What's the name of this semi-group theorem? I encountered this theorem, that for a bounded linear transform $L$ and a real parameter $t$ and initial data $u_0$, we have
$$\frac{d}{dt} \exp(Lt)[u_0] = L \exp(Lt)[u_0].$$
What is the name of this theorem though? How do I Google this if I want to learn more about it?
 A: I don't think that there is any specific name for this result; it is typically considered a standard fact for the matrix exponential function or, more generally, for the operator exponential function.
However, you ask how to find more information on it and, as Tom Copeland pointed out, this seems to be a very reasonable question. So here are a few observations that might be helpful:
Derivative of the operator exponential function
If $L$ is a quadratic finite-dimensional matrix (over $\mathbb{R}$ or $\mathbb{C}$) or, more generally, a bounded linear operator on a Banach space $X$, then one can conclude from the definition of the exponential function
$$
  \exp(tL) = \sum_{n=0}^\infty \frac{t^n L^n}{n!},
$$
which converges with respect to the operator norm (in the space $\mathcal{L}(X)$ of bounded linear operators on $X$) that the mapping $\mathbb{R} \ni t \mapsto e^{tL} \in \mathcal{L}(X)$ is differentiable with derivative
$$
  \frac{d}{dt} \exp(tL) = L \exp(tL) = \exp(tL)L.
  \qquad (*)
$$
Since for every given $u_0 \in X$ the mapping $\mathcal{L}(X) \ni T \mapsto Tu_0 \in X$ is linear and continuous, this implies readily the formula in the question for each $u_0 \in X$.
Chain rule
As pointed out in a comment by LSpice, the formula $(*)$ can also be interpreted as a consequence of the chain rule for differentiable maps between Banach spaces.
Due to the non-commutativity of the multiplication of linear operators, it is not easy to determine the derivative of the mapping $\exp: \mathcal{L}(X) \to \mathcal{L}(X)$, even for finite dimensional $X$.
But things become clearer if one considers the closed subalgebra of $\mathcal{L}(X)$ that is generated by a fixed operator $L$ and the identity operator; let's denote this algebra by $\mathcal{A}(L)$. The advantage is that this algebra is commutative. The exponential mapping is continuous on $\mathcal{L}(X)$ (even this is not completely obvious at first glance; but one can show it by using a non-commutative version of the geometric sum formula) and hence, $\exp$ leaves $\mathcal{A}(L)$ invariant.
By using the commutativity of $\mathcal{A}(L)$ one can check that the mapping $\exp: \mathcal{A}(L) \to \mathcal{A}(L)$ is differentiable, and its (Fréchet) derivative at any point $T \in \mathcal{A}(L)$ is simply given by the multiplication with $\exp(T)$ (note that the derivative at any point in $\mathcal{A}(L)$ is, by definition, a linear map from $\mathcal{A}(L)$ to $\mathcal{A}(L)$).
Moreover, the mapping $\mathbb{R} \ni t \to tL \in \mathcal{A}(L)$ is linear and thus its own derivative. But one typically likes to interprete derivatives of a path - i.e. of a differentiable mapping $\varphi$ from an interval into a Banach space $Y$ - as an element of $Y$ rather than as linear mapping $\mathbb{R} \to Y$; and this is done by identifying the derivative with its value at $1$. (Note that the formula in the question and the formula $(*)$ also implicitly use this identification.)
By using this identification, the derivative of $\mathbb{R} \ni t \to tL \in \mathcal{A}(L)$ becomes the operator $L$.
Now the chain rule, applied to the composition
$$
  \mathbb{R} \overset{t \mapsto tL}{\to} \mathcal{A}(L) \overset{\exp}{\to} \mathcal{A}(L)
$$
gives the formula $(*)$.
(I'm not sure whether this chain rule approach might also be helpful for your MathOverflow question here.
In order to determine this it would be necessary to revise your question there in order to explicitly state its precise setting .)
Functional calculus
There is also a way to derive the formula $(*)$ from the same formula for the scalar exponential function $\exp: \mathbb{C} \to \mathbb{C}$.
This can be done by representing $\exp(tL)$ by means of the holomorphic functional calculus, which means the formula
$$
  \exp(tL) = \frac{1}{2\pi i} \int_{\beta} \exp(t\lambda) (\lambda - L)^{-1} \, d\lambda,
$$
where $\beta$ is any cycle in the complex plane the encircles each point in the spectrum of $L$ exactly once, and where the integral is meant as a Riemann integral with values in the Banach space $\mathcal{L}(X)$.
To obtain $(*)$ one just needs a rule for differentiation of parameter integrals - which is similar to the case of scalar-valued integrals.
$C_0$-semigroups
In infinite dimensions, the case where $L$ is a bounded linear operator is not too useful, since most evolution equations contain differential operators, which are unbounded rather than bounded operators.
For this case, the exponential series does not converge, in general (and in fact, it is not even completely clear how the exponential series shoud be interpreted), so one needs a different definition for $\exp(tL)$, then.
This is provided by the theory of $C_0$-semigroups.
(But note that, in many case, this only yields a sensible definition of $\exp(tL)$ for $t \ge 0$, and it also depends on the properties of $L$ whether $\exp(tL)$ makes sense at all within this theory.)
A very good reference for this theory is the following book (but there are also many further good books for this topic):
[1] Engel and Nagel: One-Parameter Semigroups for Linear Evolution Equations, Springer, 2000
For a $C_0$-semigroups with generator $L$ on a Banach space $X$ one cannot expect the formula in the question to be true for all $u_0 \in X$, in general.
However, it is always true for all $u_0$ in the domain of $L$; see for instance [1, Lemma II.1.3(ii)].
For the special case of so-called analytic semigroup and, more generally, for immediately differentiable semigroups, the formula is even true for all $u_0 \in D(A)$, but only if $t > 0$ (and not for $t=0$, in general); see [1, Sections II.4(a) and (b)] for details.
