Uploaded by TheIntegralCALC on 12.02.2010

Transcript:

Hi! Welcome back!

Another product rule problem,

this time with three terms:

so, f(x) equals

x times

(2x minus 1)

times (2x plus 1).

Okay, so, this is...

The... The primary point of this is just to demonstrate

how to apply product rule when you've got three terms.

The first term is x,

the second term is (2x plus 1),

the third term is (2x plus 1).

I think I said plus here, I meant minus if I did.

So, same as before,

you take the derivative of one term,

you keep the other two constant,

then, you take the derivative of another term,

keeping the other two constant,

and so on.

You could do it with an infinite number of terms.

So,

the derivative, we're going to go and say f'(x) equals

and we're going to go in order.

So,

you take the derivative of the first term, x.

The derivative of x is just 1,

so,

that's implied. We keep the other two the same,

so (2x minus 1) times (2x plus 1).

And you know what?

I'm going to need to

write this smaller so you guys can see it.

I just don't want to have a... want to run out a room.

(2x minus 1) times (2x plus 1).

Okay...

Then,

adding (always adding), so plus,

we are going to go ahead and keep this one the same, x,

take the derivative of the second one.

The derivative of (2x minus 1) is just 2,

so we can go ahead and write that in front,

2x,

2 comes from here, x is a...

the x, remaining the same right here,

and then, we multiply

by (2x plus 1),

because this remains constant.

And then finally, we're going to add

the third term.

We're taking the derivative of the third one this time,

keeping these two the same.

So, the derivative of this term

is just 2,

so we're going to go ahead and move that out from like we did last time,

2, and then we keep these two,

there's the x and then times

(2x minus 1).

(2x minus 1)

So, that is...

that is fundamentally

what product rule is.

If you had to...

If you were asked a certain test

and you would have to find the derivative,

and it wasn't necessarily demonstrate product rule,

you could expand all of these,

multiply everything out,

and then take the derivative term by term.

If you're asked to show product rule,

you're going to have to do it this way.

And then,

I'm guessing that almost certainly you're going to have to simplify this,

so let's just go ahead and do that really quickly.

2x times 2x,

4x^2,

2x times 1 is going to be plus 2x,

these two,

minus 2x,

and then -1 times 1 is minus 1,

and then, plus.

Now, we're on to this part,

and I'm going to go ahead and multiply that out.

2x times 2x, 4x^2,

2x times 1, plus 2x,

and then, plus.

I'm going to go ahead and multiply these out as well.

2x times 2x is 4x^2,

and 2x minus...

or times -1 here

minus 2x.

So then, we can simplify.

We've got 4...

8...

12...

12x^2,

and then

2x minus 2x is 0,

plus 2x is 2x,

minus 2x is 0,

and then -1,

and that's everything,

so, 12x^2 minus 1,

And, that is our final answer,

as the derivative of this problem

showing you

how to apply product rule to this function.

Thanks!

Another product rule problem,

this time with three terms:

so, f(x) equals

x times

(2x minus 1)

times (2x plus 1).

Okay, so, this is...

The... The primary point of this is just to demonstrate

how to apply product rule when you've got three terms.

The first term is x,

the second term is (2x plus 1),

the third term is (2x plus 1).

I think I said plus here, I meant minus if I did.

So, same as before,

you take the derivative of one term,

you keep the other two constant,

then, you take the derivative of another term,

keeping the other two constant,

and so on.

You could do it with an infinite number of terms.

So,

the derivative, we're going to go and say f'(x) equals

and we're going to go in order.

So,

you take the derivative of the first term, x.

The derivative of x is just 1,

so,

that's implied. We keep the other two the same,

so (2x minus 1) times (2x plus 1).

And you know what?

I'm going to need to

write this smaller so you guys can see it.

I just don't want to have a... want to run out a room.

(2x minus 1) times (2x plus 1).

Okay...

Then,

adding (always adding), so plus,

we are going to go ahead and keep this one the same, x,

take the derivative of the second one.

The derivative of (2x minus 1) is just 2,

so we can go ahead and write that in front,

2x,

2 comes from here, x is a...

the x, remaining the same right here,

and then, we multiply

by (2x plus 1),

because this remains constant.

And then finally, we're going to add

the third term.

We're taking the derivative of the third one this time,

keeping these two the same.

So, the derivative of this term

is just 2,

so we're going to go ahead and move that out from like we did last time,

2, and then we keep these two,

there's the x and then times

(2x minus 1).

(2x minus 1)

So, that is...

that is fundamentally

what product rule is.

If you had to...

If you were asked a certain test

and you would have to find the derivative,

and it wasn't necessarily demonstrate product rule,

you could expand all of these,

multiply everything out,

and then take the derivative term by term.

If you're asked to show product rule,

you're going to have to do it this way.

And then,

I'm guessing that almost certainly you're going to have to simplify this,

so let's just go ahead and do that really quickly.

2x times 2x,

4x^2,

2x times 1 is going to be plus 2x,

these two,

minus 2x,

and then -1 times 1 is minus 1,

and then, plus.

Now, we're on to this part,

and I'm going to go ahead and multiply that out.

2x times 2x, 4x^2,

2x times 1, plus 2x,

and then, plus.

I'm going to go ahead and multiply these out as well.

2x times 2x is 4x^2,

and 2x minus...

or times -1 here

minus 2x.

So then, we can simplify.

We've got 4...

8...

12...

12x^2,

and then

2x minus 2x is 0,

plus 2x is 2x,

minus 2x is 0,

and then -1,

and that's everything,

so, 12x^2 minus 1,

And, that is our final answer,

as the derivative of this problem

showing you

how to apply product rule to this function.

Thanks!