Htykuut

Given $f:mathbb{R}^nto mathbb{R}$ and $Ain (mathbb{M_n(R}))$ such that $f(x) = x^{t}Ax.$ I want to find the derivative of this map and show that it is $C^{1}$. This problem has been partially answered before assuming that $A$ is symmetric, but note that in this case that $A$ is not necessarily symmetric. Therefore the derivative
$$df_x(h) = x^tAh+h^tAx.$$
The map is clearly linear but to show it is continuous, I am not sure whether I am using the right inequalities:
$$|df_x(h)|=|x^tAh|+|h^tAx|leq 2||x||cdot ||A||cdot ||h||$$
This shows that $df_x$ is continuous. To show that it is $C^{1}$ we do the following:
$$|df_x(h)-df_y(h)|leq 2||h||cdot ||A||cdot ||x-y||.$$
I am not sure whether these inequalities are correct, but I am using them because I have seen some exercise corrections similar to this problem. Any hints/suggestions will be much appreciated.

The directional derivative you have for $d_{x}f: mathbb{R}^{n} rightarrow mathbb{R}$ is correct, and as sort of stated in the comments that a linear map between finite-dimensional vector spaces is continuous iff it is linear (which is equivalent to boundedness with linear maps and finite-dimensional vector spaces).

More explicitly however, you can proceed as: for all $epsilon > 0$, suppose $y,z in mathbb{R}^{n}$ are such that $| y - z | < epsilon$. Then for a fixed $x in mathbb{R}^{n}$, and $A in M_{ntimes n}(mathbb{R})$,

$$
d_{x}f(y) - d_{x}f(z) = x^{T}Ay + y^{T}Ax - x^{T}Az - z^{T}Ax\
= x^{T}A(y-z) + (y-z)^{T} Ax
$$
and therefore with respect to any norm (all norms are equivalent on finite-dimensional vector spaces),
$$
| d_{x}f(y) - d_{x}f(z) | = | x^{T}A(y-z) + (y-z)^{T} Ax | leq 2|x| |A||y - z | < 2|x| |A| epsilon.
$$
So as both $xin mathbb{R}^{n}$ and $A in M_{n times n}(mathbb{R})$ are fixed, we can set $delta := 2| x | | A | epsilon$, then formally:
For any $epsilon > 0$ such that $| y - z | < epsilon$, there exists a $delta > 0$ such that
$$
|y - z|< epsilon implies | d_{x}f(y) - d_{x}f(z)| < delta
$$