The moment generating function of a normally distributed random variable is $\displaystyle M_{X}(t) = e^{\mu t + \sigma^{2} t^{2}/2} $. Find $\displaystyle E[X] $ and $\displaystyle \text{Var}(X) $.

So $\displaystyle M'_{X}(t) = (\mu + \sigma^{2}t) e^{\mu t + \sigma^{2} t^{2}/2} $ and thus $\displaystyle M'_{X}(0) = \mu $. Then find $\displaystyle E[X^2] $ by differentiating again and evaluating at $\displaystyle t = 0 $, and use shortcut formula $\displaystyle \mbox{Var}(X) = E[X^2] - (E[X])^{2} $?

Couldn't you also use $\displaystyle R_{X}(t) = \ln(e^{\mu t + \sigma^{2} t^{2}/2}) $, and the first and second derivatives would give you $\displaystyle E[X] $ and $\displaystyle \text{Var}(X) $ respectively?