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### Course: Calculus, all content (2017 edition)>Unit 2

Lesson 7: Using the formal definition of derivative

# 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.

## Want to join the conversation?

• How come the slope is -2? I didn't quite get that.
• 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
• At , if h was approaching 0 from the positive direction, would it be undefined?
• 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.
• See into 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
• 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
• 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?
• 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 as

f(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.
• I don't understand how Sal found the value of h around .. Since we are looking at value as h approaches 0, how come our h is around 7 on the x-axis?
• 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.
• 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.
• 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.
• 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?
• 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, at ), Sal said the derivative of f was f'(x) = lim (h → 0) [f(x₀ + h) - f(x₀) / h]. The function is equivalent in this example lim [f(8 + h) - f(8) / h], then why h → 8− instead of h → 0?
• 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.
• At , 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.
• at Sal 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.