Question 1

Just a philosophical digression.

If mathematicians until today don't know how to define a process like division by zero, isn't it impossible to assume that if a division by zero has any kind of definition, humanity can never surely estimate.

It seems kind of strange that the financial sector, higher analysis and modelling relies on mathematical concepts which aren't rigourously defined.

That would mean that at any given point you can estimate that a function has some
supposed behaviour but since you can never instantaneously easily access that point parametrically you are after all only speculating.

Accepted Answer

You raised an interesting topic for discussion, and to answer that, I have to get back to one of your sentences.

"It seems kind of strange that the financial sector, higher analysis and modelling relies on mathematical concepts which aren't rigourously defined."

You might get the wrong impression about division by zero. Mathematicians weren't just saying "Okay, let's define a new operation and call it division- Oh wait, it doesn't work on zero- let's say it's undefined, because any sensible answer will break the rules of math". I admit it, it might have happened in the beginning- but in the last few years, mathematicians have been making extremely careful and formal definitions of math objects, including the division operation.
Division is built as a function (of two variables), that says: given a and b, return a/b. It was built in a way that doesn't allow 0 to be the denominator,  so when we say that 1/0 is not defined, we mean that division is not built to fit in zero in there.

I'll give you an example- When I'll ask you what x to the power of a duck is, you'll think I'm crazy. Why? because exponention is not defined when talking about ducks. You say, "exponention is an operation that you put on numbers, not on animals". With the same arguement, I can tell you "division is an operation that you put on non-zero numbers as the denominator".

Not all mathematical operation need to be defined on all numbers, you just decide what numbers do you want the operation to work with.

Question 2

I understand both forms, but what is the point of the alternative form of derivative? What can the alternative form do, which the standard form can't?

Accepted Answer

The alternate form can give you the numerical value of the derivative at a particular point (where x=a), rather than a general formula for the derivative.  There are times when you might want to compute it that way (especially with more complex problems).

Question 3

What does derivative mean?

Accepted Answer

The derivative of a function is the  measure of change in that function. Consider the parabola y=x^2.

For negative x-values, on the left of the y-axis, the parabola is decreasing (falling down towards y=0), while for positive x-values, on the right of the y-axis, the parabola is increasing (shooting up from y=0). 
If you find the derivative of y=x^2 (it will be explained in later videos but the derivative of x^2 is 2x) and substitute in a negative x-value, you will get a negative number, meaning you have negative change i.e. the function is decreasing. On the other hand, if you substitute in a positive x-value,  you will get a positive number, meaning you have positive change i.e. the function is increasing.

Question 4

As far as I know, `f'(a)` is the slope of a tangent line at `x = a`. Does anyone have any idea how we can apply this definition in real life? In your example, please tell me what is *slope* and what is *tangent line*?

Accepted Answer

Slope can be thought as "How far you go out horizontally and how far you go up vertically to get back to the line". Another way of saying that is "Change in y over change in x." A tangent line is a line that intersects the graph at exactly one point. Slope is often seen as a rate of change, so differentials and derivatives are seen as a rate of change. Physics is interested often times how fast a particle is moving in which ever way, so you set up a coordinate system to find position, then if you differentiate position you obtain velocity. But let's say you want to know how fast your velocity is changing - that's acceleration. You can graph velocity versus time and if you create a tangent line to one point that would be your instantaneous acceleration or your acceleration at that instant.

Question 5

Can´t I make something like f(x+h) -f(x)/(x+h)-x -> f(x) + f(h) - f(x)/h -> f(h)/h ?

Accepted Answer

f(x+h) IS NOT equal to f(x) + f(h). So, no, you cannot evaluate it that way.
Example:
f(x) = x²
f(x+h) = (x+h)² = x² + 2hx + h²
while f(x)+f(h) = x² + h²
 
Note: there are a few types of functions where f(x+h) happens to equal f(x) + f(h), but this is the exception to the rule.

Question 6

When you draw the tangent line to point a at the beginning of the video in both graphs, would it be possible to find take another random point on that line and find the slope of the tangent line? Since the tangent line is straight, wouldnt that meant the slope is the same at every point? Thank you.

Accepted Answer

Good question – if you could draw the tangent line "nearly perfectly" then that would work.  There are, however, a number of drawbacks (so to speak) of this strategy.  I encourage you to try your proposed method on a few equations at a couple of different points and compare the answers you get with the actual answers.

Some shortcomings to a graphical strategy:
1) Tedious.
2) Error prone.
3) In situation where you need an equation it would be impractical (as well as exceedingly inaccurate) to make estimates from graphs.
 – For many applications of derivatives (e.g. optimization), we need an equation not just the derivative at a single point.  
 – Sometimes we even want the second derivative and so an equation for the derivative is doubly necessary.

Question 7

Hey, isn't a tangent line just a secant line with two points in the same place? Before I learned calculus here on Khan Academy, I taught myself to approach a tangent line with a secant line whose points get closer and closer and closer in the same way you approach an undefined point on a function with a limit.

Accepted Answer

That's a good way to look at it, and it's actually the basis of the typical definitions of the derivative. Geometrically, the derivative is the limit of the secant slope as the two points of the secant line get closer to each other, just like you described. I think Sal has a video on that entitled "Tangent slope as the limit of secant slope," or something to that effect.

Question 8

If we take a look at any of these two forms for derivative, we always will get 0/0. So it looks like any derivative is undefined. Could someone explain me this moment?

Accepted Answer

If you substitute h directly into the expressions, then yes, you get 0/0. But this does not necessarily mean that the limit does not exist, it may just mean we have to work a little more to evaluate it. For example, consider the function f(x)=x. The derivative is
lim [(x+h)-x]/h
h->0

Substituting h immediately gives 0/0, but if we first do some simplifying algebra, we get
lim [x+h-x]/h
h->0

lim [h]/h
h->0

lim 1
h->0

1

So the derivative of f(x)=x is simply 1.

Question 9

at 4:19 shouldnt  x approach zero 
then and only,'I think ' we might get closer to zero
PLEASE correct if i am wrong
I am talking about the slope at point 'a' on the second graph

Accepted Answer

Since x = a+h, when h approaches 0, x approaches a.

Question 10

at 1:23 Sal cancels the x's and the equation remaining is:

the limit (as x approaches 0) of ( f(x+h) - f(x) )/h

As x approaches 0, doesn't the whole equation approach infinity? If this is true, then why isn't the derivative of every function infinity?

Accepted Answer

The limit is when h approaches 0, not x. When h approaches 0, f(x+h) approaches f(x) and the numerator, f(x+h)-f(x), approaches 0. The numerator and denominator both approach 0, so the limit is 0/0. 0/0 is one of the inderterminate answers, so it is not necessarily infinity. The answer depends on the function we put in the limit.

AP®︎ Calculus AB (2017 edition)

Course: AP®︎ Calculus AB (2017 edition) > Unit 2

Formal and alternate form of the derivative

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