It is said that it is possible to calculate the Jacobian of a composed function by multiplying the Jacobians of each function, that is
$$ J_f = J_{f_1} \cdot J_{f_2} \cdots J_{f_nx} $$ where
$$ f = f_1 \circ f_2 \circ \cdots\circ f_n $$
I'm trying to do this on very simple examples that are not well-behaved (where the jacobians are not square matrices), and I can't get this done. I'm hoping someone can help me solve this (Any external lectures or material is also welcome. I could't find any related question).
let $$ f(x) = x^2 + x $$
This could be written as $$ f_1(x, y) = x + y $$ $$ f_2(x) = x^2 $$ $$ f(x) = f_1(f_2(x), x) $$
We now know that
$$ Jf_1 = [1, 1] $$ $$ Jf_2 = [2x] $$ And then $$ Jf = Jf_1 \cdot Jf_2 $$
Which is not a valid product. What to do in this case? Forcing both matrices as 2x2 seems to lead to an invalid solution:
Jf1 = [1, 1]
[0, 0]
Jf2 = [2x 0]
[0 0]
Jf1 * Jf2 = [2x 0]
[0 0]
(Would mean that df/dx = 2x). What am I missing?
Note that you are composing the following functions:
$$f:\mathbb{R}^2\to \mathbb{R}, \quad f(x,y)=x+y$$ and
$$g:\mathbb{R}\to \mathbb{R}^2, \quad g(x,y)=(x^2,0).$$
Now, you have $$J(f\circ g)=J(f)J(g).$$ Indeed
$$J(f\circ g)=\left(\begin{array}{cc} 1 & 1\end{array}\right)\left(\begin{array}{c} 2x\\ 0\end{array}\right)=(2x).$$