Uploaded by videosbyjulieharland on 13.10.2009

Transcript:

>> This is the third video on the derivative,

and what we are working with is the function F

of X equals X squared minus 3.

That was a parabola opening upward,

and we were trying the find the slope of the tangent line

at the point 1, negative 2, and we were using the definition

for the slope of a tangent line, which is M equals the limit

as H approaches 0 of F of C plus H minus F of C

over H. Remember this formula came from the formula for slope,

and C is the X value in the ordered pair.

So we went ahead and went through this process

and we ended up with a slope of 2, because the limit

as H approaches 0 was 2.

So remember, our function was F of X equals X squared minus 3,

the parabola opening upward, and using the definition

for the slope of the tangent line, it was the formula

for the limit as H approaches 0, et cetera.

We found that at the point 1, negative 2,

the slope of that tangent line, was positive 2.

So if we wanted to, we could use the point/slope formula

to find the equation of the tangent line.

So let's just do that.

Remember the point/slope formula is Y minus Y1 equals M times X

minus X1, and if you did want to use that formula,

then we would have Y minus Y1 is negative 2.

The slope is 2, so you put that in for M,

and X1 is positive 1 -- that comes from the ordered pair --

so we have Y plus 2 equals 2X minus 2, or Y equals 2X minus 4.

So, now we're going to go back up to our original picture

of this graph to see if it makes sense that it looks

like that would be the equation of the tangent line.

This is where we started.

We considered the function F of X equals X squared minus 3

and then we were just estimating where the tangent line might be

at the point 1, negative 2.

Remember, we said the tangent line down here, the red line

that looked like the slope at 0, negative 3,

the slope of that tangent line was 0.

And then we were trying to find out what the slope

of this green line is.

So we're trying to draw a tangent line.

And so the question, does it look

like it is what we just came up with.

We came up with Y equals 2X minus 4 for that tangent line,

and actually, that's what it looks like.

It looks to me like it did go through a Y intercept

of 4 and a slope of 2.

That was an approximation;

we could have been off a little bit, but from my picture,

you sort of get the idea.

So the slope of the tangent line was 2, and the actual equation,

if we want to do the slope/point formula --

[ Pause ]

Let's look back at the definition of tangent line

with slope M. We took this limit as H approaches 0 of F

of C plus H minus F of C over H, and if it existed,

it equaled M. Then we say that the line passing

through that ordered pair, C, F of C, with that slope M,

whatever you get when you apply the limit, that is going

to be the slope of the tangent line to the graph.

Now what's interesting is, we're going to now go

on to the formal definition of a derivative that's going

to look exactly like that.

The difference is, instead of you seeing this letter C,

we're going to have a X. So C was a particular value of,

you know, in an ordered pair.

So here's the formula you've been waiting for,

the derivative of a function.

This is the definition.

F prime of X-- that's the way we read this --

F prime of X is equal to the limit as H approaches 0

of this difference quotient.

Remember this had to do with the formula for slope.

F of X plus H minus F of X --

that was the difference of the Y values,

and at the bottom we originally would have X plus H minus X,

so we end up with just H in the denominator.

So that is the derivative.

So if you have any ordered pair X, Y, you can plug

in the X value into this F prime of X, take the limit,

and you'll be able to figure out the slope of the tangent line

to the curve at that point on the curve.

So let's do it for the problem we've been working on.

We've been working on F of X equals X squared minus 3.

And just notice, we're going to have to take a limit

as H approaches 0, and in that limit we're going to also have

to compute what F of X plus H is.

So, to make it easier, let's just compute F of X plus H.

So we're going to plug in X plus H for X, and we're going

to square the binomial.

Remember, it's X plus H times X plus H,

so that's X squared plus 2XH plus H squared minus 3.

So we know F of X, we know F of X plus H, and now we're going

to plug that into the formula to get the derivative.

So we've got F prime of X is equal to -- okay.

All right, this says we're going to plug in -- oops, I forgot.

It's got to be the limit.

Don't forget to write the limit as H approaches 0

and in the numerator, we're going to have F of X plus H,

which is this X squared plus 2XH plus H squared minus 3 minus F

of X - up here is F of X which is X squared minus 3 --

and all of that is going to be over H, all right.

And we have to keep going.

Let's write that as the same thing.

Remember you have to keep writing the limit

as H approaches 0, and let's simplify the numerator.

So X squared plus 2XH plus H squared minus 3 minus X squared

plus 3 when I distribute the minus sign.

And in the numerator, let's see,

the X squared minus X squared cancels,

the negative 3 and plus 3.

I end up with just 2XH,

XH squared over H. This should look familiar.

It's almost like what we just did a few minutes ago.

I'm going to go down the page, still.

Equals the limit as H approaches of 0 of C. I'm just going

to show the factoring here.

H times 2X plus H all over H. See how the Hs cancel?

So I really end up with just 2X plus H. Which if you plug in --

now you can plug in 0 for H-- that's the first time,

you don't have to write the limit.

So you have 2X plus 0 and the answer is 2X.

See the difference?

This time I got 2X.

All right, so what does this mean?

We had this function F of X equals X squared minus 3,

and we now just came out with F prime of X was equal to 2X.

So what that means is that any point on the curve,

if I want to find the slope, I can plug in the X value

and it'll tell me slope.

Kind of interesting, don't you think?

So, let's look back at our graph.

Here is our graph, here is our original function,

F of X equals X squared minus 3.

What it says is F prime of X equals 2X.

So at any ordered pair, we should be able to tell the slope

of the line if we know the ordered pair.

So here's an ordered pair.

This our ordered pair, 1, negative 2, and what happens?

When we plug in 1 for X and the F prime of X you get 2,

so that's why the slope of this line was 2.

How about this ordered pair down here.

It was 0, negative 3, and here the X value is 0.

If you plug in F prime of 0, you are going to get a slope of 0,

which is what it'll look like.

Now, let's say we were going to do this point over here.

This was negative 3, 6,

and I want to know the slope of this line.

That looks pretty steep, and it also looks

like a negative slope.

If you plug in negative 3 for X, you're going

to get a slope of negative 6.

So this tangent line, wherever it is,

it's going to be very steep -- hard to even see it.

It has a slope of negative 6.

And that's all about the interesting part

of finding the derivative.

and what we are working with is the function F

of X equals X squared minus 3.

That was a parabola opening upward,

and we were trying the find the slope of the tangent line

at the point 1, negative 2, and we were using the definition

for the slope of a tangent line, which is M equals the limit

as H approaches 0 of F of C plus H minus F of C

over H. Remember this formula came from the formula for slope,

and C is the X value in the ordered pair.

So we went ahead and went through this process

and we ended up with a slope of 2, because the limit

as H approaches 0 was 2.

So remember, our function was F of X equals X squared minus 3,

the parabola opening upward, and using the definition

for the slope of the tangent line, it was the formula

for the limit as H approaches 0, et cetera.

We found that at the point 1, negative 2,

the slope of that tangent line, was positive 2.

So if we wanted to, we could use the point/slope formula

to find the equation of the tangent line.

So let's just do that.

Remember the point/slope formula is Y minus Y1 equals M times X

minus X1, and if you did want to use that formula,

then we would have Y minus Y1 is negative 2.

The slope is 2, so you put that in for M,

and X1 is positive 1 -- that comes from the ordered pair --

so we have Y plus 2 equals 2X minus 2, or Y equals 2X minus 4.

So, now we're going to go back up to our original picture

of this graph to see if it makes sense that it looks

like that would be the equation of the tangent line.

This is where we started.

We considered the function F of X equals X squared minus 3

and then we were just estimating where the tangent line might be

at the point 1, negative 2.

Remember, we said the tangent line down here, the red line

that looked like the slope at 0, negative 3,

the slope of that tangent line was 0.

And then we were trying to find out what the slope

of this green line is.

So we're trying to draw a tangent line.

And so the question, does it look

like it is what we just came up with.

We came up with Y equals 2X minus 4 for that tangent line,

and actually, that's what it looks like.

It looks to me like it did go through a Y intercept

of 4 and a slope of 2.

That was an approximation;

we could have been off a little bit, but from my picture,

you sort of get the idea.

So the slope of the tangent line was 2, and the actual equation,

if we want to do the slope/point formula --

[ Pause ]

Let's look back at the definition of tangent line

with slope M. We took this limit as H approaches 0 of F

of C plus H minus F of C over H, and if it existed,

it equaled M. Then we say that the line passing

through that ordered pair, C, F of C, with that slope M,

whatever you get when you apply the limit, that is going

to be the slope of the tangent line to the graph.

Now what's interesting is, we're going to now go

on to the formal definition of a derivative that's going

to look exactly like that.

The difference is, instead of you seeing this letter C,

we're going to have a X. So C was a particular value of,

you know, in an ordered pair.

So here's the formula you've been waiting for,

the derivative of a function.

This is the definition.

F prime of X-- that's the way we read this --

F prime of X is equal to the limit as H approaches 0

of this difference quotient.

Remember this had to do with the formula for slope.

F of X plus H minus F of X --

that was the difference of the Y values,

and at the bottom we originally would have X plus H minus X,

so we end up with just H in the denominator.

So that is the derivative.

So if you have any ordered pair X, Y, you can plug

in the X value into this F prime of X, take the limit,

and you'll be able to figure out the slope of the tangent line

to the curve at that point on the curve.

So let's do it for the problem we've been working on.

We've been working on F of X equals X squared minus 3.

And just notice, we're going to have to take a limit

as H approaches 0, and in that limit we're going to also have

to compute what F of X plus H is.

So, to make it easier, let's just compute F of X plus H.

So we're going to plug in X plus H for X, and we're going

to square the binomial.

Remember, it's X plus H times X plus H,

so that's X squared plus 2XH plus H squared minus 3.

So we know F of X, we know F of X plus H, and now we're going

to plug that into the formula to get the derivative.

So we've got F prime of X is equal to -- okay.

All right, this says we're going to plug in -- oops, I forgot.

It's got to be the limit.

Don't forget to write the limit as H approaches 0

and in the numerator, we're going to have F of X plus H,

which is this X squared plus 2XH plus H squared minus 3 minus F

of X - up here is F of X which is X squared minus 3 --

and all of that is going to be over H, all right.

And we have to keep going.

Let's write that as the same thing.

Remember you have to keep writing the limit

as H approaches 0, and let's simplify the numerator.

So X squared plus 2XH plus H squared minus 3 minus X squared

plus 3 when I distribute the minus sign.

And in the numerator, let's see,

the X squared minus X squared cancels,

the negative 3 and plus 3.

I end up with just 2XH,

XH squared over H. This should look familiar.

It's almost like what we just did a few minutes ago.

I'm going to go down the page, still.

Equals the limit as H approaches of 0 of C. I'm just going

to show the factoring here.

H times 2X plus H all over H. See how the Hs cancel?

So I really end up with just 2X plus H. Which if you plug in --

now you can plug in 0 for H-- that's the first time,

you don't have to write the limit.

So you have 2X plus 0 and the answer is 2X.

See the difference?

This time I got 2X.

All right, so what does this mean?

We had this function F of X equals X squared minus 3,

and we now just came out with F prime of X was equal to 2X.

So what that means is that any point on the curve,

if I want to find the slope, I can plug in the X value

and it'll tell me slope.

Kind of interesting, don't you think?

So, let's look back at our graph.

Here is our graph, here is our original function,

F of X equals X squared minus 3.

What it says is F prime of X equals 2X.

So at any ordered pair, we should be able to tell the slope

of the line if we know the ordered pair.

So here's an ordered pair.

This our ordered pair, 1, negative 2, and what happens?

When we plug in 1 for X and the F prime of X you get 2,

so that's why the slope of this line was 2.

How about this ordered pair down here.

It was 0, negative 3, and here the X value is 0.

If you plug in F prime of 0, you are going to get a slope of 0,

which is what it'll look like.

Now, let's say we were going to do this point over here.

This was negative 3, 6,

and I want to know the slope of this line.

That looks pretty steep, and it also looks

like a negative slope.

If you plug in negative 3 for X, you're going

to get a slope of negative 6.

So this tangent line, wherever it is,

it's going to be very steep -- hard to even see it.

It has a slope of negative 6.

And that's all about the interesting part

of finding the derivative.