Strange derivation for decay in particle physics textbook. Okay so here's the derivation:
$P(x)$: probability to not interact after distance $x$
$w \space dx$: probability to interact between $x$ and $dx$
The probability of not interacting between $x+dx$ is then:
$$
P(x+dx)=P(x)(1-w \space dx) \\
P(x)+\frac{dP}{dx}dx=P-P\space w \space dx
$$
It's that last step I don't get. How does $P(x+dx)=P(x)+\frac{dP}{dx}dx$? I can see how it's like adding $dP(x)$ but is there a more rigorous way to analyse this?
 A: In the "physicist notation" you are just using $\frac{dP}{dx} dx = dP$, which is actually exact equality (there is no approximation involved). Standard analysis would make this rigorous by introducing an approximation and its error, by writing
$$P(x+\Delta x)=P(x)-wP(x) \Delta x+o(\Delta x).$$
This rearranges to
$$\frac{P(x+\Delta x)-P(x)}{\Delta x} = -wP(x) + o(1).$$
Then we take the limit, and conclude that the left side converges to $-wP(x)$ as $\Delta x \to 0$.
A: That says that the value a little distance from here ($x + dx)$ is the value here $P(x)$ plus the rate-of-change of value with distance $(dP/dx)$ times the distance moved. 
It's not exactly correct, but it's a good approximation, called the first-order Taylor approximation of $P$ around $x$. 
Suppose you're driving 30 miles per hour, and keeping track of the temperature change. In the first hour, the temp rises 2 degrees, from 28 to 30, and you expect that to keep happening. What do you expect the temp to be half an hour from now? Answer: the current 30 degrees, plus (2deg/hour) * (1/2 hour) = 30 + 1 = 31 degrees. Same idea is going on here. 
A: Using high school mathematics you should be familar with
$$
f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\approx \frac{f(x+h)-f(x)}{h}
$$
swapping in $f\to P$ and $h\to dx$ we find
$$
\frac{dP}{dx} \approx \frac{P(x+dx)-P(x)}{dx}
$$
so we get
$$
P(x+dx) = P(x) + \frac{dP}{dx}dx
$$
where we assume the approximation is valid and we can create an equality.
