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### Course: AP®︎ Calculus AB (2017 edition) > Unit 6

Lesson 9: Connecting ƒ, ƒ’, and ƒ’’- The graphical relationship between a function & its derivative (part 1)
- The graphical relationship between a function & its derivative (part 2)
- Connecting f and f' graphically
- Visualizing derivatives
- Connecting f, f', and f'' graphically
- Connecting f, f', and f'' graphically (another example)
- Connecting f, f', and f'' graphically
- Curve sketching with calculus: polynomial
- Curve sketching with calculus: logarithm

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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.

## Want to join the conversation?

- Is the tangent line the derivative of the function at the certain point?(4 votes)
- Almost...the SLOPE of the tangent line is the derivative of the function at some certain point.(27 votes)

- sin(x) is a constant function but its slope cos(x) isn't a straight line. why?(5 votes)
- 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.(13 votes)

- Is there a case in which we have a curve in the derivative, or are there only lines?(5 votes)
- 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)

- 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)
- 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)

- whats the difference between the shaded and not-shaded circles ?(5 votes)
- 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)

- Around4:24, Sal draws a open dot to start the blue line - shouldn't it be closed because we know it's zero?(4 votes)
- At2:00, 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 at2:00(4 votes)- If the original graph is of a parabola, rather than a circle, then the graph of the derivative
a straight line, since d/dx[ax² + bx + c] = 2ax + b**is**

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)

- At3:55, 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)
- Yes,

f(x) is negative, but f ' (x) (or F Prime) will be positive, since it is essentially the slope of the line and the slope at that point is positive.(2 votes)

- Why the derivative of a sharp turn is not possible?(2 votes)
- 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)

- Why do the beginning of the pink line and the end of the orange line not end in open circles? Shouldn't the derivative be undefined at those spots, because we only know the limit from one side?(2 votes)
- Since there is only one side we can approach at the endpoints of an interval, that is good enough to make the derivative legitimate at that point (as long as that limit is defined).

Yes, this does get a little strange at points. If you were to graph the derivative of the absolute value function over all real numbers, it would be undefined at 0, but if you were talking about the absolute value function defined on [0,1], then the derivative would be defined at 0 (and would be equal to 1).(2 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.