Interpretation of the slope in a probability distribution. When considering a certain probability-distribution $P(t)$, then what information does the derivative $\frac{\mathrm d P(t)}{\mathrm{dt}}$ give, in general?
Context:
I raised this question on account of a physics exercise, which I could not rigorously complete. In short the question goes like, 
assume that the probability of an electron, in an ensemble of electrons, suffering a collision in an interval $\mathrm{dt}$ is $\mathrm{dt} / \tau$. Show that an electron picked at random at a given moment had no collisions during the the next $t$ seconds with probability $e^{-t/\tau}$. 
Noting that $P(t)$ gives the probability that certain electrons have not yet suffered a collision, the difference $\frac{\mathrm d P(t)}{\mathrm{dt}} \cdot \mathrm{dt}$ gives the probabilty that those electrons have suffered a collision in an interval $\mathrm{dt}$. We now devide by $P(t)$, to treat the case for one particular electron. So we get,
$$
\frac{\frac{\mathrm d P(t)}{\mathrm{dt}}\cdot \mathrm{dt}}{P(t)}  = -\frac{\mathrm{dt}}{\tau} \implies P(t) = e^{-t/\tau}.  
$$ 
PS: Please tell me if this question is too off topic and belongs in the physics section.
 A: Your calculation is a variant of a standard approach to the derivation
of the Poisson process model.  Let us pick an electron at time $t = 0$
and ask for the probability that it suffers no collisions in the
interval $(0,T]$.  The assumptions are that
(i) in a very short interval of time $\Delta t$, the electron c
an suffer at most one collision,
and the probability of a collision is $\frac{\Delta t}{\tau}$
and 
(ii) the numbers of collisions suffered by the electron in any 
two disjoint intervals of time $(t_1,t_2]$ and $(t_3, t_4]$
(where $t_1 < t_2 \leq t_3 < t_4$) are independent random variables.
Thus, with $P(T)$ denoting the probability that the electron
suffers no collisions in the interval $(0,T]$, we have that
$$\begin{align}
P(T+\Delta T) &= P\{\text{no collisions in interval}~ (0,T+\Delta T]\}\\
&= P\{\text{no collisions in interval}~ (0,T]~\text{and}~\text{no collisions in interval}~ (T,T+\Delta T]\}\\
&= P\{\text{no collisions in interval}~ (0,T]\}\cdot P\{\text{no collisions in interval}~ (T,T+\Delta T]\}\\
&= P(T)\left(1-\frac{\Delta T}{\tau}\right)
\end{align}$$
leading to
$\displaystyle\frac{P(T+\Delta T)-P(T)}{\Delta T} = -\frac{1}{\tau}P(T) 
~\Rightarrow~ \frac{\mathrm dP(T)}{\mathrm dT} = -\frac{1}{\tau}P(T)
~\Rightarrow~ P(T)= e^{-T/\tau}$.  If the assumptions make sense
in physics (as opposed to mathematics), then you can make
the following interpretation for the derivative of $P(T)$.

The tangent to the $P(T)$ curve (whose slope is, of course,
  $\frac{\mathrm dP(T)}{\mathrm dT}$, and is a negative number), 
  crosses the horizontal axis at $T + \tau$ showing that
  at the current rate of change, the probability of no collisions
  will reduce to $0$ in a further $\tau$ seconds. Now, $\tau$ is
  the expected time till the first collision occurs after $t=0$.
  In fact, even though there have been no collisions thus far in our
  model, $\tau$ is still the expected time till the next collision.
  This is the memoryless property of the exponential distribution. 

