Calculating slope of tangent line using derivative definition Calculus-Derivative: Finding the slope (or derivative) of a curve at a particular point.
Calculating slope of tangent line using derivative definition
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- In the last video we tried to figure out 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.
- 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 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 the point 3 plus delta x squared.
- So let's figure out the slope of this secant
- 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
- 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 [? once ?], 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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