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In the last video we tried to figure out the slope of a

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point or the slope of a curve at a certain point.

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And the way we did, we said OK, well let's find the slope

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between that point and then another point that's not too

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far away from that point.

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And we got the slope of the secant

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line.

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And it looks all fancy, but this is just the y value of the

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point that's not too far away, and this is just the y value

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point of the point in question, so this is

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just your change in y.

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And then you divide that by your change in x.

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So in the example we did, h was the difference

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between our 2 x values.

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This distance was h.

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And that gave us the slope of that line.

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We said hey, what if we take the limit as this point

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right here gets closer and closer to this point.

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If this point essentially almost becomes this point, then

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our slope is going to be the slope of our tangent line.

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And we define that as the derivative of our function.

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We said that's equal to f prime of x.

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So let's if we can apply this in this video to maybe make

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things a little bit more concrete in your head.

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So let me do one.

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First I'll do a particular case where I want to find the

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slope at exactly some point.

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So let me draw my axes again.

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Let's draw some axes right there.

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Let's say I have the curve-- this is the curve-- y

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is equal to x squared.

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So this is my y-axis, this is my x-axis, and I want to know

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the slope at the point x is equal to 3.

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When I say the slope you can imagine a tangent line here.

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You can imagine a tangent line that goes just like that, and

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it would just barely graze the curve at that point.

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But what is the slope of that tangent line?

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What is the slope of that tangent line which is the same

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as the slope of the curve right at that point.

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So to do it, I'm actually going to do this exact technique that

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we did before, then we'll generalize it so you don't have

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to do it every time for a particular number.

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So let's take some other point here.

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Let's call this 3 plus delta x.

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I'm changing the notation because in some books you'll

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see an h, some books you'll see a delta x, doesn't hurt to

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be exposed to both of them.

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So this is 3 plus delta x.

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So first of all what is this point right here?

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This is a curve y is equal to x squared, so f of x is 3

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squared-- this is the point 9.

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This is the point 3,9 right here.

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And what is this point right here?

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So if we were go all the way up here, what is that point?

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Well here our x is 3 plus delta x.

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It's the same thing as this one right here,

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as x [UNINTELLIGIBLE]

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plus h.

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I could have called this 3 plus h just as easily.

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So it's 3 plus delta x up there.

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So what's the y value going to be?

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Well whatever x value is, it's on the curve, it's going

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to be that squared.

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So it's going to be the point 3 plus delta x squared.

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So let's figure out the slope of this secant

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line.

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And let me zoom in a little bit, because that might help.

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So if I zoom in on just this part of the curve,

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it might look like that.

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And then I have one point here, and then I have the

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other point is up here.

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That's the secant

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line.

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Just like that.

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This was the point over here, the point 3,9.

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And then this point up here is the point 3 plus delta x, so

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just some larger number than 3, and then it's going to

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be that number squared.

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So it's going to be 3 plus delta x squared.

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What is that?

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That's going to be 9.

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I'm just foiling this out, or you do the distribute

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property twice.

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a plus b squared is a squared plus 2 a b plus b squared, so

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it's going to be 9 plus two times the product

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of these things.

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So plus 6 delta x, and then plus delta x squared.

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That's the coordinate of the second line.

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This looks complicated, but I just took this x value and I

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squared it, because it's on the line y is equal to x squared.

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So the slope of the secant

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line is going to be the change in y divided

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by the change in x.

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So the change in y is just going to be this guy's y value,

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which is 9 plus 6 delta x plus delta x squared.

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That's this guy's y value, minus this guy's y value.

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So minus 9.

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That's your change in y.

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And you want to divide that by your change in x.

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Well what is your change in x?

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This is actually going to be pretty convenient.

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This larger x value-- we started with this point on the

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top, so we have to start with this point on the bottom.

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So it's going to be 3 plus delta x.

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And then what's this x value?

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What is minus 3?

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That's his x value.

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So what does this simplify to?

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The numerator-- this 9 and that 9 cancel out,

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we get a 9 minus 9.

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And in the denominator what happens?

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This 3 and minus 3 cancel out.

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So the change in x actually end up becoming this delta x, which

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makes sense, because this delta x is essentially how much more

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this guy is then that guy.

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So that should be the change in x, delta x.

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So the slope of my secant

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line has simplified to 6 times my change in x, plus my change

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in x squared, all of that over my change in x.

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And now we can simplify this even more.

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Let's divide the numerator and the denominator

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by our change in x.

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And I'll switch colors just to ease the monotony.

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So my slope of my tangent of my secant

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line-- the one that goes through both of these-- is

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going to be equal if you divide the numerator and

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denominator this becomes 6.

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I'm just dividing numerator and denominator by delta

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x plus six plus delta x.

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So that is the slope of this secant

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line So slope is equal to 6 plus delta x.

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That's this one right here.

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That's this reddish line that I've drawn right there.

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So this number right here, if the delta x was [? once ?], if

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these were the points 3 and 4, then my slope would be 6 plus

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1, because I'm picking a point 4 where the delta x here

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would have to be 1.

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So the slope would be 7.

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So we have a general formula for no matter what my delta

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x is, I can find the slow between 3 and 3 plus delta x.

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Between those two points.

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Now we wanted to find the slope at exactly that

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point right there.

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So let's see what happens when delta x get

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smaller and smaller.

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This is what delta x is right now.

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It's this distance.

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But if delta x got a little bit smaller, then the secant

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line would look like that.

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Got even smaller, the secant

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line would look like that, it gets even smaller.

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Then we're getting pretty close to the slope

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of the tangent line.

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The tangent line is this thing right here that I

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want to find the slope of.

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Let's find a limit as our delta x approaches 0.

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So the limit as delta x approaches 0 of our

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slope of the secant

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line of 6 plus delta x is equal to what?

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This is pretty straightforward.

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You can just set this equal to 0 and it's equal to 6.

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So the slope of our tangent line at the point x is equal to

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3 right there is equal to 6.

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And another way we could write this if we wrote that f of

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x is equal to x squared.

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We now know that the derivative or the slope of the tangent

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line of this function at the point 3-- I just only evaluated

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it at the point 3 right there-- that that is equal to 6.

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I haven't yet come up with a general formula for the slope

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of this line at any point, and I'm going to do that

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in the next video.

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