(x_1+x_2) = 0
$$

Moreover, you seem to interpret $f$ as the density function of a multidimensional uniform distribution; in that case, beware of the fact that its derivative should be taken as a distributional derivative. In consequence, in your example with $f(x_1,x_2) = x_1x_2 \chi_{[0,1]}(x_1)\chi_{[0,1]}(x_2)$, where $\chi_{[0,1]}$ is the characteristic map over the given interval, we have :
$$
\nabla f(x_1,x_2) = \begin{pmatrix} \partial f/\partial x_1 \\ \partial f/\partial x_2 \end{pmatrix}
$$
$-$ because $f$ has several variables $-$, with
$$
\frac{\partial f}{\partial x_1} = x_2 \chi_{[0,1]}(x_1)\chi_{[0,1]}(x_2) + x_1x_2 \chi_{[0,1]}'(x_1)\chi_{[0,1]}(x_2)
$$
where
$$
\chi_{[0,1]}'(x_1) = (H(x_1) - H(x_1-1))' = \delta(x_1)-\delta(x_1-1),
$$
$H$ being the Heaviside function and $\delta = H'$ the Dirac delta function, hence
$$
\begin{array}{rcl}
\displaystyle \frac{\partial f}{\partial x_1} 
   &=& \displaystyle 
   x_2 \chi_{[0,1]}(x_1)\chi_{[0,1]}(x_2) + x_1x_2(\delta(x_1)-\delta(x_1-1))\chi_{[0,1]}(x_2) \\
   &\equiv& \displaystyle 
   (1-\delta(x_1-1))x_2\chi_{[0,1]}(x_2)
\end{array}
$$
and similarly for $\frac{\partial f}{\partial x_2}$.
