Finding derivative of a definite integral I'm a self-taught, so this is possibly too basic, but I'm stuck in the last step of this segment of the derivation posted here in that I don't know whether you are supposed to solve the $\frac{d}{dx}\int_{0}^{\sqrt{x}}... du$ calculus step by thinking of a derivative of an integral leaving you where you started (intuitive enough), or if there is a change in variables, how you go about getting $\frac{d}{dx}\sqrt{x}$ in the final equation. 
\begin{align}
f(x) & = \frac{d}{dx} \Pr(X^2 \le x) = \frac{d}{dx} \Pr(-\sqrt{x}\le X\le\sqrt{x}) \\  \\
& = \frac{d}{dx} \frac{1}{\sqrt{2\pi}} \int_{-\sqrt{x}}^\sqrt{x} e^{-u^2/2} \;du = \frac{2}{\sqrt{2\pi}}\frac{d}{dx} \int_0^\sqrt{x} e^{-u^2/2} \;du \\  \\
& = \frac{2}{\sqrt{2\pi}} e^{-\sqrt{x}^2/2} \frac{d}{dx} \sqrt{x}
\end{align}
 A: There might be some confusion - I understand  that your question, and correct me please if I am mistaken, is how to differentiate the integral. This is often refered to as differentiation "under" the integral (or of), and sometimes called the Leibniz Integral Rule*. This rule is really a more general result, however, whereas what we are interested in is really an application of the fundamental theorem of calculus to differentiate an integral with variable limits. The reason there is a $\frac{d}{dx} \sqrt{x}$ in the final equation is because there is a formula for such a differentiation (the more general ones are available elsewhere including the linked wikipedia). I have included below the simple version of this formula.
$$
\frac{d}{dx} \int_a ^{f(x)} g(t) d t = g(f(x)) \cdot f'(x).
$$
For your integral, $f(x) = \sqrt{x}$, so 
$$
\frac{d}{dx} \frac{2}{\sqrt{2 \pi}} \int_0 ^{( \sqrt{x} )} e^{-t^2 / 2} d t
= \frac{2}{\sqrt{2 \pi}} \frac{d}{dx} \int_a ^{f(x)} g(t) d t 
= \frac{2}{\sqrt{2 \pi}} g(f(x)) \cdot f'(x)
$$
$$
= \frac{2}{\sqrt{2 \pi}} e^{-(\sqrt{x})^2 / 2} \cdot \frac{d}{dx}(\sqrt{x}).
$$
If your question is more along the lines of why this formula is so, perhaps I can offer a satisfactory intuitive explanation. The fundamental theorem of calculus basically says that to calculate a definite integral, you can use the antiderivative. So, suppose $G(t)$ is an antiderivative of $g(t)$. That means that 
$$
\int g(t) d t = G(t) \text{ and } G'(t) = g(t).
$$
Then, $ \int_a ^{f(x)} g(t) d t = G(f(x)) - G(a)$ by the fundamental theorem. But then, we know that
$$
\frac{d}{dx} \int_a ^{f(x)} g(t) d t 
= \frac{d}{dx} [G(f(x)) - G(a)]
= \frac{d}{dx} (G(f(x))) \text{ as } G(a) \text{ is constant.}
$$
So, by the chain rule for derivatives,
$$
\frac{d}{dx} \int_a ^{f(x)} g(t) d t 
= \frac{d}{dx} (G(f(x)))
= G'(f(x)) \cdot f'(x)
= g(f(x)) \cdot f'(x).
$$
* https://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign
A: Let $G(x)$ a primitive of $g(x)$, i.e.: 
$$
\dfrac{d}{dx} G(x)=g(x)
$$
than, by the Foundamental Theorem of Calculus we have:
$$
\int_0^{f(x)}g(t)dt= G(f(x))-G(0)
$$
and (using the chain rule)
$$
\dfrac{d}{dx}\int_0^{f(x)}g(t)dt=\dfrac{d}{dx}\left[G(f(x))-G(0) \right]=G'(f(x))f'(x)=g(f(x))f'(x)
$$
Use this result for $f(x)=\sqrt{x}$ and $g(x)=e^{-u^2/2}$ and you have the result.
