In the last video we tried to figure out the slope of a point or the slope of a curve at a certain point. And the way we did, we said OK, well let’s find the slope between that point and then another point that’s not too far away from that point.
The following video will tell you about Calculating slope of tangent line using derivative definition Differential Calculus. You can use youtube to mp3 tool to convert the video to mp3.Notice: download videos and mp3 must be licensed by the version owner and used only for study and research, and not for commercial use and dissemination.
Calculating slope of tangent line using derivative definition Differential Calculus
And we got the slope of the secant line. And it looks all fancy, but this is just the y value of the point that’s not too far away, and this is just the y value point of the point in question, so this is just your change in y. And then you divide that by your change in x.
So in the example we did, h was the difference between our 2 x values. This distance was h. And that gave us the slope of that line. We said hey, what if we take the limit as this point right here gets closer and closer to this point.
If this point essentially almost becomes this point, then our slope is going to be the slope of our tangent line. And we define that as the derivative of our function. We said that’s equal to f prime of x. So let’s if we can apply this in this video to maybe make things a little bit more concrete in your head.
So let me do one. First I’ll do a particular case where I want to find the slope at exactly some point. So let me draw my axes again.
Let’s draw some axes right there. Let’s say I have the curve– this is the curve– y is equal to x squared. So this is my y-axis, this is my x-axis, and I want to know the slope at the point x is equal to 3.
When I say the slope you can imagine a tangent line here. You can imagine a tangent line that goes just like that, and it would just barely graze the curve at that point. But what is the slope of that tangent line? What is the slope of that tangent line which is the same as the slope of the curve right at that point.
So to do it, I’m actually going to do this exact technique that we did before, then we’ll generalize it so you don’t have to do it every time for a particular number. So let’s take some other point here. Let’s call this 3 plus delta x.
I’m changing the notation because in some books you’ll see an h, some books you’ll see a delta x, doesn’t hurt to be exposed to both of them. So this is 3 plus delta x. So first of all what is this point right here? This is a curve y is equal to x squared, so f of x is 3 squared– this is the point 9. This is the point 3,9 right here.
And what is this point right here? So if we were go all the way up here, what is that point? Well here our x is 3 plus delta x. It’s the same thing as this one right here, as x naught plus h. I could have called this 3 plus h just as easily. So it’s 3 plus delta x up there.
So what’s the y value going to be? Well whatever x value is, it’s on the curve, it’s going to be that squared. So it’s going to be the point 3 plus delta x squared. So let’s figure out the slope of this secant line. And let me zoom in a little bit, because that might help.
So if I zoom in on just this part of the curve, it might look like that. And then I have one point here, and then I have the other point is up here. That’s the secant line. Just like that.
This was the point over here, the point 3,9. And then this point up here is the point 3 plus delta x, so just some larger number than 3, and then it’s going to be that number squared. So it’s going to be 3 plus delta x squared. What is that? That’s going to be 9.
I’m just foiling this out, or you do the distribute property twice. a plus b squared is a squared plus 2 a b plus b squared, so it’s going to be 9 plus two times the product of these things. So plus 6 delta x, and then plus delta x squared.
That’s the coordinate of the second line. This looks complicated, but I just took this x value and I squared it, because it’s on the line y is equal to x squared. So the slope of the secant line is going to be the change in y divided by the change in x. So the change in y is just going to be this guy’s y value, which is 9 plus 6 delta x plus delta x squared.
That’s this guy’s y value, minus this guy’s y value. So minus 9. That’s your change in y. And you want to divide that by your change in x.
Well what is your change in x? This is actually going to be pretty convenient. This larger x value– we started with this point on the top, so we have to start with this point on the bottom. So it’s going to be 3 plus delta x.
And then what’s this x value? What is minus 3? That’s his x value. So what does this simplify to? The numerator– this 9 and that 9 cancel out, we get a 9 minus 9. And in the denominator what happens? This 3 and minus 3 cancel out.
So the change in x actually end up becoming this delta x, which makes sense, because this delta x is essentially how much more this guy is then that guy. So that should be the change in x, delta x. So the slope of my secant line has simplified to 6 times my change in x, plus my change in x squared, all of that over my change in x. And now we can simplify this even more.
Let’s divide the numerator and the denominator by our change in x. And I’ll switch colors just to ease the monotony. So my slope of my tangent of my secant line– the one that goes through both of these– is going to be equal if you divide the numerator and denominator this becomes 6.
I’m just dividing numerator and denominator by delta x plus six plus delta x. So that is the slope of this secant line So slope is equal to 6 plus delta x. That’s this one right here.
That’s this reddish line that I’ve drawn right there. So this number right here, if the delta x was one, if these were the points 3 and 4, then my slope would be 6 plus 1, because I’m picking a point 4 where the delta x here would have to be 1. So the slope would be 7. So we have a general formula for no matter what my delta x is, I can find the slow between 3 and 3 plus delta x.
Between those two points. Now we wanted to find the slope at exactly that point right there. So let’s see what happens when delta x get smaller and smaller. This is what delta x is right now.
It’s this distance. But if delta x got a little bit smaller, then the secant line would look like that. Got even smaller, the secant line would look like that, it gets even smaller. Then we’re getting pretty close to the slope of the tangent line.
The tangent line is this thing right here that I want to find the slope of. Let’s find a limit as our delta x approaches 0. So the limit as delta x approaches 0 of our slope of the secant line of 6 plus delta x is equal to what? This is pretty straightforward.
You can just set this equal to 0 and it’s equal to 6. So the slope of our tangent line at the point x is equal to 3 right there is equal to 6. And another way we could write this if we wrote that f of x is equal to x squared.
We now know that the derivative or the slope of the tangent line of this function at the point 3– I just only evaluated it at the point 3 right there– that that is equal to 6. I haven’t yet come up with a general formula for the slope of this line at any point, and I’m going to do that in the next video. .
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