Let $f,g: \mathbb{R} \rightarrow \mathbb{R}^n$ be differentiable
functions and let $h: \mathbb{R} \rightarrow \mathbb{R}$ be defined by
$h(t) = f(t) \cdot g(t)$.Prove that $\frac{dh}{dt}(a) = (Df(a))^T \cdot g(a) + f(a) \cdot (Dg(a))^T$,
where $a \in \mathbb{R}$ and $M^T$ represents the transpose of a
matrix $M$ (we are considering, for example, the row matrix
$(Df(a))^T$ as an element of $\mathbb{R}^n$.
I know that this has something to do with gradient vectors and product rule, but I actually have no ideia of how to prove this. Any help, please?
Best Answer
You can track the vector components.
Let $f_i, g_i$ be the vector components of the function values. Note that $(Df)_i=\frac{d}{dt}f_i$.
$$ \frac{dh}{dt}=\frac{d}{dt}\sum_i f_i(t) g_i(t)=\sum_i f_i(t)\frac{d}{dt}g_i(t)+g_i(t)\frac{d}{dt}f_i(t) $$
which leads to the product rule formula you stated by definition.
The use of transpose in the original statement is superfluous. Transpose only matters if you phrase $h$ as the matrix-vector multiplication of $f^Tg$, where you can see the application of (the true $\mathbb{R}^m\rightarrow\mathbb{R}^n$) product rule simply retains the transpose on $f$; except that is not what your question asks.