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So this is our new p of x. So this is approximately going to be sine of x as we add more and more terms. And so the first term here, f of 0, that's just going to be 0. So we're not even going to need to include that. The next term is going to be f prime of 0, which is 1 times x. So it's going to be x. Then the next term is f prime, the second derivative at 0, which we see here is 0. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
So we're not even going to need to include that. The next term is going to be f prime of 0, which is 1 times x. So it's going to be x. Then the next term is f prime, the second derivative at 0, which we see here is 0. Let me scroll down a little bit. It is 0, so we won't have the second term. This third term right here, the third derivative of sine of x evaluated at 0 is negative 1. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
Then the next term is f prime, the second derivative at 0, which we see here is 0. Let me scroll down a little bit. It is 0, so we won't have the second term. This third term right here, the third derivative of sine of x evaluated at 0 is negative 1. So we're now going to have a negative 1. Let me scroll down so you can see this. This is negative 1 in this case, times x to the third over 3 factorial. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
This third term right here, the third derivative of sine of x evaluated at 0 is negative 1. So we're now going to have a negative 1. Let me scroll down so you can see this. This is negative 1 in this case, times x to the third over 3 factorial. So negative x to the third over 3 factorial. And then the next term is going to be 0, because that's the fourth derivative. The fourth derivative evaluated at 0 is the next coefficient. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
This is negative 1 in this case, times x to the third over 3 factorial. So negative x to the third over 3 factorial. And then the next term is going to be 0, because that's the fourth derivative. The fourth derivative evaluated at 0 is the next coefficient. We see that that is going to be 0, so it's going to drop off. And what you're going to see here, and actually maybe I haven't done enough terms for you to feel good about this. Let me do one more term right over here, just so it becomes clear. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
The fourth derivative evaluated at 0 is the next coefficient. We see that that is going to be 0, so it's going to drop off. And what you're going to see here, and actually maybe I haven't done enough terms for you to feel good about this. Let me do one more term right over here, just so it becomes clear. f of the fifth derivative of x is going to be cosine of x again. The fifth derivative, let me do it in that same color, just so it's consistent. The fifth derivative evaluated at 0 is going to be 1. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
Let me do one more term right over here, just so it becomes clear. f of the fifth derivative of x is going to be cosine of x again. The fifth derivative, let me do it in that same color, just so it's consistent. The fifth derivative evaluated at 0 is going to be 1. So the fourth derivative evaluated at 0 is 0. Then you go to the fifth derivative evaluated at 0 is going to be positive 1. And if I kept doing this, it would be positive 1 times, I don't have to write the 1 as a coefficient, times x to the fifth over 5 factorial. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
The fifth derivative evaluated at 0 is going to be 1. So the fourth derivative evaluated at 0 is 0. Then you go to the fifth derivative evaluated at 0 is going to be positive 1. And if I kept doing this, it would be positive 1 times, I don't have to write the 1 as a coefficient, times x to the fifth over 5 factorial. So there's something interesting going on here. And for cosine of x, I had 1, essentially 1 times x to the 0. Then I don't have x to the first power. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
And if I kept doing this, it would be positive 1 times, I don't have to write the 1 as a coefficient, times x to the fifth over 5 factorial. So there's something interesting going on here. And for cosine of x, I had 1, essentially 1 times x to the 0. Then I don't have x to the first power. I don't have x to the odd powers, actually. And then I just essentially have x to all of the even powers. And whatever power it is, I'm dividing it by that factorial. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
Then I don't have x to the first power. I don't have x to the odd powers, actually. And then I just essentially have x to all of the even powers. And whatever power it is, I'm dividing it by that factorial. And then the signs keep switching. And I shouldn't say this is an even power, because 0 really isn't. Well, I guess you can view it as an even number, because I won't go into all of that. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
And whatever power it is, I'm dividing it by that factorial. And then the signs keep switching. And I shouldn't say this is an even power, because 0 really isn't. Well, I guess you can view it as an even number, because I won't go into all of that. But it's essentially 0, 2, 4, 6, so on. And so forth. So this is interesting, especially when you compare it to this. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
Well, I guess you can view it as an even number, because I won't go into all of that. But it's essentially 0, 2, 4, 6, so on. And so forth. So this is interesting, especially when you compare it to this. This is all of the odd powers. This is x to the first over 1 factorial. I didn't write it here. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
So this is interesting, especially when you compare it to this. This is all of the odd powers. This is x to the first over 1 factorial. I didn't write it here. This is x to the third over 3 factorial, plus x to the fifth over 5 factorial. Yeah, 0 would be an even number. Anyway, I don't, that's almost, my brain is in a different place right now. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
I didn't write it here. This is x to the third over 3 factorial, plus x to the fifth over 5 factorial. Yeah, 0 would be an even number. Anyway, I don't, that's almost, my brain is in a different place right now. And you could keep going. If we kept this process up, you would then keep switching signs, x to the seventh over 7 factorial, plus x to the ninth over 9 factorial. So there's something interesting here. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
Anyway, I don't, that's almost, my brain is in a different place right now. And you could keep going. If we kept this process up, you would then keep switching signs, x to the seventh over 7 factorial, plus x to the ninth over 9 factorial. So there's something interesting here. You once again see this kind of complementary nature between sine and cosine here. You see almost this, they're filling each other's gaps over here. Cosine of x is all of the even powers of x, divided by that powers factorial. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
So there's something interesting here. You once again see this kind of complementary nature between sine and cosine here. You see almost this, they're filling each other's gaps over here. Cosine of x is all of the even powers of x, divided by that powers factorial. Sine of x, when you take its polynomial representation, is all of the odd powers of x, divided by its factorial, and you switch signs. In the next video, I'll do e to the x. And what's really fascinating is that e to the x starts to look like a little bit of a combination here, but not quite. | Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3 |
So I have two different expressions here that I want to take the derivative of. And what I want you to do is pause the video and think about how you would first approach taking the derivative of this expression and how that might be the same or different as your approach in taking the derivative of this expression. The goal here isn't to compute the derivatives all the way, but really to just think about how we identify what strategies to use. Okay, so let's first tackle this one. And the key when looking at a complex expression like either of these is to look at the big picture structure of the expression. So one way to think about it is let's look at the outside rather than the inside details. So if we look at the outside here, we have the sign of something. | Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 |
Okay, so let's first tackle this one. And the key when looking at a complex expression like either of these is to look at the big picture structure of the expression. So one way to think about it is let's look at the outside rather than the inside details. So if we look at the outside here, we have the sign of something. So there's a sign of something going on here that I'm going to circle in red or in this pink color. So that's how my brain thinks about it. From the outside, I'm like, okay, big picture, I'm taking the sign of some stuff. | Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 |
So if we look at the outside here, we have the sign of something. So there's a sign of something going on here that I'm going to circle in red or in this pink color. So that's how my brain thinks about it. From the outside, I'm like, okay, big picture, I'm taking the sign of some stuff. I might be taking some stuff to some exponent. In this case, I'm inputting it into a trigonometric expression. But if you have a situation like that, it's a good sign that the chain rule is in order. | Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 |
From the outside, I'm like, okay, big picture, I'm taking the sign of some stuff. I might be taking some stuff to some exponent. In this case, I'm inputting it into a trigonometric expression. But if you have a situation like that, it's a good sign that the chain rule is in order. So let me write that down. So we would want to use, in this case, the chain rule, CR for chain rule. And how would we apply it? | Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 |
But if you have a situation like that, it's a good sign that the chain rule is in order. So let me write that down. So we would want to use, in this case, the chain rule, CR for chain rule. And how would we apply it? Well, we would take the derivative of the outside with respect to this inside times the derivative of this inside with respect to x. And I'm gonna write it the way that my brain sometimes thinks about it. So we can write this as the derivative with respect to that something. | Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 |
And how would we apply it? Well, we would take the derivative of the outside with respect to this inside times the derivative of this inside with respect to x. And I'm gonna write it the way that my brain sometimes thinks about it. So we can write this as the derivative with respect to that something. I'm just gonna make that pink circle for the something rather than writing it all again, of sine of that something, sine of that something, not even thinking about what that something is just yet, times the derivative with respect to x of that something. This is just an application of the chain rule. No matter what was here in this pink colored circle, it might have been something with square roots and logarithms and whatever else, as long as it's being contained within the sine, I would move to this step. | Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 |
So we can write this as the derivative with respect to that something. I'm just gonna make that pink circle for the something rather than writing it all again, of sine of that something, sine of that something, not even thinking about what that something is just yet, times the derivative with respect to x of that something. This is just an application of the chain rule. No matter what was here in this pink colored circle, it might have been something with square roots and logarithms and whatever else, as long as it's being contained within the sine, I would move to this step. The derivative with respect to that something of sine of that something times the derivative with respect to x of the something. Now, what would that be tangibly in this case? Well, this first part, I will do it in orange, this first part would just be cosine of x squared plus five times cosine of x. | Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 |
No matter what was here in this pink colored circle, it might have been something with square roots and logarithms and whatever else, as long as it's being contained within the sine, I would move to this step. The derivative with respect to that something of sine of that something times the derivative with respect to x of the something. Now, what would that be tangibly in this case? Well, this first part, I will do it in orange, this first part would just be cosine of x squared plus five times cosine of x. So that's that circle right over there. Let me close the cosine right over there. And then times the derivative with respect to x, times the derivative with respect to x of all of this again, of x squared plus five times cosine of x. | Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 |
Well, this first part, I will do it in orange, this first part would just be cosine of x squared plus five times cosine of x. So that's that circle right over there. Let me close the cosine right over there. And then times the derivative with respect to x, times the derivative with respect to x of all of this again, of x squared plus five times cosine of x. And then I would close my brackets. And of course, I wouldn't be done yet. I have more derivative taking to do. | Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 |
And then times the derivative with respect to x, times the derivative with respect to x of all of this again, of x squared plus five times cosine of x. And then I would close my brackets. And of course, I wouldn't be done yet. I have more derivative taking to do. Here, now I would look at the big structure of what's going on. And I have two expressions being multiplied. I don't have just one big expression that's being an input into like a sine function or cosine function or one big expression that's taken to some exponent. | Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 |
I have more derivative taking to do. Here, now I would look at the big structure of what's going on. And I have two expressions being multiplied. I don't have just one big expression that's being an input into like a sine function or cosine function or one big expression that's taken to some exponent. I have two expressions being multiplied. I have this being multiplied by this. And so if I'm just multiplying two expressions, that's a pretty good clue that to compute this part, I would then use the product rule. | Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 |
I don't have just one big expression that's being an input into like a sine function or cosine function or one big expression that's taken to some exponent. I have two expressions being multiplied. I have this being multiplied by this. And so if I'm just multiplying two expressions, that's a pretty good clue that to compute this part, I would then use the product rule. And I could keep doing that and compute it, and I encourage you to do so, but this is more about the strategies and how do you recognize them. But now let's go to the other example. Well, this looks a lot more like this step of the first problem than the beginning of the original problem. | Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 |
And so if I'm just multiplying two expressions, that's a pretty good clue that to compute this part, I would then use the product rule. And I could keep doing that and compute it, and I encourage you to do so, but this is more about the strategies and how do you recognize them. But now let's go to the other example. Well, this looks a lot more like this step of the first problem than the beginning of the original problem. Here, I don't have a sine of a bunch of stuff or a bunch of stuff being raised to one exponent. Here, I have the product of two expressions, just like we saw over here. We have this expression being multiplied by this expression. | Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 |
Well, this looks a lot more like this step of the first problem than the beginning of the original problem. Here, I don't have a sine of a bunch of stuff or a bunch of stuff being raised to one exponent. Here, I have the product of two expressions, just like we saw over here. We have this expression being multiplied by this expression. So my brain just says, okay, I have two expressions. Then I'm going to use the product rule. Two expressions being multiplied, I'm going to use the product rule. | Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 |
We have this expression being multiplied by this expression. So my brain just says, okay, I have two expressions. Then I'm going to use the product rule. Two expressions being multiplied, I'm going to use the product rule. If it was one expression being divided by another expression, then I would use the quotient rule. But in this case, it's going to be the product rule. And so that tells me that this is going to be the derivative with respect to x of the first expression. | Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 |
Two expressions being multiplied, I'm going to use the product rule. If it was one expression being divided by another expression, then I would use the quotient rule. But in this case, it's going to be the product rule. And so that tells me that this is going to be the derivative with respect to x of the first expression. Just going to do that with the orange circle times the second expression. I'm going to do that with the blue circle plus the first expression, not taking its derivative, the first expression, times the derivative with respect to x of the second expression. Once again, here, this is just the product rule. | Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 |
And so that tells me that this is going to be the derivative with respect to x of the first expression. Just going to do that with the orange circle times the second expression. I'm going to do that with the blue circle plus the first expression, not taking its derivative, the first expression, times the derivative with respect to x of the second expression. Once again, here, this is just the product rule. You can substitute sine of x squared plus five where you see this orange circle. And you can substitute cosine of x where you see this blue circle. But the whole point here isn't to actually solve this or compute this, but really to just show how you identify the structures in these expressions to think about, well, do I use the chain rule first and then use the product rule here? | Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 |
Once again, here, this is just the product rule. You can substitute sine of x squared plus five where you see this orange circle. And you can substitute cosine of x where you see this blue circle. But the whole point here isn't to actually solve this or compute this, but really to just show how you identify the structures in these expressions to think about, well, do I use the chain rule first and then use the product rule here? Or in this case, do I use the product rule first? And even once you do this, you're not going to be done. Then to compute this derivative, you're going to have to use the chain rule. | Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 |
So our first reaction might just say, okay, well let's just use our limit properties a little bit. This is going to be the same thing as the limit as x approaches negative one of x plus one over, over the limit, the limit as x approaches negative one of square root of x plus five minus two. And then we could say, all right, this thing up here, x plus one, if we think about the graph y equals x plus one, it's continuous everywhere, especially at x equals negative one, and so to evaluate this limit, we just have to evaluate this expression at x equals negative one. So this numerator is just going to evaluate to negative one plus one. And then our denominator, square root of x plus five minus two isn't continuous everywhere, but it is continuous at x equals negative one, and so we can do the same thing. We can just substitute negative one for x. So this is going to be the square root of negative one plus five minus two. | Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3 |
So this numerator is just going to evaluate to negative one plus one. And then our denominator, square root of x plus five minus two isn't continuous everywhere, but it is continuous at x equals negative one, and so we can do the same thing. We can just substitute negative one for x. So this is going to be the square root of negative one plus five minus two. Now what does this evaluate to? Well, in the numerator we get a zero, and in the denominator, negative one plus five is four, take the principal root is two minus two, we get zero again. So we get, we got zero over zero. | Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3 |
So this is going to be the square root of negative one plus five minus two. Now what does this evaluate to? Well, in the numerator we get a zero, and in the denominator, negative one plus five is four, take the principal root is two minus two, we get zero again. So we get, we got zero over zero. Now when you see that, you might be tempted to give up. You say, oh look, there's a zero in the denominator, maybe this limit doesn't exist, maybe I'm done here, what do I do? And if this was non-zero up here in the numerator, if you're taking a non-zero value and dividing it by zero, that is undefined, and your limit would not exist. | Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3 |
So we get, we got zero over zero. Now when you see that, you might be tempted to give up. You say, oh look, there's a zero in the denominator, maybe this limit doesn't exist, maybe I'm done here, what do I do? And if this was non-zero up here in the numerator, if you're taking a non-zero value and dividing it by zero, that is undefined, and your limit would not exist. But when you have zero over zero, this is indeterminate form, and it doesn't mean necessarily that your limit does not exist. And as we'll see in this video and many future ones, there are tools at our disposal to address this, and we will look at one of them. Now the tool that we're gonna look at is, is there another way of rewriting this expression so that we can evaluate its limit without getting the zero over zero? | Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3 |
And if this was non-zero up here in the numerator, if you're taking a non-zero value and dividing it by zero, that is undefined, and your limit would not exist. But when you have zero over zero, this is indeterminate form, and it doesn't mean necessarily that your limit does not exist. And as we'll see in this video and many future ones, there are tools at our disposal to address this, and we will look at one of them. Now the tool that we're gonna look at is, is there another way of rewriting this expression so that we can evaluate its limit without getting the zero over zero? Well let's just rewrite, let's just take this, let me give it, so let's take this thing right over here, and let's say this is g of x. So essentially what we're trying to do is find the limit of g of x as x approaches negative one. So we could write g of x is equal to x plus one, and the whole reason why I'm defining it as g of x is just to be able to think of it more clearly as a function and manipulate the function, and then think about similar functions, over x plus five minus two, or x plus one over the square root of x plus five minus two. | Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3 |
Now the tool that we're gonna look at is, is there another way of rewriting this expression so that we can evaluate its limit without getting the zero over zero? Well let's just rewrite, let's just take this, let me give it, so let's take this thing right over here, and let's say this is g of x. So essentially what we're trying to do is find the limit of g of x as x approaches negative one. So we could write g of x is equal to x plus one, and the whole reason why I'm defining it as g of x is just to be able to think of it more clearly as a function and manipulate the function, and then think about similar functions, over x plus five minus two, or x plus one over the square root of x plus five minus two. Now the technique we're gonna use is, when you get this indeterminate form, and if you have a square root in either the numerator or the denominator, it might help to get rid of that square root. And this is often called rationalizing expression. In this case you have a square root in the denominator, so it would be rationalizing the denominator. | Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3 |
So we could write g of x is equal to x plus one, and the whole reason why I'm defining it as g of x is just to be able to think of it more clearly as a function and manipulate the function, and then think about similar functions, over x plus five minus two, or x plus one over the square root of x plus five minus two. Now the technique we're gonna use is, when you get this indeterminate form, and if you have a square root in either the numerator or the denominator, it might help to get rid of that square root. And this is often called rationalizing expression. In this case you have a square root in the denominator, so it would be rationalizing the denominator. And so this would be, the way we would do it, is we'd be leveraging our knowledge of difference of squares. We know, we know that a plus b times a minus b is equal to a squared minus b squared, you learned that in algebra a little while ago. Or, if we had the square root of a plus b, and we were to multiply that times the square root of a minus b, well that'd be the square root of a squared, which is just going to be a minus b squared. | Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3 |
In this case you have a square root in the denominator, so it would be rationalizing the denominator. And so this would be, the way we would do it, is we'd be leveraging our knowledge of difference of squares. We know, we know that a plus b times a minus b is equal to a squared minus b squared, you learned that in algebra a little while ago. Or, if we had the square root of a plus b, and we were to multiply that times the square root of a minus b, well that'd be the square root of a squared, which is just going to be a minus b squared. So we can just leverage these ideas to get rid of this radical down here. The way we're going to do it, is we're gonna multiply the numerator and the denominator by the square root of x plus five plus two, right? We have the minus two, so we're gonna multiply it times the plus two. | Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3 |
Or, if we had the square root of a plus b, and we were to multiply that times the square root of a minus b, well that'd be the square root of a squared, which is just going to be a minus b squared. So we can just leverage these ideas to get rid of this radical down here. The way we're going to do it, is we're gonna multiply the numerator and the denominator by the square root of x plus five plus two, right? We have the minus two, so we're gonna multiply it times the plus two. So let's do that. So we have square root of x plus five plus two, and we're gonna multiply the numerator times the same thing, because we don't want to change the value of the expression. This is one, so if we take the expression divided by the same expression, it's going to be one. | Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3 |
We have the minus two, so we're gonna multiply it times the plus two. So let's do that. So we have square root of x plus five plus two, and we're gonna multiply the numerator times the same thing, because we don't want to change the value of the expression. This is one, so if we take the expression divided by the same expression, it's going to be one. So this is, so square root of x plus five plus two. And so this is going to be equal to, this is going to be equal to x plus one times the square root, times the square root of x plus five plus two. And then the denominator is going to be, well, it's going to be the square root of x plus five squared, which would be just x plus five, and then minus two squared, minus four. | Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3 |
This is one, so if we take the expression divided by the same expression, it's going to be one. So this is, so square root of x plus five plus two. And so this is going to be equal to, this is going to be equal to x plus one times the square root, times the square root of x plus five plus two. And then the denominator is going to be, well, it's going to be the square root of x plus five squared, which would be just x plus five, and then minus two squared, minus four. And so this down here simplifies to x plus five minus four, it's just x plus one. So this is just, this is just x plus one. And it probably jumps out at you that both the numerator and the denominator have an x plus one in it, so maybe we can simplify. | Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3 |
And then the denominator is going to be, well, it's going to be the square root of x plus five squared, which would be just x plus five, and then minus two squared, minus four. And so this down here simplifies to x plus five minus four, it's just x plus one. So this is just, this is just x plus one. And it probably jumps out at you that both the numerator and the denominator have an x plus one in it, so maybe we can simplify. So we could simplify and just say, well, g of x is equal to the square root of x plus five plus two. Now some of you might be feeling a little off here, and you would be correct. Your spider senses would be, say, is this, is this definitely the same thing as what we originally had before we canceled out the x plus ones? | Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3 |
And it probably jumps out at you that both the numerator and the denominator have an x plus one in it, so maybe we can simplify. So we could simplify and just say, well, g of x is equal to the square root of x plus five plus two. Now some of you might be feeling a little off here, and you would be correct. Your spider senses would be, say, is this, is this definitely the same thing as what we originally had before we canceled out the x plus ones? And the answer is the way I just wrote it, it is not the exact same thing. It is the exact same thing everywhere, except at x equals negative one. This thing right over here is defined at x equals negative one. | Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3 |
Your spider senses would be, say, is this, is this definitely the same thing as what we originally had before we canceled out the x plus ones? And the answer is the way I just wrote it, it is not the exact same thing. It is the exact same thing everywhere, except at x equals negative one. This thing right over here is defined at x equals negative one. This thing right over here is not defined at x equals negative one. And g of x was not, was not, so g of x right over here, you don't get a good result when you try x equals negative one. And so in order for this to truly be the same thing as g of x, the same function, we have to say for x not equal to negative one. | Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3 |
This thing right over here is defined at x equals negative one. This thing right over here is not defined at x equals negative one. And g of x was not, was not, so g of x right over here, you don't get a good result when you try x equals negative one. And so in order for this to truly be the same thing as g of x, the same function, we have to say for x not equal to negative one. Now this is a simplified version of g of x. It is the same thing. For any input x that g of x is defined, this is going to give you the same output. | Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3 |
And so in order for this to truly be the same thing as g of x, the same function, we have to say for x not equal to negative one. Now this is a simplified version of g of x. It is the same thing. For any input x that g of x is defined, this is going to give you the same output. And this has the exact same domain now, now that we've put this constraint in, as g of x. Now you might say, okay, well, how does this help us? Because we want to find the limit as x approaches negative one. | Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3 |
For any input x that g of x is defined, this is going to give you the same output. And this has the exact same domain now, now that we've put this constraint in, as g of x. Now you might say, okay, well, how does this help us? Because we want to find the limit as x approaches negative one. And even here, I had to put this little constraint here that x cannot be equal to negative one. How do we think about this limit? Well, lucky for us, we know, lucky for us, we know that if we just take another function, f of x, if we say f of x is equal to the square root of x plus five plus two, well, then we know that f of x is equal to g of x for all x not equal to negative one, because f of x does not have that constraint. | Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3 |
Because we want to find the limit as x approaches negative one. And even here, I had to put this little constraint here that x cannot be equal to negative one. How do we think about this limit? Well, lucky for us, we know, lucky for us, we know that if we just take another function, f of x, if we say f of x is equal to the square root of x plus five plus two, well, then we know that f of x is equal to g of x for all x not equal to negative one, because f of x does not have that constraint. And we know if this is true of two, if this is true of two functions, then the limit as x approaches, the limit, let me write this down, is since we know this, because of this, we know that the limit of f of x as x approaches negative one is going to be equal to the limit of g of x as x approaches negative one. And this, of course, is what we want to figure out, what was the beginning of the problem. And but we can now use f of x here, because only at x equals negative one that they are not the same. | Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3 |
Well, lucky for us, we know, lucky for us, we know that if we just take another function, f of x, if we say f of x is equal to the square root of x plus five plus two, well, then we know that f of x is equal to g of x for all x not equal to negative one, because f of x does not have that constraint. And we know if this is true of two, if this is true of two functions, then the limit as x approaches, the limit, let me write this down, is since we know this, because of this, we know that the limit of f of x as x approaches negative one is going to be equal to the limit of g of x as x approaches negative one. And this, of course, is what we want to figure out, what was the beginning of the problem. And but we can now use f of x here, because only at x equals negative one that they are not the same. And if you were to graph g of x, it just has a point discontinuity, or a removable, or I should just say, yeah, a point discontinuity right over here at x equals negative one. And so what is the limit? And we are in the home stretch now. | Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3 |
And but we can now use f of x here, because only at x equals negative one that they are not the same. And if you were to graph g of x, it just has a point discontinuity, or a removable, or I should just say, yeah, a point discontinuity right over here at x equals negative one. And so what is the limit? And we are in the home stretch now. What is the limit of f of x, or we could say the limit of the square root of x plus five plus two as x approaches negative one? Well, this expression is continuous, or this function is continuous at x equals negative one, so we can just evaluate it at x equals negative one. So this is going to be the square root of negative one plus five plus two. | Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3 |
And we are in the home stretch now. What is the limit of f of x, or we could say the limit of the square root of x plus five plus two as x approaches negative one? Well, this expression is continuous, or this function is continuous at x equals negative one, so we can just evaluate it at x equals negative one. So this is going to be the square root of negative one plus five plus two. So this is four, square root, principle root of four is two. Two plus two is equal to four. So since the limit of f of x as x approaches negative one is four, the limit of g of x as x approaches negative one is also four. | Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3 |
So this is going to be the square root of negative one plus five plus two. So this is four, square root, principle root of four is two. Two plus two is equal to four. So since the limit of f of x as x approaches negative one is four, the limit of g of x as x approaches negative one is also four. And if this little, this little, I guess you could say, leap that I just made right over here doesn't make sense to you, think about it visually. Think about it visually. So if this is my y-axis, and this is my x-axis, g of x looked something like this. | Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3 |
So since the limit of f of x as x approaches negative one is four, the limit of g of x as x approaches negative one is also four. And if this little, this little, I guess you could say, leap that I just made right over here doesn't make sense to you, think about it visually. Think about it visually. So if this is my y-axis, and this is my x-axis, g of x looked something like this. The g of x, the g of x, let me draw it, g of x looked something, something like this. And it had a gap at negative one. So it had a gap right over there. | Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3 |
So if this is my y-axis, and this is my x-axis, g of x looked something like this. The g of x, the g of x, let me draw it, g of x looked something, something like this. And it had a gap at negative one. So it had a gap right over there. While f of x, f of x would have the same graph, except it wouldn't have, it wouldn't have the gap. And so if you're trying to find the limit, it seems completely reasonable. Well, let's just use f of x and evaluate what f of x would be to kind of fill that gap at x equals negative one. | Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3 |
The graph of r in polar coordinates consists of two loops, as shown in the figure above. So let's think about why it has two loops. So as our theta, when theta is zero, r is zero, and then as our theta increases, we start tracing out this first loop all the way until when theta is equal to pi. So we traced out this first loop from theta is equal to zero to theta is equal to pi, and then the second loop has a larger r, so these are larger r's. This is when we're going from pi to two pi, and you might say, well, why doesn't it show up down here? Well, between sine of pi and sine of two pi, this part right over here is going to be negative, so it flips it over, the r, into this side, and the magnitude of the r is larger and larger because of this three theta, and so when we go from pi to two pi, we trace out the larger circle. Fair enough. | Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3 |
So we traced out this first loop from theta is equal to zero to theta is equal to pi, and then the second loop has a larger r, so these are larger r's. This is when we're going from pi to two pi, and you might say, well, why doesn't it show up down here? Well, between sine of pi and sine of two pi, this part right over here is going to be negative, so it flips it over, the r, into this side, and the magnitude of the r is larger and larger because of this three theta, and so when we go from pi to two pi, we trace out the larger circle. Fair enough. That seems pretty straightforward. Point P is on the graph of r, right over there, and the y-axis. Find the rate of change of the x-coordinate with respect to theta at the point P. All right, so let's think about this a little bit. | Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3 |
Fair enough. That seems pretty straightforward. Point P is on the graph of r, right over there, and the y-axis. Find the rate of change of the x-coordinate with respect to theta at the point P. All right, so let's think about this a little bit. They don't give us x as a function of theta. We have to figure out that from what they've given us. So just as a bit of a polar coordinates refresher, if this is our theta right over there, this is our r, and that would be a point on our curve for this theta. | Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3 |
Find the rate of change of the x-coordinate with respect to theta at the point P. All right, so let's think about this a little bit. They don't give us x as a function of theta. We have to figure out that from what they've given us. So just as a bit of a polar coordinates refresher, if this is our theta right over there, this is our r, and that would be a point on our curve for this theta. Now, how do you convert that to x and y's? Well, you can construct a little bit of a right triangle right over here, and we know from our basic trigonometry that the length of this base right over here, this is going to be the hypotenuse. Let me just write that. | Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3 |
So just as a bit of a polar coordinates refresher, if this is our theta right over there, this is our r, and that would be a point on our curve for this theta. Now, how do you convert that to x and y's? Well, you can construct a little bit of a right triangle right over here, and we know from our basic trigonometry that the length of this base right over here, this is going to be the hypotenuse. Let me just write that. That's going to be our x-coordinate. Our x-coordinate right over here is going to be equal to our hypotenuse, which is r times the cosine of theta. If we wanted the y-coordinate as a function of r and theta, it'd be y is equal to r sine of theta, but they don't want us to worry about y here, just the x-coordinate. | Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3 |
Let me just write that. That's going to be our x-coordinate. Our x-coordinate right over here is going to be equal to our hypotenuse, which is r times the cosine of theta. If we wanted the y-coordinate as a function of r and theta, it'd be y is equal to r sine of theta, but they don't want us to worry about y here, just the x-coordinate. So we know this, but we want it purely in terms of theta. So how do we get there? Well, what we can do is take this expression for r. r itself is a function of theta, and replace it right over there. | Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3 |
If we wanted the y-coordinate as a function of r and theta, it'd be y is equal to r sine of theta, but they don't want us to worry about y here, just the x-coordinate. So we know this, but we want it purely in terms of theta. So how do we get there? Well, what we can do is take this expression for r. r itself is a function of theta, and replace it right over there. And so what we can do is we can write, well, x of theta is going to be equal to r, which itself is three theta sine of theta times cosine of theta, times cosine of theta. And now we want to find the rate of change of the x-coordinate with respect to theta at a point. So let's just find the derivative of x with respect to theta. | Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3 |
Well, what we can do is take this expression for r. r itself is a function of theta, and replace it right over there. And so what we can do is we can write, well, x of theta is going to be equal to r, which itself is three theta sine of theta times cosine of theta, times cosine of theta. And now we want to find the rate of change of the x-coordinate with respect to theta at a point. So let's just find the derivative of x with respect to theta. So x prime of theta is equal to, well, I have the product of three expressions over here. I have this first expression, three theta, then I have sine theta, and then I have cosine theta. So we can apply the product rule to find the derivative. | Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3 |
So let's just find the derivative of x with respect to theta. So x prime of theta is equal to, well, I have the product of three expressions over here. I have this first expression, three theta, then I have sine theta, and then I have cosine theta. So we can apply the product rule to find the derivative. If you're using the product rule with the expression of three things, you essentially just follow the same pattern when you're taking the product of two things. The first term is going to be the derivative of the first of the expressions, three times the other two expressions. So we're gonna have three times sine of theta, cosine of theta, plus the second term is going to be the derivative of the middle term times the other two expressions. | Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3 |
So we can apply the product rule to find the derivative. If you're using the product rule with the expression of three things, you essentially just follow the same pattern when you're taking the product of two things. The first term is going to be the derivative of the first of the expressions, three times the other two expressions. So we're gonna have three times sine of theta, cosine of theta, plus the second term is going to be the derivative of the middle term times the other two expressions. So we're gonna have three theta, and then derivative of sine theta is cosine theta, times another cosine theta, you're gonna have cosine squared of theta, or cosine of theta squared, just like that. And then you're gonna have the derivative of the last term is going to be the derivative of cosine theta times these other two expressions. Well, the derivative of cosine theta is negative sine theta. | Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3 |
So we're gonna have three times sine of theta, cosine of theta, plus the second term is going to be the derivative of the middle term times the other two expressions. So we're gonna have three theta, and then derivative of sine theta is cosine theta, times another cosine theta, you're gonna have cosine squared of theta, or cosine of theta squared, just like that. And then you're gonna have the derivative of the last term is going to be the derivative of cosine theta times these other two expressions. Well, the derivative of cosine theta is negative sine theta. So if you multiply negative sine theta times three theta sine theta, you're going to have negative three theta sine squared theta. And so we want to evaluate this at point P. So what is theta at point P? Well, point P does happen on our first pass around, and so at point P, theta is equal to, theta right over here, is equal to pi over two. | Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3 |
Well, the derivative of cosine theta is negative sine theta. So if you multiply negative sine theta times three theta sine theta, you're going to have negative three theta sine squared theta. And so we want to evaluate this at point P. So what is theta at point P? Well, point P does happen on our first pass around, and so at point P, theta is equal to, theta right over here, is equal to pi over two. So pi over two. So what we really just need to find is, well, what is x prime of pi over two? Well, that is going to be equal to three times sine of pi over two, sine of pi over two, which is one, times cosine of pi over two, which is zero, so this whole thing is zero, plus three times pi over two, this is three pi over two, times cosine squared of pi over two, or cosine of pi over two squared. | Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3 |
Well, point P does happen on our first pass around, and so at point P, theta is equal to, theta right over here, is equal to pi over two. So pi over two. So what we really just need to find is, well, what is x prime of pi over two? Well, that is going to be equal to three times sine of pi over two, sine of pi over two, which is one, times cosine of pi over two, which is zero, so this whole thing is zero, plus three times pi over two, this is three pi over two, times cosine squared of pi over two, or cosine of pi over two squared. Well, that's just zero. So so far, everything is zero. Minus three times pi over two, three pi over two, times sine of pi over two squared. | Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3 |
Well, that is going to be equal to three times sine of pi over two, sine of pi over two, which is one, times cosine of pi over two, which is zero, so this whole thing is zero, plus three times pi over two, this is three pi over two, times cosine squared of pi over two, or cosine of pi over two squared. Well, that's just zero. So so far, everything is zero. Minus three times pi over two, three pi over two, times sine of pi over two squared. Well, what's sine of pi over two? Well, that's one, you square it, you still get one. So all of this simplified to negative three pi over two. | Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3 |
Minus three times pi over two, three pi over two, times sine of pi over two squared. Well, what's sine of pi over two? Well, that's one, you square it, you still get one. So all of this simplified to negative three pi over two. Now, it's always good to get a reality check. Does this make sense, that the rate of change of x with respect to theta is negative three pi over two? Well, think about what's happening. | Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3 |
So all of this simplified to negative three pi over two. Now, it's always good to get a reality check. Does this make sense, that the rate of change of x with respect to theta is negative three pi over two? Well, think about what's happening. As theta increases a little bit, x is definitely going to decrease, so it makes sense that we have a negative out here. So right over here, rate of change of x with respect to theta, negative three pi over two. As theta increases, our x for sure is decreasing, but at least it does make intuitive sense. | Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3 |
Well, think about what's happening. As theta increases a little bit, x is definitely going to decrease, so it makes sense that we have a negative out here. So right over here, rate of change of x with respect to theta, negative three pi over two. As theta increases, our x for sure is decreasing, but at least it does make intuitive sense. | Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3 |
As theta increases, our x for sure is decreasing, but at least it does make intuitive sense. | Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3 |