An online video showed me that: $$ J(a \cdot b) = a^{T} J(b) + b^{T} J(a) $$ for a,b being vectors. But my question is: isn't the dot product a scalar? So isn't the jacobian of a scalar just 0?
Video for reference: https://www.youtube.com/watch?v=r8oIjsBBljA at 3:00
On
The source of confusion here is the meaning of 'the Jacobian.'
The most familiar instance in which one sees this term is the $n$-variable substitution rule, and there the Jacobian is often taken to be the determinant of an $n$-by-$n$ matrix of partial derivatives. This definition only makes since if a the argument of the Jacobian is an $n$-vector of scalar functions. As a consequence, the formula you cite would be clearly false under this definition.
However, you will also see the Jacobian defined as the matrix of first partial derivatives. In this case, the argument can be any $m$-vector and the result will be some rectangular $m$-by-$n$ matrix. In particular this works for a single scalar function and yields the formula cited.
(1) true, (2) false. Scalar $=$ 1D $\ne$ constant.