Suppose we have a function $x(t)$ and a transformation $t \rightarrow t'=t+ \delta t$, and want to find the change in derivative $\delta \frac{d x}{d t}$ due to that transformation. I treat it as kind of differentiation and get $\delta \frac{d x}{d t} = - \frac{d x}{d t} \frac{d \delta t}{d t} $. It is right, isn't it?
Now we want to do the same with a function $u(x^{\mu})$, transformation $x^{\mu} \rightarrow x^{\mu'} = x^{\mu} + \delta x^{\mu}$ and derivative $u_{,\mu}$. It seems that I miss something really fundamental, because the written above approach doesn't work here. Could you help me and say what is it?
I will appreciate any piece of advice.
EDIT: to ones interested in the problem: I was trying to apply this wonderful Feynman's argument to get Noether currents in classical field theory https://physics.stackexchange.com/q/19847/97253 .
EDIT 2: the original problem is to derive Noether current (without their usual derivation through looking at global transformation and making them local, but using Feynman's argument in the above link). $S[u, x^{\mu}] = \int_\Omega \mathcal{L}(x^{\mu}, u, \partial u) d^4 x$ is action functional. $x^{\mu} \rightarrow x^{\mu'} = x^{\mu} + \delta x^{\mu}$ - is a transformation under which action is in invariant. I look at the transformation of the form $\delta x^{\mu} = \epsilon(\theta(x^0-x^0_i) - \theta(x^0-x^0_f))*...$, where $\theta(x)$ is step-function. Now I need to variate action. Variation of 4-volume is obvious: $\delta d^4 x = d^4 x \frac{\partial \delta x^\mu}{\partial x^\mu}$ (just variation of Jacobian). Nonobvious for me is variation w.r.t. $\partial u$, which do change when we perform our transformation. It contains calculation of the $\frac{\partial \mathcal(L)}{\partial u_{,\mu}}\delta u_{,\mu}$, which leads to my question.