Martingal differentiation

Let $(X_n)$ be a sequence of independent, identically distributed random variables with finite moment-generating function $M(t) = \mathbb{E}\left[\exp(tX_1\right)] < \infty$ for $t \in \mathbb{R}$. Let $S_n = \sum_{i=1}^n X_i$ and $Y_n = \frac{1}{(M(t))^n}\exp(tS_n)$ for $n \geq 0$ and $t \in \mathbb{R}$.

In post Sequence of martingales it was proven that $(Y_n)_{n \geq 0}$ is a martingal. How could one show now that differentiating $Y_n$ at $t=0$ one would still obtain martingals?

What I've tried so far:

$\frac{dX_n}{dt} = -n \frac{\frac{dM(t)}{dt}}{M(t)^{n+1}}e^{tS_n} + \frac{1}{M(t)^{n}}S_ne^{tS_n}$. Taking $t=0$ in the above equation I got $\frac{dX_n}{dt}|_{t=0} = S_n - n$. Or am I missing something?

Thanks

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What did you try? What is the expression you obtain for the derivative of $Y_n(t)$ at $t=0$? – Did Jan 20 '13 at 8:24
I edited the question description with what I've tried. – eugen1806 Jan 20 '13 at 14:42
You forgot to take into account $dM(t)/dt$ at $t=0$. – Did Jan 20 '13 at 14:56
Ah, yes, indeed. Then $\frac{dX_n}{dt}|_{t=0} = S_n - n \mathbb{E}(X_1)$ and $\mathbb{E}(S_n|\mathcal{F}_{n-1}) = \mathbb{E}(X_n + S_{n-1} |\mathcal{F}_{n-1}) = \mathbb{E}(X_n) + S_{n-1}$ and as a consequence $\mathbb{E}(S_n - n \mathbb{E}(X_1) | \mathcal{F}_{n-1}) = S_{n-1} - (n-1)\mathbb{E}(X_1)$. – eugen1806 Jan 20 '13 at 15:36