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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.

## Want to join the conversation?

• Is there any difference between an anti-derivitive and an integral? if not why not just use the term integral?
• 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.
• Is there a conceptual difference between the anti-derivative shown here and the more involved concept of anti-derivatives taught in Integral Calculus?
• No. Sal is merely leading us into a better understanding of derivatives and anti-derivatives by first teaching them abstractly and intuitively.
• what does d/dx mean at ?
• Its called the "Derivative operator"! you put it infront of an expression to mean "Take the derivative of this!"
• So is anti-derivative just another term for the integral? Or is there a difference?
• Strictly speaking, the anti-derivative is one possible value, whereas the indefinite integral gives all possible values.
e.g antiderivative of x^2 is x^3/3 but the indefinite integral of x^2 is x^3/3 + C
Well that's what my calculus for dummies book says anyways.
• 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)?
• well, because of the quirks of derivatives, if there can be multiple possible slopes at a point, then we make the derivative undefined. In this case, the derivative of that 'sharp point' could be 1 if we continued it from the left, but is could also be -2 if we continued it from the right. Also, if you look at https://www.khanacademy.org/math/calculus/differential-calculus/visualizing-derivatives-tutorial/v/intuitively-drawing-the-derivative-of-a-function ()
sal also explains it. To address how the point was defined in this video, you have to note that it is given that the point is at that exact spot. When you take an antiderivative, you don't know what the value is (so you make it undefined), but in that function, they gave you the value, so it is defined there.
great question
• Would the last cusp in the antiderivative to the left of the graph have a defined slope?
• 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)
• 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...:)
• 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`.
• 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?
• 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.
• Why does the derivative of the common point between the parabola and a straight line is undefined?