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Let $\bar g(\bar x) = \bar x$

Let $g(x) = x$

let $\bar x = x + \alpha$

let $x = x$

Now taking the partial derivative w.r.t $\bar x$:

$k_1 = \frac{\partial \bar g}{\partial\bar x} = 1$

let $\alpha = 0$, then $\frac{\partial \bar g}{\partial\bar x} = 1$

Because partial differentiation only looks at explicit variables, and $\alpha$ is not explicitly in $\bar g$ can we first let $\alpha = 0$ then work out the partial? Eg:

Let$\alpha = 0$, then $k_2 = \frac{\partial \bar g}{\partial\bar x} = \frac{\partial g}{\partial x} = 1$

This makes $k_2 = k_1$

Is this allowed?

I appreciate any help you can offer.

Incubator Harcore
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    "Because partial differentiation only looks at explicit variables, and $\alpha$ is not explicitly in $g$ can we first let $\alpha=0$ then work out the partial? ": yes. – kevinkayaks Nov 16 '22 at 22:38
  • @kevinkayaks , thank you so much. – Incubator Harcore Nov 16 '22 at 22:41
  • @kevinkayaks, Did they use this method for https://math.stackexchange.com/questions/4578133/help-with-deriving-the-euler-lagrange-equationpossible-misunderstanding-leibniz?noredirect=1#comment9636412_4578133

    did they remove the $\varepsilon$ because it was implicitly used?

     – Incubator Harcore Nov 17 '22 at 05:04

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