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I would like to do, in every calculation that I perform, the following substitution:

$e^x -> (1+x)$.

That is, every term with an exponential should become the Taylor series to the first-order evaluated at zero ($e^{f(x)} = 1+f`(0) x$). Surely my attempt was to try:

$e^x = (1+x)$

however it returns me that it is Protected.

I must address that I make several calculations and this term may appear in several different contexts. As examples:

$e^{2(x+y)}$ and this should return $(1+2x)e^{2y}$;

or

$e^{-x}-1$ which should become $-x$.

Is it possible to generalize this to any function following a structure, like: every exponential following a $\delta$ will be simplified, but the ones without it will not. As an example:

$e^{2(x+\delta x) + y-3\delta c} = (1+2\delta x) (1-3\delta c) e^{2x+y}$.

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    $\begingroup$ Exp[expr] /. {Exp[x_] :> 1 + x} will not do what you want but Exp[expr] /. {Exp[ xpr_] :> (1 + D[xpr, x] x)*(Exp[xpr /. {x :> 0}])} comes closer. $\endgroup$ – Alan Sep 23 at 21:17
  • $\begingroup$ Generally substitutions work better if they involve a variable rather than an expression. Try the inverse of your expression, x->Log[1+x]. You will probably still need to make some simplifications afterword. $\endgroup$ – Bill Watts Sep 23 at 22:21
  • $\begingroup$ You can introduce an additional parameter in the exponents and perform a series expansion with respect to it: $e^xe^{2(x+y)}\rightarrow e^{\alpha x}e^{2\alpha x+2y}$ $\endgroup$ – yarchik Sep 24 at 7:23
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One way to do it.

erule[a_] = {a -> Log[1 + a]}

expr = Exp[-x] - 1

Series[expr /. erule[x], {x, 0, 1}] // Normal
(*-x*)

expr = Exp[2 (x + y)]

Series[expr /. erule[x], {x, 0, 1}] // Normal // Simplify
(*(2 x + 1) E^(2 y)*)

expr = E^(2 (x + δx) + y - 3 δc)

Series[expr /. erule[δx] /. erule[δc], {δx, 0, 1}, {δc, 0, 1}] // Normal // Simplify
(*(3 δc - 1) (2 δx + 1) (-E^(2 x + y))*)

While this method is not automated, it does work, at least for these examples.

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