Edit: Show that $f$ is differentiable on $\Bbb{R}^n$ and compute $f'$

Question

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!

Answer 1

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 $x$.

Answer 2

Consider the rule for differentiating inner products described here, and use the usual derivative properties (addition, chain rule, differentiating constant and linear functions).

Answer 3

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