Prove Variance of a normal distribution is (sigma)^2 (using its moment generating function) Prove that the  Variance of a normal distribution is (sigma)^2 (using its moment generating function).
What I did so far:
$Var(X) = E(X^2) - (E(X))^2$
$$E(X^2) = Mx'(0) = \frac r{\sqrt{2\pi}*\sigma} * exp(-[(x-\mu)/\sigma]^2/2)$$
$E(X) = Mx''(0) = \frac {r^2}{\sqrt{2\pi}*\sigma} * exp(-[(x-\mu)/\sigma]^2/2)$
So $Var(X) = \frac r{\sqrt{2\pi}*\sigma} * exp(-[(x-\mu)/\sigma]^2/2) -(r^4)/[(2\pi)*(\sigma^2)] * exp(-[(x-\mu)/\sigma]^2)$
However, I dont see how I can proceed from here. How can I take these terms together? Could anyone please help me out?
 A: It is not clear from the wording of the question whether you have been given the mgf or not. If you have not, then we need to calculate it.  A sketch of how to do this is given in the remark at the end. But for now let us assume we know the mgf.
Suppose we know that $M_X(t)$, the mgf of the  random variable $X$, is given by $M_X(t)=\exp(a t+\frac{1}{2}b t^2)$. We want to show that $a$ is the mean of $X$ and $b$ is the variance of $X$.
Calculate $M_X'(t)$. We get
$$M_X'(t)=(a+bt)\exp(a t+\frac{1}{2}b t^2).$$
So $M_X'(0)=a$. That means that $a$ is the mean $\mu$ of $X$, that is, $a=E(X)$.
To find $E(X^2)$, we find the second derivative of $M_X(t)$ at $t=0$. We get
$$M_X''(t)=(a+bt)^2 \exp(a t+\frac{1}{2}b t^2)+b\exp(a t+\frac{1}{2}b t^2).$$
Set $t=0$. We get $M_X''(0)=a^2+b$. So $E(X^2)=a^2+b$.
Finally, the variance of $X$ is $E(X^2)-(E(X))^2$. (Please note: it is not $E(X^2)-E(X)$.) We have
$$E(X^2)-(E(X))^2=(a^2+b)-a^2=b.$$ 
Remark:  We sketch how to find the mgf of the general normal. Recall that $M_X(t)=E(e^{tX})$. Thus
$$M_X(t)=\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma} \exp(tx)\exp(-(x-\mu)^2/(2\sigma^2))\,dx.$$
The product of the exponentials is $\exp\left(-\frac{x^2-2x(\mu+t\sigma^2)+\mu^2}{2\sigma^2}\right)$.  Now make the change of variable $\sigma u=x-(\mu +t\sigma^2)$. After some mildly tedious algebra, and using the fact that $\int_{-\infty}^\infty e^{-w^2/2}\,dw=\sqrt{2\pi}$, we get the mgf.  
