= The derivative is the generator of the translation group
Take the group of all <Translation (geometry)> in <\R^1>.
Let's see how the <generator of a Lie algebra>[generator] of this group is the <derivative> <operator>:
$$
\pdv{}{x}
$$
The way to think about this is:
* the translation group operates on the argument of a function $f(x)$
* the generator is an <operator> that operates on $f$ itself
So let's take the <exponential map (Lie theory)>:
$$
e^{x_0\pdv{}{x}}f(x) = \left( 1 + x_0 \pdv{}{x} + x_0^2 \pdv{^2}{x^2} + \ldots\right)f(x)
$$
and we notice that this is exactly the <Taylor series> of $f(x)$ around the identity element of the translation group, which is 0! Therefore, if $f(x)$ behaves nicely enough, within some <radius of convergence> around the origin we have for finite $x_0$:
$$
e^{x_0\pdv{}{x}}f(x) = f(x + x_0)
$$
This example shows clearly how the <exponential map (Lie theory)> applied to a (differential) <operator> can generate finite (non-infinitesimal) <Translation (geometry)>!
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