Differential of normal distribution Let
$$f(x)=\frac{\exp\left(-\frac{x^2}{2\sigma^2}\right)}{\sigma\sqrt{2\pi}}$$
(Normal distribution curve) Where $\sigma$ is constant. Is my derivative correct and can it be simplified further? 
$$\begin{align} f'(x)
&=\frac d{dx}\left(\frac{\exp\left(-\frac{x^2}{2\sigma^2}\right)}{\sigma\sqrt{2\pi}}\right)\\
&=\frac{1}{\sigma\sqrt{2\pi}}\frac d{dx}\left(\exp\left(-\frac{x^2}{2\sigma^2}\right)\right)\\
\end{align}$$
$$\begin{align}
\implies\frac d{dx}\exp\left(-\frac{x^2}{2\sigma^2}\right)&=\frac d{dx}\sum_{n=0}^\infty\frac{\left(-\frac{x^2}{2\sigma^2}\right)^n}{n!}
=\sum_{n=0}^\infty\frac d{dx}\frac{\left(-\frac{x^2}{2\sigma^2}\right)^n}{n!}\\
&=\sum_{n=0}^\infty \frac 1{n!}\frac d{dx}\left(-\frac{x^2}{2\sigma^2}\right)^n=\sum_{n=0}^\infty -\frac 1{n!}\frac{1}{2\sigma^{2n}}\frac d{dx}x^{2n}\\
&=\sum_{n=0}^\infty -\frac 1{n!}\frac{1}{2\sigma^{2n}}2nx^{2n-1}=\sum_{n=0}^\infty -\frac {1}{n!}\frac{2nx^{2n-1}}{2\sigma^{2n}}\\
\end{align}$$
$$\implies f'(x)=\frac{-\sum_{n=0}^\infty \left(\frac {1}{n!}\frac{2nx^{2n-1}}{2\sigma^{2n}}\right)}{\sigma\sqrt{2\pi}}$$
I attempted to simplify the infinite summation to the exponential function but I am failing. How would you simplify this further?

Okay so it simplifies to:
$$f'(x)=-\frac{x\exp\left(-\frac{x^2}{2\sigma^2}\right)}{\sigma^3\sqrt{2\pi}}$$
