TITLE: Edit: Show that $f$ is differentiable on $\Bbb{R}^n$ and compute $f'$
QUESTION [5 upvotes]: Good day all!
Edit: I'm currently doing a personal study on differentiation on $\Bbb{R}^n$ but I have this challenging problem. Although, some answers have been provided on how to show that show that $f$ is differentiable on $\Bbb{R}^n$ but I would further like to compute $f'$ on $\Bbb{R}^n$.
There is this function
$$f:\Bbb{R}^n\to \Bbb{R}$$ 
$$x\mapsto f(x)=\frac{1}{2}\langle x,u(x)\rangle+\langle x,b\rangle$$
where $u:\Bbb{R}^n\to\Bbb{R}^n$ is linear and symmetric $:$ $(\forall\;x,y\in \Bbb{R}^n,\langle x,u(y)\rangle=\langle u(x), y \rangle)$ and $b\in \Bbb{R}^n.$
Honestly, I am just coming across this kind of function. I want to know what name it's called. How do I show that $f$ is differentiable on $\Bbb{R}^n$ and how do I compute $f'$?
Thanks for your help!

REPLY [3 votes]: Using the symmetry and the linearity of $u$ you can easily check that
$$
f(x+h) = f(x) + \langle u(x) , h\rangle + \langle h, b \rangle + O(|h|^2),
$$
i.e.
$$
f(x+h) = f(x) + L_x(h) + o(|h|),
$$
with $L_x(h) := \langle u(x) ,h\rangle + \langle h, b \rangle$.
Hence, by definition, the linear map $L_x$ is the differential of $f$ at $x$.

REPLY [2 votes]: sorry but I dont have enough score to write this as a comment
I  think the best way is to passe by the defintion 
Let $P$ be a point in $\mathbb{R}^n$ and $w$ a vector in $\mathbb{R}^n$
you can define a curve $\alpha :(-\epsilon , \epsilon)\rightarrow \mathbb{R}^n $
 such that $\alpha(0)=p$ and $\alpha^{'}(0)=w$ 
now we can define a curve $\beta = F o \alpha\; $ and the answer is 
$dF_p(w)=\beta^{'}(o)$ this is the idea you have just to write the calculations 
Good luck