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?

If you want to use the method that you started, then you should note that we can only "pull out the negative sign" from $$\left(-\frac{x^2}{2\sigma^2}\right)^n$$ when $n$ is odd. It would be better to keep a $(-1)^n$ term around, because then, $$\begin{align}\frac d{dx}\exp\left(-\frac{x^2}{2\sigma^2}\right) &= \sum_{n=0}^\infty \frac 1{n!}\frac d{dx}\left(-\frac{x^2}{2\sigma^2}\right)^n\qquad[\text{as you showed}]\\ &= \sum_{n=0}^\infty \frac 1{n!}\frac{(-1)^n}{2^n\sigma^{2n}}\frac d{dx}\left[x^{2n}\right]\\ &= \sum_{n=0}^\infty \frac 1{n!}\frac{(-1)^n}{2^n\sigma^{2n}}\cdot 2nx^{2n-1}\\ &= \sum_{n=1}^\infty \frac 1{n!}\frac{(-1)^n}{2^n\sigma^{2n}}\cdot 2nx^{2n-1}\qquad[\text{the }n=0\text{ term vanishes}]\\ &= \sum_{n=1}^\infty \frac 1{(n-1)!}\frac{(-1)^n}{2^{n-1}\sigma^{2n}} x^{2n-1}\qquad[\text{cancellation}]\\ &= \sum_{n=0}^\infty \frac 1{n!}\frac{(-1)^{n+1}}{2^n\sigma^{2(n+1)}} x^{2(n+1)-1}\qquad[\text{reindex by }n\mapsto n+1]\\ &= \sum_{n=0}^\infty \frac 1{n!}\frac{(-1)^{n+1}}{2^n\sigma^{2n+2}} x^{2n+1}\\ &= -\frac{x}{\sigma^2}\sum_{n=0}^\infty \frac 1{n!}\frac{(-1)^n}{2^n\sigma^{2n}} x^{2n}\\ &= -\frac{x}{\sigma^2}\sum_{n=0}^\infty \frac 1{n!}\left(-\frac{x^2}{2\sigma^2}\right)^n\\ &= -\frac{x}{\sigma^2}\exp\left(-\frac{x^2}{2\sigma^2}\right).\end{align}$$
There are far too many places to go wrong, here, though. Better just to use the chain rule.
