Say $X \sim N(\mu, \sigma^2)$ with pdf $f(x, \mu)$. We are interested in expectation of $g(x)$. Then $$E[g(x, \mu)] = \int_{-\infty}^{\infty} g(x, \mu) f(x, \mu) dx$$
Now I want partial derivative of this. Why does the following hold? $$\frac{d}{d\mu}E[g(x, \mu)]= \int_{-\infty}^{\infty} \frac{dg(x, \mu)}{d\mu} f(x, \mu) dx = \int_{-\infty}^{\infty} g(x, \mu) \frac{df(x)}{d\mu} dx$$
I mean, why not the chain rule inside the integral and only one function is partially differentiated?
It is easy to understand the rationale of it, but I am struggling with the proof...
This question has such messy notation and mixes up so many different things that it is difficult to figure out what exactly is being asked.
Let $g(x,\mu)$ denote an ordinary two-variable real-valued function of two real variables. We will assume that the function is integrable with respect to both variables. Let $X$ denote a normal random variable with mean $\mu$ and variance $\sigma^2$. Thus, $X$ has a probability density function given by $$f_X(t) = f(t,\mu) = \frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{1}{2} \left(\frac{t-\mu}{\sigma}\right)^2\right), ~~-\infty < t < \infty.\tag{1}$$ Now, there is nothing random about the function $g(x,\mu)$ and so its expected value is just $g(x,\mu)$ 1tself. Indeed the same notion,
is a fundamental notion of probability theory (as well as of real life). More formally, $$E[g(x,\mu)] = \int_{-\infty}^\infty g(x,\mu)f_X(t)\,\mathrm dt = g(x,\mu)\int_{-\infty}^\infty f(t,\mu)\,\mathrm dt = g(x,\mu). \tag{2}$$ and it is worth noting in passing that we would have obtained the same answer even if we had used a different density function for $X$, or indeed a discrete mass function, etc. From $(2)$, we have that $$\frac{\partial}{\partial\mu}E[g(x,\mu)] = \frac{\partial}{\partial\mu}g(x,\mu)\tag{3}$$ and that's all there is to it. You could, if you like, write $(3)$ as $$\frac{\partial}{\partial\mu}E[g(x,\mu)] = \int_{-\infty}^{\infty}\frac{\partial}{\partial\mu}g(x,\mu) f(t,\mu)\,\mathrm dt\tag{4}$$ which sort of looks like what you want to prove, but actually isn't at all the same.
"But, but, but,..." you splutter indignantly, "All this is pure hooey. Everybody, with the possible exception of Old Harry And His Old Aunt, knows that $(1)$ is incorrect: the density of $X$ is $f_X(x)$ and not $f_X(t)$ as you have written in $(1$). Thus $(2)$ is not at all what I know $E[g(x,\mu)]$ to be. It's got to be $$E[g(x,\mu)] = \int_{-\infty}^\infty g(x,\mu)f(x,\mu)\,\mathrm dx \tag{5}$$ the way I wrote it in my question."
All right, have it your way.. If $Y = g(X,\mu)$ is a random variable that is a function of the random variable $X$, then the law of the unconscious statistician gives us that $$E[g(X,\mu)] = \int_{-\infty}^\infty g(x,\mu)f(x,\mu)\,\mathrm dx. \tag{6}$$ If you cannot see any difference between $(5)$ and $(6)$, compare the 5th character in each equation; that is a fundamental difference.
The equation $(6)$ can be differentiated on both sides with respect to $\mu$ giving \begin{align} \frac{\partial}{\partial\mu}E[g(X,\mu)] &= \frac{\partial}{\partial\mu}\int_{-\infty}^\infty g(x,\mu)f(x,\mu)\,\mathrm dx\\ &=\int_{-\infty}^\infty \left(f(x,\mu)\frac{\partial}{\partial\mu}g(x,\mu)+g(x,\mu)\frac{\partial}{\partial\mu}f(x,\mu)\right)\,\mathrm dx\\ &= \int_{-\infty}^\infty f(x,\mu)\frac{\partial}{\partial\mu}g(x,\mu)\,\mathrm dx + \int_{-\infty}^\infty g(x,\mu)\frac{\partial}{\partial\mu}f(x,\mu)\,\mathrm dx\tag{7} \end{align} which more nearly resembles what you want to prove if you replace that second $=$ sign in your question with a $+$ sign. But note that $$\frac{\partial}{\partial\mu}f(x,\mu) = \frac{x-\mu}{\sigma^2}f(x,\mu)$$ and so if $g(x,\mu)$ is an even function of $(x-\mu)$, then the integrand in the second integral in $(7)$ is a odd function of $(x-\mu)$, the value of that integral is $0$ and so $$\frac{\partial}{\partial\mu}E[g(X,\mu)] = \int_{-\infty}^\infty f(x,\mu)\frac{\partial}{\partial\mu}g(x,\mu)\,\mathrm dx$$ just the way you want it.