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The paradox of the derivative | Chapter 2, Essence of calculus
3Blue1Brown
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May 12, 2026
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Transcript
0:15
The goal here is simple, explain what a derivative is.
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The thing is though, there's some subtlety to this topic,
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and a lot of potential for paradoxes if you're not careful.
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So a secondary goal is that you have an appreciation
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for what those paradoxes are and how to avoid them.
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You see, it's common for people to say that the derivative measures an instantaneous
0:35
rate of change, but when you think about it, that phrase is actually an oxymoron.
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Change is something that happens between separate points in time,
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and when you blind yourself to all but just a single instant,
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there's not really any room for change.
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You'll see what I mean more as we get into it,
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but when you appreciate that a phrase like instantaneous rate of change is actually
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nonsense, I think it makes you appreciate just how clever the fathers of calculus
1:00
were in capturing the idea that phrase is meant to evoke,
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but with a perfectly sensible piece of math, the derivative.
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As our central example, I want you to imagine a car that starts at some point A,
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speeds up, and then slows down to a stop at some point B 100 meters away,
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and let's say it all happens over the course of 10 seconds.
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That's the setup to have in mind as we lay out what the derivative is.
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Well, we could graph this motion, letting the vertical axis represent the
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distance traveled, and the horizontal axis represent time, so at each time t,
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represented with a point somewhere on the horizontal axis,
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the height of the graph tells us how far the car has traveled in total after
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that amount of time.
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It's pretty common to name a distance function like this s of t.
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I would use the letter d for distance, but that
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guy already has another full time job in calculus.
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Initially, the curve is quite shallow, since the car is slow to start.
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During that first second, the distance it travels doesn't change that much.
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For the next few seconds, as the car speeds up,
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the distance traveled in a given second gets larger,
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which corresponds to a steeper slope in this graph.
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Then towards the end, when it slows down, that curve shallows out again.
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If we were to plot the car's velocity in meters per second as a function of time,
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it might look like this bump.
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At early times, the velocity is very small.
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Up to the middle of the journey, the car builds up to some maximum velocity,
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covering a relatively large distance each second.
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Then it slows back down towards a speed of zero.
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These two curves are definitely related to each other.
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If you change the specific distance vs.
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time function, you'll have some different velocity vs.
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time function.
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What we want to understand is the specifics of that relationship.
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Exactly how does velocity depend on a distance vs.
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time function?
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To do that, it's worth taking a moment to think
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critically about what exactly velocity means here.
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Intuitively, we all might know what velocity at a given moment means,
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it's just whatever the car's speedometer shows in that moment.
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Intuitively, it might make sense that the car's velocity should be higher at times when
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this distance function is steeper, when the car traverses more distance per unit time.
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But the funny thing is, velocity at a single moment makes no sense.
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If I show you a picture of a car, just a snapshot in an instant,
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and I ask you how fast it's going, you'd have no way of telling me.
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What you'd need are two separate points in time to compare.
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That way you can compute whatever the change in distance across those times is,
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divided by the change in time.
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Right?
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I mean, that's what velocity is, it's the distance traveled per unit time.
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So how is it that we're looking at a function for velocity that
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only takes in a single value of t, a single snapshot in time?
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It's weird, isn't it?
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We want to associate individual points in time with a velocity,
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but actually computing velocity requires comparing two separate points in time.
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If that feels strange and paradoxical, good!
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You're grappling with the same conflicts that the fathers of calculus did.
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And if you want a deep understanding for rates of change, not just for a moving car,
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but for all sorts of things in science, you're going to need to resolve this apparent
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paradox.
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First, I think it's best to talk about the real world,
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and then we'll go into a purely mathematical one.
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Let's think about what the car's speedometer is probably doing.
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At some point, say 3 seconds into the journey,
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the speedometer might measure how far the car goes in a very small amount of time,
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maybe the distance traveled between 3 seconds and 3.01 seconds.
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Then it could compute the speed in meters per second as that tiny
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distance traversed in meters divided by that tiny time, 0.01 seconds.
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That is, a physical car just side-steps the paradox and
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doesn't actually compute speed at a single point in time.
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It computes speed during a very small amount of time.
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So let's call that difference in time dt, which you might think of as 0.01 seconds,
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and let's call that resulting difference in distance ds.
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So the velocity at some point in time is ds divided by dt,
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the tiny change in distance over the tiny change in time.
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Graphically, you can imagine zooming in on some point of this distance vs.
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time graph above t equals 3.
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That dt is a small step to the right, since time is on the horizontal axis,
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and that ds is the resulting change in the height of the graph,
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since the vertical axis represents the distance traveled.
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So ds divided by dt is something you can think of as the rise
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over run slope between two very close points on this graph.
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Of course, there's nothing special about the value t equals 3.
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We could apply this to any other point in time,
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so we consider this expression ds over dt to be a function of t,
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something where I can give you a time t and you can give me back the value of this
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ratio at that time, the velocity as a function of time.
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For example, when I had the computer draw this bump curve here,
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the one representing the velocity function, here's what I had the computer actually do.
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First, I chose a small value for dt, I think in this case it was 0.01.
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Then I had the computer look at a whole bunch of times t between 0 and 10,
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and compute the distance function s at t plus dt,
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and then subtract off the value of that function at t.
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In other words, that's the difference in the distance traveled between the given time,
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t, and the time 0.01 seconds after that.
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Then you can just divide that difference by the change in time, dt,
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and that gives you velocity in meters per second around each point in time.
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So with a formula like this, you could give the computer any curve representing any
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distance function s of t, and it could figure out the curve representing velocity.
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Now would be a good time to pause, reflect, and make sure this idea
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of relating distance to velocity by looking at tiny changes makes sense,
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because we're going to tackle the paradox of the derivative head on.
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This idea of ds over dt, a tiny change in the value of the function s divided by
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the tiny change in the input that caused it, that's almost what a derivative is.
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And even though a car's speedometer will actually look at a concrete change in time,
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like 0.01 seconds, and even though the drawing program here is looking at an actual
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concrete change in time, in pure math the derivative is not this ratio ds over dt for a
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specific choice of dt. Instead, it's whatever that ratio approaches as your choice for dt
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approaches 0.
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Luckily there is a really nice visual understanding for what it means to ask what
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this ratio approaches, Remember, for any specific choice of dt,
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this ratio ds over dt is the slope of a line passing through two separate points
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on the graph, right?
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Well as dt approaches 0, and as those two points approach each other,
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the slope of the line approaches the slope of a line that's
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tangent to the graph at whatever point t we're looking at.
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So the true honest-to-goodness pure math derivative is not the
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rise over run slope between two nearby points on the graph,
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it's equal to the slope of a line tangent to the graph at a single point.
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Now notice what I'm not saying, I'm not saying that the derivative is
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whatever happens when dt is infinitely small, whatever that would mean.
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Nor am I saying that you plug in 0 for dt.
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This dt is always a finitely small non-zero value, it's just that it approaches 0 is all.
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I think that's really clever.
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Even though change in an instant makes no sense,
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this idea of letting dt approach 0 is a really sneaky backdoor
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way to talk reasonably about the rate of change at a single point in time.
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Isn't that neat?
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It's kind of flirting with the paradox of change in
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an instant without ever needing to actually touch it.
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And it comes with such a nice visual intuition too,
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as the slope of a tangent line to a single point on the graph.
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And because change in an instant still makes no sense,
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I think it's healthiest for you to think of this slope not as some instantaneous
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rate of change, but instead as the best constant approximation for a rate of
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change around a point.
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By the way, it's worth saying a couple words on notation here.
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Throughout this video I've been using dt to refer to a tiny change in t with
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some actual size, and ds to refer to the resulting change in s,
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which again has an actual size, and this is because that's how I want you to
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think about them.
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But the convention in calculus is that whenever you're using the letter d like this,
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you're kind of announcing your intention that eventually you're
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going to see what happens as dt approaches 0.
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For example, the honest-to-goodness pure math derivative is written as ds divided by dt,
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even though it's technically not a fraction per se,
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but whatever that fraction approaches for smaller and smaller nudges in t.
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I think a specific example should help here.
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You might think that asking about what this ratio approaches
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for smaller and smaller values would make it much more difficult to compute,
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but weirdly it kind of makes things easier.
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Let's say you have a given distance vs time function that happens to be exactly t cubed.
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So after 1 second the car has traveled 1 cubed equals 1 meters,
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after 2 seconds it's traveled 2 cubed, or 8 meters, and so on.
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Now what I'm about to do might seem somewhat complicated,
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but once the dust settles it really is simpler,
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and more importantly it's the kind of thing you only ever have to do once in calculus.
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Let's say you wanted to compute the velocity, ds divided by dt,
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at some specific time, like t equals 2.
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For right now let's think of dt as having an actual size,
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some concrete nudge, we'll let it go to 0 in just a bit.
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The tiny change in distance between 2 seconds and 2 plus dt
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seconds is s of 2 plus dt minus s of 2, and we divide that by dt.
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Since our function is t cubed, that numerator looks like 2 plus dt cubed minus 2 cubed.
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And this is something we can work out algebraically.
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Again, bear with me, there's a reason I'm showing you the details here.
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When you expand that top, what you get is 2 cubed plus 3 times 2 squared dt
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plus 3 times 2 times dt squared plus dt cubed, and all of that is minus 2 cubed.
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Now there's a lot of terms, and I want you to remember that it looks like a mess,
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but it does simplify.
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Those 2 cubed terms cancel out.
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Everything remaining here has a dt in it, and since there's a dt on the bottom there,
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many of those cancel out as well.
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What this means is that the ratio ds divided by dt has boiled down into
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3 times 2 squared plus 2 different terms that each have a dt in them.
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So if we ask what happens as dt approaches 0, representing the idea of looking at a
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smaller and smaller change in time, we can just completely ignore those other terms.
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By eliminating the need to think about a specific dt,
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we've eliminated a lot of the complication in the full expression.
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So what we're left with is this nice clean 3 times 2 squared.
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You can think of that as meaning that the slope of a line tangent to
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the point at t equals 2 of this graph is exactly 3 times 2 squared, or 12.
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And of course, there's nothing special about the time t equals 2.
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We could more generally say that the derivative
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of t cubed as a function of t is 3 times t squared.
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Now take a step back, because that's beautiful.
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The derivative is this crazy complicated idea.
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We've got tiny changes in distance over tiny changes in time,
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but instead of looking at any specific one of those,
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we're talking about what that thing approaches.
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I mean, that's a lot to think about.
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And yet what we've come out with is such a simple expression, 3 times t squared.
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And in practice, you wouldn't go through all this algebra each time.
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Knowing that the derivative of t cubed is 3t squared is one of those things that all
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calculus students learn how to do immediately without having to re-derive it each time.
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And in the next video, I'm going to show you a nice way to think about
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this and a couple other derivative formulas in really nice geometric ways.
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But the point I want to make by showing you all of the algebraic guts
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here is that when you consider the tiny change in distance caused by a
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tiny change in time for some specific value of dt, you'd have kind of a mess.
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But when you consider what that ratio approaches as dt approaches 0,
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it lets you ignore much of that mess, and it really does simplify the problem.
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That right there is kind of the heart of why calculus becomes useful.
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Another reason to show you a concrete derivative like this is that it
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sets the stage for an example of the kind of paradoxes that come about
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if you believe too much in the illusion of instantaneous rate of change.
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So think about the actual car traveling according to this t cubed distance function,
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and consider its motion at the moment t equals 0, right at the start.
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Now ask yourself whether or not the car is moving at that time.
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On the one hand, we can compute its speed at that point using the derivative,
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3t squared, which for time t equals 0 works out to be 0.
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Visually, this means that the tangent line to the graph at that point is perfectly flat,
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so the car's quote-unquote instantaneous velocity is 0,
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and that suggests that obviously it's not moving.
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But on the other hand, if it doesn't start moving at time 0, when does it start moving?
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Really, pause and ponder that for a moment.
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Is the car moving at time t equals 0?
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Do you see the paradox?
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The issue is that the question makes no sense.
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It references the idea of change in a moment, but that doesn't actually exist.
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That's just not what the derivative measures.
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What it means for the derivative of a distance function to be 0 is that the best
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constant approximation for the car's velocity around that point is 0 m per second.
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For example, if you look at an actual change in time,
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say between time 0 and 0.1 seconds, the car does move.
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It moves 0.001 m.
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That's very small, and importantly, it's very small compared to the change in time,
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giving an average speed of only 0.01 m per second.
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And remember, what it means for the derivative of this motion to be 0 is that
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for smaller and smaller nudges in time, this ratio of m per second approaches 0.
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But that's not to say that the car is static.
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Approximating its movement with a constant velocity of 0 is,
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after all, just an approximation.
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So whenever you hear people refer to the derivative as an instantaneous rate of change,
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a phrase which is intrinsically oxymoronic, I want you to think of that as a
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conceptual shorthand for the best constant approximation for rate of change.
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In the next couple videos, I'll be talking more about the derivative,
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what it looks like in different contexts, how do you actually compute it,
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why is it useful, things like that, focusing on visual intuition as always.
— end of transcript —
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