The price of an action follow the function $f(t)=e^{-t}$. The question is, what is the rate of change of the price ? For me, the rate of change is the rate of change by unit of time $h$, i.e. $$\frac{f(t+h)-f(t)}{(t+h)-t}=\frac{e^{-t-h}-e^{-t}}{h}=e^{-t}\left(\frac{e^{-h}-1}{h}\right).$$
In other word, during an interval of time $[x,x+h]$, the price of the action increased by $\left(\frac{e^{-h}+1}{h}\right)he^{-t}$. I understand it as : the rate of change of the price is $\left(\frac{e^{-h}+1}{h}\right)$ multiplicate by a quantity that depend on the position only (here is $e^{-t}$). But the most important is $\frac{e^{-h}-1}{h}$ that really describe the rate of increasing independently on the position.
In my solution, they say that it's the derivative, i.e. $e^{-t}$. I really don't understand how to interpret this result. What does it mean exactly ? If the rate of change of the time is $h$, then the rate of change of the price is $e^{-t}$ ? It doesn't really make sense for me... Could someone explain ?
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$\begingroup$It should be $e^{-h}-1$ in the numerator.
This goes back to fundamental calculus. For a nonlinear function, the rate of change doesn't really make sense, because the function can wildly change between $t$ and $t+h$. So what you've calculated is some approximate rate of change over an interval. Presumably the question is after the instantaneous rate of change, which is your result under the limit $\lim_{h\rightarrow 0}$. A careful calculation of the limit will yield the answer, that it's just the derivative of the original function.
$\endgroup$ 2 $\begingroup$I'm going to take a different tack (and one that I wish I had seen when I was a student studying this material).
Let's plot $\frac{\mathrm{e}^{-h} - 1}{h}$ versus $h$.
This graph contains all the information of the rate of change as we vary $h$. Notice, there is a hole in the graph at $h = 0$, because division by zero is undefined. However, it is quite clear what value the continuous extension of the function to include $h = 0$ is, $-1$.
If I ask you for an average rate of change, I have to tell you $h$ so that you know which point from this graph to report. If I ask you for the average rate of change when $h=0$, that is undefined (again, because division by zero is undefined). But if I ask you what it would be, the answer is $-1$, as we can easily see. So the average rate of change when $h=0$ is undefined, but the instantaneous rate of change can be defined, using the limit as $h \rightarrow 0$ of the average rate(s)${}^*$ of change, and has a definite value.
${}^*$ When we take a limit, we imagine a sequence of $h$ values and a sequence of average rates of change. However, grammatically, the use of a plural is incorrect, since we are taking the limit of one thing, the average rate of change versus $h$, not several different things, because we only have one thing.
$\endgroup$ $\begingroup$The average rate of change over some interval of length $h$ starting at time $t$ is given by$$ e^{-t}\left(\frac{e^{-h}-1}h\right) $$The point of the derivative is to see what happens to this rate when this interval becomes very, very small. It's called "momentary rate of change" for a reason. And when $h$ comes very close to $0$, the expression in the bracket comes very close to $-1$. So the derivative is $-e^{-t}$, and it describes the rate of change at the exact moment $t$.
$\endgroup$ 4 $\begingroup$You probably know slope, or the average rate of change between two points. Let’s assume there is a function $f(x)$. The average rate of change is interpreted as the slope of a secant passing through those two points. In other words, the ratio of the change in the dependent variable to the change in the independent variable:
$$\overline{m} = \frac{\Delta f(x)}{\Delta x} = \frac{f(x+h)-f(x)}{h}$$
Which in this case, as you’ve mentioned, is
$$\overline{m} = \frac{e^{-x-h}-e^{-x}}{h} = e^{-x}\cdot\frac{e^{-h}-1}{h}$$
This would give you the average rate in change of the price. If you were to plot this function and take any two points, $\overline{m}$ would give the slope of the straight secant line passing through those two points.
However, what if you want the rate of change at one instant in time, or rather, the slope at one point? You would have to take those two points and shrink $\Delta x$ down to $0$, so you get
$$m = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$$
Which is basically taking two points infinitely (might not really be the proper term here, but it gives you the idea) close to each other to give the slope, or rate of change, at that particular point. In your case,
$$m = e^{-x}\cdot\lim_{h \to 0}\frac{e^{-h}-1}{h} \to -e^{-x}$$
which would give the instantaneous rate of change, or slope, at point $x$.
$\endgroup$ 4 $\begingroup$Just adding a point that none of the other answers have mentioned yet: when calculus was first developed, the derivative was seen as the result of dividing an infinitely small change by an infinitely small interval—basically a sophisticated way of calculating $\frac00$.
This was an embarrassment: although calculus worked, it wasn't at all clear that the final step of dividing one infinitely small quantity by another was actually valid. It wasn't until the nineteenth century that the solution we use today was found, which is basically to define everything in terms of limits so we nicely sidestep things like $\frac00$ and $0\cdot\infty$.
Until then, everyone had to pretend that it did make sense to, say, calculate a speed by dividing a distance of $0$ by a time of $0$.
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