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# The derivative of x² at x=3 using the formal definition

AP.CALC:
CHA‑2 (EU)
,
CHA‑2.B (LO)
,
CHA‑2.B.2 (EK)
,
CHA‑2.B.3 (EK)
,
CHA‑2.B.4 (EK)

## Video transcript

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 it we said okay well let's find the slope between that point and then another point that's not too far away from that point and we got the slope of the secant line 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 of your 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 two x-values this distance was H and that gave us the point that the slope of that line we said hey what if we take the limit as this as this point right here gets closer and closer at 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 see if we can apply this in this video to 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's say let me draw my axes again let's draw some axes right there and let's say I have the curve 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 and when I say the slope you can imagine a tangent line here do it in a light yellow you can imagine a tangent line that goes just like that and it would just 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 going to actually do this exact technique that we did before and 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 take some other point up out here let's call this 3 plus Delta X I'll 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 what is this point right here this is a curve Y is equal to x squared so f of X f of X is 3 squared this is the point 9 this is the point 3 comma 9 right here let me draw that out well I'll draw it out in a second and what is this point right here so if we were to go all the way up here what is that point well here our X is now our X is 3 plus Delta X it's the same thing as this one right here is X not 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 this is going to be the point 3 plus Delta X 3 plus Delta x squared so let's figure out the slope of this secant line let's figure out the slope of the secant line 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 you see if I can draw that that's the secant line secant secant line just like that this was the point over here this at this point is the point 3 comma 9 and then this point up here is the point 3 plus Delta X so it's 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 boiling this out or you do the distributive property twice if you have a plus B squared is a squared plus 2 a B plus B squared so it's going to be 9 plus 2 times the product of these things so plus 6 Delta X and then plus Delta x squared plus Delta X 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 this line 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 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 guys Y value I'll do it in green so minus 9 minus 9 that's your change in Y and you want to divide that by your change in X 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 3 plus Delta X and then what's this x-value well it's minus 3 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 that's this Delta X is essentially how much more this guy is than that guy so that should be the change in X Delta X so the slope of my secant line so my slope has simplified to 6 times my change in X plus my change in x squared all of that over my change in X 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 will switch colors just to ease the monotony so of 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 denominators becomes six right I'm just dividing numerator denominator by Delta X plus 6 plus Delta X so that is the slope of the secant line so slope slope is equal to six plus Delta X that's this one right here that's this reddish line that I've drawn right there so if this number right here if the Delta X was one this was the point if these were the points three and four then my slope would be six plus one because I'm picking a point four where the Delta X here would have to be one so the slope would be seven because so we have a general formula for no matter what my Delta X is I can find the slope between three and three plus Delta X between those two points now we wanted to find the point the slope at exactly that point right there so let's see what happens when we take when Delta X gets 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 it would start the secant line would look like that we've 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 so let's find the limit as our Delta X approaches zero so the limit as Delta X approaches zero of our slope of the secant line of six plus Delta X is equal to what this is pretty straightforward you can just set this equal to zero and it is equal to six so the slope of our tangent line the slope of our tangent line at the point X is equal to three right there is equal to six 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 three I just only evaluated it at the point three right there that that is equal to six 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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