Note that $S$ is an open subset of $\mathbb{R^n}$.
We have $f, g: S \to \mathbb{R^m}$ are functions in $S$ that is differentiable at $a \in S$.
$\cdot$ between $f$ and $g$ represents the dot-product below:
Claim: $\phi:= f \cdot g: S \to \mathbb{R}$ is differentiable at $a$
I am interested in a proof by definition of differentiability of a vector-valued function; no theorems please, I am seeking an alternative method. I have already proved using other theorems. Oh, except the theorem that states since $f,g$ is differentiable at $a$ then we know that $Df(a)$ is the matrix with components of the matching partial derivatives, evaluated at $a$.
Attempt: I'm not exactly sure which version of the differentiability definition of a vector-valued function to use but I decided to go with the following:
We will show that there exists a matrix $M$ such that $\phi(a+h)= \phi(a)+Mh +E(h)$ where limit of $\frac{E(h)}{|h|} \to 0 \in \mathbb{R^m}$ as $h \to 0$
I wrote down $\phi(a+h) = (f_1g_1+...+f_mg_m)(a_1+h_1,...,a_n+h_n)$ got stuck here.
Any direction or a new approach of definitions would be appreciated. Thanks in advance.
Edit#1:
So we know that $\phi(a+h)=f_1(a+h)g_1(a+h)+...+f_m(a+h)g_m(a+h)$ by properties of functions $f,g$.
Since we are given that $f,g$ are differentiable at $a$ we know that this means each of the component $f_j,g_j$ are differentiable at $a$.
i.e. There exists a vector $m_j$ such that $f_j(a+h)=f_j(a)+ m \cdot h+E_j(h)$ where $\frac{E_j(h)}{|h|} \to 0$ as $ h \to 0$. This applies the same way for $g_j$.
Below I denote $E_j^f$ representing the error term at the $jth$ index of the function $f$. Also denote $m_{jf}$ representing the vector of the function $f$ at the $jth$ index.
This means that $\phi(a+h)=(f_1(a)+m_{1f} \cdot h +E_1^f(h)) (g_1(a)+m_{1g} \cdot h + E_1^g(h))$ $+...+$ $(f_m(a)+m_{mf} \cdot h +E_m^f(h)) (g_m(a)+m_{mg} \cdot h + E_m^g(h))$
This is the step I got up to, but I am not sure if I should expand the terms above. It seems very very messy. And If @Mathemetica was trying to accomplish the same thing in a different way I couldn't understand you. Thanks.