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Worked example: Derivative from limit expression

AP.CALC:
CHA‑2 (EU)
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CHA‑2.B (LO)
,
CHA‑2.B.2 (EK)
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CHA‑2.B.3 (EK)
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CHA‑2.B.4 (EK)
Sal interprets a limit expression to find that it describes the derivative of f(x)=x³ at the point x=5. Created by Sal Khan.

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  • piceratops ultimate style avatar for user Danny
    It would be nice if the video focused more on finding the derivative at an indicated point or a separate video for it. I'm great at the Formal form but the alternate form is more difficult for me. Even just a simple f(x)= 1/x, a=2 explanation would be great. If someone could help me out that would be great as well.
    (18 votes)
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    • aqualine ultimate style avatar for user Linka
      If I understand you're question correctly, you'd like to have an explanation of how to use alternate form to get the derivative at a point. Hopefully my explanation below isn't too long >n<. I also tried to explain how formal and alternate form are different ways of saying the same thing.

      Let's do this example you have provided, f(x)=1/x, with a=2.

      For the sake of demonstration, let's do the formal form first. Formal form is saying "I'm going to find the derivative by decreasing a distance from my point at 2 down to 0".

      You would probably write it out like this:
      f'(x)=lim (h->0) ((1/(2+h)-1/2)/(h)
      I'll let you crack the algebra to save space, but unless I made an error you should eventually boil it down to this:
      -1/2(2+h)
      and as h approaches 0, you get -1/4.

      Ok, now let's talk alternate form. Instead of taking a point h distance away from 2 and reducing that distance to 0, I'm going to say it another way- "I'm going to just take a point x, and bring it closer to 2." If you think about it, it's the same thing. Once x approaches 2, the distance between the two will be 0-just like with the distance h from 2 in formal form. It's just an alternate way of finding the same thing. To put it more succinctly, we're substituting h with x, and just saying that the closer we get x to 2 is the same thing as saying we're moving a point 2+h closer to 2 by reducing h to 0.

      So let's write it out:
      f'(x)=lim (x->a) (1/x-1/2)/(x-2)

      If you do the algebra (hint: Get the denominators at the top to play nice first), you'll eventually get:

      -1/2x

      Plug in 2 for x (remember, we're having x approach 2), and you'll get the same thing as formal form- -1/4.

      Hope that was what you were looking for and it wasn't too long. Let me know if this helps.
      (33 votes)
  • ohnoes default style avatar for user Cyan Wind
    Any tip to understand the difference(s) between the formal form and the alternate one clearer and to apply them better? In what case(s) we should use the prior, and in what case(s) the latter is better?
    (14 votes)
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    • marcimus pink style avatar for user Alex Tran
      Generally, we will use the formal form of a derivative when trying to find elementary derivatives like finding the derivative of y = x, or the derivative of y = x^2. The alternate form is still useful, although it is primarily used in finding slopes of lines, not curves.
      Differences between formal form and alternate form are in how the limits are defined. In both, we are letting two points approach each other, but one uses lim h --> 0 and the other uses lim x --> a.
      (13 votes)
  • piceratops ultimate style avatar for user Travis Petersen
    I calculated the slope of the tangent line of f(x)=x^3 at x=5 to be 75. Can anyone confirm this for me?
    (4 votes)
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  • male robot hal style avatar for user mikevez
    The secant line is found by (Y2-Y1 / X2-X1) I always thought the (X2 , Y2) were the point furthest to the right? But in this video () he is coming from the left. How do we know what values to put in ( f(x)-f(a) / x-a ) ? or does it matter what order they are in?
    (4 votes)
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    • duskpin sapling style avatar for user Vu
      The secant line is basically the slope of the 2 points. In other words, changes in Y over changes in X. It doesn't matter from which direction you approach it from as long as you are consistent with your order pair, meaning, let (x,f(x)) and (a,f(a)) be the two points, you can use f(x)-f(a)/x-a or f(a)-f(x)/a-x.

      Generally, in the book it will teach you that f(x) is on the order pair from the right and > f(a) this is because it's more intuitive when doing the subtraction. But just remember that the secant line is just the slope between any two points on the graph. And if you know how to find the slope of any two points, then you can find the secant line.
      (4 votes)
  • orange juice squid orange style avatar for user Kevin Au
    Knowing only the formal and the alternative form of the derivative, are we supposed to know how to evaluate f'(x) or f'(a)? Specifically, I am having troubles simplifying (x^3 - 125)/(x-5) so that I can take the limit of it as x approaches 5. If so, I might have just missed something and I will go back some videos to revise.
    (2 votes)
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  • leaf green style avatar for user Stephen Vallejos
    what is a prime function with the symbol ' such as f'(a) ?
    (1 vote)
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  • blobby green style avatar for user Luke Schep
    How do you find f(x)= 14x+13, if f'(1) or f prime of 1
    (0 votes)
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    • leaf green style avatar for user fosterz
      I'm not sure I completely understand your question, but since your function is linear, the slope is constant, so the derivative is also constant. Your equation is of the linear form: y=mx+b where m is the slope. In your case it is 14, so the derivative is f'(x)=14.
      f'(1)=14 too
      (5 votes)
  • starky sapling style avatar for user 20leunge
    I thought the derivative was going to be an actual value. Why was it kept in that form?
    (1 vote)
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    • mr pants teal style avatar for user Moon Bears
      Normally if a function is differentiable, then you'd want to write down what that resulting function is so that you can compute it at any point you'd like - instead of a numerical value. For instance if f(x) = x^2 then f'(x) = 2x. This is true for any value of x.
      (2 votes)
  • blobby green style avatar for user 😊
    I am needing help with a problem.

    Use the 4 step process to find f'(x) and then find f'(1) f'(2) f'(3)

    f(x)=3x-7
    (1 vote)
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    • cacteye blue style avatar for user Jerry Nilsson
      𝑓(𝑥) = 3𝑥 − 7

      Step 1: 𝑓(𝑥 + ℎ) = 3(𝑥 + ℎ) − 7 = 3𝑥 − 7 + 3ℎ

      Step 2: 𝑓(𝑥 + ℎ) − 𝑓(𝑥) = 3𝑥 − 7 + 3ℎ − (3𝑥 − 7) = 3ℎ

      Step 3: (𝑓(𝑥 + ℎ) − 𝑓(𝑥))∕ℎ = 3ℎ∕ℎ = 3

      Step 4: lim(ℎ → 0) (𝑓(𝑥 + ℎ) − 𝑓(𝑥))∕ℎ = lim(ℎ → 0) 3 = 3

      So, 𝑓 '(𝑥) = 3 ⇒ 𝑓 '(1) = 𝑓 '(2) = 𝑓 '(3) = 3
      (2 votes)
  • blobby blue style avatar for user Isabella Mathews
    Why is the derivative of all these functions just algebraic expressions? Don't they have any numerical value? Because I remember in the intro of calculus that we can use it to find the instantaneous value of a curve. If they do have a value, how can we find it? For e.g., what is the numerical value of f'(5) in the above problem?
    (1 vote)
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    • aqualine seedling style avatar for user Seongjoo
      Because you are solving for the general derivative of the functions.To find the particular solution for a X-value, all you have to do is plug in the X-value into the derivative.

      For your example of f'(5), as f(x) = x^3
      f'(x) = 3x^2
      So you plug in 5 for the X-value
      f'(5) = 3(5)^2 = 75

      Also, this video solves for the derivative using the formal definition but there is another way that you will learn later which is much faster and easier (in my opinion).
      (2 votes)

Video transcript

The alternate form of the derivative of the function f, at a number a, denoted by f prime of a, is given by this stuff. Now this might look a little strange to you, but if you really think about what it's saying, it's really just taking the slope of the tangent line between a comma f of a. So let's imagine some arbitrary function like this. Let's say that that is-- well I'll just write that's our function f. And so you could have the point when x is equal to a-- this is our x-axis-- when x is equal to a, this is the point a, f of a. You notice a, f of a. And then we could take the slope between that and some arbitrary point, let's call that x. So this is the point x, f of x. And notice, the numerator right here, this is just our change in the value of our function. Or you could view that as the change in the vertical axis. So that would give you this distance right over here. That's what we're doing up here in the numerator. And then in the denominator, we're finding the change in our horizontal values, horizontal coordinates. Let me do that in a different color. So the change in the horizontal, that's this right over here. And then they're trying to find the limit as x approaches a. So as x gets closer and closer and closer and closer to a, what's going to happen is, is that when x is out here, we have this secant line. We're finding the slope of this secant line. But as x gets closer and closer, the secant lines better and better and better approximate the slope of the tangent line. Where the limit, as x approaches a, but doesn't quite equal a, is going to be-- this is actually our definition of our derivative. Or I guess the alternate form of the derivative definition. And this would be the slope of the tangent line, if it exists. So with that all that out the way, let's try to answer their question. With the Alternative Form of the Derivative as an aid, make sense of the following limit expression by identifying the function f and the number a. So right here, they want to find the slope of the tangent line at 5. Here they wanted to find the slope of the tangent line at a. So it's pretty clear that a is equal to 5. And that f of a is equal to 125. Now what about f of x? Well here, it's a limit of f of x minus f of a. Well here it's the limit as x to the third minus 125. And this makes sense. If f of x is equal to x to the third, then it makes sense that f of 5 is going to be 5 to the third, is going to be 125. And we're also taking up here the limit as x approaches a. Here we're taking the limit as x approaches 5. So this is the derivative of the function f of x is equal to x to the third. Let me write that down in the green color. x to the third at the number a is equal to 5. And so we can imagine this. Let's try to actually graph it, just so that we can imagine it. Actually, I'll do it out here, where I have a little bit better contrast with the colors. So let's say that is my y-axis. Let's say that this is my x-axis. I'm not going to quite draw it to scale. Let's say this right over here is the 125. Or y, this is when y equals 125. This is when x is equal to 5, so they're clearly not at the same scale. But the function is going to look something like this. We know what x to the third looks like, it looks something like this. So here, our a is equal to 5. This point right over here is 5, 125. And then we're taking the slope between that point and an arbitrary x-value. Or I should say an arbitrary other point on the curve. So this right over here would be the point, we could call that x, x to the third. We know that f of x is equal to x to the third. And let me make it clear. This is a graph of y is equal to x to the third. And so this expression, right over here, all of this, this is the slope between these two points. And as we take the limit as x approaches 5, so right now this is our x, as x gets closer and closer and closer to 5, the secant lines are going to better and better approximate the slope of the tangent line at x equals 5. So the slope of a tangent line would look something like that.