Here is a picture of my problem
Basically, given that $R_X(t) = \log(M_X(t))$, I need to show that $\text{Var}(X) = R′′(0)$.
As an attempt, I know that $\text{Var}[X] = E[X^2] - (E[X])^2$ and that $R'_X(0) = E(X)$, but I'm not sure how to get $E[X^2]$.
$R_X(t)$ is called the cumulant of $X$, which is the natural logarithm of the moment generating function $M_X(t) = E[e^{tX}]$. It is well known that if $M_X$ is defined on an open interval containing $0$, then $$E[X] = M'(0), \quad E[X^2] = M''(0).$$ It therefore follows that \begin{align} & R'(0) = \frac{M'(0)}{M(0)} = E[X] \\ & R''(0) = \frac{M''(0)M(0) - (M'(0))^2}{M(0)^2} = E[X^2] - (E[X])^2 = \text{Var}(X). \end{align}