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The graphical relationship between a function & its derivative (part 1)

Given the graph of a function, Sal sketches the graph of its derivative. Created by Sal Khan.

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  • leaf green style avatar for user avneesh.muralitharan
    Is the tangent line the derivative of the function at the certain point?
    (4 votes)
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  • piceratops sapling style avatar for user Meghna
    sin(x) is a constant function but its slope cos(x) isn't a straight line. why?
    (4 votes)
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    • piceratops ultimate style avatar for user Christian Wen
      no sinx is not a constant function. Do you mean that it is a function that cycles? Because it does do that. But it isn't constant. A function that has a slope of a straight line is a parabola. But sinx isn't a parabola. It's slope increases and then decreases and then increases and then decreases. That's why the slope of sinx is cosx.
      (14 votes)
  • marcimus pink style avatar for user Wafae EL MAHJOUBI
    Is there a case in which we have a curve in the derivative, or are there only lines?
    (5 votes)
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    • leafers ultimate style avatar for user Marco Merlini
      A linear function is a function that has degree one (as in the highest power of the independent variable is 1). If the derivative (which lowers the degree of the starting function by 1) ends up with 1 or lower as the degree, it is linear. If the derivative gives you a degree higher than 1, it is a curve.
      (8 votes)
  • piceratops ultimate style avatar for user Matěj Krátký
    I think, at the "first point" of f(x), if I can say it like that, the derivative also shouldn't be defined, because we just cannot make a tangent line over there.
    (5 votes)
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    • leaf grey style avatar for user Qeeko
      I presume you mean the point x = 0? Most texts on elementary calculus would not define the derivative at such a point. Such texts usually only define the derivative of a function at an interior point of said function's domain. Informally speaking, a point is an interior point if we may "approach" it from both sides, while still being in the domain of the function.

      More formally: let A be a nonempty set of real numbers, and suppose a ∈ A. We call a an interior point of A if and only if there exists some open interval, containing a, which is entirely contained in A. If a is such a point, we may approach it from both sides from within A.

      So, for most elementary purposes, you are correct, the derivative is not defined there. However, more advanced texts may define what is called a one-sided derivative (see, e.g., http://en.wikipedia.org/wiki/Left_and_right_derivative for a brief overview). Other texts, such as Analysis I (Tao, 2nd ed., 2009, p. 250) does not make the distinction between the "ordinary" derivative and the one-sided derivative (this is an honours level text on introductory real analysis - not well suited for a beginner).
      (5 votes)
  • blobby green style avatar for user weirdmind1
    whats the difference between the shaded and not-shaded circles ?
    (4 votes)
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    • male robot hal style avatar for user Yamanqui García Rosales
      Shaded circles mean that the function reaches that loin and it's defined there, while open circles mean that the function approaches infinitely close to that point but never reach it, so the function is undefined on the open point.

      This is related to the difference between the minus-than ( < ) and greater-than (>) and the minus-or-equal-than (<=), and the greater-or-equal-than (>=).

      For example, if a function is defined in the domain 2 < x <= 5 then the value on the 2 would be represented by an open circle (since the function approaches it but never actually reaches it), while the 5 would be a filled circle, since the function is defined there.
      (4 votes)
  • piceratops tree style avatar for user aakanksha.j.saxena
    Around , Sal draws a open dot to start the blue line - shouldn't it be closed because we know it's zero?
    (4 votes)
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  • piceratops seedling style avatar for user stephenwist
    At , should the graph of the derivative resemble something like the graph of tangent? i.e. this form http://www.purplemath.com/modules/trig/graphs22.gif
    The slope of the semicircle is extremely positive at first and gradually decreases to zero, and does not decrease at a constant value as is shown at
    (4 votes)
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    • leaf blue style avatar for user Stefen
      If the original graph is of a parabola, rather than a circle, then the graph of the derivative is a straight line, since d/dx[ax² + bx + c] = 2ax + b
      If the original graph is a circle, then the graph of the derivative will be similar (but opposite) to the purple math image you linked to. The graph will look like this: https://www.desmos.com/calculator/uoe1bollo2
      There will be vertical asymptotes at the left and right edges of the circle. As we move along x from x=0, the derivative will be very positive, gradually reducing to zero at x=<circle radius>, (where the slope is parallel to the x axis), and then the graph of the derivative will get more and more negative.
      (2 votes)
  • spunky sam blue style avatar for user Adeyemi47
    At , the red colored line is increasing, but is in f(x)<0. So, I get that it will have a constant f'(x) slope, but shouldn't it be in f'(x)<0.
    (3 votes)
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  • orange juice squid orange style avatar for user sadiafaruque2000
    Why the derivative of a sharp turn is not possible?
    (2 votes)
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    • starky ultimate style avatar for user Felicia L.
      The derivative is basically a tangent line. Recall the limit definition of a tangent line. As the two points making a secant line get closer to each other, they approach the tangent line. With a sharp turn like a cusp, there is no point that the secant line approaches. I hope that makes sense!
      (3 votes)
  • old spice man green style avatar for user Akash Pandey
    Why is the derivative of second and third function a horizontal line? I mean I don't understand how to represent derivative of a function graphically.
    (2 votes)
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    • spunky sam blue style avatar for user Ethan Dlugie
      I'm not quite sure that I understand your question. Are you asking what the second and third derivatives look like (i.e. d^2/dx^2, d^3/dx^3)?
      In that case, look at your horizontal line. What is its slope? Zero, right? That is the first derivative. The first derivative of a horizontal line is 0 everywhere. What is the slope of that function (the second derivative)? It is a flat line, y=0, so the slope must still be 0 which means that the second derivative is everywhere zero.
      Continue this train of thought, and it is clear that any derivative (first, second, third...17th...37th) of a horizontal line must always be 0.
      (3 votes)

Video transcript

So I've got this crazy discontinuous function here, which we'll call f of x. And my goal is to try to draw its derivative right over here. So what I'm going to need to think about is the slope of the tangent line, or the slope at each point in this curve, and then try my best to draw that slope. So let's try to tackle it. So right over here at this point, the slope is positive. And actually, it's a good bit positive. And then as we get larger and larger x's, the slope is still positive, but it's less positive-- and all the way up to this point right over here, where it becomes 0. So let's see how I could draw that over here. So over here we know that the slope must be equal to 0-- right over here. Remember over here, I'm going to try to draw y is equal to f prime of x. And I'm going to assume that this is some type of a parabola. And you'll learn shortly why I had to make that assumption. But let's say that, so let's see, here the slope is quite positive. So let's say the slope is right over here. And then it gets less and less and less positive. And I'll assume it does it in a linear fashion. That's why I had to assume that it's some type of a parabola. So it gets less and less and less positive. Notice here, for example, the slope is still positive. And so when you look at the derivative, the slope is still a positive value. But as we get larger and larger x's up to this point, the slope is getting less and less positive, all the way to 0. And then the slope is getting more and more negative. And at this point, it seems like the slope is just as negative as it was positive there. So at this point right over here, the slope is just as negative as it was positive right over there. So it seems like this would be a reasonable view of the slope of the tangent line over this interval. Now let's think about as we get to this point. Here the slope seems constant. Our slope is a constant positive value. So once again, our slope here is a constant positive line. Let me be careful here because at this point, our slope won't really be defined, because our slope, you could draw multiple tangent lines at this little pointy point. So let me just draw a circle right over there. But then as we get right over here, the slope seems to be positive. So let's draw that. The slope seems to be positive, although it's not as positive as it was there. So the slope looks like it is-- I'm just trying to eyeball it-- so the slope is a constant positive this entire time. We have a line with a constant positive slope. So it might look something like this. And let me make it clear what interval I am talking about. I want these things to match up. So let me do my best. So this matches up to that. This matches up over here. And we just said we have a constant positive slope. So let's say it looks something like that over this interval. And then we look at this point right over here. So right at this point, our slope is going to be undefined. There's no way that you could find the slope over-- or this point of discontinuity. But then when we go over here, even though the value of our function has gone down, we still have a constant positive slope. In fact, the slope of this line looks identical to the slope of this line. Let me do that in a different color. The slope of this line looks identical. So we're going to continue at that same slope. It was undefined at that point, but we're going to continue at that same slope. And once again, it's undefined here at this point of discontinuity. So the slope will look something like that. And then we go up here. The value of the function goes up, but now the function is flat. So the slope over that interval is 0. The slope over this interval, right over here, is 0. So we could say-- let me make it clear what interval I'm talking about-- the slope over this interval is 0. And then finally, in this last section-- let me do this in orange-- the slope becomes negative. But it's a constant negative. And it seems actually a little bit more negative than these were positive. So I would draw it right over there. So it's a weird looking function. But the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point. And by doing so, we have essentially drawn the derivative over that interval.