In Spivak's calculus of manifolds there is a product rule given as below.
if $f,g:\mathbb{R}^n \rightarrow \mathbb{R}$ are differentiable at a, then
$D(f*g)(a)=g(a)Df(a)+f(a)Dg(a).$
My question is, if this condition of '$f,g:\mathbb{R}^n \rightarrow \mathbb{R}$' is always necessary? What if we have vector valued output for each of the functions? Is there a generalized product rule? If not, how is this answer valid because $x^T$ and $Ax$ are surely not satisfying above condition.
Best Answer
If you're still interested, you can define a "generalised product rule" even when the target space of your functions is a vector space (it can even be infinite dimensional).
As a remark, Spivak would use the notation $DF(\beta)$ to mean the derivative of $F$ at $\beta$, but I used $dF_{\beta}$ (as in Loomis and Sternberg), simply to avoid a lot of brackets when evaluating this linear transformation at a certain point $x$. But, aside from this slight notation change, in this formula, we like to think of $\omega$ as "multiplication" so that $F(\xi)$ is the "product" of $g(\xi)$ and $h(\xi)$. Notice how the derivative formula (*) takes the nice form "differentiate the first keep the second $+$ keep the first, differentiate the second". To read more about this, I HIGHLY recommend the book Advanced Calculus by Loomis and Sternberg. This theorem is in fact Theorem $8.4$ of Chapter $3$ (with slightly different notation).
Addition in Response to OP's comment:
Let $A \in M_{n \times n}(\mathbb{R})$. Then, for $x \in \mathbb{R^n}$, the quantity $x^t A x$ can be written using the standard inner product of $\mathbb{R^n}$ as $\langle Ax, x \rangle$. So, what we're doing is considering the following functions (mimicing the notation above):
where in the last line, I made use of symmetry and bilinearity of the inner product. Now, you may be wondering, since $F$ maps $\mathbb{R^n}$ to $\mathbb{R}$, how might we calculate its partial derivatives? The relationship is very simple (see the book I linked above, Section 3.8 for more details): \begin{align} (\partial_i F)(x) &= dF_x(e_i) \\ &= \langle (A^t + A)x, e_i \rangle \end{align} where $\partial_i F(x)$ is what you might normally write as $\dfrac{\partial F}{\partial x_i}(x)$, and $e_i = (0, \dots, 1, \dots, 0) \in \mathbb{R^n}$, with $1$ in the $i^{th}$ place.