Sentence stringlengths 92 3.31k | video_title stringlengths 24 106 |
|---|---|
This is definitely going to be positive. So P prime prime of 0.4725 is greater than zero. So the fact that the second derivative is less than zero, that means that my derivative is decreasing, my derivative, my first derivative is decreasing when x is equal to this value, which means that our graph, our function is con... | Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3 |
Concave, concave downwards. And concave downwards means it looks something like this. And so, and you can see when it looks something like that, the slope is constantly decreasing. So if you have an interval where the slope is decreasing and you know the point where the slope is exactly zero, which is where x is equal ... | Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3 |
So if you have an interval where the slope is decreasing and you know the point where the slope is exactly zero, which is where x is equal to 3.528, it must be a maximum. It must be a maximum. So we actually do take on a maximum value when x is 3.528. On the other side, we see that over here we are concave upwards, con... | Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3 |
On the other side, we see that over here we are concave upwards, concave upwards. Concave upwards, the graph will look something like this over here. And if the slope is zero where the graph looks like that, we see that that is a local minimum. That is a local minimum. And so we definitely don't want to do this. We wou... | Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3 |
That is a local minimum. And so we definitely don't want to do this. We would produce 472 and a half units if we were looking to minimize our profit, maximize our loss. So we definitely don't want to do this. But let's actually think about what our profit is going to be if we produce 3.528 thousands of shoes, or 3,528 ... | Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3 |
So we definitely don't want to do this. But let's actually think about what our profit is going to be if we produce 3.528 thousands of shoes, or 3,528 shoes. Well, to do that, we just have to input it back into our original profit function right over here. So let's do that. So get my calculator out. So my original prof... | Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3 |
So let's do that. So get my calculator out. So my original profit function is right over there. So I want to be able to see that and that. So I get negative 3.528 to the third power plus 6 times 3.528 squared minus 5 times 3.528 gives me, and we get a drum roll now, gives me a profit of 13.128. So let me write this dow... | Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3 |
So I want to be able to see that and that. So I get negative 3.528 to the third power plus 6 times 3.528 squared minus 5 times 3.528 gives me, and we get a drum roll now, gives me a profit of 13.128. So let me write this down. The profit when I produce 3,528 shoes is approximately equal to, or it is equal to, if I prod... | Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3 |
The profit when I produce 3,528 shoes is approximately equal to, or it is equal to, if I produce exactly that many shoes, it's equal to 13.128. Or actually, it's approximately because I'm still rounding. 13.128. So if I produce 3,528 shoes in a given period, I am going to have a profit of $13,128. Remember, this right ... | Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3 |
Differentiability. So let's think about that first. It's always helpful to draw ourselves a function. So that's our y-axis. This is our x-axis. And let's just draw some function here. So let's say my function looks like this. | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
So that's our y-axis. This is our x-axis. And let's just draw some function here. So let's say my function looks like this. And we care about the point x equals C, which is right over here. So that's the point x equals C. And then this value, of course, is going to be f of C. And one way that we can find the derivative... | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
So let's say my function looks like this. And we care about the point x equals C, which is right over here. So that's the point x equals C. And then this value, of course, is going to be f of C. And one way that we can find the derivative at x equals C, or the slope of the tangent line at x equals C, is we could start ... | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
So let's say this is some arbitrary x out here. So then this point right over there, this value, this y value, would be f of x. This graph, of course, is a graph of y equals f of x. And we can think about finding the slope of this line, this secant line between these two points. But then we can find the limit as x appr... | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
And we can think about finding the slope of this line, this secant line between these two points. But then we can find the limit as x approaches C. And as x approaches C, this secant, the slope of the secant line is going to approach the slope of the tangent line, or it's going to be the derivative. And so we could tak... | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
So what's the slope? Well, it's going to be change in y over change in x. The change in y is f of x minus f of C. That's our change in y right over here. And this is all a review. This is just one definition of the derivative, or one way to think about the derivative. So it's going to be f of x minus f of C, that's our... | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
And this is all a review. This is just one definition of the derivative, or one way to think about the derivative. So it's going to be f of x minus f of C, that's our change in y, over our change in x, which is x minus C. It is x minus C. So if this limit exists, then we're able to find the slope of the tangent line at... | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
And we call that slope of the tangent line, we call that the derivative at x equals C. We say that this is going to be equal to f prime of C. All of this is review. So if we're saying, one way to think about it, if we're saying that the function f is differentiable at x equals C, we're really just saying that this limi... | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
Now let's give ourselves a review of continuity. Con-ti-nuity. So the definition for continuity is if the limit as x approaches C of f of x is equal to f of C. Now this might seem a little bit, you know, well, it might pop out to you as being intuitive, or it might seem a little, well, where did this come from? Well, l... | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
Well, let's visualize it, and then hopefully it'll make some intuitive sense. So if you have a function, so let's actually look at some cases where you're not continuous. And that actually might make it a little bit more clear. So if you had a point discontinuity at x equals C, so this is x equals C. So if you had a po... | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
So if you had a point discontinuity at x equals C, so this is x equals C. So if you had a point discontinuity, so let me draw it like this actually. So you have a gap here, and x equals, when x equals C, f of C is actually way up here. So this is f of C, and then the function continues like this. The limit as x approac... | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
The limit as x approaches C of f of x is going to be this value, which is clearly different than f of C, this value right over here. If you take the limit, if you take the limit as x approaches C of f of x, you're approaching this value. This right over here is the limit as x approaches C of f of x, which is different ... | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
You have a point discontinuity. So for at least in this case, this definition of continuity would properly identify this as not a continuous function. Now you could also think about a jump discontinuity. You could also think about a jump discontinuity. So let's look at this. And all of this is hopefully a little bit of... | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
You could also think about a jump discontinuity. So let's look at this. And all of this is hopefully a little bit of review. So a jump discontinuity at C, at x equals C, might look like this. Might look like this. So this is at x equals C. So this is x equals C right over here. This would be f of C. But if you tried to... | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
So a jump discontinuity at C, at x equals C, might look like this. Might look like this. So this is at x equals C. So this is x equals C right over here. This would be f of C. But if you tried to evaluate the limit as x approaches C of f of x, you'd get a different value as you approach C from the negative side. You wo... | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
This would be f of C. But if you tried to evaluate the limit as x approaches C of f of x, you'd get a different value as you approach C from the negative side. You would approach this value. And as you approach C from the positive side, you would approach f of C. And so the limit wouldn't exist. So this limit right ove... | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
So this limit right over here wouldn't exist in the case of this type of a jump discontinuity. So once again, this definition would properly say that this is not, this one right over here is not continuous. This limit actually would not even exist. And then you could even look at a, you could look at a function that is... | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
And then you could even look at a, you could look at a function that is truly continuous. If you look at a function that is truly continuous, so something like this. Something like this. That is x equals C. Well, this is f of C. This is f of C. And if you were to take the limit as x approaches C, as x approaches C from... | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
That is x equals C. Well, this is f of C. This is f of C. And if you were to take the limit as x approaches C, as x approaches C from either side of f of x, you're going to approach f of C. So here you have the limit as x approaches C of f of x indeed is equal to f of C. So it's what you would expect for a continuous f... | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
So differentiability implies this, this limit right over here exists. So let's start with a slightly different limit. Let me draw a line here actually. Let me draw a line just so we're doing something different. So let's take, let us take the limit as x approaches C of f of x, of f of x minus f of C. Of f of x minus f ... | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
Let me draw a line just so we're doing something different. So let's take, let us take the limit as x approaches C of f of x, of f of x minus f of C. Of f of x minus f of C. Well can we rewrite this? Well we could rewrite this as the limit as x approaches C. And we can essentially take this expression and multiply and ... | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
This is going to be equal to, it's going to be the limit, and I'm just applying the property of limit, property, I'm applying a property of limits here. So the limit of the product is equal to the same thing as the product of the limits. So it's the limit as x approaches C of x minus C times the limit, let me write it ... | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
Well if we assume that f is differentiable at C, and we're going to do that, actually I should have started off there. Let's assume, let's assume, because we wanted to show that differentiability improves continuity. If we assume f differentiable, differentiable at C, well then this right over here is just going to be ... | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
This is f prime, f prime of C. And what is this thing right over here? The limit as x approaches C of x minus C? Well that's just going to be zero. As x approaches C, it's going to approach C minus C, it's just going to be zero. So what's zero times f prime of C? Well f prime of C is just going to be some value, so zer... | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
As x approaches C, it's going to approach C minus C, it's just going to be zero. So what's zero times f prime of C? Well f prime of C is just going to be some value, so zero times anything is just going to be zero. So I did all that work to get a zero. Now why is this interesting? Well we just said, we just assumed tha... | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
So I did all that work to get a zero. Now why is this interesting? Well we just said, we just assumed that if f is differentiable at C, and we evaluate this limit, we get zero. So if we assume f is differentiable at C, we can write, we can write the limit, I'm just rewriting it, the limit as x approaches C of f of x mi... | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
So if we assume f is differentiable at C, we can write, we can write the limit, I'm just rewriting it, the limit as x approaches C of f of x minus f of C, and I could even put parentheses around it like that, which I already did up here, is equal to zero. Well this is the same thing, I could use limit properties again,... | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
The limit of the difference is the same thing as the difference of the limits. Well what's this thing over here going to be? Well f of C is just a number, it's not a function of x anymore, it's just f of C is going to evaluate to something. So this is just going to be f of C. This is just going to be f of C. So if the ... | Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3 |
So we're told this table gives select values of the differentiable function f. So it gives us the value of the function at a few values for x, in particular five different values for x. It tells us what the corresponding f of x is. And they say what is the best estimate for f prime of four? So this is the derivative of... | Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3 |
So this is the derivative of our function f when x is equal to four. Or another way to think about it, what is the slope of the tangent line when x is equal to four for f of x? So what is the best estimate for f prime of four we can make based on this table? So let's just visualize what's going on before we even look a... | Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3 |
So let's just visualize what's going on before we even look at the choices. So let me draw some axes here. And let me plot these points. We know that these would sit on the curve of y is equal to f of x. When x is zero, f of x is 72. So this is the point zero, 72. This is the point three, 95. | Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3 |
We know that these would sit on the curve of y is equal to f of x. When x is zero, f of x is 72. So this is the point zero, 72. This is the point three, 95. Clearly two different scales on the x and y-axis. This is the point five, 112. This is 677. | Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3 |
This is the point three, 95. Clearly two different scales on the x and y-axis. This is the point five, 112. This is 677. This is 954. Actually, let me write out the, this is one, two, three, four, five, six, seven, eight, nine, and 10. Now they want us to know, they wanna know what is the derivative of our function whe... | Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3 |
This is 677. This is 954. Actually, let me write out the, this is one, two, three, four, five, six, seven, eight, nine, and 10. Now they want us to know, they wanna know what is the derivative of our function when f is equal to four. Well, they haven't told us even what the value of f is at four. We don't know what tha... | Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3 |
Now they want us to know, they wanna know what is the derivative of our function when f is equal to four. Well, they haven't told us even what the value of f is at four. We don't know what that point is. But what they're trying to do is, well, we're trying to make a best estimate. And using these points, we don't even ... | Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3 |
But what they're trying to do is, well, we're trying to make a best estimate. And using these points, we don't even know exactly what the curve looks like. It could look like all sorts of things. We could try to fit a reasonably smooth curve. The curve might look something like that. But it might be wackier. It might d... | Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3 |
We could try to fit a reasonably smooth curve. The curve might look something like that. But it might be wackier. It might do something like this. Well, let me try to do it. It might look something like this. So we don't know for sure. | Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3 |
It might do something like this. Well, let me try to do it. It might look something like this. So we don't know for sure. All we know is that it needs to go through those points because they've just sampled the function at those points. But let's just, for the sake of this exercise, let's assume the simplest, let's say... | Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3 |
So we don't know for sure. All we know is that it needs to go through those points because they've just sampled the function at those points. But let's just, for the sake of this exercise, let's assume the simplest, let's say it's a nice, smooth curve without too many twists and turns that goes through these points jus... | Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3 |
So what they're asking, okay, when x is equal to four, if this yellow curve were the actual curve, then what is the slope of the tangent line at that point? So we would be visualizing that. Now to be clear, this tangent line that I just drew, this would be for this version of our function that I did connecting these po... | Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3 |
That does not have to be the actual function. We know that the actual function has to go through those points. But I'm just doing this for visualization purposes. One of the whole ideas here is that all we do have is a sample and we're trying to get a best estimate. We don't know if it's even gonna be a good estimate. ... | Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3 |
One of the whole ideas here is that all we do have is a sample and we're trying to get a best estimate. We don't know if it's even gonna be a good estimate. It's just going to be a best estimate. So what we generally do when we just have some data around a point is let's use the data points that are closest to that poi... | Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3 |
So what we generally do when we just have some data around a point is let's use the data points that are closest to that point and find slopes of secant lines pretty close around that point. And that's going to give us our best estimate for the slope of the tangent line. So what points do we have near F of four or near... | Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3 |
Well, they give us what F is equal to when x is equal to three. They give us this point right over here. Let me do this in another color. So three comma 95, that is that right over there. And they also give us five comma 112. That is that point right over there. And so what we could do, we could say, well, what is the ... | Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3 |
So three comma 95, that is that right over there. And they also give us five comma 112. That is that point right over there. And so what we could do, we could say, well, what is the average rate of change between these two points? Another way to think about it is what is the slope of the secant line between those two p... | Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3 |
And so what we could do, we could say, well, what is the average rate of change between these two points? Another way to think about it is what is the slope of the secant line between those two points? And that would be our best estimate for the slope of the tangent line at x equals four. Do we know that it's a good es... | Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3 |
Do we know that it's a good estimate? Do we know that it's even close? No, we don't know for sure, but that would be the best estimate. It would be better than trying to take the average rate of change between when x equals three and x equals six, or between when x equals zero and x equals nine. These are pretty close ... | Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3 |
It would be better than trying to take the average rate of change between when x equals three and x equals six, or between when x equals zero and x equals nine. These are pretty close around four. And so let's do that. Let's find the average rate of change between when x goes from three to five. So we can see here our ... | Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3 |
Let's find the average rate of change between when x goes from three to five. So we can see here our change in x, let me do this in a new color. So our change in x here is equal to plus two, and I can draw that out. My change in x here is plus two, and my change in y is going to be, when my x increased by two, my chang... | Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3 |
My change in x here is plus two, and my change in y is going to be, when my x increased by two, my change in y is plus, let's see, this is, if I add 10, I get to 105, and then I add another seven, so this is plus 17. So this is plus 17 right over here. Plus 17. And so my change in y over change in x, change in y over m... | Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3 |
And so my change in y over change in x, change in y over my change in x, for this secant line between when x is equaling three and x is equaling five, is going to be equal to 17 over two. 17 over two, which is equal to 8.5. So the slope of this green line here is 8.5, and that would be our best estimate for the slope o... | Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3 |
This one is actually pretty straightforward to define explicitly in terms of x, to solve for y. So if we divide both sides by x, we get square root of y is equal to 1 over x. And then if you square both sides, you get y is equal to 1 over x squared, which is the same thing as x to the negative 2 power. And so if you wa... | Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3 |
And so if you want the derivative of y with respect to x, this is pretty straightforward. This is just an application of the chain rule. We get dy dx is equal to negative 2 x to the negative 2 minus 1, x to the negative 3 power. So that's pretty straightforward. But what I want to see is if we get the same exact result... | Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3 |
So that's pretty straightforward. But what I want to see is if we get the same exact result when we differentiate implicitly. So let's apply our derivative operator to both sides of this equation. And so let me make it clear what we're doing. x times the square root of y and 1 right over there. When you apply the deriv... | Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3 |
And so let me make it clear what we're doing. x times the square root of y and 1 right over there. When you apply the derivative operator to the expression on the left-hand side, we are just going to have to apply, well, actually we're going to apply both the product rule and the chain rule. The product rule tells us, ... | Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3 |
The product rule tells us, so we have the product of two functions of x. You could view it that way. So the product rule tells us this is going to be the derivative with respect to x of x times the square root of y plus x, not taking its derivative, plus x times the derivative with respect to x of the square root of y.... | Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3 |
Let me make it clear this bracket, of the square root of y. And on the right-hand side, right over here, the derivative with respect to x of this constant, that's just going to be equal to 0. So what does this simplify to? Well, the derivative with respect to x of x is just 1. So we're just going to be left, this simpl... | Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3 |
Well, the derivative with respect to x of x is just 1. So we're just going to be left, this simplifies to 1, so we're just going to be left with the square root of y right over here. So we're just going to be, this is going to simplify to a square root of y. And what does this over here simplify to? Well, the derivativ... | Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3 |
And what does this over here simplify to? Well, the derivative with respect to x of the square root of y, here we want to apply the chain rule, so let me make it clear. So we have plus this x, plus this x, plus whatever business this is. And I'm going to do this in blue. Well, it's going to be the derivative of square ... | Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3 |
And I'm going to do this in blue. Well, it's going to be the derivative of square root of something with respect to that something. Well, the derivative of square root of something with respect to that something, or the derivative of something to the one half with respect to that something is going to be one half times... | Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3 |
Once again, this right over here is the derivative of the square root of y with respect to y. We've seen this multiple times. If I were to say the derivative of the square root of x with respect to x, you would get 1 half x to the negative 1 half. Now I'm just doing it with y's. But we're not done yet. Remember, our de... | Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3 |
Now I'm just doing it with y's. But we're not done yet. Remember, our derivative operator wasn't to say with respect to y. It's with respect to x. So this only gets us with respect to y. We need to apply the entire chain rule. We have to multiply that. | Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3 |
It's with respect to x. So this only gets us with respect to y. We need to apply the entire chain rule. We have to multiply that. We have to multiply that times the derivative of y with respect to x in order to get the real derivative of this expression with respect to x. So let's multiply times the derivative of y wit... | Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3 |
We have to multiply that. We have to multiply that times the derivative of y with respect to x in order to get the real derivative of this expression with respect to x. So let's multiply times the derivative of y with respect to x. We don't know what that is. That's actually what we're trying to solve for. But to use t... | Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3 |
We don't know what that is. That's actually what we're trying to solve for. But to use the chain rule, we just have to say it's the derivative of the square root of y with respect to y times the derivative of y with respect to x. This is the derivative of this thing with respect to x. So we get this on the left-hand si... | Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3 |
This is the derivative of this thing with respect to x. So we get this on the left-hand side, on the right-hand side, we just have a 0. And now once again, we can attempt to solve for the derivative of y with respect to x. And maybe the easiest first step is to subtract the square root of y from both sides of this equa... | Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3 |
And maybe the easiest first step is to subtract the square root of y from both sides of this equation. And actually, let me move all of this stuff over so I have, once again, more room to work with. Let me cut it, actually. And then let me paste it. Let me move it over right over here. So we went from there to there. I... | Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3 |
And then let me paste it. Let me move it over right over here. So we went from there to there. I didn't gain a lot of real estate, but hopefully this helps a little bit. And actually, I don't even like that. Let me leave it where it was before. So then if we subtract the square root of y from both sides, we get, and I'... | Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3 |
I didn't gain a lot of real estate, but hopefully this helps a little bit. And actually, I don't even like that. Let me leave it where it was before. So then if we subtract the square root of y from both sides, we get, and I'll try to simplify as I go, we get this thing, which I can rewrite as x times, well, it's just ... | Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3 |
So then if we subtract the square root of y from both sides, we get, and I'll try to simplify as I go, we get this thing, which I can rewrite as x times, well, it's just going to be x in the numerator, divided by 2 times the square root of y. y to the negative 1 half is just the square root of y in the denominator. And... | Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3 |
So copy and then paste. So let's go back up here just to continue our simplification solving for dy dx. Well, to solve for dy dx, we just have to divide both sides by x over 2 times the square root of y. So we're left with dy dx is equal to, or dividing both sides by this is the same thing as multiplying by the recipro... | Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3 |
So we're left with dy dx is equal to, or dividing both sides by this is the same thing as multiplying by the reciprocal of this, is equal to 2 times the square root of y over x, over my yellow x, times the negative square root of y. Well, what's this going to simplify to? This is going to be equal to the square root of... | Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3 |
The negative times the 2, you get negative 2. So you get negative 2y over x is equal to the derivative of y with respect to x. Now, you might be saying, look, we just figured out the derivative implicitly, and it looks very different than what we did right over here. When we just used the power rule, we got negative 2x... | Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3 |
So this car right over here is approaching an intersection at 60 miles per hour. And right now, right at this moment, it is 0.8 miles from the intersection. Now we have this truck over here. It's approaching the same intersection on a street that is perpendicular to the street that the car is on. And right now it is 0.... | Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3 |
It's approaching the same intersection on a street that is perpendicular to the street that the car is on. And right now it is 0.6 miles. So that is 0.6 miles from the intersection and is approaching the intersection at 30 miles per hour. Now my question to you is, what is the rate at which the distance between the car... | Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3 |
Now my question to you is, what is the rate at which the distance between the car and the truck is changing? Well, to think about that, let's first just think about what we're asking. So we're asking about the distance between the car and the truck. So right at this moment when the car is 0.8 miles from the intersectio... | Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3 |
So right at this moment when the car is 0.8 miles from the intersection, the truck is 0.6 miles from the intersection, the truck is traveling at 30 miles per hour towards the intersection, the car is traveling 60 miles per hour towards the intersection, right at this moment, what is the rate at which this distance righ... | Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3 |
Well, we know the distance of the car and the intersection. And let's just call that distance y. So y is equal to 0.8 miles. We also know that d, so let me write this. We know that y is 0.8 miles right now. 0.8 miles. We also know that dy dt, the rate at which y is changing with respect to time, is what? | Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3 |
We also know that d, so let me write this. We know that y is 0.8 miles right now. 0.8 miles. We also know that dy dt, the rate at which y is changing with respect to time, is what? Well, y is decreasing by 60 miles per hour. So let me write it as negative 60 miles per hour. Now, similarly, let's say that this distance ... | Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3 |
We also know that dy dt, the rate at which y is changing with respect to time, is what? Well, y is decreasing by 60 miles per hour. So let me write it as negative 60 miles per hour. Now, similarly, let's say that this distance right over here is x. x is 0.6 miles right at this moment. So we know that x is equal to 0.6 ... | Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3 |
Now, similarly, let's say that this distance right over here is x. x is 0.6 miles right at this moment. So we know that x is equal to 0.6 miles. What is the rate at which x is changing with respect to time? Well, we know it's 30 miles per hour is how fast we're approaching the intersection. But x is decreasing by 30 mi... | Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3 |
Well, we know it's 30 miles per hour is how fast we're approaching the intersection. But x is decreasing by 30 miles every hour. So we should say it's negative 30 miles per hour. So we know what y is. We know what x is. We know how fast y is changing, how fast x is changing with respect to time. So what we could try to... | Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3 |
So we know what y is. We know what x is. We know how fast y is changing, how fast x is changing with respect to time. So what we could try to do here is come up with a relationship between x, y, and s, and then differentiate that relationship with respect to time. And it seems like we have pretty much everything we nee... | Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3 |
So what we could try to do here is come up with a relationship between x, y, and s, and then differentiate that relationship with respect to time. And it seems like we have pretty much everything we need to solve for this. So what's the relationship between x, y, and s? Well, we know that this is a right triangle. The ... | Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3 |
Well, we know that this is a right triangle. The streets are perpendicular to each other. So we can use the Pythagorean theorem. We know that x squared plus y squared is going to be equal to s squared. And then we can take the derivative of both sides of this with respect to time to get a relationship between all the t... | Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3 |
We know that x squared plus y squared is going to be equal to s squared. And then we can take the derivative of both sides of this with respect to time to get a relationship between all the things that we care about. So what's the derivative of x squared with respect to time? Well, it's going to be the derivative of x ... | Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3 |
Well, it's going to be the derivative of x squared with respect to x, which is just 2x, times the derivative of x with respect to time, times dx dt. Once again, just a chain rule. Derivative of something squared with respect to the something times the derivative of the something with respect to time. And we use similar... | Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3 |
And we use similar logic right over here when we want to take the derivative of y squared with respect to time. Derivative of y squared with respect to y times the derivative of y with respect to time. Now on the right-hand side of this equation, we, once again, take the derivative with respect to time, so it's the der... | Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3 |
Once again, this is all just an application of the chain rule. Now it looks like we know what x is, we know what dx dt is, we know what y is, we know what dy dt is. All we need to figure out is what s and then what ds dt is, the rate at which this distance is changing with respect to time. Well, what's s right now? Wel... | Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3 |
Well, what's s right now? Well, we can actually use the Pythagorean theorem at this exact moment. We know that x squared, so x is 0.6, we know 0.6 squared plus y squared, 0.8 squared, is equal to s squared. Well, this is 0.36 plus 0.64 is equal to s squared. This is 1 is equal to s squared. And we only care about posit... | Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3 |
Well, this is 0.36 plus 0.64 is equal to s squared. This is 1 is equal to s squared. And we only care about positive distances, so we have s is equal to 1 right now. So we also know what s is. So let's substitute all of these numbers in and then try to solve for what we came here to do, solve for ds dt. So the rate at ... | Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.