How is the derivative of the CDF of a random variable $X$ its PDF? For a continuos random variable $X$, if we have its p.d.f. $f(x)$, then the cumulative densitity function (c.d.f.) of $X$, $F(x)$, is
$$F(x) = \int_{-\infty}^{x}f(t)dt$$
We also have $$F'(x) = \frac{d}{dx}F(x)= f(x)$$

Why does the last step in the following equation yields $f(x)$?
$$\frac{d}{dx}F(x) = \frac{d}{dx} \int_{-\infty}^{x}f(t)dt = \int_{-\infty}^{x} \frac{\partial}{\partial x} f(t)dt $$
This is just the Fundamental Theorem of Calculus.

A PDF (of a univariate distribution) is a function defined such that it is 1.) everywhere non-negative and 2.) integrates to 1 over $\Bbb R$.

If we define $F(x) = \int_{-\infty}^x f(t)\ dt$, then the Fundamental Theorem of Calculus gives you the desired result.

This function, $F(x)$, is called the "cumulative distribution function," or CDF. It is defined in this manner, so the relationship between CDF and PDF is not coincidental -- it is by design.

Note that your last step is incorrect -- $x$ is the independent variable of the derivative there, and it is also the upper limit of the integral (so the resulting integral will be a function in terms of $x$). You can't move the $d/dx$ inside the integral.
http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
read the 1st part of proof, then you can understand but you are assumed to know mean value theorem and squezee theorem.
hope it helps.