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The graphical relationship between a function & its derivative (part 2)

Given the graph of a function, Sal sketches the graph of its antiderivative. In other words, he sketches the graph of the function whose derivative is the given function. Created by Sal Khan.

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  • marcimus pink style avatar for user hana
    Is there any difference between an anti-derivitive and an integral? if not why not just use the term integral?
    (50 votes)
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    • spunky sam blue style avatar for user Niema Moshiri
      No difference: they mean the same thing. Usually, you'll just use the word "integral." Some people like the word "anti-derivative" since an integral is literally taking the reverse of a derivative. Also, since these are instructional videos, the term "anti-derivative" is helpful because it kind of explicitly tells you what's going on (you have some derivative, now you're going back, hence "anti"), while the word "integral" itself doesn't tell you anything about what's going on unless you already know what it means.

      In short, it doesn't matter at all. I personally like the term "integral" more, but it's all up to personal preference.
      (86 votes)
  • mr pants teal style avatar for user box 0f rox
    Is there a conceptual difference between the anti-derivative shown here and the more involved concept of anti-derivatives taught in Integral Calculus?
    (5 votes)
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  • aqualine ultimate style avatar for user Katie
    So is anti-derivative just another term for the integral? Or is there a difference?
    (3 votes)
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  • blobby green style avatar for user akibshahjahan
    what does d/dx mean at ?
    (5 votes)
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  • leaf green style avatar for user archers2009
    So why could it be defined in the antiderivative (at ) but undefined in the derivative function (when Sal was drawing the derivative of a function in the previous video, he made it the same)?
    (4 votes)
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  • piceratops ultimate style avatar for user Everest Witman
    Would the last cusp in the antiderivative to the left of the graph have a defined slope?
    (4 votes)
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    • piceratops ultimate style avatar for user Michael McCafferty
      If you are referring to the part of F(x) where the orange and pink lines meet, then yes, the derivative is undefined at this point.

      f(x), the derivative of F(x), is discontinuous from the blue part to the pink part and the function does not exist at that point (it has open circles at both the end of the blue part and beginning of the pink part)
      (3 votes)
  • mr pants teal style avatar for user box 0f rox
    Just making sure I'm understanding this correctly...

    We don't know the Original Function, but we know it's derivative. The derivative of the Unknown Function's derivative is the anti-derivative, which is also the Original Function.

    Do I have that right?
    (3 votes)
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  • piceratops ultimate style avatar for user Kevin George Joe
    in this example Sal drew the anti-derivative as one big continuous function...is it okay if I draw the anti-derivative even if the function is not continuous at certain points and has "gaps" like in the derivative function?
    I'd really like some help on this...:)
    (3 votes)
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    • leaf grey style avatar for user Qeeko
      Since differentiability implies continuity, it is not possible for an anti-derivative to be discontinuous at a point where the "derivative-function" is defined, as this would imply that the anti-derivative would not be differentiable at this point (a contradiction).

      An anti-derivative may be discontinuous at points where the "derivative-function" is undefined, however, but this is a rather trivial observation.

      Observe that a function may have a discontinuous derivative, though. As an example, consider the function ƒ defined on all of R by ƒ(x) = x²sin(1/x) when x ≠ 0, and let ƒ(0) = 0. Then the following holds (see if you can prove all of these claims. In particular, see if you can prove claims III) and IV)):

      I) ƒ is differentiable everywhere, i.e., differentiable on all of R;
      II) ƒ'(x) = 2xsin(1/x) - cos(1/x) for x ≠ 0;
      III) ƒ'(0) = 0;
      IV) ƒ' is not continuous at 0.
      (3 votes)
  • aqualine ultimate style avatar for user Al.Dumbrava
    Ha! I finally get why we always put + C after calculating the integral of a function. It's because you can shift it up or down by and y-value C and the graph would still be right.
    (4 votes)
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  • male robot hal style avatar for user Tasin Hoque
    Why does the derivative of the common point between the parabola and a straight line is undefined?
    (3 votes)
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    • leaf green style avatar for user Surf
      Because a sharp angle doesn't have a slope.

      Think about what is happening at that point for f(x). Coming from the left, the slope is a constant negative value, and then suddenly becomes positive. But if you were to define it at the common point, what value would you choose? The value coming from the left, or the right, or somewhere in between? Because there's no one value (or any value) that you can assign to the common point, the slope and therefore the derivative is left undefined.

      That's my intuition of why a slope can't be defined at a sharp point.

      A more mathematically rigorous way to look at it is that f'(x) is a function defined by a limit: the limit of the slope of the secant line between f(x) and f(x+h) as h ->0. At a sharp point, you get two different values depending on whether h approaches 0 from the positive or negative side. Therefore the limit (by definition) doesn't exist at that point, so neither does f'(x)
      (1 vote)

Video transcript

In the last video we looked at a function and tried to draw its derivative. Now in this video, we're going to look at a function and try to draw its antiderivative. Which sounds like a very fancy word, but it's just saying the antiderivative of a function is a function whose derivative is that function. So for example, if we have f of x, and let's say that the antiderivative of f of x is capital F of x. And this tends to be the notation, when you're talking about an antiderivative. This just means that the derivative of capital F of x, which is equal to, you could say capital f prime of x, is equal to f of x. So we're going to try to do here is, we have our f of x. And we're going to try to think about what's a possible function that this could be the derivative of? And you're going to study this in much more depth when you start looking at integral calculus. But there's actually many possible functions that this could be the derivative of. And our goal in this video is just to draw a reasonable possibility. So let's think about it a little bit. So let's, over here on the top, draw y is equal to capital F of x. So what we're going to try to draw is a function where its derivative could look like this. So what we're essentially doing is, when we go from what we're draw up here to this, we're taking the derivative. So let's think about what this function could look like. So when we look at this derivative, it says over this interval over this first interval right over here, let me do this in purple, it says over this interval from x is equal to 0 all the way to whatever value of x, this right over here, it says that the slope is a constant positive 1. So let me draw a line with a slope of a constant positive 1. And I could shift that line up and down. Once again, there's many possible antiderivatives. I will just pick a reasonable one. So I could have a line that looks something like this. I want to draw a slope of positive 1 as best as I can. So let's say it looks something like this. And I could make the function defined here or undefined here. The derivative is undefined at this point. I could make the function defined or undefined as I see fit. This will probably be a point of discontinuity on the original function. It doesn't have to be, but I'm just trying to draw a possible function. So let's actually just say it actually is defined at that point right over there. But since this is going to be discontinuous, the derivative is going to be undefined at that point. So that's that first interval. Now let's look at the second interval. The second interval, from where that first interval ended, all the way to right over here. The derivative is a constant negative 2. So that means over here, I'm going to have a line of constant negative slope, or constant negative 2 slope. So it's going to be twice as steep as this one right over here. So I actually could just draw it. I could make it a continuous function, I could just make a negative 2 slope, just like this. And it looks like this interval is about half as long as this interval. So it maybe gets to the exact same point. So it could look something-- let me draw it a little bit neater-- like this. The slope right over here is equal to 1, we see that right over there in the derivative. And then the slope right over here is equal to negative 2. We see that in the derivative. Now things get interesting. Once again, I could have shifted this blue line up and down. I did not have to construct a continuous function like this. But I'm doing it just for fun. There's many possible antiderivatives of this function. Now what's going on over the next interval? And I'll do it do it in orange. The slope starts off at a very high value. It starts off at positive 2. Then it keeps decreasing, and it gets to 0, right over here. The slope gets to 0 right over there. And then it starts becoming more and more and more and more negative. So I'll just try to make this a continuous function, just for fun. Once again, it does not have to be. So over here the slope is very positive. It's a positive 2. So our slope is going to be like, this is negative 2, so it's going to be a positive 2. And then it gets more and more and more negative up to this, or, it becomes less and less and less positive I should say. Even here the slope is positive. And it gets to 0 right over there. So maybe it gets to 0 maybe right over here. And so what we have, we could have some type of a parabola. So the downward facing parabola. So notice, the slope is a very positive value. It's a positive 2 right over here, then it becomes less and less and less and less positive, all the way to 0 right over there. And then the slope starts turning negative. And so our function could look something like that over the interval. Let me draw it a little bit neater. This is symmetric. How ever positive our slope was here, it's equally negative here. So our curve should probably be symmetric as well. So let me draw it like this over that interval. It could look like that. And then finally, over this last interval, or I guess we could say, it keeps going, our slope is 0. And once again I don't have to draw a continuous function. But when the slope is 0, that just means that I have a line with slope 0. I have a horizontal line. And I could draw that horizontal line up here, in which case I'd have to say its discontinuous. Or I could draw that horizontal line right over here, and try to make my antiderivative a continuous function. So once again, I could have shifted any of these segments up or down, and gotten the exact same derivative. But then I would not have gotten a continuous function like this. But we have been able to construct a possible antiderivative for f of x. And just as a reminder, that's just saying this f of x is-- the antiderivative is a function that f of x could be the derivative of.