Let $a$ element of $\mathbb{R^n}$, a $f$ functiom from $\mathbb{R^n}$ to $\mathbb{R}$ such that $$f(x)=e^{||x||^2-2<a,x>}$$. Find $Df(x)$ and local max and min.
My solution:
First of all I want to get $Df(x)$ soo the function $f$ sholud be written as a composition of functions. But than I see that $||x||^2=<x,x>$, so our function can be written as $f(x)=e^{<x,x>-2<a,x>}=e^{<x-2a,x>}$. Now our function looks a little better.
Let's $s(x)=<x-2a,x>$ s is linnar operator so we know how $Ds$ looks. $$Df(x)(h)=Des(x)[Ds(x)(h)]=e^{<x-2a,x>}*<h-2a,h>$$ My question is it this ok for first part of task?
Your $s$ isn't linear. Rather, write $s(x) = \|x\|^2 - 2\langle a, x \rangle = q(x) -2L(x)$.
$q = B\circ(id,id)$, where $B: (x,y)\mapsto \langle x, y\rangle$ is bilinear. By the chain rule, $Dq(x)h = 2\langle x, h\rangle$. Now $L$ is linear, so $DL(x) = L$. Thus: $Ds(x)h = 2\left( \langle x,h\rangle - \langle a,h\rangle \right) = 2\langle h, x - a\rangle$.
We have $f = \exp \circ s$, so:
$$Df(x)h = D(\exp)(s(x)) Ds(x)h = 2e^{s(x)} \langle h, x-a \rangle$$
From this, it's clear that $Df(x)h=0$ for all $h$ iff $x = a$.