PROFESSOR: OK.
This video is about
derivatives.
Two rules for finding
new derivatives.
If we know the derivative
of a function f--
say we've found that--
and we know the derivative
of g--
we've found that--
then there are functions that
we can build out of those.
And two important and
straightforward ones are the
product, f of x times g of x,
and the quotient, the ratio f
of x over g of x.
So those are the two
rules we need.
If we know df dx and we know dg
dx, what's the derivative
of the product?
Well, it is not df
dx times dg dx.
And let me reduce the suspense
by writing down what it is.
It's the first one times the
derivative of the second, we
know that, plus another term,
the second one times the
derivative of the first. OK.
So that's the rule to learn.
Two terms, you see
the pattern.
And maybe I ought to use it,
give you some examples, see
what it's good for, and
also some idea of
where it comes from.
And then go on to the
quotient rule,
which is a little messier.
OK.
So let me just start by using
this in some examples.
Right underneath, here.
OK.
So let me take, as a first
example, f of x equals x
squared and g of x equals x.
So then what is p of x?
It's x squared times x.
I'm multiplying the functions.
So I've got x cubed, and I want
to know its derivative.
And I know the derivatives
of these guys.
OK, so what does the
rule tell me?
It tells me that the derivative
of p, dp dx--
so p is x cubed.
So I'm looking for the
derivative of x cubed.
And if you know that, it's OK.
Let's just see it
come out here.
So the derivative of x cubed,
by my formula there, is the
first one, x squared, times the
derivative of the second,
which is 1, plus the second one,
x, times the derivative
of the first, which is 2x.
So what do we get?
x squared, two more x
squared, 3x squared.
The derivative of x cubed
is 3x squared.
x cubed goes up faster than
x squared, and this
is a steeper slope.
Oh, let's do x to the fourth.
So x to the fourth-- now I'll
take f to be x cubed, times x.
Because x cubed, I just found.
x, its derivative is 1, so I
can do the derivative of x
fourth the same way.
It'll be f.
So practicing that formula again
with x cubed and x, it's
x cubed times 1 plus this guy
times the derivative of f.
Right?
I'm always going back
to that formula.
So the derivative of f, x
cubed, we just found--
3x squared-- so I'll
put it in.
And what do we have?
x cubed here, three more
x cubeds here.
That's a total of 4x cubed.
OK.
We got another one.
Big deal.
What is important is--
and it's really what
math is about--
is the pattern, which we can
probably guess from those two
examples and the one we
already knew, that the
derivative of x squared
was 2x.
So everybody sees a 2 here and
a 3 here and a 4 here, coming
from 2, 3, and 4 there.
And everybody also sees that
the power dropped by one.
The derivative of x
squared was an x.
The derivative of x cubed
involved an x squared.
Well, let's express this
pattern in algebra.
It's looking like the derivative
of x to the n--
we hope for any n.
We've got it for n equals 2,
3, 4, probably 0 and 1.
And if the pattern continues,
what do we think?
This 4, this n shows up there,
and the power drops by 1.
So that'll be x to the n minus
1, the same power minus 1, one
power below.
So that's a highly important
formula.
And actually it's important
to know it, not--
right now, well, we've done
two or three examples.
I guess the right way for me
to get this for n equals--
so we really could check
1, 2, 3, and so on.
All the positive integers.
We could complete the proof.
We could establish
the pattern.
Actually, induction would
be one way to do it.
If we know it for, as we did
here, for n equals 3, then
we've got it for 4.
If we know it for 4, the same
product formula would get it
for 5 and onwards, and would
give us that answer.
Good.
Even better is the fact that
this formula is also true if n
is a fraction.
If we're doing the square root
of x, you recognize the square
root of x is x to the--
what's the exponent there
for square root?
1/2.
So I would like to
know for 1/2.
OK, let me take a couple of
steps to get to that one.
All right.
The steps I'm going to take are
going to look just like
this, but this was powers of x,
and it'll be very handy if
I can do powers of f of x.
I'd like to know--
I want to find--
So here's what I'm headed for.
I'd like to know the derivative
of f of x to the
n-th power equals what?
That's what I'd like to know.
So let me do f of x.
Let me do it just
as I did before.
Take n equals 2,
f of x squared.
So what's the derivative of
f of x squared, like sine
squared or whatever
we're squaring.
Cosine squared.
Well, for f of x squared, all
I'm doing is I'm taking f to
be the same as g.
I'll use the product rule.
If g and f are the same, then
I've got something squared.
And my product rule says that
the derivative-- and I just
copy this rule.
Now I'm taking p is going
to be f squared, right?
Can I just write f
squared equals--
so it's f times--
f is the same as g.
Are you with me?
I'm just using the rule in a
very special case when the two
functions are the same.
The derivative of
f squared is f.
What do I have? f times the
derivative of f, df dx.
That's the first term.
And then what's the
second term?
Notice I wrote f instead of g,
because they're the same.
And the second term is,
again, a copy of that.
So I have 2 of these.
Times 2, just the way
I had a 2 up there.
This was the case
of x squared.
This is the case of
f of x squared.
Let me go one more
step to f cubed.
What am I going to
do for f cubed?
The derivative of--
hold on.
I have to show you what to
pay attention to here.
To pay attention to is--
the 2 we're familiar with.
This would have been the x,
that's not a big deal.
But there's something new.
A df dx factor is coming in.
It's going to stay with us.
Let me see it here.
The derivative of
f of x cubed.
Now let's practice
with this one.
OK.
So now what am I
going to take?
How do I get f of x cubed?
Well, I've got f, so I'd better
take g to be f squared.
Then when I multiply,
I've got cubed.
So g is now going to be f
squared for this case.
Can I take my product rule
with f times f squared?
My product rule of f times
f squared is--
I'm doing this now with g equals
f squared, just the way
I did it over there at
some point with one
of them as a square.
OK.
I'm near the end of
this calculation.
OK.
So what do I have. If this
thing is cubed, I
have f times f squared.
That's f cubed.
And I take its derivative
by the rule.
So I take f times the derivative
of f squared, which
I just figured out
as 2f df dx.
That's the f dg dx.
And now I have g, which is
f squared, times df dx.
What are you seeing there?
You're seeing--
well, again, these combine.
That's what's nice about
this example.
Here I have one f squared df dx,
and here I have two more.
That's, all together, three.
So the total was 3 times
f squared times df dx.
And let me write down what
that pattern is saying.
Here it will be n.
Because here it was a 2.
Here it's going to be
2 plus 1-- that's 3.
And now if I have the n-th
power, I'm expecting an n
times the next lower power
of f, f to the n
minus 1, times what?
Times this guy that's hanging
around, df dx.
That's my--
you could call that
the power rule.
The derivative of a power.
This would be the power rule
for just x to the n-th, and
this is the derivative of a
function of x to the n-th.
There's something special
here that we're
going to see more of.
This will be, also, an example
of what's coming as maybe the
most important rule,
the chain rule.
And typical of it is that when
I take this derivative, I
follow that same pattern--
n, this thing, to one lower
power, but then the derivative
of what's inside.
Can I use those words?
Because I'll use it again
for the chain rule.
n times one lower
power, times the
derivative of what's inside.
And why do I want to
do such a thing?
Because I'd like to find out
the derivative of the
square root of x.
OK.
Can we do that?
I want to use this, now.
So I want to use this to find
the derivative of the
square root of x.
OK.
So that will be my function.
f of x will be the
square root of x.
So this is a good example.
That's x to the 1/2 power.
What would I love
to have happen?
I would like this formula to
continue with n equals 1/2,
but no change in the formula.
And that does happen.
How can I do that?
OK, well, square root of
x is what I'm tackling.
The easy thing would
be, if I square
that, I'll get x, right?
The square of the square root.
Well, square root
of x squared--
so there's f of x.
I'm just going to use the fact
that the square root of x
squared is x.
Such is mathematics.
You can write down really
straightforward ideas, but it
had to come from somewhere.
And now what am I going to do?
I'm going to take
the derivative.
Well, the derivative on
the right side is a 1.
The derivative of x is 1.
What is the derivative of
that left-hand side?
Well, that fits my pattern.
You see, here is my
f of x, squared.
And I had a little
formula for the
derivative of f of x squared.
So the derivative of this
is 2 times the thing
to one lower power--
square root of x just
to the first power--
times the derivative of what's
inside, if you allow me to use
those words.
It's this, df dx.
And that's of course what I
actually wanted, the square
root of x, dx.
This lecture is not going
to have too many more
calculations, but this
is a good one to see.
That's clear.
I take the derivative
of both sides.
That's clear.
This is the 2 square
root of x.
And now I've got what I want,
as soon as I move these over
to the other side.
So I divide by that.
Can I now just do that with an
eraser, or maybe just X it
out, and put it here.
1 over 2 square root of x.
Am I seeing what I want
for the derivative of
square root of x?
I hope so.
I'm certainly seeing the 1/2.
So the 1/2--
that's the n.
It's supposed to show up here.
And then what do I
look for here?
One lower power than 1/2,
which will be x
to the minus 1/2.
And is that what I have?
Yes.
You see the 1/2.
And that square root of x,
that's x to the 1/2, but it's
down in the denominator.
And things in the
denominator--
the exponent for those,
there's a minus sign.
We'll come back to that.
That's a crucial fact, going
back to algebra.
But, you know, calculus is
now using all that--
I won't say stuff.
All those good things
that we learned in
algebra, like exponents.
So that was a good example.
OK.
So my pattern held
for n equals 1/2.
And maybe I'll just say that
it also would hold for cube
roots, and any root,
and other powers.
In other words, I get
this formula.
This is the handy formula that
we're trying to get.
We got it very directly for
positive whole numbers.
Now I'm getting it for
n equals 1 over any--
now I'm getting it for capital
Nth roots, like 1/2.
Then I could go on
to get it for--
I could take then the n-th
power of the n-th root.
I could even stretch this
to get up to m over n.
Any fraction, I can get to.
But I can't get to negative
exponents yet, because those
are divisions.
Negative exponent is a division,
and I'm going to
need the quotient rule,
which is right
now still a big blank.
OK.
Pause for a moment.
We've used the product rule.
I haven't explained
it, though.
Let me, so, explain
the product rule.
Where did it come from?
I'm going back before the
examples, and before that
board full of chalk, back to
that formula and just think,
where did it come from?
How did we find the derivative
of f times g,
of the product p?
So we needed delta p, right?
And then I'm going to
divide by delta x.
OK.
So let me try to make--
what's the delta p when p is--
remember, p is f times g.
Thinking about f times g, maybe
let's make it visual.
Let's make it like a rectangle,
where this side is
f of x and this side
is g of x.
Then this area is f
times g, right?
The area of a rectangle.
And that's our p.
OK, that's sitting there at x.
Now move it a little.
Move x a little bit.
Move x a little and figure out,
how much does p change?
That's our goal.
We need the change in p.
If I move x by a little bit,
then f changes a little, by a
little amount, delta f, right?
And g changes a little, by
a little amount, delta g.
You remember those deltas?
So it's the change in f.
There's a delta x in here.
x is the starting point.
It's the thing we
move a little.
When we move x a little, by
delta x, f will move a little,
g will move a little, and their
product will move a little.
And now, can you see, in the
picture, where is the product?
Well, this is where
f moved to.
This is where g moved to.
The product is this,
that bigger area.
So where is delta p?
Where is the change between
the bigger area and the
smaller area?
It's this.
I have to figure out, what's
that new area?
The delta p is in here.
OK, can you see what
that area--
well, look, here's
the way to do it.
Cut it up into little
three pieces.
Because now they're little
rectangles, and we know the
area of rectangles.
Right?
So help me out here.
What is the area of
that rectangle?
Well, its base is f, and
its height is delta g.
So that is f times delta g.
What about this one?
That has height g and
base delta f.
So here I'm seeing a g times
delta f, for that area.
And what about this little
corner piece?
Well, its height is just delta
g, its width is delta f.
This is delta g times delta f.
And it's going to disappear.
This is like a perfect place
to recognize that an
expression--
that's sort of like
second order.
Let me use words without
trying to
pin them down perfectly.
Here is a zero-order, an
f, a real number, times
a small delta g.
So that's first order.
That's going to show up--
you'll see it disappear.
These three pieces, remember,
were the delta p.
So what have I got here?
I've got this piece, f delta
g, and I'm always
dividing by delta x.
And then I have this piece,
which is the g times the delta
f, and I divide by
the delta x.
And then this piece that I'm
claiming I don't have to worry
much about, because I divide
that by delta x.
So that was the third piece.
This is it, now.
The picture has led to the
algebra, the formula for delta
p, the change in the product
divided by delta x.
That's what calculus says--
OK, look at that, and then
take the tricky step, the
calculus step, which is let
delta x get smaller and
smaller and smaller,
approaching 0.
So what do those three terms
do as delta x gets smaller?
Well, all the deltas
get smaller.
So what happens to this term
as delta x goes to 0?
As the change in x is just
tiny, tiny, tiny?
That term is the one that gives
the delta g over delta
x, in the limit when delta x
goes to 0, is that one, right?
And this guy is giving my g.
That ratio is familiar, df dx.
You see, the cool thing about
splitting it into these pieces
was that we got this piece
by itself, which was
just the f delta g.
And we know what that does.
It goes here.
And this piece--
we know what that does.
And now, what about
this dumb piece?
Well, as delta x goes
to 0, this would go
to df dx, all right.
But what would delta g do?
It'll go to 0.
You see, we have two little
things divided by only one
little thing.
This ratio is sensible, it gives
df dx, but this ratio is
going to 0.
So forget it.
And now the two pieces that we
have are the two pieces of the
product rule.
OK.
Product rule sort of visually
makes sense.
OK.
I'm ready to go to the
quotient rule.
OK, so how am I going to deal,
now, with a ratio of
f divided by g?
OK.
Let's put that on
a fourth board.
How to deal then with the
ratio of f over g.
Well, what I know is the
product rule, right?
So let me multiply both sides
by g of x and get a product.
There, that looks better.
Of course the part that I don't
know is in here, but
just fire away.
Take the derivative
of both sides.
OK.
The derivative of the left
side is df dx, of course.
Now I can use the
product rule.
It's g of x, dq dx.
That's the very, very
thing I'm wanting.
dq dx--
that's my big empty space.
That's going to be the
quotient rule.
And then the second one
is q of x times dg dx.
That's the product rule
applied to this.
Now I have it.
I've got dq dx.
Well, I've got to get
it by itself.
I want to get dq dx by itself.
So I'm going to move this
part over there.
Let me, even, multiply
both sides--
this q, of course, I recognize
as f times g.
This is f of x times g of x.
That's what q was.
Now I'm going to--
oh, was not.
It was f of x over g of x.
Good Lord.
You would never have allowed
me to go on.
OK.
Good.
This is came from the product
rule, and now my final job is
just to isolate dq dx and
see what I've got.
What I'll have will be
the quotient rule.
One good way is if I multiply
both sides by g.
So I multiply everything by
g, so here's a g, df dx.
And now this guy I'm going to
bring over to the other side.
When I multiply that by g, that
just knocks that out.
When I bring it over,
it comes over with a
minus sign, f dg dx.
And this one got multiplied by
g, so right now I'm looking at
g squared, dq dx.
The guy I want.
Again, just algebra.
Moving stuff from one
side to the other
produced the minus sign.
Multiplying by g, you
see what happened.
So what do I now finally do?
I'm ready to write
this formula in.
I've got it there.
I've got dq dx, just
as soon as I divide
both sides by g squared.
So let me write that
left-hand side.
g df dx minus f dg dx, and I
have to divide everything--
this g squared has got
to come down here.
It's a little bit messier
formula but
you get used to it.
g squared.
That's the quotient rule.
Can I say it in words?
Because I actually say
those words to myself
every time I use it.
So here are the words I say,
because that's a kind of
messy-looking expression.
But if you just think
about words--
so for me, remember we're
dealing with f over g. f is
the top, g at the bottom.
So I say to myself, the bottom
times the derivative of the
top minus the top times the
derivative of the bottom,
divided by the bottom squared.
That wasn't brilliant,
but anyway, I
remember it that way.
OK.
so now, finally, I'm ready to go
further with this pattern.
I still like that pattern.
We've got the quotient rule, so
the two rules are now set,
and I want to do one last
example before stopping.
And that example is going to
be a quotient, of course.
And it might as well be
a negative power of x.
So now my example--
last example for today--
my quotient is going to be 1.
The f of x will be 1 and the
g of x-- so this is my f.
This is my g.
I have a ratio of two things.
And as I've said, this
is x to the minus n.
Right?
That's what we mean.
We can think again
about exponents.
A negative exponent becomes
positive when it's in the
denominator.
And we want it in the
denominator so we can use this
crazy quotient rule.
All right.
So let me think through
the quotient rule.
So the derivative of this ratio,
which is x to the minus
n That's the q, is 1
over x to the n.
The derivative is--
OK, ready for the
quotient rule?
Bottom times the derivative
of the top--
ah, but the top's just
a constant, so its
derivative is 0--
minus--
remembering that minus-- the
top times the derivative of
the bottom.
Ha.
Now we have a chance to use our
pattern with a plus exponent.
The derivative of the bottom
is nx to the n minus 1.
So it's two terms, again,
but with a minus sign.
And then the other thing I must
remember is, divide by g
squared, x to the
n twice squared.
OK.
That's it.
Of course, I'm going
to simplify it,
and then I'm done.
So this is 0.
Gone.
This is minus n, which I like.
I like to see minus
n come down.
That's my pattern, that this
exponent should come down.
Minus n, and then I want
to see-- oh, what
else do I have here?
What's the power of x?
Well, here I have an
x to the n-th.
And here I have, twice, so can
I cancel this one and just
keep this one?
So I still have an
x to the minus 1.
I don't let him go.
Actually the pattern's here.
The answer is minus n minus
capital N, which was the
exponent, times x to
one smaller power.
This is x to the minus n, and
then there's another x
to the minus 1.
The final result was that the
derivative is minus nx to the
minus n, minus 1.
And that's the good pattern
that matches here.
When little n matches minus
big N, that pattern is the
same as that.
So we now have the derivatives
of powers of x as an example
from the quotient rule
and the product rule.
Well, I just have to
say one thing.
We haven't got--
We've fractions, we've got
negative numbers, but we don't
have a whole lot of other
numbers, like pi.
We don't know what is,
for example, the
derivative of x to the pi.
Because pi isn't--
pi is positive, so we're OK in
the product rule, but it's not
a fraction and we haven't
got it yet.
What do you think it is?
You're right--
it is pi x to the pi minus 1.
Well, actually I never met x
to the pi in my life, until
just there, but I've certainly
met all kinds of powers of x
and this is just one
more example.
OK.
So that's quotient rule--
first came product rule, power
rule, and then quotient rule,
leading to this calculation.
Now, the quotient rule I can
use for other things, like
sine x over cosine x.
We're far along, and one
more big rule will
be the chain rule.
OK, that's for another time.
Thank you.
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