Derivative of moment generating function

Question

If the moment generating function of $X$ exists, i.e., $$M_X(t)=E[e^{tX}],$$ then the derivative with respect to $t$ is usually taken as $$\frac{dM_X(t)}{dt}=E[Xe^{tX}].$$ Usually, if we want to change the order of derivative and calculus, there are some conditions need to verified. Why the derivative goes inside for the moment generating function?

Answer 1

Lame, but what I could find:

(1) "If a moment generating function exists, then $m(t)$ is continuously differentiable in some neighborhood of the origin." Mood, Graybill, Boes (1974) An Intro. to the Theory of Statistics, 3e, p78.

If you make differentiability part of the definition of MGF, then there is nothing to prove.

(2) Similarly, shortly after the definition and informal proofs of properties of MGFs, "If the MGF of $X$ exists, then $E(X^r) = M_X^{(r)}(0)$ for all $r = 1,2, \dots$ and $M_X(t) = 1 + \sum_{i=1}^\infty \frac{E(X^r)t^r}{r!}.$" Bain, Englehardt (1992): Intro. to Probability and Mathematical Statistics, 2e, p79.

(3) However, many other authors use phrases such as "assuming differentiation is allowed [inside the expectation]..." (always without elaboration in sources I found).

Because characteristic functions (which exist for all distributions) are used in more formal and advanced texts, I'm wondering if there are slightly different definitions of MGFs in various mid-level texts.

The relevant Wikipedia article (as of 12 Sept '17) seems skeletal--and unhelpful on this issue.