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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?
• Almost...the SLOPE of the tangent line is the derivative of the function at some certain point.
• sin(x) is a constant function but its slope cos(x) isn't a straight line. why?
• 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.
• Is there a case in which we have a curve in the derivative, or are there only lines?
• 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.
• 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.
• 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).
• 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.
• Around , Sal draws a open dot to start the blue line - shouldn't it be closed because we know it's zero?
• 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
• 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.
• 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.
• 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.
• Why the derivative of a sharp turn is not possible?