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Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3 | All right, let's see if we can find the indefinite integral of one over five x squared minus 30x plus 65 dx. Pause this video and see if you can figure it out. All right, so this is going to be an interesting one. It'll be a little bit hairy, but we're gonna work through it together. So immediately, you might try multiple integration techniques and be hitting some walls. And what we're going to do here is actually try to complete the square in this denominator right over here. And then by completing the square, we're gonna get it in the form that it looks like the derivative of arctan. |
Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3 | It'll be a little bit hairy, but we're gonna work through it together. So immediately, you might try multiple integration techniques and be hitting some walls. And what we're going to do here is actually try to complete the square in this denominator right over here. And then by completing the square, we're gonna get it in the form that it looks like the derivative of arctan. And if that's a big hint to you, once again, pause the video and try to move forward. All right, now let's do this together. So I'm just gonna try to simplify this denominator so that my coefficient on my x squared term is a one. |
Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3 | And then by completing the square, we're gonna get it in the form that it looks like the derivative of arctan. And if that's a big hint to you, once again, pause the video and try to move forward. All right, now let's do this together. So I'm just gonna try to simplify this denominator so that my coefficient on my x squared term is a one. And so I can just factor a five out of the denominator. And if I did that, then this integral will become 1 5th times the integral of one over, so I factored a five out of the denominator. So it is x squared minus six x plus 13 dx. |
Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3 | So I'm just gonna try to simplify this denominator so that my coefficient on my x squared term is a one. And so I can just factor a five out of the denominator. And if I did that, then this integral will become 1 5th times the integral of one over, so I factored a five out of the denominator. So it is x squared minus six x plus 13 dx. And then as I mentioned, I'm gonna complete the square down here. So let me rewrite it. So this is equal to 1 5th times the integral of one over, and so x squared minus six x. |
Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3 | So it is x squared minus six x plus 13 dx. And then as I mentioned, I'm gonna complete the square down here. So let me rewrite it. So this is equal to 1 5th times the integral of one over, and so x squared minus six x. It's clearly not a perfect square the way it's written. Let me write this plus 13 out here. Now what could I add, and then I'm gonna have to subtract if I don't wanna change the value of the denominator, in order to make this part right over here a perfect square? |
Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3 | So this is equal to 1 5th times the integral of one over, and so x squared minus six x. It's clearly not a perfect square the way it's written. Let me write this plus 13 out here. Now what could I add, and then I'm gonna have to subtract if I don't wanna change the value of the denominator, in order to make this part right over here a perfect square? Well, we've done this before. You take half of your coefficient here, which is negative three, and you square that. So you wanna add a nine here. |
Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3 | Now what could I add, and then I'm gonna have to subtract if I don't wanna change the value of the denominator, in order to make this part right over here a perfect square? Well, we've done this before. You take half of your coefficient here, which is negative three, and you square that. So you wanna add a nine here. But if you add a nine, then you have to subtract a nine as well. And so this part is going to be x minus three squared. And then this part right over here is going to be equal to a positive four. |
Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3 | So you wanna add a nine here. But if you add a nine, then you have to subtract a nine as well. And so this part is going to be x minus three squared. And then this part right over here is going to be equal to a positive four. And we of course don't wanna forget our dx out here. And so let me write it in this form. So this is going to be equal to 1 5th times the integral of one over, get myself some space, x minus three squared plus four, which could also write as plus two squared. |
Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3 | And then this part right over here is going to be equal to a positive four. And we of course don't wanna forget our dx out here. And so let me write it in this form. So this is going to be equal to 1 5th times the integral of one over, get myself some space, x minus three squared plus four, which could also write as plus two squared. Actually, let me do it that way. Plus two squared dx. Now many of y'all might already be saying, hey, this looks a lot like arctangent, but I'm gonna try to simplify it even more so it becomes very clear that it looks like arctangent is going to be involved. |
Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3 | So this is going to be equal to 1 5th times the integral of one over, get myself some space, x minus three squared plus four, which could also write as plus two squared. Actually, let me do it that way. Plus two squared dx. Now many of y'all might already be saying, hey, this looks a lot like arctangent, but I'm gonna try to simplify it even more so it becomes very clear that it looks like arctangent is going to be involved. I'm actually gonna do some u substitution in order to do it. So the first thing I'm gonna do is, let's factor a four out of the denominator here. So if we do that, then this is going to become 1 5th times 1 4th, which is going to be 1 20th times the integral of one over x minus three squared over two squared. |
Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3 | Now many of y'all might already be saying, hey, this looks a lot like arctangent, but I'm gonna try to simplify it even more so it becomes very clear that it looks like arctangent is going to be involved. I'm actually gonna do some u substitution in order to do it. So the first thing I'm gonna do is, let's factor a four out of the denominator here. So if we do that, then this is going to become 1 5th times 1 4th, which is going to be 1 20th times the integral of one over x minus three squared over two squared. And then this is going to be a plus one. And of course, we have our dx. And then we could write this as, and I'm trying to just do every step here. |
Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3 | So if we do that, then this is going to become 1 5th times 1 4th, which is going to be 1 20th times the integral of one over x minus three squared over two squared. And then this is going to be a plus one. And of course, we have our dx. And then we could write this as, and I'm trying to just do every step here. A lot of these you might have been able to do in your head. One over, and I'll just write this as x minus three over two squared plus one. Plus one, and then dx. |
Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3 | And then we could write this as, and I'm trying to just do every step here. A lot of these you might have been able to do in your head. One over, and I'll just write this as x minus three over two squared plus one. Plus one, and then dx. And now the u substitution is pretty clear. I am just going to make the substitution that u is equal to x minus three over two. Or we could even say that's u is equal to 1 1 2 x minus 3 1 2, that's just x minus three over two. |
Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3 | Plus one, and then dx. And now the u substitution is pretty clear. I am just going to make the substitution that u is equal to x minus three over two. Or we could even say that's u is equal to 1 1 2 x minus 3 1 2, that's just x minus three over two. And du is going to be equal to 1 1 2 dx. And so what I can do here is, actually let me start to re-engineer this integral a little bit, so that we see a 1 1 2 here. So if I make this a 1 1 2, and then I multiply the outside by two, so I divide by two, multiply by two is one way to think about it. |
Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3 | Or we could even say that's u is equal to 1 1 2 x minus 3 1 2, that's just x minus three over two. And du is going to be equal to 1 1 2 dx. And so what I can do here is, actually let me start to re-engineer this integral a little bit, so that we see a 1 1 2 here. So if I make this a 1 1 2, and then I multiply the outside by two, so I divide by two, multiply by two is one way to think about it. This becomes 1 1. And so doing my u substitution, I get 1 1, that's that 1 1 there, times the integral of, well, I have 1 1 2 dx right over here, which is the same thing as du. So I could put the du either in the numerator, I could put it out here. |
Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3 | So if I make this a 1 1 2, and then I multiply the outside by two, so I divide by two, multiply by two is one way to think about it. This becomes 1 1. And so doing my u substitution, I get 1 1, that's that 1 1 there, times the integral of, well, I have 1 1 2 dx right over here, which is the same thing as du. So I could put the du either in the numerator, I could put it out here. And then I have one over, this is u squared, u squared plus one. Now you might immediately recognize, what's the derivative of arc tan of u? Well, that would be one over u squared plus one. |
Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3 | So I could put the du either in the numerator, I could put it out here. And then I have one over, this is u squared, u squared plus one. Now you might immediately recognize, what's the derivative of arc tan of u? Well, that would be one over u squared plus one. So this is going to be equal to 1 1 0 times the arc tangent of u. And of course, we can't forget our big constant c, because we're taking an indefinite integral. And now we just wanna do the reverse substitution. |
Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3 | Well, that would be one over u squared plus one. So this is going to be equal to 1 1 0 times the arc tangent of u. And of course, we can't forget our big constant c, because we're taking an indefinite integral. And now we just wanna do the reverse substitution. We know that u is equal to this business right over here, so we deserve a little bit of a drum roll. This is going to be equal to 1 1 0 times the arc tangent of u. Well, u is just x minus three over two, which could also be written like this. |
Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3 | It has a vertical tangent at the point three comma zero. So three comma zero has a vertical tangent. Let me draw that. So it has a vertical tangent right over there. And a horizontal tangent at the point zero comma negative three, zero comma negative three, so it has a horizontal tangent right over there. And also has a horizontal tangent at six comma three. Six comma three, let me draw the horizontal tangent. |
Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3 | So it has a vertical tangent right over there. And a horizontal tangent at the point zero comma negative three, zero comma negative three, so it has a horizontal tangent right over there. And also has a horizontal tangent at six comma three. Six comma three, let me draw the horizontal tangent. Just like that. Select all the x values for which f is not differentiable. Select all that apply. |
Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3 | Six comma three, let me draw the horizontal tangent. Just like that. Select all the x values for which f is not differentiable. Select all that apply. So f prime, f prime, I'll write it in shorthand. So we say no f prime under, it's going to happen under three conditions. The first condition, you could say, well we have a vertical tangent. |
Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3 | Select all that apply. So f prime, f prime, I'll write it in shorthand. So we say no f prime under, it's going to happen under three conditions. The first condition, you could say, well we have a vertical tangent. Vertical tangent. Why is a vertical tangent a place where it's hard to define our derivative? Well remember, our derivative, we're really trying to find our rate of change of y with respect to x. |
Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3 | The first condition, you could say, well we have a vertical tangent. Vertical tangent. Why is a vertical tangent a place where it's hard to define our derivative? Well remember, our derivative, we're really trying to find our rate of change of y with respect to x. When you have a vertical tangent, you change your x a very small amount, you have an infinite change in y, either in the positive or the negative direction. So that's one situation where you have no derivative. And they tell us where we have a vertical tangent in here, where x is equal to three. |
Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3 | Well remember, our derivative, we're really trying to find our rate of change of y with respect to x. When you have a vertical tangent, you change your x a very small amount, you have an infinite change in y, either in the positive or the negative direction. So that's one situation where you have no derivative. And they tell us where we have a vertical tangent in here, where x is equal to three. So we have no, f is not differentiable at x equals three because of the vertical tangent. You might say, what about horizontal tangents? No, horizontal tangents are completely fine. |
Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3 | And they tell us where we have a vertical tangent in here, where x is equal to three. So we have no, f is not differentiable at x equals three because of the vertical tangent. You might say, what about horizontal tangents? No, horizontal tangents are completely fine. Horizontal tangents are places where the derivative is equal to zero. So f prime of six is equal to zero. F prime of zero is equal to zero. |
Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3 | No, horizontal tangents are completely fine. Horizontal tangents are places where the derivative is equal to zero. So f prime of six is equal to zero. F prime of zero is equal to zero. What are other scenarios? Well another scenario where you're not going to have a defined derivative is where the graph is not continuous. It's not continuous. |
Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3 | F prime of zero is equal to zero. What are other scenarios? Well another scenario where you're not going to have a defined derivative is where the graph is not continuous. It's not continuous. And we see right over here at x equals negative three, our graph is not continuous. So x equals negative three, it's not continuous. And those are the only places where f is not differentiable that they're giving us options on. |
Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3 | It's not continuous. And we see right over here at x equals negative three, our graph is not continuous. So x equals negative three, it's not continuous. And those are the only places where f is not differentiable that they're giving us options on. We don't know what the graph is doing to the left or the right. These I guess would be interesting cases, but they haven't given us those choices here. And we already said, at x equals zero, the derivative is zero, it's defined, it's differentiable there. |
Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3 | And those are the only places where f is not differentiable that they're giving us options on. We don't know what the graph is doing to the left or the right. These I guess would be interesting cases, but they haven't given us those choices here. And we already said, at x equals zero, the derivative is zero, it's defined, it's differentiable there. And at x equals six, the derivative is zero. We have a flat, flat tangent. So once again, it's defined there as well. |
Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3 | And we already said, at x equals zero, the derivative is zero, it's defined, it's differentiable there. And at x equals six, the derivative is zero. We have a flat, flat tangent. So once again, it's defined there as well. Let's do another one of these. Oh, and actually I didn't include, I think that this takes care of this problem, but there's a third scenario in which we have, I'll call it a sharp turn. A sharp turn. |
Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3 | So once again, it's defined there as well. Let's do another one of these. Oh, and actually I didn't include, I think that this takes care of this problem, but there's a third scenario in which we have, I'll call it a sharp turn. A sharp turn. And this isn't the most mathy definition right over here, but it's easy to recognize. A sharp turn is something like that. Or like, or like, well no, that doesn't look too sharp. |
Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3 | A sharp turn. And this isn't the most mathy definition right over here, but it's easy to recognize. A sharp turn is something like that. Or like, or like, well no, that doesn't look too sharp. Or like this. And the reason why I think where you have these sharp bends or sharp turns as opposed to something that looks more smooth like that, the reason why we're not differentiable there is as we approach this point, as we approach this point from either side, we have different slopes. Notice our slope is positive right over here. |
Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3 | Or like, or like, well no, that doesn't look too sharp. Or like this. And the reason why I think where you have these sharp bends or sharp turns as opposed to something that looks more smooth like that, the reason why we're not differentiable there is as we approach this point, as we approach this point from either side, we have different slopes. Notice our slope is positive right over here. As x increases, y is increasing, while our slope is negative here. So as you try to find the limit of our slope as we approach this point, it's not going to exist because it's different on the left-hand side and the right-hand side. So that's why the sharp turns, I don't see any sharp turns here, so it doesn't apply to this example. |
Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3 | Notice our slope is positive right over here. As x increases, y is increasing, while our slope is negative here. So as you try to find the limit of our slope as we approach this point, it's not going to exist because it's different on the left-hand side and the right-hand side. So that's why the sharp turns, I don't see any sharp turns here, so it doesn't apply to this example. Let's do one more examples. And actually this one does have some sharp turns, so this could be interesting. The graph of function f is given to the left right here. |
Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3 | So that's why the sharp turns, I don't see any sharp turns here, so it doesn't apply to this example. Let's do one more examples. And actually this one does have some sharp turns, so this could be interesting. The graph of function f is given to the left right here. It has a vertical asymptote at x equals negative three. We see that. And horizontal asymptotes at y equals zero. |
Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3 | The graph of function f is given to the left right here. It has a vertical asymptote at x equals negative three. We see that. And horizontal asymptotes at y equals zero. Yep, this end of the curve, as x approaches negative infinity, it looks like y is approaching zero. And it has another horizontal asymptote at y equals four. As x approaches infinity, it looks like our graph is trending down to y is equal to four. |
Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3 | And horizontal asymptotes at y equals zero. Yep, this end of the curve, as x approaches negative infinity, it looks like y is approaching zero. And it has another horizontal asymptote at y equals four. As x approaches infinity, it looks like our graph is trending down to y is equal to four. Select the x values for which f is not differentiable. So first of all, we could think about vertical tangents. Doesn't seem to have any vertical tangents. |
Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3 | As x approaches infinity, it looks like our graph is trending down to y is equal to four. Select the x values for which f is not differentiable. So first of all, we could think about vertical tangents. Doesn't seem to have any vertical tangents. Then we could think about where we are not continuous. Well, we're definitely not continuous where we have this vertical asymptote right over here. So we're not continuous at x equals negative three. |
Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3 | Doesn't seem to have any vertical tangents. Then we could think about where we are not continuous. Well, we're definitely not continuous where we have this vertical asymptote right over here. So we're not continuous at x equals negative three. We're also not continuous at x is equal to one. And then the last situation where we are not going to be differentiable is where we have a sharp turn, or you could kind of view it as a sharp point on our graph. And I see a sharp point right over there. |
Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3 | So we're not continuous at x equals negative three. We're also not continuous at x is equal to one. And then the last situation where we are not going to be differentiable is where we have a sharp turn, or you could kind of view it as a sharp point on our graph. And I see a sharp point right over there. Notice, as we approach from the left-hand side, the slope looks like a constant, I don't know, it looks like a positive 3 1⁄2. Well, as we go to the right side of that, it looks like our slope turns negative. And so if you were to try to find the limit of the slope as we approach from either side, which is essentially what you're trying to do when you try to find the derivative, well, it's not going to be defined because it's different on either side. |
Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3 | And I see a sharp point right over there. Notice, as we approach from the left-hand side, the slope looks like a constant, I don't know, it looks like a positive 3 1⁄2. Well, as we go to the right side of that, it looks like our slope turns negative. And so if you were to try to find the limit of the slope as we approach from either side, which is essentially what you're trying to do when you try to find the derivative, well, it's not going to be defined because it's different on either side. So f is also not differentiable at the x value that gives us that little sharp point right over there. And if you were to graph the derivative, which we will do in future videos, you will see that the derivative is not continuous at that point. So let me mark that off. |
Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3 | And so if you were to try to find the limit of the slope as we approach from either side, which is essentially what you're trying to do when you try to find the derivative, well, it's not going to be defined because it's different on either side. So f is also not differentiable at the x value that gives us that little sharp point right over there. And if you were to graph the derivative, which we will do in future videos, you will see that the derivative is not continuous at that point. So let me mark that off. And then we could check x equals zero. X equals zero is completely cool. We're at a point that our tangent line is definitely not vertical. |
Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3 | What I want to do in this video is a few examples that test our intuition of the derivative as a rate of change or the steepness of a curve or the slope of a curve or the slope of a tangent line of a curve, depending on how you actually want to think about it. So here it says f prime of five. So this notation, prime, this is another way of saying, well, what's the derivative? Let's estimate the derivative of our function at five. And when we say f prime of five, this is the slope, slope of tangent line, tangent line at five. Or you could view it as the, you could view it as the rate of change of y with respect to x, which is really how we define slope, with respect to x of our function f. So let's think about that a little bit. We see they put the point, the point five comma f of five right over here. |
Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3 | Let's estimate the derivative of our function at five. And when we say f prime of five, this is the slope, slope of tangent line, tangent line at five. Or you could view it as the, you could view it as the rate of change of y with respect to x, which is really how we define slope, with respect to x of our function f. So let's think about that a little bit. We see they put the point, the point five comma f of five right over here. And so if we want to estimate the slope of the tangent line, if we want to estimate the steepness of this curve, we could try to draw a line that is tangent right at that point. And so let me see if I can do that. So if I were to draw a line starting there, if I just wanted to make it tangent, it looks like it would do something like that, that right at that point, that looks to be about how steep that curve is. |
Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3 | We see they put the point, the point five comma f of five right over here. And so if we want to estimate the slope of the tangent line, if we want to estimate the steepness of this curve, we could try to draw a line that is tangent right at that point. And so let me see if I can do that. So if I were to draw a line starting there, if I just wanted to make it tangent, it looks like it would do something like that, that right at that point, that looks to be about how steep that curve is. Now what makes this an interesting thing in nonlinear is that it's constantly changing. The steepness, it's very low here, and it gets steeper and steeper and steeper as we move to the right for larger and larger x values. But if we look at the point in question, when x is equal to five, remember, f prime of five would be, if we were estimating it, this would be the slope of this line here. |
Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3 | So if I were to draw a line starting there, if I just wanted to make it tangent, it looks like it would do something like that, that right at that point, that looks to be about how steep that curve is. Now what makes this an interesting thing in nonlinear is that it's constantly changing. The steepness, it's very low here, and it gets steeper and steeper and steeper as we move to the right for larger and larger x values. But if we look at the point in question, when x is equal to five, remember, f prime of five would be, if we were estimating it, this would be the slope of this line here. And the slope of this line, it looks like for every time we move one in the x direction, we're moving two in the y direction. Delta y is equal to two when delta x is equal to one. So our change in y with respect to x, at least for this tangent line here, which would represent our change in y with respect to x right at that point, is going to be equal to two over one, or two. |
Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3 | But if we look at the point in question, when x is equal to five, remember, f prime of five would be, if we were estimating it, this would be the slope of this line here. And the slope of this line, it looks like for every time we move one in the x direction, we're moving two in the y direction. Delta y is equal to two when delta x is equal to one. So our change in y with respect to x, at least for this tangent line here, which would represent our change in y with respect to x right at that point, is going to be equal to two over one, or two. And they told us to estimate it, but all of these are way off. Having a negative two derivative would mean that as we increase our x, our y is decreasing. So if our curve looks something like this, we would have a slope of negative two. |
Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3 | So our change in y with respect to x, at least for this tangent line here, which would represent our change in y with respect to x right at that point, is going to be equal to two over one, or two. And they told us to estimate it, but all of these are way off. Having a negative two derivative would mean that as we increase our x, our y is decreasing. So if our curve looks something like this, we would have a slope of negative two. If having slopes in this, a positive of.1, that would be very flat. Down here we might have a slope closer to.1. Negative.1, that might be closer on this side, where it's downward sloping, but very close to flat. |
Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3 | So if our curve looks something like this, we would have a slope of negative two. If having slopes in this, a positive of.1, that would be very flat. Down here we might have a slope closer to.1. Negative.1, that might be closer on this side, where it's downward sloping, but very close to flat. A slope of zero, that would be right over here at the bottom where right at that moment, as we change x, y is not increasing or decreasing. The slope of the tangent line right at that bottom point would have a slope of zero. So I feel really good about that response. |
Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3 | Negative.1, that might be closer on this side, where it's downward sloping, but very close to flat. A slope of zero, that would be right over here at the bottom where right at that moment, as we change x, y is not increasing or decreasing. The slope of the tangent line right at that bottom point would have a slope of zero. So I feel really good about that response. Let's do one more of these. All right, so they're telling us to compare the derivative of g at four to the derivative of g at six. And which one of these is greater? |
Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3 | So I feel really good about that response. Let's do one more of these. All right, so they're telling us to compare the derivative of g at four to the derivative of g at six. And which one of these is greater? And like always, pause the video and see if you could figure this out. Well, this is just an exercise. Let's see if we were to make a line that indicates the slope there. |
Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3 | And which one of these is greater? And like always, pause the video and see if you could figure this out. Well, this is just an exercise. Let's see if we were to make a line that indicates the slope there. And you could view this as a tangent line. So let me try to do that. So, no, that doesn't do a good job. |
Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3 | Let's see if we were to make a line that indicates the slope there. And you could view this as a tangent line. So let me try to do that. So, no, that doesn't do a good job. So right over here at, so that looks like a pretty, I think I can do a better job than that. No, that's too shallow. Let's see, not shallow, that's too flat. |
Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3 | So, no, that doesn't do a good job. So right over here at, so that looks like a pretty, I think I can do a better job than that. No, that's too shallow. Let's see, not shallow, that's too flat. So let me try to really, okay, that looks pretty good. So that line that I just drew seems to be indicative of the rate of change of y with respect to x or the slope of that curve or that line, you could view it as a tangent line so that we can think about what its slope is going to be. And then if we go further down over here, this one is, it looks like it is steeper, but in the negative direction. |
Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3 | Let's see, not shallow, that's too flat. So let me try to really, okay, that looks pretty good. So that line that I just drew seems to be indicative of the rate of change of y with respect to x or the slope of that curve or that line, you could view it as a tangent line so that we can think about what its slope is going to be. And then if we go further down over here, this one is, it looks like it is steeper, but in the negative direction. So it looks like it is steeper for sure, but it's in the negative direction. As we increase, think of it this way, as we increase x one here, it looks like we are decreasing y by about one. So it looks like g prime of four, g prime of four, the derivative when x is equal to four, is approximately, I'm estimating it, negative one. |
Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3 | And then if we go further down over here, this one is, it looks like it is steeper, but in the negative direction. So it looks like it is steeper for sure, but it's in the negative direction. As we increase, think of it this way, as we increase x one here, it looks like we are decreasing y by about one. So it looks like g prime of four, g prime of four, the derivative when x is equal to four, is approximately, I'm estimating it, negative one. While the derivative here, when we increase x, if we increase x by, if we increase x by one, it looks like we're decreasing y by close to three. So g prime of six looks like it's closer to negative three. So which one of these is larger? |
Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3 | So it looks like g prime of four, g prime of four, the derivative when x is equal to four, is approximately, I'm estimating it, negative one. While the derivative here, when we increase x, if we increase x by, if we increase x by one, it looks like we're decreasing y by close to three. So g prime of six looks like it's closer to negative three. So which one of these is larger? Well, this one is less negative, so it's gonna be greater than the other one. And you could have done this intuitively. If you just look at the curve, this is some type of a sinusoid here. |
Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3 | So which one of these is larger? Well, this one is less negative, so it's gonna be greater than the other one. And you could have done this intuitively. If you just look at the curve, this is some type of a sinusoid here. You have right over here, the curve is flat. It's, you have right at that moment, you have no change in y with respect to x. Then it starts to decrease at a, then it decreases at an even faster rate, then it decreases at a faster rate. |
Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3 | If you just look at the curve, this is some type of a sinusoid here. You have right over here, the curve is flat. It's, you have right at that moment, you have no change in y with respect to x. Then it starts to decrease at a, then it decreases at an even faster rate, then it decreases at a faster rate. Then it starts, it's still decreasing, but it's decreasing at slower and slower rates, decreasing at slower rates. And right at that moment, it's not, you have your slope of your tangent line is zero. Then it starts to increase, increase, so on and so forth. |
Interpreting the meaning of the derivative in context AP Calculus AB Khan Academy.mp3 | We're told that Eddie drove from New York City to Philadelphia. The function d gives the total distance Eddie has driven in kilometers t hours after he left. What is the best interpretation for the following statement? D prime of two is equal to 100. So pause this video and I encourage you to write it out. What do you think this means? And be sure to include the appropriate units. |
Interpreting the meaning of the derivative in context AP Calculus AB Khan Academy.mp3 | D prime of two is equal to 100. So pause this video and I encourage you to write it out. What do you think this means? And be sure to include the appropriate units. All right, now let's do this together. If d is equal to the distance driven, then to get d prime, you're taking the derivative with respect to time. So one way to think about it is, it is the rate of change of d. So we could view this as d prime is going to give you the instantaneous rate. |
Interpreting the meaning of the derivative in context AP Calculus AB Khan Academy.mp3 | And be sure to include the appropriate units. All right, now let's do this together. If d is equal to the distance driven, then to get d prime, you're taking the derivative with respect to time. So one way to think about it is, it is the rate of change of d. So we could view this as d prime is going to give you the instantaneous rate. And they are both functions of t. So one way to view d prime of two is equal to 100, that would mean, well, what is our time now? Well, that is our t, and that's in hours. So two hours, actually, let me color code it. |
Interpreting the meaning of the derivative in context AP Calculus AB Khan Academy.mp3 | So one way to think about it is, it is the rate of change of d. So we could view this as d prime is going to give you the instantaneous rate. And they are both functions of t. So one way to view d prime of two is equal to 100, that would mean, well, what is our time now? Well, that is our t, and that's in hours. So two hours, actually, let me color code it. So two hours after leaving, after leaving, Eddie drove, and this means, let me be grammatically correct, drove at an instantaneous, instantaneous, instantaneous rate of, and let me use a different color now for this part, of 100, and what are the units? Well, the distance was given in kilometers, and now we're gonna be thinking about kilometers per unit time, kilometers per hour. So this is 100 kilometers, kilometers per hour. |
Interpreting the meaning of the derivative in context AP Calculus AB Khan Academy.mp3 | So two hours, actually, let me color code it. So two hours after leaving, after leaving, Eddie drove, and this means, let me be grammatically correct, drove at an instantaneous, instantaneous, instantaneous rate of, and let me use a different color now for this part, of 100, and what are the units? Well, the distance was given in kilometers, and now we're gonna be thinking about kilometers per unit time, kilometers per hour. So this is 100 kilometers, kilometers per hour. So that's the interpretation there. Let's do another example. Here we are told a tank is being drained of water. |
Interpreting the meaning of the derivative in context AP Calculus AB Khan Academy.mp3 | So this is 100 kilometers, kilometers per hour. So that's the interpretation there. Let's do another example. Here we are told a tank is being drained of water. The function v gives the volume of liquid in the tank in liters after t minutes. What is the best interpretation for the following statement? The slope of the line tangent to the graph of v at t equals seven is equal to negative three. |
Interpreting the meaning of the derivative in context AP Calculus AB Khan Academy.mp3 | Here we are told a tank is being drained of water. The function v gives the volume of liquid in the tank in liters after t minutes. What is the best interpretation for the following statement? The slope of the line tangent to the graph of v at t equals seven is equal to negative three. So pause this video again and try to do what we just did with the previous example. Write out that interpretation, and make sure to get the units right. All right, so let's just remind ourselves what's going on. |
Interpreting the meaning of the derivative in context AP Calculus AB Khan Academy.mp3 | The slope of the line tangent to the graph of v at t equals seven is equal to negative three. So pause this video again and try to do what we just did with the previous example. Write out that interpretation, and make sure to get the units right. All right, so let's just remind ourselves what's going on. V is going to give us the volume as a function of time. Volume is in liters, and time is in minutes. And so if they're talking about the slope of the tangent line to the graph, the slope of the tangent line to the graph of v, that's just v prime. |
Interpreting the meaning of the derivative in context AP Calculus AB Khan Academy.mp3 | All right, so let's just remind ourselves what's going on. V is going to give us the volume as a function of time. Volume is in liters, and time is in minutes. And so if they're talking about the slope of the tangent line to the graph, the slope of the tangent line to the graph of v, that's just v prime. So if you take the derivative with respect to time, that's going to give you v prime, and these are all functions of t. These are all functions of t. And they say at t equals seven, it's equal to negative three. So this, which is the same thing as the slope of tangent line slope of tangent, tangent line, and they tell us that v prime of, at time equals seven minutes, our rate of change of volume with respect to time is equal to negative three. And so you could say, if we were to write it out, this means that after, after seven minutes, seven minutes, the tank is being drained at an instantaneous, instantaneous. |
Interpreting the meaning of the derivative in context AP Calculus AB Khan Academy.mp3 | And so if they're talking about the slope of the tangent line to the graph, the slope of the tangent line to the graph of v, that's just v prime. So if you take the derivative with respect to time, that's going to give you v prime, and these are all functions of t. These are all functions of t. And they say at t equals seven, it's equal to negative three. So this, which is the same thing as the slope of tangent line slope of tangent, tangent line, and they tell us that v prime of, at time equals seven minutes, our rate of change of volume with respect to time is equal to negative three. And so you could say, if we were to write it out, this means that after, after seven minutes, seven minutes, the tank is being drained at an instantaneous, instantaneous. That's why we need that calculus for that instantaneous rate. At an instantaneous rate of, now, you might be tempted to say it's being drained at an instantaneous rate of negative three liters per minute. But remember, the negative three just shows that the volume is decreasing. |
Interpreting the meaning of the derivative in context AP Calculus AB Khan Academy.mp3 | And so you could say, if we were to write it out, this means that after, after seven minutes, seven minutes, the tank is being drained at an instantaneous, instantaneous. That's why we need that calculus for that instantaneous rate. At an instantaneous rate of, now, you might be tempted to say it's being drained at an instantaneous rate of negative three liters per minute. But remember, the negative three just shows that the volume is decreasing. So one way to think about it is, this negative is already being accounted for when you're saying it's being drained. If this was positive, that means it is being filled. So it is being drained at an instantaneous rate of three liters per minute. |
Interpreting the meaning of the derivative in context AP Calculus AB Khan Academy.mp3 | But remember, the negative three just shows that the volume is decreasing. So one way to think about it is, this negative is already being accounted for when you're saying it's being drained. If this was positive, that means it is being filled. So it is being drained at an instantaneous rate of three liters per minute. Three liters per minute. And how did I know the units were liters per minute? Well, the volume function is in terms of liters, and the time is in terms of minutes, and then I'm taking the derivative with respect to time, so now it's going to be liters per minute, and we are done. |
Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3 | And like always, I encourage you to pause this video and try to figure this out on your own. And I will give you two hints. First hint is, well, we don't know what the derivative of sine inverse of x is, but we do know what the derivative of the sine of something is. And so maybe if you rearrange this and use some implicit differentiation, maybe you can figure out what dy dx is. Remember, this right over here, this right over here is our goal. We essentially want to figure out the derivative of this with respect to x. So I'm assuming you've had a go at it, so let's work through this together. |
Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3 | And so maybe if you rearrange this and use some implicit differentiation, maybe you can figure out what dy dx is. Remember, this right over here, this right over here is our goal. We essentially want to figure out the derivative of this with respect to x. So I'm assuming you've had a go at it, so let's work through this together. So if y is the inverse sine of x, that's just like saying that, that's equivalent to saying that sine of y is equal to x. Sine of y is equal to x. So now we have things that we're a little bit more familiar with, and now we can do a little bit of implicit differentiation. |
Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3 | So I'm assuming you've had a go at it, so let's work through this together. So if y is the inverse sine of x, that's just like saying that, that's equivalent to saying that sine of y is equal to x. Sine of y is equal to x. So now we have things that we're a little bit more familiar with, and now we can do a little bit of implicit differentiation. We could take the derivative of both sides with respect to x. So derivative of the left-hand side with respect to x and the derivative of the right-hand side with respect to x. Well, what's the derivative of the left-hand side with respect to x going to be? |
Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3 | So now we have things that we're a little bit more familiar with, and now we can do a little bit of implicit differentiation. We could take the derivative of both sides with respect to x. So derivative of the left-hand side with respect to x and the derivative of the right-hand side with respect to x. Well, what's the derivative of the left-hand side with respect to x going to be? And here we just apply the chain rule. It's going to be the derivative of sine of y with respect to y, which is going to be cosine of y times the derivative of y with respect to x. So times dy dx. |
Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3 | Well, what's the derivative of the left-hand side with respect to x going to be? And here we just apply the chain rule. It's going to be the derivative of sine of y with respect to y, which is going to be cosine of y times the derivative of y with respect to x. So times dy dx. Times dy dx. And the right-hand side, what's the derivative of x with respect to x? Well, that's obviously just going to be equal to 1. |
Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3 | So times dy dx. Times dy dx. And the right-hand side, what's the derivative of x with respect to x? Well, that's obviously just going to be equal to 1. And so we could solve for dy dx, divide both sides by cosine of y, and we get the derivative of y with respect to x is equal to 1 over cosine of y. Now, this still isn't that satisfying because I have the derivative in terms of y. So let's see if we can re-express it in terms of x. |
Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3 | Well, that's obviously just going to be equal to 1. And so we could solve for dy dx, divide both sides by cosine of y, and we get the derivative of y with respect to x is equal to 1 over cosine of y. Now, this still isn't that satisfying because I have the derivative in terms of y. So let's see if we can re-express it in terms of x. So how could we do that? Well, we already know that x is equal to sine of y. Let me rewrite it. |
Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3 | So let's see if we can re-express it in terms of x. So how could we do that? Well, we already know that x is equal to sine of y. Let me rewrite it. We already know that x is equal to sine of y. So if we could rewrite this bottom expression in terms, instead of cosine of y, if we could use our trigonometric identities to rewrite it in terms of sine of y, then we'll be in good shape because x is equal to sine of y. Well, how can we do that? |
Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3 | Let me rewrite it. We already know that x is equal to sine of y. So if we could rewrite this bottom expression in terms, instead of cosine of y, if we could use our trigonometric identities to rewrite it in terms of sine of y, then we'll be in good shape because x is equal to sine of y. Well, how can we do that? Well, we know from our trigonometric identities, we know that sine squared of y plus cosine squared of y is equal to 1. Or if we want to solve for cosine of y, subtract sine squared of y from both sides, we know that cosine squared of y is equal to 1 minus sine squared of y, or that cosine of y, just take the principal root of both sides, is equal to the principal root of 1 minus sine squared of y. So we can rewrite this as being equal to 1 over, 1 over, instead of cosine of y, we could rewrite it as 1 minus sine squared of y. |
Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3 | Well, how can we do that? Well, we know from our trigonometric identities, we know that sine squared of y plus cosine squared of y is equal to 1. Or if we want to solve for cosine of y, subtract sine squared of y from both sides, we know that cosine squared of y is equal to 1 minus sine squared of y, or that cosine of y, just take the principal root of both sides, is equal to the principal root of 1 minus sine squared of y. So we can rewrite this as being equal to 1 over, 1 over, instead of cosine of y, we could rewrite it as 1 minus sine squared of y. Now why is this useful? Well, sine of y is just x. So this is the same, if we just substitute back in, and let me just write it that way so it's a little bit clearer. |
Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3 | So we can rewrite this as being equal to 1 over, 1 over, instead of cosine of y, we could rewrite it as 1 minus sine squared of y. Now why is this useful? Well, sine of y is just x. So this is the same, if we just substitute back in, and let me just write it that way so it's a little bit clearer. I could write it as sine y squared. We know that this thing right over here is x. So this is going to be equal to, and we deserve a little bit of a drum roll, 1 over the square root of 1 minus, instead of sine of y, we know that x is equal to sine of y. |
Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3 | So this is the same, if we just substitute back in, and let me just write it that way so it's a little bit clearer. I could write it as sine y squared. We know that this thing right over here is x. So this is going to be equal to, and we deserve a little bit of a drum roll, 1 over the square root of 1 minus, instead of sine of y, we know that x is equal to sine of y. So 1 minus x squared. So there you have it. The derivative with respect to x of the inverse sine of x is equal to 1 over the square root of 1 minus x squared. |
Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3 | So this is going to be equal to, and we deserve a little bit of a drum roll, 1 over the square root of 1 minus, instead of sine of y, we know that x is equal to sine of y. So 1 minus x squared. So there you have it. The derivative with respect to x of the inverse sine of x is equal to 1 over the square root of 1 minus x squared. So let me just make that very clear. If you were to take the derivative with respect to x of both sides of this, you would get dy dx is equal to this on the right-hand side. Or we could say the derivative with respect to x of the inverse sine of x is equal to 1 over the square root of 1 minus x squared. |
Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3 | The derivative with respect to x of the inverse sine of x is equal to 1 over the square root of 1 minus x squared. So let me just make that very clear. If you were to take the derivative with respect to x of both sides of this, you would get dy dx is equal to this on the right-hand side. Or we could say the derivative with respect to x of the inverse sine of x is equal to 1 over the square root of 1 minus x squared. Now you could always reprove this if your memory starts to fail you, and actually that is the best way to really internalize this. But this is also just a good thing to know, especially as we go into more and more calculus and you might see this in expression. You might say, oh, okay, that's the derivative of the inverse sine of x, and that might prove to be useful. |
_-substitution intro AP Calculus AB Khan Academy.mp3 | So first when you look at it, it seems like a really complicated integral. We have this polynomial right over here being multiplied by this exponential expression and over here in the exponent, we essentially have another polynomial. It seems kind of crazy. And the key intuition here, the key insight, is that you might want to use a technique here called u-substitution. Substitution. And I'll tell you in a second how I would recognize that we have to use u-substitution. And then over time, you might even be able to do this type of thing in your head. |
_-substitution intro AP Calculus AB Khan Academy.mp3 | And the key intuition here, the key insight, is that you might want to use a technique here called u-substitution. Substitution. And I'll tell you in a second how I would recognize that we have to use u-substitution. And then over time, you might even be able to do this type of thing in your head. U-substitution is essentially unwinding the chain rule. In the chain, well, I'll go in more depth in another video where I really talk about that intuition. But the way I would think about it is, well, I have this crazy exponent right over here. |
_-substitution intro AP Calculus AB Khan Academy.mp3 | And then over time, you might even be able to do this type of thing in your head. U-substitution is essentially unwinding the chain rule. In the chain, well, I'll go in more depth in another video where I really talk about that intuition. But the way I would think about it is, well, I have this crazy exponent right over here. I have the x to the third plus x squared. And this thing right over here happens to be the derivative of x to the third plus x squared. The derivative of x to the third is three x squared. |
_-substitution intro AP Calculus AB Khan Academy.mp3 | But the way I would think about it is, well, I have this crazy exponent right over here. I have the x to the third plus x squared. And this thing right over here happens to be the derivative of x to the third plus x squared. The derivative of x to the third is three x squared. Derivative of x squared is two x, which is a huge clue to me that I could use u-substitution. So what I do here is this thing, or this little expression here, where I also see its derivative being multiplied, I can set that equal to u. So I can say u is equal to x to the third plus x squared. |
_-substitution intro AP Calculus AB Khan Academy.mp3 | The derivative of x to the third is three x squared. Derivative of x squared is two x, which is a huge clue to me that I could use u-substitution. So what I do here is this thing, or this little expression here, where I also see its derivative being multiplied, I can set that equal to u. So I can say u is equal to x to the third plus x squared. Now what is going to be the derivative of u with respect to x? du dx, well, we've done this multiple times. It's going to be three x squared plus two x. |
_-substitution intro AP Calculus AB Khan Academy.mp3 | So I can say u is equal to x to the third plus x squared. Now what is going to be the derivative of u with respect to x? du dx, well, we've done this multiple times. It's going to be three x squared plus two x. And now we can write this in differential form. And du dx, this isn't really a fraction of a differential of du divided by a differential of dx. It really is a form of notation. |
_-substitution intro AP Calculus AB Khan Academy.mp3 | It's going to be three x squared plus two x. And now we can write this in differential form. And du dx, this isn't really a fraction of a differential of du divided by a differential of dx. It really is a form of notation. But it is often useful to kind of pretend that it is a fraction. And you could kind of view this if you wanted to just get a du, if you just wanted to get a differential form over here. How much does u change for a given change in x? |
_-substitution intro AP Calculus AB Khan Academy.mp3 | It really is a form of notation. But it is often useful to kind of pretend that it is a fraction. And you could kind of view this if you wanted to just get a du, if you just wanted to get a differential form over here. How much does u change for a given change in x? You could multiply both sides times dx. So both sides times dx. And so if we were to pretend that they were fractions, and it will give you the correct differential form, you're going to be left with du is equal to, du is equal to three x squared plus two x dx. |
_-substitution intro AP Calculus AB Khan Academy.mp3 | How much does u change for a given change in x? You could multiply both sides times dx. So both sides times dx. And so if we were to pretend that they were fractions, and it will give you the correct differential form, you're going to be left with du is equal to, du is equal to three x squared plus two x dx. Now why is this over here? Why did I go through the trouble of doing that? Well, we see we have a three x squared plus two x, and then it's being multiplied by a dx right over here. |
_-substitution intro AP Calculus AB Khan Academy.mp3 | And so if we were to pretend that they were fractions, and it will give you the correct differential form, you're going to be left with du is equal to, du is equal to three x squared plus two x dx. Now why is this over here? Why did I go through the trouble of doing that? Well, we see we have a three x squared plus two x, and then it's being multiplied by a dx right over here. I could rewrite this original integral. I could rewrite this as the integral of, let me do it in that color, of three x squared plus two x times dx times e, let me do that in that other color, times e to the x to the third plus x squared. Now, what's interesting about this? |
_-substitution intro AP Calculus AB Khan Academy.mp3 | Well, we see we have a three x squared plus two x, and then it's being multiplied by a dx right over here. I could rewrite this original integral. I could rewrite this as the integral of, let me do it in that color, of three x squared plus two x times dx times e, let me do that in that other color, times e to the x to the third plus x squared. Now, what's interesting about this? Well, the stuff that I have in magenta here is exactly equal to du. This is exactly equal to du. And then this stuff I have up here, x to the third plus x squared, that is what I set u equal to. |
_-substitution intro AP Calculus AB Khan Academy.mp3 | Now, what's interesting about this? Well, the stuff that I have in magenta here is exactly equal to du. This is exactly equal to du. And then this stuff I have up here, x to the third plus x squared, that is what I set u equal to. That is going to be equal to u. So I can rewrite my entire integral, and now you might recognize why this is going to simplify things a good bit. It's going to be equal to, and what I'm gonna do is I'm gonna change the order. |
_-substitution intro AP Calculus AB Khan Academy.mp3 | And then this stuff I have up here, x to the third plus x squared, that is what I set u equal to. That is going to be equal to u. So I can rewrite my entire integral, and now you might recognize why this is going to simplify things a good bit. It's going to be equal to, and what I'm gonna do is I'm gonna change the order. I'm gonna put the du, this entire du, I'm gonna stick it on the other side here so it looks like more of the standard form that we're used to seeing our indefinite integrals in. So it's going to be, we're gonna have our du, our du, at times e, times e to the u, times e to the u. And so what would the antiderivative of this be in terms of u? |
_-substitution intro AP Calculus AB Khan Academy.mp3 | It's going to be equal to, and what I'm gonna do is I'm gonna change the order. I'm gonna put the du, this entire du, I'm gonna stick it on the other side here so it looks like more of the standard form that we're used to seeing our indefinite integrals in. So it's going to be, we're gonna have our du, our du, at times e, times e to the u, times e to the u. And so what would the antiderivative of this be in terms of u? Well, the derivative of e to the u is e to the u. The antiderivative of e to the u is e to the u. So it's going to be equal to e to the u, e to the u. |
_-substitution intro AP Calculus AB Khan Academy.mp3 | And so what would the antiderivative of this be in terms of u? Well, the derivative of e to the u is e to the u. The antiderivative of e to the u is e to the u. So it's going to be equal to e to the u, e to the u. Now, there's a possibility that there was some type of a constant factor here, so let me write that. So plus c. And now, to get it in terms of x, we just have to unsubstitute the u. We know what u is equal to, so we could say that this is going to be equal to e. Instead of writing u, we could say u is x to the third plus x squared. |
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