CDF derivative of pdf? How does this work?

For a random variable $X$ if we have a pdf $f(x)$, then Continuous Random Variable is

$$F(x) = \int_{-\infty}^{x}f(t)dt$$ Next $F'(x) = \frac{d}{dx}F(x)= f(x)$

I don't follow this, $$\frac{d}{dx}F(x) = \frac{d}{dx} \int_{-\infty}^{x}f(t)dt = \int_{-\infty}^{x} \frac{\partial}{\partial x} f(t)dt$$

I don't see how that last step yields $f(x)$

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you differentiate w.r.t. the upper limit as it is not an integral with a parameter. Also, the passage "... then Continuous Random Variable is" is remarkable by itself. Is it from the book? –  Ilya Nov 30 '12 at 20:42
Oh okay woops thank you –  sidht Nov 30 '12 at 20:43

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.