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

Lesson 7: Using the formal definition of derivative- The derivative of x² at x=3 using the formal definition
- The derivative of x² at any point using the formal definition
- Limit expression for the derivative of a linear function
- Limit expression for the derivative of cos(x) at a minimum point
- Limit expression for the derivative of function (graphical)
- Tangent lines and rates of change

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# 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!!(16 votes)
- 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.(52 votes)

- At1:27of the video, Sal says "an arbitrary point". What does that mean?(10 votes)
- That means he just picked a point at random.(16 votes)

- @0:51How He Knows That Slope is -4 by just seeing the Equation of Function?(6 votes)
- The equation of a line is
`y = mx + b`

. m is the slope of the line. In the video the equation is`g(x) = -4x + 7`

, so -4 is the slope of the line.(18 votes)

- 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!(2 votes)
- 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.(6 votes)

- at0:40I 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 :/(3 votes)
- The slope of a line can be found from its coefficient before x in its equasion.

For example, the line y = -4x + 7 has a slope of -4 because it is the coefficient before x. It also represents the tangent of the angle created by the line and the positive direction of the X axis.

However, most curves don't behave like that.(2 votes)

- 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(3 votes)- u can cancel out (x+1) and (x+1) because you are not putting x= -1 ....you are putting x very close to -1 .(2 votes)

- At4:04, shouldn't the simplified form just be 6 + h, rather than h squared? The answer would be the same though.(3 votes)
- Yeah however you proceed with the problem, you are supposed to get the same answer, Anne.. But i don't understand what u're mentioning to in this problem. I don't see any h or h squared in the problem... Can you be a bit more clear please?(2 votes)

- so where the limit approaches -1 do i just plug all values of -1 in for x?

Cheers :D(1 vote)- Yes, but only after manipulating the formula is such a way that plugging in the limit value doesn't give us 0/0. We start with a formula that gives us this meaningless result if we simply plug in the limit value, and we have to work with it to create a formula where that's no longer the case.(5 votes)

- 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(2 votes)
- Maybe I've been awake too long but I'm having an aneurism on this one but at4:40, we're going from 4x + 7 - 11 / x + 1 to -4 ( x + 1 ) / x + 1. How do we get from substituting 4x + -4 in the numerator to -4(x + 1). Is it that "x" in the numerator is substituted by the denominator " x + 1 "?(1 vote)
- You're missing a minus sign at the beginning of the numerator, which was -4x + 7 - 11. Combining the two constants we get -4x - 4 in the numerator, and then factoring out -4 we get -4(x + 1).(3 votes)

## Video transcript

Let g of x equal
negative 4x plus 7. What is the value of the limit
as x approaches negative 1 of all of this? So before we think about this,
let's just visualize the line. And then we can think about
what they're asking here. So let me draw some axes here. So this is my vertical axis
and this is my horizontal axis. And let's say this is my x-axis. We'll label that the x-axis. I'll graph g of
x. g of x is going to have a positive-- I guess
you would say y-intercept. or vertical axis intercept. It's going to have a
slope of negative 4, so it's going to look
something like this. Let me draw my best. So it's going to look
something like that. And we already know the slope
here is going to be negative 4. We get that right from
this slope intercept form of the equation, slope
is equal to negative 4. And they ask us,
what is the limit as x approaches negative 1
of all of this kind of stuff? So let's plot the
point negative 1. So when x is
negative 1, so that's this point right over here. And this point right over here
would be the point negative 1, g of negative 1. Let me label everything else. So I could call this my y-axis. I could call this graph. This is the graph of
y is equal to g of x. So what they're
doing right over here is they're finding the slope
between an arbitrary point x, g of x, and this point
right over here. So let's do that. So let's take another x. So let's say this is x. This would be the x, g of x. And this expression
right over here, notice it is your change
in the vertical axis. That would be your g of x. Let me make it this way. So this would be your
change in the vertical axis. That would be g of x
minus g of negative 1. And then that's over-- actually,
let me write it this way so you can keep track of the
colors-- minus g of negative 1, all of that over your change
in the horizontal axis. Well, your change in
the horizontal axis is this distance, which is the
same thing as this distance. Notice your change in vertical
over change in horizontal, change in vertical over
change in horizontal, reviewing the green
point as the endpoint. So it's going to be
x minus negative 1. And this is the exact
same expression. These are the exact
same expression. You can simplify them,
minus negative 1, and this becomes plus. This could become a plus 1. But these are the
exact same expression. So this is the
expression, really, for the slope between
negative 1 and g of negative 1 and an arbitrary x. Well, we already know that
no matter what x you pick, the slope between x, g of , and
this point right over here is going to be constant. It's going to be the
slope of the line. It's going to be
equal to negative 4. This thing is going to
be equal to negative 4. It's going to be
equal to negative 4. Doesn't matter how close
x gets, and weather x comes from the right or
whether x comes from the left. So this thing, taking
the limit of this, this just gets
you to negative 4. It's really just the
slope of the line. So even if you were to take
the limit as x approaches negative 1, as x gets
closer and closer and closer to negative 1, well
then, these points are just going to get closer
and closer and closer. But every time you
calculate the slope, it's just going to be the
slope of the line, which is negative 4. Now, you could also
do this algebraically. And let's try to do
it algebraically. So let's actually
just take the limit as x approaches
negative 1 of g of x. Well, they already
told us what g of x is. It is negative 4x plus
7, minus g of negative 1. So that's minus, what
is g of negative 1? Negative 1 times
4 is positive 4. Positive 4 plus 7 is 11. All of that over x plus 1,
all of that over x plus 1. And that's really
x minus negative 1, is you want to think
of it that way. But I'll just write x
plus 1 this way here. So this is going to
be equal to the limit as x approaches negative 1 of,
in our numerator-- let's see. 7 minus 11 is negative 4. We can factor out a negative 4. It's a negative 4 times x plus
1, all of that over x plus 1. And then since we're just
trying to find the limit as x approaches negative 1, so
we can cancel those out. And this is going to be
non-zero for any x value other than negative 1. And so this is going to
be equal to negative 4. So either way, we
get negative 4. But if you just realize,
hey, this is a line. It's going to have
a constant slope. This is just the slope
of between some arbitrary point on the line and the
point negative 1, 11, really. Negative 1, 11, you say, well
that's just going to the same as the slope of a line. It's negative 4.