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If $X$ follows a distribution with p.d.f. $f(x)$ and its moment-generating function is $$M_{X}(t) = E(e^{tX}).$$ What should the derivative of $M_{X}(t)$ , $\frac{d}{dt}M_{X}(t)$ be?

Thanks for your time.

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    $\begingroup$ $E(Xe^{tX})$? The definition of $e^{tX}$ uses the exponential's power series. $\endgroup$
    – Chappers
    Jul 17 '17 at 9:23
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You have $$\mathbb{E}(e^{tX}) = \int e^{tx}f(x) ~\mathrm{d}x$$ so we obtain $$\frac{d}{dt} \mathbb{E}(e^{tX}) = \int xe^{tx}f(x)~\mathrm{d}x = \mathbb{E}(Xe^{tX})$$ at least if this integral exists finitely for all $s$ in some neighborhood of $t$.

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