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Differential Calculus
Course: Differential Calculus > Unit 2
Lesson 3: Derivative definition- Formal definition of the derivative as a limit
- Formal and alternate form of the derivative
- Worked example: Derivative as a limit
- Worked example: Derivative from limit expression
- Derivative as a limit
- The derivative of x² at x=3 using the formal definition
- The derivative of x² at any point using the formal definition
- Finding tangent line equations using the formal definition of a limit
- Limit expression for the derivative of function (graphical)
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Limit expression for the derivative of function (graphical)
Sal interprets limit expressions as the derivatives of a function given graphically, and evaluates them. Created by Sal Khan.
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How come the slope is -2? I didn't quite get that.(43 votes)
Just look at the graph around x=3. If you move one to the right, the f(x) moves two down. Change in f(x) = -2 for a change in x of 1. -2 / 1 = -2(30 votes)
At3:43, if h was approaching 0 from the positive direction, would it be undefined?(20 votes)
There is a jump discontinuity between the two points so a limit can not exist there because for a limit to exist values approaching a limit from both sides have to be approaching the same number. I think.(0 votes)
- See1:12into the video. I have watched this video now three times now and I kinda stuck. The tutor (Mr Khan) mentioned that the value is -2. Yes because it is declining I agree, however the value 2 I am not in agreement. The value seems to be 1 and not 2. Can someone help me understand this part. Tx(4 votes)
- Slope = (change in y) / (change in x)
Just look at the graph between x=3 and x=4, where y changes from 1 to -1.
The change in x is 1, the change in y is -2. So the slope is -2/1 = -2(14 votes)
I don't quite understand the logic here. In previous sections, the limit corresponds to the corresponding y-value, i.e. f(x), but now it corresponds to the slope. What did I miss in between?(5 votes)
Now we aren't just taking the limit of the function; instead, we're taking the limit of an expression that represents the average slope between two points, and as we take the limit, those two points come closer and closer together until they are equal, at which point we get the instantaneous slope.
A key difference here is that the limit now applies in more than one place (it is in both the numerator and the denominator); thus, we're taking the limit of a ratio dy/dx -- in other words, the slope -- as the function approaches a given point.
As shown in the videos, the expression for slope between an arbitrary point (x) and another point arbitrarily close to it (x+h) can be written asf(x+h) - f(x)
---------------
(x+h) - x
As we take the limit of this expression as h approaches 0, we approximate the instantaneous slope of the function (that is, the slope at exactly one point, expressed as the limit of the average slope between two points arbitrarily far from each other).
Hope that helps.(9 votes)
I don't understand how Sal found the value of h around2:27.. Since we are looking at value as h approaches 0, how come our h is around 7 on the x-axis?(5 votes)
Sal did not explain that very clearly, and did misspeak. It was not the value of h, but the value of x+h (before applying the limit h→0).
Sal was attempting to explain how a secant line becomes a tangent line when the two points the secant line connects are infinitesimally close to each other.(3 votes)
I wish I could be more specific, but this video overall was really confusing. It would help if he specified what we would have to review to even know what h is in the first place. Same for the previous video featuring y = cos(x). I have reviewed algebra, graphing functions, limits, saw the videos in the previous section (which the first video in this section makes redundant IMO), and I still don't get it.
Saying all of that, my question is what is h and how is Sal deciding which arbitrary points to choose? This isn't clear to me at all.
Edit: I still want to know the answer to my questions, but for other people who have the same concerns here, this is explained much clearer in this video https://www.khanacademy.org/math/calculus/differential-calculus/derivative_intro/v/alternate-form-of-the-derivative
Where an alternate method is given that's a little more intuitive, I think.(4 votes)- h is any number. We have to find out the limit as h assumes values near 0. Arbitrary points are random points. You can choose any one.(2 votes)
I didn't understand the fact that the limit as h approaches 0 from the negative direction was 1 where at 0 the f(x) 2. Why is this?(4 votes)
In this video Derivative as slope of a tangent line (https://www.khanacademy.org/math/differential-calculus/taking-derivatives/derivative_intro/v/calculus--derivatives-1--new-hd-version, at15:10), Sal said the derivative of f wasf'(x) = lim (h → 0) [f(x₀ + h) - f(x₀) / h]. The function is equivalent in this examplelim [f(8 + h) - f(8) / h], then whyh → 8−instead ofh → 0?(3 votes)- It was h → 0⁻, not 8⁻
Remember that h is a different variable than x. He did mention what happened when x was near 8, but unless I missed something, he consistently spoke of h approaching 0.
Also, the function was not differentiable at x=8, so we had to use one-sided limits in that case.(2 votes)
At0:43, why is the arbitrary X ahead, not behind?(1 vote)- It's only for sake of convenience when using the formula that we choose a point that is ahead. You certainly can choose a point that is behind if you want.(6 votes)
- at3:30Sal says the slope for the limit from the positive direction of 8 is infinite. how does he know that?
I thought the slop was -1.
can someone please help me?(2 votes)- You're correct that the slope on the positive side of 8 is clearly -1.
I'm not sure if Sal worded that 100% correct. I would say the limit as h approaches 0+ would be -1.
The limit at 8 does not exist because limit as h ---> 0- does not equal h---> 0+
So the "infinite slope" he is drawing applies only AT 8.(2 votes)
3:43Video transcript
With the graph of the
function f as an aid, evaluate the following limits. So the first one is the
limit as x approaches 3 of f of x minus f
of 3 over x minus 3. So let's think about x minus--
x equals 3 is right over here. This right over here is f
of 3, or we could say f of 3 is 1 right over here. That's the point 3 comma f of 3. And they're essentially
trying to find the slope between an
arbitrary x and that point as that x gets
closer and closer to 3. So we can imagine an
x that is above 3, that is, say, right over here. Well, if we're trying to find
the slope between this x comma f of x and 3 comma
f of 3, we see that it gets this
exact same form. Your end point is f of x. So it's f of x minus
f of 3 is your change in the vertical axis. That's this distance
right over here. And we would divide
by your change in the horizontal axis,
which is your change in x. And that's going
to be x minus 3. So that's the exact expression
that we have up here when I picked this as an arbitrary x. And we see that that
slope, just by looking at the line between
those two intervals, seems to be negative 2. And the slope was the same
thing if we go the other side. If x was less than
3, then we also would have a slope
of negative 2. Either way, we have a
slope of negative 2. And that's important
because this limit is just the limit
as x approaches 3. So it can be as x approaches
3 from the positive direction or from the negative direction. But in either case, the slope,
as we get closer and closer to this point right over
here, is negative 2. Now let's think about what
they're asking us here. So we have 8, f of 8. So let's think. We have 8. This is 8 comma f of 8. So that's 8 comma f
of 8 right over there. And they have f of 8 plus h. So our temptation
might be to say, hey, 8 plus h is going to
be someplace out here. It's going to be
something larger than 8. But notice, they
have the limit as h approaches 0 from the
negative direction. So approaching 0 from
the negative direction means you're coming
to 0 from below. You're at negative 1,
negative 0.5, negative 0.1, negative 0.0001. So h is actually going
to be a negative number. So 8 plus h would actually
be-- we could just pick an arbitrary point. It could be something
like this right over here. So this might be the
value of 8 plus h. And this would be the
value of f of 8 plus h. So once again, they're
finding-- or this expression is the slope between
these two points. And then we are taking the
limit as h approaches 0 from the negative direction. So as h gets closer
and closer to 0, this down here moves further
and further to the right. And these points move closer
and closer and closer together. So this is really
just an expression of the slope of
the line, roughly-- and we see that it's constant. So what's the slope of the
line over this interval? Well, you can just eyeball
it and see, well, look. Every time x changes by 1,
our f of x changes by 1. So the slope of the
line there is 1. It would have been a
completely different thing if this said limit
as h approaches 0 from the positive direction. Then we would be looking
at points over here. And we would see that we would
slowly approach, essentially, a vertical slope, kind
of an infinite slope.