Derivation of Noether's theorem by Gateaux derivative Noether's theorem states that if:
$\ \int_{a}^{b} F(x, y, y') \,dx  = \ \int_{a_{new}}^{b_{new}} F(x_{new}, y_{new}, y_{new}') \,dx_{new} $
for any $a$, $b$ and $y(x)$, and when $x$ and $x_{new}$ and $y$ and $y_{new}$ are of transformations:
$x_{new} = \Phi(x, y, y', \epsilon)$
$y_{new} = \Psi(x, y, y', \epsilon)$
$x = \Phi(x, y, y', \epsilon = 0)$
$y = \Psi(x, y, y', \epsilon = 0)$
$\epsilon$ is a variable scalar parameter, and if $y^*(x)$ is an extremal of the functional $\ \int_{a}^{b} F(x, y, y') \,dx$, meaning that $y^*(x)$ satisfies:
$\frac{\partial F(x, y^*, y^{*'})}{\partial y^*} = \frac{d}{dx} \frac{\partial F(x, y^*, y^{*'})}{\partial y^{*'}}$
then there is
$\frac{\partial F(x, y^*, y^{*'})}{\partial y^{*'}} \psi(x, y^*, y^{*'}) + (F(x, y^*, y^{*'}) - y^{*'}\frac{\partial F(x, y^*, y^{*'})}{\partial y^{*'}})\phi(x, y^*, y^{*'}) = \text{constant}$, with
$\phi(x, y, y') = \frac{\partial \Phi(x, y, y', \epsilon)}{\partial \epsilon}\vert{_{\epsilon = 0}}$
$\psi(x, y, y') = \frac{\partial \Psi(x, y, y', \epsilon)}{\partial \epsilon}\vert{_{\epsilon = 0}}$
My question is: is it possible to derive
$\frac{\partial F(x, y^*, y^{*'})}{\partial y^{*'}} \psi(x, y^*, y^{*'}) + (F(x, y^*, y^{*'}) - y^{*'}\frac{\partial F(x, y^*, y^{*'})}{\partial y^{*'}})\phi(x, y^*, y^{*'}) = \text{constant}$
from
$\frac{d{\ \int_{a_{new}}^{b_{new}} F(x_{new}, y_{new}, y_{new}') \,dx_{new} }}{d\epsilon}\vert{_{\epsilon = 0}}$ ?
I read some proof for example, https://drchristianphsalas.com/2020/08/06/proving-noethers-theorem/, but I don't understand the procedures such as ignoring the high order terms, so that only the linear part is used, bla bla.... I want to know if it is possible to only use total derivatives and/or partial derivatives to derive the equations. If so, could you demonstrate that derivation? Thank you all.
 A: Let me expand on my comment. Consider a function $f\colon \mathbb R^n\to \mathbb R$, for simplicity. The Gateaux derivative reduces to the directional derivative, that is
$$
D_vf(x)=\left.\frac{\partial}{\partial \epsilon}f(x+\epsilon v)\right|_{\epsilon=0}. $$
This can be obtained by Taylor expansion, as hinted by the note you are citing; indeed,
$$
f(x+\epsilon v)=f(x)+D_v f(x) \epsilon + o(\epsilon), $$
so to compute $D_v f(x)$ you have to "ignore the higher order terms" (that is, ignore that $o(\epsilon)$) and "use only the linear part" (that is, use only the coefficient of $\epsilon$). The quotation marks refer to the words from the notes that you mentioned.
Example. Consider the functional
$$ 
\Phi(y)=\int_a^b y(x)^4\, dx.$$
This is a function of $y$, which itself is a function of $x$. This is why we call $\Phi$ a "functional".
We want to compute the following directional derivative, or Gateaux derivative;
$$
D_y \Phi(y^\star)=\left.\frac{\partial }{\partial \epsilon}\Phi(y^\star +\epsilon y)\right|_{\epsilon=0}.$$
To do so we want to use the method just introduced. So, we compute the Taylor expansion via the binomial theorem;
$$
\Phi(y^\star +\epsilon y)=\int_a^b(y^\star(x))^4\, dx +4\epsilon \int_a^b (y^\star(x))^3y(x)\, dx+ o(\epsilon).$$
Note that we did not write the terms in $\epsilon^2, \epsilon^3, \epsilon^4$, since we are not going to need them. Instead, we sweeped all those terms under the $o(\epsilon)$. (We could have used $O(\epsilon^2)$ instead of $o(\epsilon)$. It would have been the same). And now we are ready to read our result;
$$
D_y \Phi(y^\star)=4 \int_a^b (y^\star(x))^3y(x)\, dx.$$
