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The derivative of x² at x=3 using the formal definition

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)
,
CHA‑2.B.4 (EK)
Sal finds the limit expression for the derivative of f(x)=x² at the point x=3 and evaluates it. Created by Sal Khan.

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  • old spice man green style avatar for user Skywalker94
    It's probably a dumb question, but what's the difference between the tangent line and the secant line? I don't really get where the whole secant line thing is coming from.
    (199 votes)
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  • blobby green style avatar for user Diego Agostini
    At about , shouldn't it simplify to "6+ (delta x) squared"?
    (28 votes)
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    • blobby green style avatar for user Josh
      No because it's 6 * (delta)x / (delta)x = 6, and (delta)x^2 / (delta)x, which simplifies to ( (delta)x * (delta)x ) / (delta)x, which is equal to (delta)x. You have to divide both sides of the + sign by (delta)x, if you're going to simplify.
      (44 votes)
  • blobby green style avatar for user liaborrelli
    I don't understand why a derivative is written as F'(x). Why is that the notation we use? Thanks for any help!
    (18 votes)
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    • old spice man green style avatar for user Hector
      Simple notation:
      1. Lagrange introduced the prime notation f'(x). We use it because is one of the most common modern notations and is most useful when we wish to talk about the derivative as being a function itself.
      2. Newton introduced the dot notation ẏ, used in physics to denote time derivatives.
      3. Leibniz introduced the Leibniz's notation dy/dx, useful for partial differentiation, and
      4. Euler introduced the Euler's notation which uses a differential operator Df(x) useful for solving linear differential equations.
      (51 votes)
  • mr pink red style avatar for user edominguezarzac
    Aren't we looking for the limit as x approaches 3 ???
    (18 votes)
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  • male robot hal style avatar for user anthony
    What is the use of a derivative? Any real-life applications?
    (9 votes)
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    • spunky sam blue style avatar for user James Pearson
      The most obvious application quoted is usually for Speed, Acceleration and Distance. They are all derivatives/integrals of each other. E.g. Acceleration is the derivative of Speed.

      However there are any number of possibilities since it can be applied to pretty much anything where an analysis of a rate of change is helpful. Actually calculating a best fit function in the first place is an expertise in itself though.
      (22 votes)
  • female robot grace style avatar for user Emily
    At , Sal simplifies (6(delta)x + (delta)x^2)/(delta)x to 6+(delta)x. However, later he substitutes (delta)x for 0 to get a slope of 6. If he'd put that in the original expression it would have been division by 0. Why does this work? Aren't you not allowed to divide by 0?
    (11 votes)
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    • male robot hal style avatar for user Sid
      That's not exactly what he does. He makes dx very very very small, but not zero. Close to zero, but not zero.

      6+0.0000000000000000000000001 is pretty close to 6, and the smaller you make dx the less significant it gets. It will not become exactly zero, but it will become so small that we can ignore it.
      (9 votes)
  • blobby green style avatar for user Edmund Yong
    so a derivative is a function that shows the slope of the original function (curve) ?
    (8 votes)
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  • leaf blue style avatar for user Owen Z.
    I keep seeing the term "limit of difference quotient" in my worksheets. Is this just another term for what this video uses or is it an entirely separate concept?
    (3 votes)
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    • spunky sam blue style avatar for user Ethan Dlugie
      Indeed it is the same thing. The difference quotient (in calculus) is used to refer to the slope between two points, the average rate of change between two points, rise over run, ∆y/∆x, or whatever other name you might have for it. The limit of the difference quotient then becomes the instantaneous rate of change aka the derivative.
      (3 votes)
  • aqualine ultimate style avatar for user ahamed.zoha
    Hey, we can solve a curve using a tangent line. Why use a secant line then?
    (4 votes)
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  • blobby green style avatar for user Utkarsh Jetly
    So we can write f'(x) as dy/dx?
    (3 votes)
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    • female robot grace style avatar for user loumast17
      yep another way of writing it is d/dx y or the derivative of y with respect to x.

      I say that because when you take the derivative of the derivative it can be written as f''(x) or (d^2)y/(dx)^2 because basically you are doing d/dx d/dx y. It might look like things are being multiplied and you'll see that kind of notation used a lot in calculus.

      Hopefully that last part made sense, but if not, to make sure I answer your question, Yes those are two ways of writing the derivative.
      (1 vote)

Video transcript

In the last video we tried to figure out the slope of a point or the slope of a curve at a certain point. And the way we did, we said OK, well let's find the slope between that point and then another point that's not too far away from that point. And we got the slope of the secant line. And it looks all fancy, but this is just the y value of the point that's not too far away, and this is just the y value point of the point in question, so this is just your change in y. And then you divide that by your change in x. So in the example we did, h was the difference between our 2 x values. This distance was h. And that gave us the slope of that line. We said hey, what if we take the limit as this point right here gets closer and closer to this point. If this point essentially almost becomes this point, then our slope is going to be the slope of our tangent line. And we define that as the derivative of our function. We said that's equal to f prime of x. So let's if we can apply this in this video to maybe make things a little bit more concrete in your head. So let me do one. First I'll do a particular case where I want to find the slope at exactly some point. So let me draw my axes again. Let's draw some axes right there. Let's say I have the curve-- this is the curve-- y is equal to x squared. So this is my y-axis, this is my x-axis, and I want to know the slope at the point x is equal to 3. When I say the slope you can imagine a tangent line here. You can imagine a tangent line that goes just like that, and it would just barely graze the curve at that point. But what is the slope of that tangent line? What is the slope of that tangent line which is the same as the slope of the curve right at that point. So to do it, I'm actually going to do this exact technique that we did before, then we'll generalize it so you don't have to do it every time for a particular number. So let's take some other point here. Let's call this 3 plus delta x. I'm changing the notation because in some books you'll see an h, some books you'll see a delta x, doesn't hurt to be exposed to both of them. So this is 3 plus delta x. So first of all what is this point right here? This is a curve y is equal to x squared, so f of x is 3 squared-- this is the point 9. This is the point 3,9 right here. And what is this point right here? So if we were go all the way up here, what is that point? Well here our x is 3 plus delta x. It's the same thing as this one right here, as x naught plus h. I could have called this 3 plus h just as easily. So it's 3 plus delta x up there. So what's the y value going to be? Well whatever x value is, it's on the curve, it's going to be that squared. So it's going to be the point 3 plus delta x squared. So let's figure out the slope of this secant line. And let me zoom in a little bit, because that might help. So if I zoom in on just this part of the curve, it might look like that. And then I have one point here, and then I have the other point is up here. That's the secant line. Just like that. This was the point over here, the point 3,9. And then this point up here is the point 3 plus delta x, so just some larger number than 3, and then it's going to be that number squared. So it's going to be 3 plus delta x squared. What is that? That's going to be 9. I'm just foiling this out, or you do the distribute property twice. a plus b squared is a squared plus 2 a b plus b squared, so it's going to be 9 plus two times the product of these things. So plus 6 delta x, and then plus delta x squared. That's the coordinate of the second line. This looks complicated, but I just took this x value and I squared it, because it's on the line y is equal to x squared. So the slope of the secant line is going to be the change in y divided by the change in x. So the change in y is just going to be this guy's y value, which is 9 plus 6 delta x plus delta x squared. That's this guy's y value, minus this guy's y value. So minus 9. That's your change in y. And you want to divide that by your change in x. Well what is your change in x? This is actually going to be pretty convenient. This larger x value-- we started with this point on the top, so we have to start with this point on the bottom. So it's going to be 3 plus delta x. And then what's this x value? What is minus 3? That's his x value. So what does this simplify to? The numerator-- this 9 and that 9 cancel out, we get a 9 minus 9. And in the denominator what happens? This 3 and minus 3 cancel out. So the change in x actually end up becoming this delta x, which makes sense, because this delta x is essentially how much more this guy is then that guy. So that should be the change in x, delta x. So the slope of my secant line has simplified to 6 times my change in x, plus my change in x squared, all of that over my change in x. And now we can simplify this even more. Let's divide the numerator and the denominator by our change in x. And I'll switch colors just to ease the monotony. So my slope of my tangent of my secant line-- the one that goes through both of these-- is going to be equal if you divide the numerator and denominator this becomes 6. I'm just dividing numerator and denominator by delta x plus six plus delta x. So that is the slope of this secant line So slope is equal to 6 plus delta x. That's this one right here. That's this reddish line that I've drawn right there. So this number right here, if the delta x was one, if these were the points 3 and 4, then my slope would be 6 plus 1, because I'm picking a point 4 where the delta x here would have to be 1. So the slope would be 7. So we have a general formula for no matter what my delta x is, I can find the slope between 3 and 3 plus delta x. Between those two points. Now we wanted to find the slope at exactly that point right there. So let's see what happens when delta x get smaller and smaller. This is what delta x is right now. It's this distance. But if delta x got a little bit smaller, then the secant line would look like that. Got even smaller, the secant line would look like that, it gets even smaller. Then we're getting pretty close to the slope of the tangent line. The tangent line is this thing right here that I want to find the slope of. Let's find a limit as our delta x approaches 0. So the limit as delta x approaches 0 of our slope of the secant line of 6 plus delta x is equal to what? This is pretty straightforward. You can just set this equal to 0 and it's equal to 6. So the slope of our tangent line at the point x is equal to 3 right there is equal to 6. And another way we could write this if we wrote that f of x is equal to x squared. We now know that the derivative or the slope of the tangent line of this function at the point 3-- I just only evaluated it at the point 3 right there-- that that is equal to 6. I haven't yet come up with a general formula for the slope of this line at any point, and I'm going to do that in the next video.