Product rule of differentiation

For vector functions:

Theorem (Product Rule). Suppose

$f: {\bf R}^n \rightarrow {\bf R}^m$ is differentiable at $x = a$ and $k: {\bf R}^n \rightarrow {\bf R}$ is differentiable at $x = a$.

Then

$kf: {\bf R}^n \rightarrow {\bf R}^m$ is differentiable at $x = a$ and $( kf )'(a) = f(a) k'(a) + k(a) f'(a).$

Note that since $f(a)$ is of dimension $m \times 1$, $k'(a)$ is of dimension $1 \times n$ (and hence $f(a) k'(a)$ is of dimension $m \times n$),

$k(a)$ is a scalar, $f'(a)$ is of dimension $m \times n$ (and hence $k(a) f'(a)$ is of dimension $m \times n$), the derivative $( kf )'(a)$ is a matrix of dimension $m \times n$.

Further keep in mind that here

1. derivatives are matrices,
2. functions are vectors.