Calculate the differentiate of $ Q: \mathbb{R}^n \rightarrow \mathbb{R}$ defined by $Q(x)=B(x;x)$

Question

I have a question on the differentiable of a function and I want to be sure that my understanding of this concept is correct.

Question:

Calculate the differentiate of $ Q: \mathbb{R}^n \rightarrow \mathbb{R}$ defined by $Q(x)=B(x;x)$ where $B: \mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R} $ is bilinear.

Answer:

1- It is given that $B$ is a bilinear mapping thus $B(x;x)$ is continuous and $ \forall x \in \mathbb{R}^n, \exists c > 0 $ s.t. $ |B(x;x) - B(0;0)| \leq c \| x -0 \| $. And as $B(0;0)=0$ because $B$ is linear mapping the last inequality can be re write $ |B(x;x)| \leq c \| x\| $

2- Now according to the definition; for any given point $x \in \mathbb{R}^n$ the differential in this point will be the linear mapping $L(h)$ verifying that $ \forall h \in \mathbb{R}^n $, $Q(x+h)$ can be writen as follow: $ Q(x+h)= Q(x) + L(h) + O(h)$ with $O(h)$ verifying $ \lim_{h \to 0} || \frac{O(h)}{h}|| = 0 $
Rem: $h$ can not appear in the exepression $Q(x)$ but $x$ can appear in the expression of $L(h)$ and $O(h)$. More over $h$ can appear in the expression of $L(h)$ only as a power of $1$ that means $h$ but not $h^2 , h^3,...$

3- First by definition we can write $Q(x+h) = B(x+h;x+h)$. After as $B(.;.)$ is a bilinear mapping (given) we can keep on and continue to write $ Q(x+h) = B(x+h;x+h) = B(x;x+h) + B(h;x+h)=B(x;x)+B(x;h)+ B(h;x+h) = B(x;x)+B(x;h)+ B(h;x) + B(h;h) $
Rem: We used the following property of bilinear mapping: $\phi(a+v;b)= \phi(a;b) + \phi(v;b) $

4- First lets note that according to "1-" we have that if we take $O(h) = B(h;h)$ that $ \lim_{h \to 0} || \frac{O(h)}{h}|| = 0 $.
Now $L(h) = B(x;h)+ B(h;x) = D[Q(x)] h $ is the differentiate of $Q(x)$ as it is a the sum of two linear mapping in $h$ by assumption.

Is it correct? I have mostly a question on my part "2-" concerning my understanding of the differentiate definition.

Thank you.

Answer 1

-According to the comment of https://math.stackexchange.com/users/167818/j%c3%bcrgen-sukumaran this is correct.

-More over to make the things even more clearer pay attention to the following fact.
$L(h)$ cannot be equal to $B(x;h) + B(h;x) + B(h;h)$ because in such a case we will have $L( \alpha h ) = B(x; \alpha h) + B( \alpha h;x) + B(\alpha h;\alpha h) = \alpha B(x; h) + \alpha B( h;x) + \alpha^2 B( h;h) \neq \alpha L(h) $. That means that $L(.)$ will not be linear.