Mixing
Why the derivative is the rate of change of the function?
$begingroup$
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 ?
real-analysis derivatives
asked Jan 3 at 22:21
user623855user623855
1287
$endgroup$
add a comment |
$begingroup$
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 ?
real-analysis derivatives
asked Jan 3 at 22:21
user623855user623855
1287
$endgroup$
$begingroup$
Related (very) math.stackexchange.com/questions/1781642/…
$endgroup$
– Ethan Bolker
Jan 4 at 1:25
add a comment |
$begingroup$
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 ?
real-analysis derivatives
asked Jan 3 at 22:21
user623855user623855
1287
$endgroup$
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 ?
real-analysis derivatives
real-analysis derivatives
asked Jan 3 at 22:21
user623855user623855
1287
asked Jan 3 at 22:21
user623855user623855
1287
edited Jan 3 at 22:30
user623855
asked Jan 3 at 22:21
user623855user623855
1287
asked Jan 3 at 22:21
user623855user623855
1287
asked Jan 3 at 22:21
user623855user623855
1287
1287
$begingroup$
Related (very) math.stackexchange.com/questions/1781642/…
$endgroup$
– Ethan Bolker
Jan 4 at 1:25
add a comment |
$begingroup$
Related (very) math.stackexchange.com/questions/1781642/…
$endgroup$
– Ethan Bolker
Jan 4 at 1:25
$begingroup$
Related (very) math.stackexchange.com/questions/1781642/…
$endgroup$
– Ethan Bolker
Jan 4 at 1:25
$begingroup$
Related (very) math.stackexchange.com/questions/1781642/…
$endgroup$
– Ethan Bolker
Jan 4 at 1:25
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
The average rate of change over some interval of length $h$ starting at time $t$ is given by
$$
e^{-t}left(frac{e^{-h}-1}hright)
$$
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 $t$.
answered Jan 3 at 22:30
ArthurArthur
112k7109191
$endgroup$
$begingroup$
Thank you for the reply. Does it make sense to talk about "rate of change at the exact moment $t$ ? Also, what is the interest if we want the rate of change per unit time for example ?
$endgroup$
– user623855
Jan 3 at 22:32
3
$begingroup$
@user623855 No, technically it doesn't really make sense. Which is why the derivative isn't defined from just a point but from a limit. We call it "rate of change at a point", but what we really mean is "what the rate of change approaches as we shrink the interval down toward zero width". That just doesn't roll of the tongue as nicely. And often the derivative is easier to calculate and work with than changes over actual intervals, so it's used as an approximation to the change over, say, a unit interval.
$endgroup$
– Arthur
Jan 3 at 22:41
$begingroup$
@user623855 Maybe for a moment imagine the slope as the direction of the curve. Imagine a small vehicle driving along the curve. At every point, the vehicle is pointing in a defined direction and we can calculate what direction that is, even though a single point can't have a direction.
$endgroup$
– timtfj
Jan 3 at 23:41
$begingroup$
@timtfj: great picture :) thank you.
$endgroup$
– user623855
Jan 3 at 23:42
add a comment |
$begingroup$
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 $lim_{hrightarrow 0}$. A careful calculation of the limit will yield the answer, that it's just the derivative of the original function.
answered Jan 3 at 22:25
Alex R.Alex R.
24.8k12452
$endgroup$
$begingroup$
"Presumably the question is after the instantaneous rate of change" : Does it really make sense to talk about a rate of change at an instantaneous time ?
$endgroup$
– user623855
Jan 3 at 22:34
1
$begingroup$
@user623855: Yes, this is the basis of all of calculus. Explicitely, $f(x+h)approx f(x)+f'(x)h$, where the approximation gets better and better as $h$ tends to 0, meaning that the instantaneous rate of change is a good approximation for how the function will jump in a short interval.
$endgroup$
– Alex R.
Jan 3 at 22:38
add a comment |
$begingroup$
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.
answered Jan 4 at 1:30
Eric TowersEric Towers
32.4k22268
$endgroup$
add a comment |
$begingroup$
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
$$overline{m} = frac{e^{-x-h}-e^{-x}}{h} = e^{-x}cdotfrac{e^{-h}-1}{h}$$
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,
$$m = e^{-x}cdotlim_{h to 0}frac{e^{-h}-1}{h} to -e^{-x}$$
which would give the instantaneous rate of change, or slope, at point $x$.
answered Jan 3 at 22:37
KM101KM101
5,9431523
$endgroup$
$begingroup$
thank you. "However, what if you want the rate of change at one instant in time", this is just $f(t)$, no need of the derivative...
$endgroup$
– user623855
Jan 3 at 22:41
$begingroup$
In your question, you said “$f(t)$ shows the price.” It’s not the its rate of change.
$endgroup$
– KM101
Jan 3 at 22:43
$begingroup$
Yes :) But you talk about instantaneous rate of change. It's $0$, and thus $f(t)$ is just $f(t)$ :)
$endgroup$
– user623855
Jan 3 at 22:44
$begingroup$
Instantaneous rate of change is not $0$, it’s the $f’(t)$ evaluated at point $t$.
$endgroup$
– KM101
Jan 3 at 22:45
add a comment |
$begingroup$
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 $0cdotinfty$.
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 $0$.
answered Jan 4 at 1:11
timtfjtimtfj
1,318318
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061077%2fwhy-the-derivative-is-the-rate-of-change-of-the-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The average rate of change over some interval of length $h$ starting at time $t$ is given by
$$
e^{-t}left(frac{e^{-h}-1}hright)
$$
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 $t$.
answered Jan 3 at 22:30
ArthurArthur
112k7109191
$endgroup$
$begingroup$
Thank you for the reply. Does it make sense to talk about "rate of change at the exact moment $t$ ? Also, what is the interest if we want the rate of change per unit time for example ?
$endgroup$
– user623855
Jan 3 at 22:32
3
$begingroup$
@user623855 No, technically it doesn't really make sense. Which is why the derivative isn't defined from just a point but from a limit. We call it "rate of change at a point", but what we really mean is "what the rate of change approaches as we shrink the interval down toward zero width". That just doesn't roll of the tongue as nicely. And often the derivative is easier to calculate and work with than changes over actual intervals, so it's used as an approximation to the change over, say, a unit interval.
$endgroup$
– Arthur
Jan 3 at 22:41
$begingroup$
@user623855 Maybe for a moment imagine the slope as the direction of the curve. Imagine a small vehicle driving along the curve. At every point, the vehicle is pointing in a defined direction and we can calculate what direction that is, even though a single point can't have a direction.
$endgroup$
– timtfj
Jan 3 at 23:41
$begingroup$
@timtfj: great picture :) thank you.
$endgroup$
– user623855
Jan 3 at 23:42
add a comment |
$begingroup$
The average rate of change over some interval of length $h$ starting at time $t$ is given by
$$
e^{-t}left(frac{e^{-h}-1}hright)
$$
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 $t$.
answered Jan 3 at 22:30
ArthurArthur
112k7109191
$endgroup$
$begingroup$
Thank you for the reply. Does it make sense to talk about "rate of change at the exact moment $t$ ? Also, what is the interest if we want the rate of change per unit time for example ?
$endgroup$
– user623855
Jan 3 at 22:32
3
$begingroup$
@user623855 No, technically it doesn't really make sense. Which is why the derivative isn't defined from just a point but from a limit. We call it "rate of change at a point", but what we really mean is "what the rate of change approaches as we shrink the interval down toward zero width". That just doesn't roll of the tongue as nicely. And often the derivative is easier to calculate and work with than changes over actual intervals, so it's used as an approximation to the change over, say, a unit interval.
$endgroup$
– Arthur
Jan 3 at 22:41
$begingroup$
@user623855 Maybe for a moment imagine the slope as the direction of the curve. Imagine a small vehicle driving along the curve. At every point, the vehicle is pointing in a defined direction and we can calculate what direction that is, even though a single point can't have a direction.
$endgroup$
– timtfj
Jan 3 at 23:41
$begingroup$
@timtfj: great picture :) thank you.
$endgroup$
– user623855
Jan 3 at 23:42
add a comment |
$begingroup$
The average rate of change over some interval of length $h$ starting at time $t$ is given by
$$
e^{-t}left(frac{e^{-h}-1}hright)
$$
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 $t$.
answered Jan 3 at 22:30
ArthurArthur
112k7109191
$endgroup$
The average rate of change over some interval of length $h$ starting at time $t$ is given by
$$
e^{-t}left(frac{e^{-h}-1}hright)
$$
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 $t$.
answered Jan 3 at 22:30
ArthurArthur
112k7109191
answered Jan 3 at 22:30
ArthurArthur
112k7109191
answered Jan 3 at 22:30
ArthurArthur
112k7109191
answered Jan 3 at 22:30
ArthurArthur
112k7109191
112k7109191
$begingroup$
Thank you for the reply. Does it make sense to talk about "rate of change at the exact moment $t$ ? Also, what is the interest if we want the rate of change per unit time for example ?
$endgroup$
– user623855
Jan 3 at 22:32
3
$begingroup$
@user623855 No, technically it doesn't really make sense. Which is why the derivative isn't defined from just a point but from a limit. We call it "rate of change at a point", but what we really mean is "what the rate of change approaches as we shrink the interval down toward zero width". That just doesn't roll of the tongue as nicely. And often the derivative is easier to calculate and work with than changes over actual intervals, so it's used as an approximation to the change over, say, a unit interval.
$endgroup$
– Arthur
Jan 3 at 22:41
$begingroup$
@user623855 Maybe for a moment imagine the slope as the direction of the curve. Imagine a small vehicle driving along the curve. At every point, the vehicle is pointing in a defined direction and we can calculate what direction that is, even though a single point can't have a direction.
$endgroup$
– timtfj
Jan 3 at 23:41
$begingroup$
@timtfj: great picture :) thank you.
$endgroup$
– user623855
Jan 3 at 23:42
add a comment |
$begingroup$
Thank you for the reply. Does it make sense to talk about "rate of change at the exact moment $t$ ? Also, what is the interest if we want the rate of change per unit time for example ?
$endgroup$
– user623855
Jan 3 at 22:32
3
$begingroup$
@user623855 No, technically it doesn't really make sense. Which is why the derivative isn't defined from just a point but from a limit. We call it "rate of change at a point", but what we really mean is "what the rate of change approaches as we shrink the interval down toward zero width". That just doesn't roll of the tongue as nicely. And often the derivative is easier to calculate and work with than changes over actual intervals, so it's used as an approximation to the change over, say, a unit interval.
$endgroup$
– Arthur
Jan 3 at 22:41
$begingroup$
@user623855 Maybe for a moment imagine the slope as the direction of the curve. Imagine a small vehicle driving along the curve. At every point, the vehicle is pointing in a defined direction and we can calculate what direction that is, even though a single point can't have a direction.
$endgroup$
– timtfj
Jan 3 at 23:41
$begingroup$
@timtfj: great picture :) thank you.
$endgroup$
– user623855
Jan 3 at 23:42
$begingroup$
Thank you for the reply. Does it make sense to talk about "rate of change at the exact moment $t$ ? Also, what is the interest if we want the rate of change per unit time for example ?
$endgroup$
– user623855
Jan 3 at 22:32
$begingroup$
Thank you for the reply. Does it make sense to talk about "rate of change at the exact moment $t$ ? Also, what is the interest if we want the rate of change per unit time for example ?
$endgroup$
– user623855
Jan 3 at 22:32
3
3
$begingroup$
@user623855 No, technically it doesn't really make sense. Which is why the derivative isn't defined from just a point but from a limit. We call it "rate of change at a point", but what we really mean is "what the rate of change approaches as we shrink the interval down toward zero width". That just doesn't roll of the tongue as nicely. And often the derivative is easier to calculate and work with than changes over actual intervals, so it's used as an approximation to the change over, say, a unit interval.
$endgroup$
– Arthur
Jan 3 at 22:41
$begingroup$
@user623855 No, technically it doesn't really make sense. Which is why the derivative isn't defined from just a point but from a limit. We call it "rate of change at a point", but what we really mean is "what the rate of change approaches as we shrink the interval down toward zero width". That just doesn't roll of the tongue as nicely. And often the derivative is easier to calculate and work with than changes over actual intervals, so it's used as an approximation to the change over, say, a unit interval.
$endgroup$
– Arthur
Jan 3 at 22:41
$begingroup$
@user623855 Maybe for a moment imagine the slope as the direction of the curve. Imagine a small vehicle driving along the curve. At every point, the vehicle is pointing in a defined direction and we can calculate what direction that is, even though a single point can't have a direction.
$endgroup$
– timtfj
Jan 3 at 23:41
$begingroup$
@user623855 Maybe for a moment imagine the slope as the direction of the curve. Imagine a small vehicle driving along the curve. At every point, the vehicle is pointing in a defined direction and we can calculate what direction that is, even though a single point can't have a direction.
$endgroup$
– timtfj
Jan 3 at 23:41
$begingroup$
@timtfj: great picture :) thank you.
$endgroup$
– user623855
Jan 3 at 23:42
$begingroup$
@timtfj: great picture :) thank you.
$endgroup$
– user623855
Jan 3 at 23:42
add a comment |
$begingroup$
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 $lim_{hrightarrow 0}$. A careful calculation of the limit will yield the answer, that it's just the derivative of the original function.
answered Jan 3 at 22:25
Alex R.Alex R.
24.8k12452
$endgroup$
$begingroup$
"Presumably the question is after the instantaneous rate of change" : Does it really make sense to talk about a rate of change at an instantaneous time ?
$endgroup$
– user623855
Jan 3 at 22:34
1
$begingroup$
@user623855: Yes, this is the basis of all of calculus. Explicitely, $f(x+h)approx f(x)+f'(x)h$, where the approximation gets better and better as $h$ tends to 0, meaning that the instantaneous rate of change is a good approximation for how the function will jump in a short interval.
$endgroup$
– Alex R.
Jan 3 at 22:38
add a comment |
$begingroup$
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 $lim_{hrightarrow 0}$. A careful calculation of the limit will yield the answer, that it's just the derivative of the original function.
answered Jan 3 at 22:25
Alex R.Alex R.
24.8k12452
$endgroup$
$begingroup$
"Presumably the question is after the instantaneous rate of change" : Does it really make sense to talk about a rate of change at an instantaneous time ?
$endgroup$
– user623855
Jan 3 at 22:34
1
$begingroup$
@user623855: Yes, this is the basis of all of calculus. Explicitely, $f(x+h)approx f(x)+f'(x)h$, where the approximation gets better and better as $h$ tends to 0, meaning that the instantaneous rate of change is a good approximation for how the function will jump in a short interval.
$endgroup$
– Alex R.
Jan 3 at 22:38
add a comment |
$begingroup$
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 $lim_{hrightarrow 0}$. A careful calculation of the limit will yield the answer, that it's just the derivative of the original function.
answered Jan 3 at 22:25
Alex R.Alex R.
24.8k12452
$endgroup$
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 $lim_{hrightarrow 0}$. A careful calculation of the limit will yield the answer, that it's just the derivative of the original function.
answered Jan 3 at 22:25
Alex R.Alex R.
24.8k12452
answered Jan 3 at 22:25
Alex R.Alex R.
24.8k12452
answered Jan 3 at 22:25
Alex R.Alex R.
24.8k12452
answered Jan 3 at 22:25
Alex R.Alex R.
24.8k12452
24.8k12452
$begingroup$
"Presumably the question is after the instantaneous rate of change" : Does it really make sense to talk about a rate of change at an instantaneous time ?
$endgroup$
– user623855
Jan 3 at 22:34
1
$begingroup$
@user623855: Yes, this is the basis of all of calculus. Explicitely, $f(x+h)approx f(x)+f'(x)h$, where the approximation gets better and better as $h$ tends to 0, meaning that the instantaneous rate of change is a good approximation for how the function will jump in a short interval.
$endgroup$
– Alex R.
Jan 3 at 22:38
add a comment |
$begingroup$
"Presumably the question is after the instantaneous rate of change" : Does it really make sense to talk about a rate of change at an instantaneous time ?
$endgroup$
– user623855
Jan 3 at 22:34
1
$begingroup$
@user623855: Yes, this is the basis of all of calculus. Explicitely, $f(x+h)approx f(x)+f'(x)h$, where the approximation gets better and better as $h$ tends to 0, meaning that the instantaneous rate of change is a good approximation for how the function will jump in a short interval.
$endgroup$
– Alex R.
Jan 3 at 22:38
$begingroup$
"Presumably the question is after the instantaneous rate of change" : Does it really make sense to talk about a rate of change at an instantaneous time ?
$endgroup$
– user623855
Jan 3 at 22:34
$begingroup$
"Presumably the question is after the instantaneous rate of change" : Does it really make sense to talk about a rate of change at an instantaneous time ?
$endgroup$
– user623855
Jan 3 at 22:34
1
1
$begingroup$
@user623855: Yes, this is the basis of all of calculus. Explicitely, $f(x+h)approx f(x)+f'(x)h$, where the approximation gets better and better as $h$ tends to 0, meaning that the instantaneous rate of change is a good approximation for how the function will jump in a short interval.
$endgroup$
– Alex R.
Jan 3 at 22:38
$begingroup$
@user623855: Yes, this is the basis of all of calculus. Explicitely, $f(x+h)approx f(x)+f'(x)h$, where the approximation gets better and better as $h$ tends to 0, meaning that the instantaneous rate of change is a good approximation for how the function will jump in a short interval.
$endgroup$
– Alex R.
Jan 3 at 22:38
add a comment |
$begingroup$
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.
answered Jan 4 at 1:30
Eric TowersEric Towers
32.4k22268
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 4 at 1:30
Eric TowersEric Towers
32.4k22268
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 4 at 1:30
Eric TowersEric Towers
32.4k22268
$endgroup$
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.
answered Jan 4 at 1:30
Eric TowersEric Towers
32.4k22268
answered Jan 4 at 1:30
Eric TowersEric Towers
32.4k22268
answered Jan 4 at 1:30
Eric TowersEric Towers
32.4k22268
answered Jan 4 at 1:30
Eric TowersEric Towers
32.4k22268
32.4k22268
add a comment |
add a comment |
$begingroup$
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
$$overline{m} = frac{e^{-x-h}-e^{-x}}{h} = e^{-x}cdotfrac{e^{-h}-1}{h}$$
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,
$$m = e^{-x}cdotlim_{h to 0}frac{e^{-h}-1}{h} to -e^{-x}$$
which would give the instantaneous rate of change, or slope, at point $x$.
answered Jan 3 at 22:37
KM101KM101
5,9431523
$endgroup$
$begingroup$
thank you. "However, what if you want the rate of change at one instant in time", this is just $f(t)$, no need of the derivative...
$endgroup$
– user623855
Jan 3 at 22:41
$begingroup$
In your question, you said “$f(t)$ shows the price.” It’s not the its rate of change.
$endgroup$
– KM101
Jan 3 at 22:43
$begingroup$
Yes :) But you talk about instantaneous rate of change. It's $0$, and thus $f(t)$ is just $f(t)$ :)
$endgroup$
– user623855
Jan 3 at 22:44
$begingroup$
Instantaneous rate of change is not $0$, it’s the $f’(t)$ evaluated at point $t$.
$endgroup$
– KM101
Jan 3 at 22:45
add a comment |
$begingroup$
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
$$overline{m} = frac{e^{-x-h}-e^{-x}}{h} = e^{-x}cdotfrac{e^{-h}-1}{h}$$
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,
$$m = e^{-x}cdotlim_{h to 0}frac{e^{-h}-1}{h} to -e^{-x}$$
which would give the instantaneous rate of change, or slope, at point $x$.
answered Jan 3 at 22:37
KM101KM101
5,9431523
$endgroup$
$begingroup$
thank you. "However, what if you want the rate of change at one instant in time", this is just $f(t)$, no need of the derivative...
$endgroup$
– user623855
Jan 3 at 22:41
$begingroup$
In your question, you said “$f(t)$ shows the price.” It’s not the its rate of change.
$endgroup$
– KM101
Jan 3 at 22:43
$begingroup$
Yes :) But you talk about instantaneous rate of change. It's $0$, and thus $f(t)$ is just $f(t)$ :)
$endgroup$
– user623855
Jan 3 at 22:44
$begingroup$
Instantaneous rate of change is not $0$, it’s the $f’(t)$ evaluated at point $t$.
$endgroup$
– KM101
Jan 3 at 22:45
add a comment |
$begingroup$
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
$$overline{m} = frac{e^{-x-h}-e^{-x}}{h} = e^{-x}cdotfrac{e^{-h}-1}{h}$$
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,
$$m = e^{-x}cdotlim_{h to 0}frac{e^{-h}-1}{h} to -e^{-x}$$
which would give the instantaneous rate of change, or slope, at point $x$.
answered Jan 3 at 22:37
KM101KM101
5,9431523
$endgroup$
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
$$overline{m} = frac{e^{-x-h}-e^{-x}}{h} = e^{-x}cdotfrac{e^{-h}-1}{h}$$
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,
$$m = e^{-x}cdotlim_{h to 0}frac{e^{-h}-1}{h} to -e^{-x}$$
which would give the instantaneous rate of change, or slope, at point $x$.
answered Jan 3 at 22:37
KM101KM101
5,9431523
answered Jan 3 at 22:37
KM101KM101
5,9431523
answered Jan 3 at 22:37
KM101KM101
5,9431523
answered Jan 3 at 22:37
KM101KM101
5,9431523
5,9431523
$begingroup$
thank you. "However, what if you want the rate of change at one instant in time", this is just $f(t)$, no need of the derivative...
$endgroup$
– user623855
Jan 3 at 22:41
$begingroup$
In your question, you said “$f(t)$ shows the price.” It’s not the its rate of change.
$endgroup$
– KM101
Jan 3 at 22:43
$begingroup$
Yes :) But you talk about instantaneous rate of change. It's $0$, and thus $f(t)$ is just $f(t)$ :)
$endgroup$
– user623855
Jan 3 at 22:44
$begingroup$
Instantaneous rate of change is not $0$, it’s the $f’(t)$ evaluated at point $t$.
$endgroup$
– KM101
Jan 3 at 22:45
add a comment |
$begingroup$
thank you. "However, what if you want the rate of change at one instant in time", this is just $f(t)$, no need of the derivative...
$endgroup$
– user623855
Jan 3 at 22:41
$begingroup$
In your question, you said “$f(t)$ shows the price.” It’s not the its rate of change.
$endgroup$
– KM101
Jan 3 at 22:43
$begingroup$
Yes :) But you talk about instantaneous rate of change. It's $0$, and thus $f(t)$ is just $f(t)$ :)
$endgroup$
– user623855
Jan 3 at 22:44
$begingroup$
Instantaneous rate of change is not $0$, it’s the $f’(t)$ evaluated at point $t$.
$endgroup$
– KM101
Jan 3 at 22:45
$begingroup$
thank you. "However, what if you want the rate of change at one instant in time", this is just $f(t)$, no need of the derivative...
$endgroup$
– user623855
Jan 3 at 22:41
$begingroup$
thank you. "However, what if you want the rate of change at one instant in time", this is just $f(t)$, no need of the derivative...
$endgroup$
– user623855
Jan 3 at 22:41
$begingroup$
In your question, you said “$f(t)$ shows the price.” It’s not the its rate of change.
$endgroup$
– KM101
Jan 3 at 22:43
$begingroup$
In your question, you said “$f(t)$ shows the price.” It’s not the its rate of change.
$endgroup$
– KM101
Jan 3 at 22:43
$begingroup$
Yes :) But you talk about instantaneous rate of change. It's $0$, and thus $f(t)$ is just $f(t)$ :)
$endgroup$
– user623855
Jan 3 at 22:44
$begingroup$
Yes :) But you talk about instantaneous rate of change. It's $0$, and thus $f(t)$ is just $f(t)$ :)
$endgroup$
– user623855
Jan 3 at 22:44
$begingroup$
Instantaneous rate of change is not $0$, it’s the $f’(t)$ evaluated at point $t$.
$endgroup$
– KM101
Jan 3 at 22:45
$begingroup$
Instantaneous rate of change is not $0$, it’s the $f’(t)$ evaluated at point $t$.
$endgroup$
– KM101
Jan 3 at 22:45
add a comment |
$begingroup$
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 $0cdotinfty$.
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 $0$.
answered Jan 4 at 1:11
timtfjtimtfj
1,318318
$endgroup$
add a comment |
$begingroup$
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 $0cdotinfty$.
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 $0$.
answered Jan 4 at 1:11
timtfjtimtfj
1,318318
$endgroup$
add a comment |
$begingroup$
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 $0cdotinfty$.
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 $0$.
answered Jan 4 at 1:11
timtfjtimtfj
1,318318
$endgroup$
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 $0cdotinfty$.
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 $0$.
answered Jan 4 at 1:11
timtfjtimtfj
1,318318
answered Jan 4 at 1:11
timtfjtimtfj
1,318318
answered Jan 4 at 1:11
timtfjtimtfj
1,318318
answered Jan 4 at 1:11
timtfjtimtfj
1,318318
1,318318
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061077%2fwhy-the-derivative-is-the-rate-of-change-of-the-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Related (very) math.stackexchange.com/questions/1781642/…
$endgroup$
– Ethan Bolker
Jan 4 at 1:25