Partial derivative of normal cdf w.r.t sigma I am trying to find the partial derivative of univariate normal cdf w.r.t $\sigma$. I just need some direction. So far I have gotten this:
$\frac{\partial }{\partial \sigma}\Phi(x,\mu,\sigma^2) = \frac{\partial }{\partial \sigma} \displaystyle\int_{-\infty}^x \frac{1}{\sqrt{2\pi}}\exp\left(-\ln(\sigma)-\frac{(x-\mu)^2}{2\sigma^2}\right)dx $
$ = \displaystyle\int_{-\infty}^x\frac{\partial }{\partial \sigma}\frac{1}{\sqrt{2\pi}}\exp\left(-\ln(\sigma)-\frac{(x-\mu)^2}{2\sigma^2}\right)dx$
$ = \displaystyle\int_{-\infty}^x\frac{1}{\sqrt{2\pi}}\exp\left(-\ln(\sigma)-\frac{(x-\mu)^2}{2\sigma^2}\right)\times\left( -\frac{1}{\sigma}+\frac{(x-\mu)^2}{\sigma^3}\right)dx$
$ = \displaystyle\int_{-\infty}^x -\frac{1}{\sqrt{2\pi}\sigma^2}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right) + \frac{(x-\mu)^2}{\sqrt{2\pi}\sigma^4}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right) dx$
I don't know how to proceed from there.
 A: There is a much simpler way of proceeding. We know that:
$$N(\mu,\sigma^{2})$$
has cumulative distribution function:
$$F_{X}(x;\mu,\sigma^{2})=\Phi\bigg(\frac{x-\mu}{\sigma}\bigg)$$
where $\Phi$ is the cumulative distribution function of the standard normal. We can proceed as follows:
$$\begin{align}
\frac{\partial}{\partial \sigma}F_{X}(x;\mu,\sigma^{2})&=\frac{\partial}{\partial \sigma}\Phi\bigg(\frac{x-\mu}{\sigma}\bigg)
\end{align}$$
Using the chain rule, this should set you up.

Using your approach is a little bit more involved. This was my attempt at it. According to Leibniz's rule, we know that:
$$\frac{\partial}{\partial\sigma}\int_{-\infty}^{x}f_{X}(x;\mu,\sigma^{2})\,dx=\int_{-\infty}^{x}\frac{\partial}{\partial\sigma}f_{X}(x;\mu,\sigma^{2})\,dx$$
Now, as you've already calculated:
$$\frac{\partial}{\partial\sigma}f_{X}(x;\mu,\sigma^{2})=-\frac{1}{\sqrt{2\pi}\sigma^{2}}e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}+\frac{(x-\mu)^{2}}{\sqrt{2\pi}\sigma^{4}}e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}$$
Now, the first of the terms above is simple enough to integrate:
$$\int_{-\infty}^{x}-\frac{1}{\sqrt{2\pi}\sigma^{2}}e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}\,dx=-\frac{1}{\sigma}\Phi\bigg(\frac{x-\mu}{\sigma}\bigg)$$
Let's focus now on the second term. For the sake of simple notation, let's allow:
$$z=\frac{x-\mu}{\sigma}$$
This gives:
$$\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{x}z^{2}e^{-\frac{z^{2}}{2}}\,dz$$
To solve, we'll use integration by parts. Let:
$$\begin{align}
u&=z,\quad dv=ze^{-\frac{z^{2}}{2}}\\
du&=1,\quad\,\,\, v=-e^{-\frac{z^{2}}{2}}
\end{align}$$
So,
$$\begin{align}
\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{x}z^{2}e^{-\frac{z^{2}}{2}}\,dz&=\frac{1}{\sqrt{2\pi}\sigma}\bigg[-ze^{-\frac{z^{2}}{2}}\bigg]_{-\infty}^{x}+\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{x}e^{-\frac{z^{2}}{2}}\,dz\\
&=\frac{1}{\sqrt{2\pi}\sigma}\bigg[-\bigg(\frac{x-\mu}{\sigma}\bigg)e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}-0\bigg]+\frac{1}{\sigma}\Phi\bigg(\frac{x-\mu}{\sigma}\bigg)\\
&=-\bigg(\frac{x-\mu}{\sigma^{2}}\bigg)\frac{1}{\sqrt{2\pi}}e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}+\frac{1}{\sigma}\Phi\bigg(\frac{x-\mu}{\sigma}\bigg)\\
&=-\bigg(\frac{x-\mu}{\sigma^{2}}\bigg)\phi\bigg(\frac{x-\mu}{\sigma}\bigg)+\frac{1}{\sigma}\Phi\bigg(\frac{x-\mu}{\sigma}\bigg)
\end{align}$$
Thus,
$$\begin{align}
\frac{\partial}{\partial\sigma}\int_{-\infty}^{x}f_{X}(x;\mu,\sigma^{2})\,dx&=-\frac{1}{\sigma}\Phi\bigg(\frac{x-\mu}{\sigma}\bigg)-\bigg(\frac{x-\mu}{\sigma^{2}}\bigg)\phi\bigg(\frac{x-\mu}{\sigma}\bigg)+\frac{1}{\sigma}\Phi\bigg(\frac{x-\mu}{\sigma}\bigg)\\
&=-\bigg(\frac{x-\mu}{\sigma^{2}}\bigg)\phi\bigg(\frac{x-\mu}{\sigma}\bigg)
\end{align}$$
