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## Calculus, all content (2017 edition)

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

Lesson 7: Using the formal definition of derivative

# Limit expression for the derivative of a linear function

Sal interprets a limit expression as the derivative of a linear function at a point, and evaluates it. Created by Sal Khan.

## Want to join the conversation?

• I don't even understand the title! (Actually the video's perfectly clear.) Can someone please say more about how tangent slope limits the value of secant slope, and how that's illustrated here? (It seems to me they're just talking about the slope of a line, one which could be tangent, secant or could exist on its own. Is it because of the g? because that makes it a function?) Thank you!! •   Envision a curve. Any curve -- it could represent a function, but let's just focus on the shape of it for now. Now, imagine picking two points on that curve (with a nice open space in-between), and draw a line that passes through them. That's your secant. Now, move one of your points a little closer to the other along the curve. Draw your secant line again. Keep doing this. As your distance between the two points gets to be hugely small (if that makes any sense), your secant line will start to change less and less.
Now, envision a line that is tangent to the curve, just touching it at one of your points. If your two points are close enough together, you couldn't tell the difference between your tangent and your secant! That's what it means to say that the tangent limits the secant -- you could never make zero distance between the points, but as you bring them closer together, the secant looks more like the tangent.
• At of the video, Sal says "an arbitrary point". What does that mean? • @ How He Knows That Slope is -4 by just seeing the Equation of Function? • he keeps talking about secant lines and lines tangent to the curve. where can i learn about these things? is there a video because i'm at a loss as to what he's referring to. thank you! • A secant line is a line that passes (locally) through the curve exactly twice (the secant line may or may not pass through the curve again somewhere else if you extend the line out, but we're only interested in the local area of the curve).

Now, let the two points that the secant line passes through get closer and closer. Take the limit as those to points converge into one point. That would be a tangent line. The slope of the tangent line will be equal to the instantaneous slope of the curve at that same point.
• at I dont get how did he draw that line...he says the line has a slope of -4 but i dont get how that works...pls help meunderstand this :/ • at the last statement sal writes
-4(x+1)/(x+1) at lim x approaches -1
how does he cancel out x+1 and x+1 is that simple cancelling as they are similar ?
how does x approaches -1 impact on the statement • At , shouldn't the simplified form just be 6 + h, rather than h squared? The answer would be the same though. • so where the limit approaches -1 do i just plug all values of -1 in for x?

Cheers :D
(1 vote) • I have a question, how do I compute the slope of a curve without knowing other values besides (9,8) ? There are no x values in the exercise that our teacher gave us  