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

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

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