Find the differential of the following map. Given $f:\mathbb{R}^n\to \mathbb{R}$  and $A\in (\mathbb{M_n(R}))$ 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.
 A: 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 $A \in M_{n\times n}(\mathbb{R})$,
$$
d_{x}f(y) - d_{x}f(z) = x^{T}Ay + y^{T}Ax - x^{T}Az - z^{T}Ax\\
= x^{T}A(y-z) + (y-z)^{T} Ax
$$
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 $x\in \mathbb{R}^{n}$ 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
$$
\|y - z\|< \epsilon \implies \| d_{x}f(y) - d_{x}f(z)\| < \delta
$$
