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It has half of the wavelength of A. Notice, when it finishes its second full waveform, it ends again right at the same time that the wavelength of A ends. And then another harmonic would be something that has a third the wavelength of an A, and a fourth of the wavelength of A. And if you look at a lot of musical instruments or what sounds good to our ears, they're playing not just the fundamental tone, but a lot of the harmonics. But anyway, that was a long-winded way of justifying why this is called the harmonic series. Harmonic series. And in a future video, we will prove that, and I don't wanna ruin the punchline, but this actually diverges. | Harmonic series and 𝑝-series AP®︎ Calculus BC Khan Academy.mp3 |
And if you look at a lot of musical instruments or what sounds good to our ears, they're playing not just the fundamental tone, but a lot of the harmonics. But anyway, that was a long-winded way of justifying why this is called the harmonic series. Harmonic series. And in a future video, we will prove that, and I don't wanna ruin the punchline, but this actually diverges. And I will come up with general rules for when things that look like this might converge or diverge. But the harmonic series in particular diverges. So if we were to write it, so in sigma form, we would write it like this. | Harmonic series and 𝑝-series AP®︎ Calculus BC Khan Academy.mp3 |
And in a future video, we will prove that, and I don't wanna ruin the punchline, but this actually diverges. And I will come up with general rules for when things that look like this might converge or diverge. But the harmonic series in particular diverges. So if we were to write it, so in sigma form, we would write it like this. We're going from n equals one to infinity of one over n. Now, another interesting thing is, well, what if we were to throw in some exponents here? So we already said, and I'll just rewrite it. It doesn't hurt to rewrite it and get more familiar with it. | Harmonic series and 𝑝-series AP®︎ Calculus BC Khan Academy.mp3 |
So if we were to write it, so in sigma form, we would write it like this. We're going from n equals one to infinity of one over n. Now, another interesting thing is, well, what if we were to throw in some exponents here? So we already said, and I'll just rewrite it. It doesn't hurt to rewrite it and get more familiar with it. This right over here is the harmonic series. One over one, which is just one, plus one over two, plus one over three, so on and so forth. But what if we were to raise each of these denominators to, say, the second power? | Harmonic series and 𝑝-series AP®︎ Calculus BC Khan Academy.mp3 |
It doesn't hurt to rewrite it and get more familiar with it. This right over here is the harmonic series. One over one, which is just one, plus one over two, plus one over three, so on and so forth. But what if we were to raise each of these denominators to, say, the second power? So you might have something that looks like this, where you have from n equals one to infinity of one over n to the second power. Well, then it would look like this. It would be one over one squared, which is one. | Harmonic series and 𝑝-series AP®︎ Calculus BC Khan Academy.mp3 |
But what if we were to raise each of these denominators to, say, the second power? So you might have something that looks like this, where you have from n equals one to infinity of one over n to the second power. Well, then it would look like this. It would be one over one squared, which is one. And we could just write that first term as one, plus one over two squared, which would be 1 1⁴, plus one over three squared, which is 1 9th. And then you could go on and on forever, forever. And then you could generalize it. | Harmonic series and 𝑝-series AP®︎ Calculus BC Khan Academy.mp3 |
It would be one over one squared, which is one. And we could just write that first term as one, plus one over two squared, which would be 1 1⁴, plus one over three squared, which is 1 9th. And then you could go on and on forever, forever. And then you could generalize it. You could say, hey, all right, what if we wanted to have a general class of series that we were to describe like this, going from n equals one to infinity of one over n to the p, where p could be any exponent. So for example, well, the way this would play out is this would be one plus one over two to the p plus one over three to the p plus one over four to the p. And it doesn't just have to be an integer value. It could be, some p could be 1⁄2, in which case you would have one plus one over the square root of two plus one over the square root of three. | Harmonic series and 𝑝-series AP®︎ Calculus BC Khan Academy.mp3 |
And then you could generalize it. You could say, hey, all right, what if we wanted to have a general class of series that we were to describe like this, going from n equals one to infinity of one over n to the p, where p could be any exponent. So for example, well, the way this would play out is this would be one plus one over two to the p plus one over three to the p plus one over four to the p. And it doesn't just have to be an integer value. It could be, some p could be 1⁄2, in which case you would have one plus one over the square root of two plus one over the square root of three. This entire class of series, and of course harmonic series is a special case where p is equal to one, this is known as p-series. So these are known as p-series. And I tried to remember it because it's p for the power that you are raising this denominator to. | Harmonic series and 𝑝-series AP®︎ Calculus BC Khan Academy.mp3 |
It could be, some p could be 1⁄2, in which case you would have one plus one over the square root of two plus one over the square root of three. This entire class of series, and of course harmonic series is a special case where p is equal to one, this is known as p-series. So these are known as p-series. And I tried to remember it because it's p for the power that you are raising this denominator to. You could also view it as you're raising the whole expression to it because one to any exponent is still going to be one. But I hinted a little bit that maybe some of these converge and some of these diverge. And we're going to prove it in future videos, but the general principle is if p is greater than one, then we are going to converge. | Harmonic series and 𝑝-series AP®︎ Calculus BC Khan Academy.mp3 |
And I tried to remember it because it's p for the power that you are raising this denominator to. You could also view it as you're raising the whole expression to it because one to any exponent is still going to be one. But I hinted a little bit that maybe some of these converge and some of these diverge. And we're going to prove it in future videos, but the general principle is if p is greater than one, then we are going to converge. And that makes sense intuitively because that means that the terms are getting smaller and smaller fast enough because the larger the exponent for that denominator, that means that the denominator is going to get bigger faster, which means that the fraction is going to get smaller faster. And if p is less than or equal to one, and of course when p is equal to one, we're dealing with the famous harmonic series, that's a situation in which we diverge. And we will prove these things in future videos. | Harmonic series and 𝑝-series AP®︎ Calculus BC Khan Academy.mp3 |
For each point on the graph, is the object speeding up, slowing down, or neither? So pause this video and see if you can figure that out. All right, now let's do it together. And first, we need to make sure we're reading this carefully, because they're not asking is the velocity increasing, decreasing, or neither. They're saying is the object speeding up, slowing down, or neither? So they're talking about speed, which is the magnitude of velocity. You could think of it as the absolute value of velocity, especially when we're thinking about it in one dimension here. | Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3 |
And first, we need to make sure we're reading this carefully, because they're not asking is the velocity increasing, decreasing, or neither. They're saying is the object speeding up, slowing down, or neither? So they're talking about speed, which is the magnitude of velocity. You could think of it as the absolute value of velocity, especially when we're thinking about it in one dimension here. So even though they're not asking about velocity, I'm actually going to answer both, so that we can see how sometimes they move together, velocity and speed, but sometimes one might be increasing while the other might be decreasing. So if we look at this point right over here, where our velocity is two meters per second, the speed is the absolute value of velocity, which would also be two meters per second, and we can see that the slope of the velocity-time graph is positive, and so our velocity is increasing, and the absolute value of our velocity, which is speed, is also increasing. A moment later, our velocity might be 2.1 meters per second, and our speed would also be 2.1 meters per second. | Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3 |
You could think of it as the absolute value of velocity, especially when we're thinking about it in one dimension here. So even though they're not asking about velocity, I'm actually going to answer both, so that we can see how sometimes they move together, velocity and speed, but sometimes one might be increasing while the other might be decreasing. So if we look at this point right over here, where our velocity is two meters per second, the speed is the absolute value of velocity, which would also be two meters per second, and we can see that the slope of the velocity-time graph is positive, and so our velocity is increasing, and the absolute value of our velocity, which is speed, is also increasing. A moment later, our velocity might be 2.1 meters per second, and our speed would also be 2.1 meters per second. That seems intuitive enough. And we get the other scenario if we go to this point right over here. Our velocity is still positive, but we see that our velocity-time graph is now downward sloping. | Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3 |
A moment later, our velocity might be 2.1 meters per second, and our speed would also be 2.1 meters per second. That seems intuitive enough. And we get the other scenario if we go to this point right over here. Our velocity is still positive, but we see that our velocity-time graph is now downward sloping. So our velocity is decreasing because of that downward slope, and the absolute value of our velocity is also decreasing. Right at that moment, our speed is two meters per second, and then a moment later, it might be 1.9 meters per second. All right, now let's go to this point. | Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3 |
Our velocity is still positive, but we see that our velocity-time graph is now downward sloping. So our velocity is decreasing because of that downward slope, and the absolute value of our velocity is also decreasing. Right at that moment, our speed is two meters per second, and then a moment later, it might be 1.9 meters per second. All right, now let's go to this point. So this point is really interesting. Here we see that our velocity, the slope of the tangent line, is still negative, so our velocity is still decreasing. But what about the absolute value of our velocity, which is speed? | Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3 |
All right, now let's go to this point. So this point is really interesting. Here we see that our velocity, the slope of the tangent line, is still negative, so our velocity is still decreasing. But what about the absolute value of our velocity, which is speed? Well, if you think about it, a moment before this, we were slowing down to get to a zero velocity, and a moment after this, we're going to be speeding up to start having negative velocity. You might say, wait, speeding up for negative velocity? Remember, speed is the absolute value. | Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3 |
But what about the absolute value of our velocity, which is speed? Well, if you think about it, a moment before this, we were slowing down to get to a zero velocity, and a moment after this, we're going to be speeding up to start having negative velocity. You might say, wait, speeding up for negative velocity? Remember, speed is the absolute value. So if your velocity goes from zero to negative one meters per second, your speed just went from zero to one meter per second. So we're slowing down here, and we're speeding up here, but right at this moment, neither is happening. We are neither speeding up nor slowing down. | Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3 |
Remember, speed is the absolute value. So if your velocity goes from zero to negative one meters per second, your speed just went from zero to one meter per second. So we're slowing down here, and we're speeding up here, but right at this moment, neither is happening. We are neither speeding up nor slowing down. Now what about this point? Here, the slope of our velocity time graph, or the slope of the tangent line, is still negative, so our velocity is still decreasing. But what about speed? | Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3 |
We are neither speeding up nor slowing down. Now what about this point? Here, the slope of our velocity time graph, or the slope of the tangent line, is still negative, so our velocity is still decreasing. But what about speed? Well, our velocity is already negative, and it's becoming more negative. So the absolute value of velocity, which is two meters per second, that is increasing at that moment in time. So our speed is actually increasing. | Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3 |
But what about speed? Well, our velocity is already negative, and it's becoming more negative. So the absolute value of velocity, which is two meters per second, that is increasing at that moment in time. So our speed is actually increasing. So notice here, you see a difference. Now what about this point? Well, the slope of the tangent line here, of our velocity time graph, is zero right at that point. | Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3 |
So our speed is actually increasing. So notice here, you see a difference. Now what about this point? Well, the slope of the tangent line here, of our velocity time graph, is zero right at that point. So that means that our velocity is not changing. So you could say velocity not changing. And if speed is the absolute value, or the magnitude of velocity, well, that will also be not changing. | Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3 |
Well, the slope of the tangent line here, of our velocity time graph, is zero right at that point. So that means that our velocity is not changing. So you could say velocity not changing. And if speed is the absolute value, or the magnitude of velocity, well, that will also be not changing. So we would say speed is, I'll just say, neither slowing down nor speeding up. Last but not least, this point right over here, the slope of the tangent line is positive, so our velocity is increasing. What about speed? | Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3 |
And if speed is the absolute value, or the magnitude of velocity, well, that will also be not changing. So we would say speed is, I'll just say, neither slowing down nor speeding up. Last but not least, this point right over here, the slope of the tangent line is positive, so our velocity is increasing. What about speed? Well, the speed here is two meters per second. Remember, it would be the absolute value of the velocity. And the absolute value is actually going down if we forward in time a little bit. | Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3 |
We've already seen that if we were to start with a differential equation, the derivative of y with respect to x is equal to y, and we had the initial condition that y of zero is equal to one, that the particular solution to this, given this initial conditions, is y of x is equal to e to the x, or I guess we could just say y is equal to e to the x if we didn't want to write it with the function notation. And that's all fair and well, and this works out well. This is a separable differential equation, and we can integrate things quite easily. But as you will see as you go further in the world of differential equations, most differential equations are not so easy to solve. In fact, many of them are impossible to solve using analytic methods. And so given that, what do you do? You've nicely described some phenomena, modeled some phenomena using differential equations, but if you can't solve it analytically, do you just give up? | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
But as you will see as you go further in the world of differential equations, most differential equations are not so easy to solve. In fact, many of them are impossible to solve using analytic methods. And so given that, what do you do? You've nicely described some phenomena, modeled some phenomena using differential equations, but if you can't solve it analytically, do you just give up? And the answer to that question is no. You do not just give up, because we now have computers, and computers are really good at numerical methods, numerical methods for approximating and giving us a sense of what the solution to a differential equation might look like. And so how do we do that? | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
You've nicely described some phenomena, modeled some phenomena using differential equations, but if you can't solve it analytically, do you just give up? And the answer to that question is no. You do not just give up, because we now have computers, and computers are really good at numerical methods, numerical methods for approximating and giving us a sense of what the solution to a differential equation might look like. And so how do we do that? Well, in this video, we can explore one of the most straightforward numerical methods for approximating a particular solution. So what we do is, so let me draw a little table here. So a little table here. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
And so how do we do that? Well, in this video, we can explore one of the most straightforward numerical methods for approximating a particular solution. So what we do is, so let me draw a little table here. So a little table here. And so, actually let me give myself some, I'm gonna do it over here on the left-hand side. So a little table. So x and then y, with x, y, and then dy, dx. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
So a little table here. And so, actually let me give myself some, I'm gonna do it over here on the left-hand side. So a little table. So x and then y, with x, y, and then dy, dx. And you could set up a table like this to create a slope field. You could just pick all the, you could sample x's and y's in the xy plane, and then figure out, for a first-order differential equation like this, what is the slope going to be at that point, and you could construct a slope field. And we're gonna do something kind of related, but instead of trying to construct a slope field, we're gonna start with this initial condition. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
So x and then y, with x, y, and then dy, dx. And you could set up a table like this to create a slope field. You could just pick all the, you could sample x's and y's in the xy plane, and then figure out, for a first-order differential equation like this, what is the slope going to be at that point, and you could construct a slope field. And we're gonna do something kind of related, but instead of trying to construct a slope field, we're gonna start with this initial condition. We know that y of zero is equal to one. We know that the particular solution of this differential equation contains this point. So we're gonna start with that point. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
And we're gonna do something kind of related, but instead of trying to construct a slope field, we're gonna start with this initial condition. We know that y of zero is equal to one. We know that the particular solution of this differential equation contains this point. So we're gonna start with that point. So we're gonna start with x is equal to zero, and let me do this in a different color. We're gonna start with x is equal to zero, y is equal to one, which is that point right over there. And we're gonna say, well, okay, what is the derivative at that point? | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
So we're gonna start with that point. So we're gonna start with x is equal to zero, and let me do this in a different color. We're gonna start with x is equal to zero, y is equal to one, which is that point right over there. And we're gonna say, well, okay, what is the derivative at that point? Well, we know the derivative at any point or any solution to this differential equation, the derivative is going to be equal to the y value. So in this case, the derivative is going to be equal to y. It's going to be equal to one. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
And we're gonna say, well, okay, what is the derivative at that point? Well, we know the derivative at any point or any solution to this differential equation, the derivative is going to be equal to the y value. So in this case, the derivative is going to be equal to y. It's going to be equal to one. And in general, if the derivative, just like what we saw in the case of slope fields, as long as the derivative is expressed as a function of x's and y of x's, then you can figure out what the slope of the tangent line will be at that point. And so you say, okay, there's a slope of one at that point. So I can depict it like that. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
It's going to be equal to one. And in general, if the derivative, just like what we saw in the case of slope fields, as long as the derivative is expressed as a function of x's and y of x's, then you can figure out what the slope of the tangent line will be at that point. And so you say, okay, there's a slope of one at that point. So I can depict it like that. And instead of just keep doing that at a bunch of points, you say, okay, well, let's just, we know that the slope is changing, or it's probably changing for most cases, but let's just assume it's fixed until our next x, and then use that assumption to estimate what the next y would be. So what am I talking about here? So when I talk about the next x, we're talking about, well, let's just step, let's say for the sake of simplicity, we're gonna have a delta x of one, a change in x of one. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
So I can depict it like that. And instead of just keep doing that at a bunch of points, you say, okay, well, let's just, we know that the slope is changing, or it's probably changing for most cases, but let's just assume it's fixed until our next x, and then use that assumption to estimate what the next y would be. So what am I talking about here? So when I talk about the next x, we're talking about, well, let's just step, let's say for the sake of simplicity, we're gonna have a delta x of one, a change in x of one. So we're gonna step from x equals zero now. We're gonna now step from that to x is equal to one. So we're now gonna go to, actually, let me not use that, I used that yellow color already for the actual graph, or for the actual e to the x. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
So when I talk about the next x, we're talking about, well, let's just step, let's say for the sake of simplicity, we're gonna have a delta x of one, a change in x of one. So we're gonna step from x equals zero now. We're gonna now step from that to x is equal to one. So we're now gonna go to, actually, let me not use that, I used that yellow color already for the actual graph, or for the actual e to the x. So now let's say x is equal to one. So we've, our delta x is one. So we've just added one here. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
So we're now gonna go to, actually, let me not use that, I used that yellow color already for the actual graph, or for the actual e to the x. So now let's say x is equal to one. So we've, our delta x is one. So we've just added one here. And what we can do in our little approximation scheme here is let's just assume that that slope was constant over that interval. So where does that get us to? Well, if y was at one, and if I have a slope of one, for one more, for one increase in x, I'm gonna increase by y by one. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
So we've just added one here. And what we can do in our little approximation scheme here is let's just assume that that slope was constant over that interval. So where does that get us to? Well, if y was at one, and if I have a slope of one, for one more, for one increase in x, I'm gonna increase by y by one. So then y is going to increase by one, and is going to get to two. And we see that point right over there. And you already might see where this is going. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
Well, if y was at one, and if I have a slope of one, for one more, for one increase in x, I'm gonna increase by y by one. So then y is going to increase by one, and is going to get to two. And we see that point right over there. And you already might see where this is going. Now, if this were actually a point on the curve, on the solution, and if it was satisfying this, what would then the derivative be? Well, the derivative is equal to y. The slope of the tangent line is going to be equal to y. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
And you already might see where this is going. Now, if this were actually a point on the curve, on the solution, and if it was satisfying this, what would then the derivative be? Well, the derivative is equal to y. The slope of the tangent line is going to be equal to y. So in this case, the slope of the tangent line is now going to be equal to two. And we could depict that, let me depict that in magenta here. So it is going to be, it is going to be two. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
The slope of the tangent line is going to be equal to y. So in this case, the slope of the tangent line is now going to be equal to two. And we could depict that, let me depict that in magenta here. So it is going to be, it is going to be two. So it's gonna look, the slope of the tangent line there is going to be two. And so what does that tell us? Well, if we step, if we step by our delta x one more, so now our x is equal to two, what should the corresponding, what should the corresponding y be? | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
So it is going to be, it is going to be two. So it's gonna look, the slope of the tangent line there is going to be two. And so what does that tell us? Well, if we step, if we step by our delta x one more, so now our x is equal to two, what should the corresponding, what should the corresponding y be? Well, let's see. Now, for every one that we increase in the x direction, we should increase two in the y direction, because the slope is two. So the very next one should be four. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
Well, if we step, if we step by our delta x one more, so now our x is equal to two, what should the corresponding, what should the corresponding y be? Well, let's see. Now, for every one that we increase in the x direction, we should increase two in the y direction, because the slope is two. So the very next one should be four. Y is equal to four. So we could imagine, we have now kind of had a constant slope, and we get to that point right over there. And now we can do the same thing. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
So the very next one should be four. Y is equal to four. So we could imagine, we have now kind of had a constant slope, and we get to that point right over there. And now we can do the same thing. Well, if we assume dy dx, based on the differential equation, has to be equal to y, we say, okay, the slope of the tangent line there is going to be the same thing as y. It's going to be four. And so if we step our x up by one, if we increment our x by one again, and once again, we just decided to increment by one. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
And now we can do the same thing. Well, if we assume dy dx, based on the differential equation, has to be equal to y, we say, okay, the slope of the tangent line there is going to be the same thing as y. It's going to be four. And so if we step our x up by one, if we increment our x by one again, and once again, we just decided to increment by one. We could have incremented by 10. We could have incremented by.01. And you could guess which one's going to give you a more accurate result. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
And so if we step our x up by one, if we increment our x by one again, and once again, we just decided to increment by one. We could have incremented by 10. We could have incremented by.01. And you could guess which one's going to give you a more accurate result. But if we step up by one now, and our slope is four, well, we're going to increase by, if we increased x by one, we're going to increase y by four. So we are going to get, we are going to get to eight. And so we are at the.3,8, which is right over here. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
And you could guess which one's going to give you a more accurate result. But if we step up by one now, and our slope is four, well, we're going to increase by, if we increased x by one, we're going to increase y by four. So we are going to get, we are going to get to eight. And so we are at the.3,8, which is right over here. And so for this next stretch, the next stretch is going to look like that. And as you can see, just by doing this, we have been able to approximate what the particular solution looks like. And you might say, hey, Sal, well, you know, that's not so good of an approximation. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
And so we are at the.3,8, which is right over here. And so for this next stretch, the next stretch is going to look like that. And as you can see, just by doing this, we have been able to approximate what the particular solution looks like. And you might say, hey, Sal, well, you know, that's not so good of an approximation. And my reply to you is, well, yeah, I mean, depends on what your goals are. But I did this by hand. I didn't even do this using a computer. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
And you might say, hey, Sal, well, you know, that's not so good of an approximation. And my reply to you is, well, yeah, I mean, depends on what your goals are. But I did this by hand. I didn't even do this using a computer. And because I wanted to do it by hand, I took fairly large delta x steps. If I wanted a better approximation, I could have lowered the delta x. And let's do that. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
I didn't even do this using a computer. And because I wanted to do it by hand, I took fairly large delta x steps. If I wanted a better approximation, I could have lowered the delta x. And let's do that. So let's take another scenario. So let's do another scenario where instead of delta x equals one, let's say delta x equals 1 1.5. So once again, x, y, and the derivative of y with respect to x. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
And let's do that. So let's take another scenario. So let's do another scenario where instead of delta x equals one, let's say delta x equals 1 1.5. So once again, x, y, and the derivative of y with respect to x. So now let's say I want to take, so we know this first point, we're given this initial condition. When x is zero, y is one. And so the slope of the tangent line is going to be one. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
So once again, x, y, and the derivative of y with respect to x. So now let's say I want to take, so we know this first point, we're given this initial condition. When x is zero, y is one. And so the slope of the tangent line is going to be one. But then if we're incrementing by 1 1.5, so then when x is, I'll just write it as 0.5, 0.5, what is our new y going to be? Well, we're gonna assume that our slope from this to this is this slope right over here. So our slope is one. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
And so the slope of the tangent line is going to be one. But then if we're incrementing by 1 1.5, so then when x is, I'll just write it as 0.5, 0.5, what is our new y going to be? Well, we're gonna assume that our slope from this to this is this slope right over here. So our slope is one. So if we increase x by 0.5, we're gonna increase y by 0.5. And we're going to get to 1.5. So we're gonna, we get 0.5, 1.5. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
So our slope is one. So if we increase x by 0.5, we're gonna increase y by 0.5. And we're going to get to 1.5. So we're gonna, we get 0.5, 1.5. We get to that point right over there. Actually, you're having trouble seeing that. This stuff right over here is this point right over here. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
So we're gonna, we get 0.5, 1.5. We get to that point right over there. Actually, you're having trouble seeing that. This stuff right over here is this point right over here. And now our new slope is going to be 1.5, which is going to look, which is going to look like, which is going to look like, actually not quite that steep. I don't want to overstate how good of an approximation it is and it's starting to get a little bit messy. But it's gonna look something like that. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
This stuff right over here is this point right over here. And now our new slope is going to be 1.5, which is going to look, which is going to look like, which is going to look like, actually not quite that steep. I don't want to overstate how good of an approximation it is and it's starting to get a little bit messy. But it's gonna look something like that. And what you would see if you kept doing this process, so if your slope is now 1.5, when you increment x by another 0.5, we get to one. So now if you increment by, if you increment by 0.5, and your slope is 1.5, your y is gonna increment by half of that, by 0.75. And so you're gonna get to 2.25. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
But it's gonna look something like that. And what you would see if you kept doing this process, so if your slope is now 1.5, when you increment x by another 0.5, we get to one. So now if you increment by, if you increment by 0.5, and your slope is 1.5, your y is gonna increment by half of that, by 0.75. And so you're gonna get to 2.25. So now you get to one, 2.25, which is this point right over here. And once again, this is a better approximation. Remember in the original one, y of one, you know, should be equal to e. y of one in the actual solution should be equal to e, 2.7, on and on and on and on and on. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
And so you're gonna get to 2.25. So now you get to one, 2.25, which is this point right over here. And once again, this is a better approximation. Remember in the original one, y of one, you know, should be equal to e. y of one in the actual solution should be equal to e, 2.7, on and on and on and on and on. Now in this one, y of, y of, y of one got us to two. In this one, y of one got us to 2.25. Once again, closer to the actual reality, closer to e. Instead of stepping by 0.5, if we stepped by 0.1, we would get even closer. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
Remember in the original one, y of one, you know, should be equal to e. y of one in the actual solution should be equal to e, 2.7, on and on and on and on and on. Now in this one, y of, y of, y of one got us to two. In this one, y of one got us to 2.25. Once again, closer to the actual reality, closer to e. Instead of stepping by 0.5, if we stepped by 0.1, we would get even closer. If we stepped by, if we stepped by 0.0001, we would get even closer and closer and closer. So there's a bunch of interesting things here. This is actually how most differential equations or techniques that are derived from this or that are based on numerical methods similar to this are how most differential equations get solved. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
Once again, closer to the actual reality, closer to e. Instead of stepping by 0.5, if we stepped by 0.1, we would get even closer. If we stepped by, if we stepped by 0.0001, we would get even closer and closer and closer. So there's a bunch of interesting things here. This is actually how most differential equations or techniques that are derived from this or that are based on numerical methods similar to this are how most differential equations get solved. And even if it's not the exact same solution or the same method, the idea that most differential equations are actually solved or, I guess you could say simulated with a numerical method because most of them actually cannot be solved in analytical form. Now you might be saying, hey, well what method is this one right over here called? Well this right over here is called Euler's, Euler's method after the famous Leonard Euler. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
This is actually how most differential equations or techniques that are derived from this or that are based on numerical methods similar to this are how most differential equations get solved. And even if it's not the exact same solution or the same method, the idea that most differential equations are actually solved or, I guess you could say simulated with a numerical method because most of them actually cannot be solved in analytical form. Now you might be saying, hey, well what method is this one right over here called? Well this right over here is called Euler's, Euler's method after the famous Leonard Euler. Euler's, Euler's method. And not only is actually this one a good way of approximating what the solution to this or any differential equation is, but actually for this differential equation in particular, you can actually even use this to find, to find E with more and more and more precision. Anyway, hopefully you found that exciting. | Euler's method Differential equations AP Calculus BC Khan Academy.mp3 |
If we were to take the derivative with respect to time, so if we were to take the derivative with respect to time of this function s, what are we going to get? Well, we're going to get ds dt, or the rate at which position changes with respect to time, and what's another word for that? The rate at which position changes with respect to time? Well, that's just velocity. So that we could write as velocity as a function of time. Now, what if we were to take the derivative of that with respect to time? So we could either view this as the second derivative, we're taking the derivative not once, but twice of our position function, or you could say that we're taking the derivative with respect to time of our velocity function. | Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3 |
Well, that's just velocity. So that we could write as velocity as a function of time. Now, what if we were to take the derivative of that with respect to time? So we could either view this as the second derivative, we're taking the derivative not once, but twice of our position function, or you could say that we're taking the derivative with respect to time of our velocity function. So this is going to be, we can write this as dv dt, the rate at which velocity is changing with respect to time, and what's another word for that? Well, that's also called acceleration. This is going to be our acceleration as a function of time. | Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3 |
So we could either view this as the second derivative, we're taking the derivative not once, but twice of our position function, or you could say that we're taking the derivative with respect to time of our velocity function. So this is going to be, we can write this as dv dt, the rate at which velocity is changing with respect to time, and what's another word for that? Well, that's also called acceleration. This is going to be our acceleration as a function of time. So you start with the position as a function of time, take its derivative with respect to time, you get velocity. Take that derivative with respect to time, you get acceleration. Well, you could go the other way around. | Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3 |
This is going to be our acceleration as a function of time. So you start with the position as a function of time, take its derivative with respect to time, you get velocity. Take that derivative with respect to time, you get acceleration. Well, you could go the other way around. If you started with acceleration, and you were to take the antiderivative of it, an antiderivative of it is going to be, actually, let me just write it this way. So an antiderivative, I'll just use the integral symbol to show that I'm taking the antiderivative, is going to be the integral of the antiderivative of a of t, and this is going to give you some expression with a plus c, and we could say, well, that's a general form of our velocity function. This is going to be equal to our velocity function, and to find the particular velocity function, we would have to know what the velocity is at a particular time that we could solve for our c. But then if we're able to do that, and we were to take the antiderivative again, then now we're taking the antiderivative of our velocity function, which would give us some expression as a function of t, and then some other constant, and if we could solve for that constant, then we know what the position is going to be. | Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3 |
Well, you could go the other way around. If you started with acceleration, and you were to take the antiderivative of it, an antiderivative of it is going to be, actually, let me just write it this way. So an antiderivative, I'll just use the integral symbol to show that I'm taking the antiderivative, is going to be the integral of the antiderivative of a of t, and this is going to give you some expression with a plus c, and we could say, well, that's a general form of our velocity function. This is going to be equal to our velocity function, and to find the particular velocity function, we would have to know what the velocity is at a particular time that we could solve for our c. But then if we're able to do that, and we were to take the antiderivative again, then now we're taking the antiderivative of our velocity function, which would give us some expression as a function of t, and then some other constant, and if we could solve for that constant, then we know what the position is going to be. The position is a function of time. Just like this, we would have some plus c here, but if we know our position at a given time, we could solve for that c. Now that we've reviewed it a little bit, but we've rewritten it in, I guess you could say, thinking of it not just from the differential point of view or from the derivative point of view, but thinking of it from the antiderivative point of view, let's see if we can solve an interesting problem. Let's say that we know that the acceleration of a particle as a function of time is equal to 1. | Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3 |
This is going to be equal to our velocity function, and to find the particular velocity function, we would have to know what the velocity is at a particular time that we could solve for our c. But then if we're able to do that, and we were to take the antiderivative again, then now we're taking the antiderivative of our velocity function, which would give us some expression as a function of t, and then some other constant, and if we could solve for that constant, then we know what the position is going to be. The position is a function of time. Just like this, we would have some plus c here, but if we know our position at a given time, we could solve for that c. Now that we've reviewed it a little bit, but we've rewritten it in, I guess you could say, thinking of it not just from the differential point of view or from the derivative point of view, but thinking of it from the antiderivative point of view, let's see if we can solve an interesting problem. Let's say that we know that the acceleration of a particle as a function of time is equal to 1. It's always accelerating at 1 unit per, and I'm not giving you time, but let's just say that we're thinking in terms of meters and seconds. This is 1 meter per second squared right over here. That's our acceleration as a function of time. | Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3 |
Let's say that we know that the acceleration of a particle as a function of time is equal to 1. It's always accelerating at 1 unit per, and I'm not giving you time, but let's just say that we're thinking in terms of meters and seconds. This is 1 meter per second squared right over here. That's our acceleration as a function of time. Let's say we don't know the velocity expressions, but we know the velocity at a particular time, and we don't know the position expressions, but we know the position at a particular time. Let's say we know that the velocity at time 3, let's say 3 seconds, is negative 3 meters per second. Actually, I'm going to write the units here and make it a little bit. | Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3 |
That's our acceleration as a function of time. Let's say we don't know the velocity expressions, but we know the velocity at a particular time, and we don't know the position expressions, but we know the position at a particular time. Let's say we know that the velocity at time 3, let's say 3 seconds, is negative 3 meters per second. Actually, I'm going to write the units here and make it a little bit. This is meters per second squared. That's going to be our unit for acceleration. This is our unit for velocity. | Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3 |
Actually, I'm going to write the units here and make it a little bit. This is meters per second squared. That's going to be our unit for acceleration. This is our unit for velocity. Let's say that we know that the position at time 2, at 2 seconds, is equal to negative 10 meters. If we're thinking in one dimension, if we're just moving along the number line, it's 10 to the left of the origin. Given this information right over here and everything that I wrote up here, can we figure out the actual expressions for velocity as a function of time? | Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3 |
This is our unit for velocity. Let's say that we know that the position at time 2, at 2 seconds, is equal to negative 10 meters. If we're thinking in one dimension, if we're just moving along the number line, it's 10 to the left of the origin. Given this information right over here and everything that I wrote up here, can we figure out the actual expressions for velocity as a function of time? Not just velocity at time 3, but velocity generally as a function of time and position as a function of time. I encourage you to pause this video right now and try to figure it out on your own. Let's just work through this. | Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3 |
Given this information right over here and everything that I wrote up here, can we figure out the actual expressions for velocity as a function of time? Not just velocity at time 3, but velocity generally as a function of time and position as a function of time. I encourage you to pause this video right now and try to figure it out on your own. Let's just work through this. What is, we know that velocity as a function of time is going to be the antiderivative of our acceleration as a function of time. Our acceleration is just 1. This is going to be, the antiderivative of this right over here is going to be T, and then we can't forget our constant, plus C. Now we can solve for C because we know V of 3 is negative 3. | Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3 |
Let's just work through this. What is, we know that velocity as a function of time is going to be the antiderivative of our acceleration as a function of time. Our acceleration is just 1. This is going to be, the antiderivative of this right over here is going to be T, and then we can't forget our constant, plus C. Now we can solve for C because we know V of 3 is negative 3. Let's just write that down. V of 3 is going to be equal to 3, 3 plus C. Every place where I saw the T, or every place where I have the T, I replace it with this 3 right over here. Actually, let me make it a little bit clearer. | Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3 |
This is going to be, the antiderivative of this right over here is going to be T, and then we can't forget our constant, plus C. Now we can solve for C because we know V of 3 is negative 3. Let's just write that down. V of 3 is going to be equal to 3, 3 plus C. Every place where I saw the T, or every place where I have the T, I replace it with this 3 right over here. Actually, let me make it a little bit clearer. V of 3 is equal to 3 plus C, and they tell us that that's equal to negative 3. That is equal to negative 3. What's C going to be? | Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3 |
Actually, let me make it a little bit clearer. V of 3 is equal to 3 plus C, and they tell us that that's equal to negative 3. That is equal to negative 3. What's C going to be? If we just look at this part of the equation, just this equation right over here, if you subtract 3 from both sides, you get C is equal to negative 6. Now we know the exact expression that defines velocity as a function of time. V of T is equal to T, T plus negative 6, or T minus 6. | Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3 |
What's C going to be? If we just look at this part of the equation, just this equation right over here, if you subtract 3 from both sides, you get C is equal to negative 6. Now we know the exact expression that defines velocity as a function of time. V of T is equal to T, T plus negative 6, or T minus 6. We can verify that. The derivative of this with respect to time is just 1, and when time is equal to 3, 3 minus 6 is indeed negative 3. We've been able to figure out velocity as a function of time. | Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3 |
V of T is equal to T, T plus negative 6, or T minus 6. We can verify that. The derivative of this with respect to time is just 1, and when time is equal to 3, 3 minus 6 is indeed negative 3. We've been able to figure out velocity as a function of time. Now let's do a similar thing to figure out position as a function of time. We know that position is going to be an antiderivative of the velocity function. Let's write that down. | Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3 |
We've been able to figure out velocity as a function of time. Now let's do a similar thing to figure out position as a function of time. We know that position is going to be an antiderivative of the velocity function. Let's write that down. Position as a function of time is going to be equal to the antiderivative of V of T, dT, which is equal to the antiderivative of T minus 6, dT, which is equal to, well, the antiderivative of T is T squared over 2. T squared over 2, we've seen that before. The antiderivative of negative 6 is negative 6T. | Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3 |
Let's write that down. Position as a function of time is going to be equal to the antiderivative of V of T, dT, which is equal to the antiderivative of T minus 6, dT, which is equal to, well, the antiderivative of T is T squared over 2. T squared over 2, we've seen that before. The antiderivative of negative 6 is negative 6T. Of course, we can't forget our constant. So plus C. This is what S of T is equal to. S of T is equal to all of this business right over here. | Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3 |
The antiderivative of negative 6 is negative 6T. Of course, we can't forget our constant. So plus C. This is what S of T is equal to. S of T is equal to all of this business right over here. Now we can try to solve for our constant. We do that using this information right over here. At 2 seconds, our position is negative 10 meters. | Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3 |
S of T is equal to all of this business right over here. Now we can try to solve for our constant. We do that using this information right over here. At 2 seconds, our position is negative 10 meters. S of 2, or I could just write it this way. Let me write it this way. S of 2 at 2 seconds is going to be equal to 2 squared over 2. | Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3 |
At 2 seconds, our position is negative 10 meters. S of 2, or I could just write it this way. Let me write it this way. S of 2 at 2 seconds is going to be equal to 2 squared over 2. That is, let's see, that's 4 over 2. That's going to be 2 minus 6 times 2. So minus 12 plus C is equal to negative 10. | Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3 |
S of 2 at 2 seconds is going to be equal to 2 squared over 2. That is, let's see, that's 4 over 2. That's going to be 2 minus 6 times 2. So minus 12 plus C is equal to negative 10. Is equal to negative 10. Let's see, we get 2 minus 12 is negative 10, plus C is equal to negative 10. You add 10 to both sides, you get C, in this case, is equal to 0. | Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3 |
So minus 12 plus C is equal to negative 10. Is equal to negative 10. Let's see, we get 2 minus 12 is negative 10, plus C is equal to negative 10. You add 10 to both sides, you get C, in this case, is equal to 0. So we've figured out what our position function is as well. This C right over here is just going to be 0. So our position, as a function of time, is equal to T squared over 2 minus 6T. | Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3 |
You add 10 to both sides, you get C, in this case, is equal to 0. So we've figured out what our position function is as well. This C right over here is just going to be 0. So our position, as a function of time, is equal to T squared over 2 minus 6T. And you can verify. When T is equal to 2, 2 squared over 2 is 2, minus 12 is negative 10. You take the derivative here, you get T minus 6, and you can see, and we already verified, that V of 3 is negative 3. | Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3 |
What we're going to do in this video is try to find the derivative with respect to x of x squared sine of x, all of that to the third power. And what's going to be interesting is that there's multiple ways to tackle it, and I encourage you to pause the video and see if you can work through it on your own. So there's actually multiple techniques. One path is to do the chain rule first. So I'll just say CR for chain rule first. And so I have, I'm taking the derivative with respect to x of something to the third power. So if I take the derivative, it would be the derivative with respect to that something, so that would be three times that something squared times the derivative with respect to x of that something, where the something in this case is x squared sine of x, x squared sine of x. | Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3 |
One path is to do the chain rule first. So I'll just say CR for chain rule first. And so I have, I'm taking the derivative with respect to x of something to the third power. So if I take the derivative, it would be the derivative with respect to that something, so that would be three times that something squared times the derivative with respect to x of that something, where the something in this case is x squared sine of x, x squared sine of x. This is just an application of the chain rule. Now, the second part, what would that be? The second part here, this is another color, in orange. | Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3 |
So if I take the derivative, it would be the derivative with respect to that something, so that would be three times that something squared times the derivative with respect to x of that something, where the something in this case is x squared sine of x, x squared sine of x. This is just an application of the chain rule. Now, the second part, what would that be? The second part here, this is another color, in orange. Well, here I would apply the product rule. I have the product of two expressions, so I would take the derivative of, let me write this down. So this is gonna be the product rule, product rule. | Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3 |
The second part here, this is another color, in orange. Well, here I would apply the product rule. I have the product of two expressions, so I would take the derivative of, let me write this down. So this is gonna be the product rule, product rule. I would take the derivative of the first expression, so x, derivative of x squared is two x, let me write a little bit to the right. This is gonna be two x times the second expression, sine of x, plus the first expression, x squared, times the derivative of the second one, cosine of x. That's just the product rule as applied to this part right over here. | Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3 |
So this is gonna be the product rule, product rule. I would take the derivative of the first expression, so x, derivative of x squared is two x, let me write a little bit to the right. This is gonna be two x times the second expression, sine of x, plus the first expression, x squared, times the derivative of the second one, cosine of x. That's just the product rule as applied to this part right over here. And all of that, of course, is being multiplied by this up front, which actually, let me just rewrite that. So all of this I could rewrite as, let's see. This would be three times, if I have the product of things raised to the second power, I could take each of them to the second power and then take their product. | Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3 |
That's just the product rule as applied to this part right over here. And all of that, of course, is being multiplied by this up front, which actually, let me just rewrite that. So all of this I could rewrite as, let's see. This would be three times, if I have the product of things raised to the second power, I could take each of them to the second power and then take their product. So x squared squared is x to the fourth, and then sine of x squared is sine squared of x, and then all of that is being multiplied by that. And if we want, we can algebraically simplify, we can distribute everything out, in which case, what would we get? Well, let's see. | Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3 |
This would be three times, if I have the product of things raised to the second power, I could take each of them to the second power and then take their product. So x squared squared is x to the fourth, and then sine of x squared is sine squared of x, and then all of that is being multiplied by that. And if we want, we can algebraically simplify, we can distribute everything out, in which case, what would we get? Well, let's see. Three times two is six. X to the fourth times x is x to the fifth. Sine squared of x times sine of x is sine of x to the third power. | Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3 |
Well, let's see. Three times two is six. X to the fourth times x is x to the fifth. Sine squared of x times sine of x is sine of x to the third power. And then, let's see, three, so plus three, x to the fourth times x squared is x to the sixth power, and then I have sine squared of x, sine squared of x, cosine of x. So there you have it. That's one strategy, chain rule first and then product rule. | Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3 |
Sine squared of x times sine of x is sine of x to the third power. And then, let's see, three, so plus three, x to the fourth times x squared is x to the sixth power, and then I have sine squared of x, sine squared of x, cosine of x. So there you have it. That's one strategy, chain rule first and then product rule. What would be another strategy? Pause the video and try to think of it. Well, we could just algebraically use our exponent properties first, in which case, this is going to be equal to the derivative with respect to x of, if I'm taking x squared times sine of x to the third power, instead I could say x to the third to the third power, which is going to be x to the sixth, and then sine of x to the third power, sine of x to the third power. | Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3 |
That's one strategy, chain rule first and then product rule. What would be another strategy? Pause the video and try to think of it. Well, we could just algebraically use our exponent properties first, in which case, this is going to be equal to the derivative with respect to x of, if I'm taking x squared times sine of x to the third power, instead I could say x to the third to the third power, which is going to be x to the sixth, and then sine of x to the third power, sine of x to the third power. I'm using the same exponent property that we used right over here to simplify this. If I have, if I'm taking the product things to some exponent, well, that's the same thing of each of them raised to the exponent and then the product of the two. Now, how would we tackle this? | Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3 |
Well, we could just algebraically use our exponent properties first, in which case, this is going to be equal to the derivative with respect to x of, if I'm taking x squared times sine of x to the third power, instead I could say x to the third to the third power, which is going to be x to the sixth, and then sine of x to the third power, sine of x to the third power. I'm using the same exponent property that we used right over here to simplify this. If I have, if I'm taking the product things to some exponent, well, that's the same thing of each of them raised to the exponent and then the product of the two. Now, how would we tackle this? Well, I, here, I would do the product rule first. So let's do that. So let's do the product rule. | Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3 |
Now, how would we tackle this? Well, I, here, I would do the product rule first. So let's do that. So let's do the product rule. So we're gonna take the derivative of the first expression. So derivative of x to the sixth is six x to the fifth times the second expression, sine to the third of x, or sine of x to the third power, plus the first x to the sixth times the derivative of the second, and I'm just gonna write that, d dx of sine of x to the third power. Now, to evaluate this right over here, it does definitely make sense to use the chain rule, use the chain rule. | Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3 |
So let's do the product rule. So we're gonna take the derivative of the first expression. So derivative of x to the sixth is six x to the fifth times the second expression, sine to the third of x, or sine of x to the third power, plus the first x to the sixth times the derivative of the second, and I'm just gonna write that, d dx of sine of x to the third power. Now, to evaluate this right over here, it does definitely make sense to use the chain rule, use the chain rule. And so what is this going to be? Well, I have the derivative of something to the third power, so that's going to be three times that something squared times the derivative of that something. So in this case, the something is sine of x, and the derivative of sine of x is cosine of x. | Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3 |
Now, to evaluate this right over here, it does definitely make sense to use the chain rule, use the chain rule. And so what is this going to be? Well, I have the derivative of something to the third power, so that's going to be three times that something squared times the derivative of that something. So in this case, the something is sine of x, and the derivative of sine of x is cosine of x. And then I have all of this business over here. I have six x to the fifth, sine to the third, or sine of x to the third power, plus x to the sixth. And if I were to just simplify this a little bit, in fact, you see it very clearly, these two things are equivalent. | Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3 |
So in this case, the something is sine of x, and the derivative of sine of x is cosine of x. And then I have all of this business over here. I have six x to the fifth, sine to the third, or sine of x to the third power, plus x to the sixth. And if I were to just simplify this a little bit, in fact, you see it very clearly, these two things are equivalent. This term is exactly equivalent to this term, the way it's written. And then this is exactly, if you multiply three times x to the sixth, sine of x squared cosine of x. So the nice thing about math, if we're doing things that make logical sense, we should get to the same end point. | Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3 |
And if I were to just simplify this a little bit, in fact, you see it very clearly, these two things are equivalent. This term is exactly equivalent to this term, the way it's written. And then this is exactly, if you multiply three times x to the sixth, sine of x squared cosine of x. So the nice thing about math, if we're doing things that make logical sense, we should get to the same end point. But the point here is that there's multiple strategies. You could use the chain rule first and then the product rule or you could use the product rule first and then the chain rule. In this case, you could debate which one is faster. | Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3 |