Question 1

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

Accepted Answer

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.

Question 2

Is there a case in which we have a curve in the derivative, or are there only lines?

Accepted Answer

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.

Question 3

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.

Accepted Answer

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).

Question 4

whats the difference between the shaded and not-shaded circles ?

Accepted Answer

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.

Question 5

Is the  tangent line the derivative of the function at the certain point?

Accepted Answer

Almost...the SLOPE of the tangent line is the derivative of the function at some certain point.

Question 6

At 2:00, 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 at 2:00

Accepted Answer

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.

Question 7

At 3:55, 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.

Accepted Answer

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.

Question 8

Why the derivative of a sharp turn is not possible?

Accepted Answer

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!

Question 9

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?

Accepted Answer

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).

AP®︎ Calculus AB (2017 edition)

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

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

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