$\begingroup$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 $f(x)$?
$$\frac{d}{dx}F(x) = \frac{d}{dx} \int_{-\infty}^{x}f(t)dt = \int_{-\infty}^{x} \frac{\partial}{\partial x} f(t)dt $$
$\endgroup$05 Answers
$\begingroup$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.
$\endgroup$1$\begingroup$You can see this by differentiating under the integral sign, which follows from the fundamental theorem of calculus:
$$ \frac{d}{dx} F(x) =\lim_{c\to-\infty} \frac{d}{dx} \int^{x}_{c} f(t) dt = f(x).1 -\lim_{c\to-\infty} f(c).\frac{dc}{dx} + \lim_{c\to-\infty}\int^{x}_{c} \frac{d}{dx} f(t) dt $$
Since $c\to-\infty$ is a constant, the second term disappears, and since $f$ is a function of $t$, $\frac{d}{dx} f(t)$ also disappears.
$\endgroup$2$\begingroup$Since the accepted answer doesn't explicitly write out the proof, and considering that the other derivation uses Leibniz's rule, here's the answer in an explicit form: $$F(x) = \int_{-\infty}^{x}{f(t)dt}$$ From the fundamental theorem of calculus, if $P(x)$ is the indefinite integral of $f(x)$ : $f(x) = \frac{dP}{dx}(x)$, then $P(b) - P(a) = \int_{a}^{b}{f(t)dt}$. Then $F(x) = P(x) - \lim_{a \to -\infty}P(a)$. $$\frac{dF}{dx}(x) = \frac{d}{dx}[P(x) - \lim_{a \to -\infty}P(a)] = \frac{dP}{dx}(x) = f(x) $$
$\endgroup$$\begingroup$Here is an even more detailed answer, since the last one does not explicitly state that it made use of the simple property: $\frac{d}{dy} f(x)=0$.
Let:
$$F(x) = \int_{-\infty}^{x}{f(t)dt}$$
Then, by the Fundamental Theorem of Calculus, it holds that:
$$F(x) = F_f(x) - F_f(-\infty) \tag{1}$$
where $F_f(x)$ is the antiderivative (indefinite integral) of $f(x)$, i.e.:
$$F_f(x) = \int{f(x)dx} \tag{2}$$
$\rm (1)$ is inserted into $F'(x)$:
$$\begin{align} F'(x) &= \frac{d}{dx}F(x) \\ &\overset{\rm{(1)}}{=} \frac{d}{dx}(F_f(x) - F_f(-\infty)) \\ &= \frac{d}{dx}F_f(x) - \frac{d}{dx}F_f(-\infty) \\ & \overset{\rm{(2)}}{=} f(x) - 0 \\ \end{align}$$
The last equality holds because the argument $x$, that the derivation is done with respect to, is not present in $F_f(-\infty)$. Thus $F'(x) = f(x)$, regardless of the lower bound of the integral that $F(x)$ is defined over (here: $-\infty$).
$\endgroup$$\begingroup$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.
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