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Professor: So, we're ready
to begin the fifth lecture.
00:00:24.720 --> 00:00:25.790
I'm glad to be back.
00:00:25.790 --> 00:00:33.240
Thank you for entertaining
my colleague, Haynes Miller.
00:00:33.240 --> 00:00:35.600
So, today we're
going to continue
00:00:35.600 --> 00:00:40.740
where he started, namely what
he talked about was the chain
00:00:40.740 --> 00:00:43.690
rule, which is probably
the most powerful technique
00:00:43.690 --> 00:00:45.860
for extending the
kinds of functions
00:00:45.860 --> 00:00:47.650
that you can differentiate.
00:00:47.650 --> 00:00:50.920
And we're going to use the
chain rule in some rather clever
00:00:50.920 --> 00:00:54.410
algebraic ways today.
00:00:54.410 --> 00:00:57.420
So the topic for
today is what's known
00:00:57.420 --> 00:01:10.670
as implicit differentiation.
00:01:10.670 --> 00:01:15.040
So implicit differentiation
is a technique
00:01:15.040 --> 00:01:17.690
that allows you to differentiate
a lot of functions you didn't
00:01:17.690 --> 00:01:20.970
even know how to find before.
00:01:20.970 --> 00:01:24.940
And it's a technique -
let's wait for a few people
00:01:24.940 --> 00:01:28.260
to sit down here.
00:01:28.260 --> 00:01:29.210
Physics, huh?
00:01:29.210 --> 00:01:36.419
Okay, more Physics.
00:01:36.419 --> 00:01:37.210
Let's take a break.
00:01:37.210 --> 00:01:40.980
You can get those after class.
00:01:40.980 --> 00:01:46.220
All right, so we're talking
about implicit differentiation,
00:01:46.220 --> 00:01:53.760
and I'm going to illustrate
it by several examples.
00:01:53.760 --> 00:01:57.610
So this is one of the most
important and basic formulas
00:01:57.610 --> 00:01:59.770
that we've already
covered part way.
00:01:59.770 --> 00:02:06.950
Namely, the derivative of
x to a power is ax^(a-1).
00:02:06.950 --> 00:02:15.780
Now, what we've got so far is
the exponents, 0, plus or minus
00:02:15.780 --> 00:02:19.010
1, plus or minus 2, etc.
00:02:19.010 --> 00:02:24.440
You did the positive integer
powers in the first lecture,
00:02:24.440 --> 00:02:30.710
and then yesterday
Professor Miller
00:02:30.710 --> 00:02:32.470
told you about the
negative powers.
00:02:32.470 --> 00:02:35.470
So what we're going
to do right now,
00:02:35.470 --> 00:02:39.500
today, is we're
going to consider
00:02:39.500 --> 00:02:44.420
the exponents which are rational
numbers, ratios of integers.
00:02:44.420 --> 00:02:46.540
So a is m/n.
00:02:46.540 --> 00:02:53.380
m and n are integers.
00:02:53.380 --> 00:02:55.414
All right, so that's
our goal for right now,
00:02:55.414 --> 00:02:56.830
and we're going
to use this method
00:02:56.830 --> 00:02:58.150
of implicit differentiation.
00:02:58.150 --> 00:03:01.010
In particular, it's important
to realize that this
00:03:01.010 --> 00:03:03.260
covers the case m = 1.
00:03:03.260 --> 00:03:04.890
And those are the nth roots.
00:03:04.890 --> 00:03:07.370
So when we take the
one over n power,
00:03:07.370 --> 00:03:09.680
we're going to cover
that right now,
00:03:09.680 --> 00:03:13.110
along with many other examples.
00:03:13.110 --> 00:03:16.360
So this is our first example.
00:03:16.360 --> 00:03:17.700
So how do we get started?
00:03:17.700 --> 00:03:20.510
Well we just write down a
formula for the function.
00:03:20.510 --> 00:03:24.620
The function is y = x^(m/n).
00:03:24.620 --> 00:03:26.620
That's what we're
trying to deal with.
00:03:26.620 --> 00:03:30.610
And now there's
really only two steps.
00:03:30.610 --> 00:03:38.120
The first step is to take this
equation to the nth power,
00:03:38.120 --> 00:03:42.900
so write it y^n = x^m.
00:03:42.900 --> 00:03:46.170
Alright, so that's just the
same equation re-written.
00:03:46.170 --> 00:03:50.180
And now, what we're
going to do is
00:03:50.180 --> 00:03:52.170
we're going to differentiate.
00:03:52.170 --> 00:04:01.600
So we're going to apply
d/dx to the equation.
00:04:01.600 --> 00:04:05.730
Now why is it that we can apply
it to the second equation, not
00:04:05.730 --> 00:04:06.670
the first equation?
00:04:06.670 --> 00:04:10.220
So maybe I should call these
equation 1 and equation 2.
00:04:10.220 --> 00:04:13.150
So, the point is, we can
apply it to equation 2.
00:04:13.150 --> 00:04:17.860
Now, the reason is that we
don't know how to differentiate
00:04:17.860 --> 00:04:18.860
x^(m/n).
00:04:18.860 --> 00:04:21.320
That's something we
just don't know yet.
00:04:21.320 --> 00:04:24.630
But we do know how to
differentiate integer powers.
00:04:24.630 --> 00:04:29.080
Those are the things that
we took care of before.
00:04:29.080 --> 00:04:32.980
So now we're in shape to be
able to do the differentiation.
00:04:32.980 --> 00:04:34.900
So I'm going to write
it out explicitly
00:04:34.900 --> 00:04:37.930
over here, without
carrying it out just yet.
00:04:37.930 --> 00:04:46.460
That's d/dx of
y^n = d/dx of x^m.
00:04:46.460 --> 00:04:51.920
And now you see
this expression here
00:04:51.920 --> 00:04:55.020
requires us to do something we
couldn't do before yesterday.
00:04:55.020 --> 00:04:58.780
Namely, this y is
a function of x.
00:04:58.780 --> 00:05:01.610
So we have to apply
the chain rule here.
00:05:01.610 --> 00:05:06.710
So this is the same as - this
is by the chain rule now -
00:05:06.710 --> 00:05:13.332
d/dy of y^n times dy/dx.
00:05:13.332 --> 00:05:15.790
And then, on the right hand
side, we can just carry it out.
00:05:15.790 --> 00:05:17.280
We know the formula.
00:05:17.280 --> 00:05:18.420
It's mx^(m-1).
00:05:21.550 --> 00:05:24.960
Right, now this is our scheme.
00:05:24.960 --> 00:05:29.360
And you'll see in a minute
why we win with this.
00:05:29.360 --> 00:05:32.390
So, first of all, there
are two factors here.
00:05:32.390 --> 00:05:33.870
One of them is unknown.
00:05:33.870 --> 00:05:35.870
In fact, it's what
we're looking for.
00:05:35.870 --> 00:05:38.570
But the other one is going
to be a known quantity,
00:05:38.570 --> 00:05:40.507
because we know how
to differentiate y
00:05:40.507 --> 00:05:42.507
to the n with respect to y.
00:05:42.507 --> 00:05:44.340
That's the same formula,
although the letter
00:05:44.340 --> 00:05:46.330
has been changed.
00:05:46.330 --> 00:05:53.070
And so this is the same as -
I'll write it underneath here -
00:05:53.070 --> 00:06:07.040
n y^(n-1) dy/dx = m x^(m-1).
00:06:07.040 --> 00:06:14.565
Okay, now comes, if you
like, the non-calculus part
00:06:14.565 --> 00:06:15.190
of the problem.
00:06:15.190 --> 00:06:17.140
Remember the non-calculus
part of the problem
00:06:17.140 --> 00:06:20.030
is always the messier
part of the problem.
00:06:20.030 --> 00:06:22.210
So we want to figure
out this formula.
00:06:22.210 --> 00:06:25.600
This formula, the
answer over here,
00:06:25.600 --> 00:06:29.850
which maybe I'll
put in a box now,
00:06:29.850 --> 00:06:33.420
has this expressed much more
simply, only in terms of x.
00:06:33.420 --> 00:06:36.140
And what we have to do now
is just solve for dy/dx
00:06:36.140 --> 00:06:39.730
using algebra, and then solve
all the way in terms of x.
00:06:39.730 --> 00:06:41.700
So, first of all,
we solve for dy/dx.
00:06:44.230 --> 00:06:47.930
So I do that by dividing the
factor on the left-hand side.
00:06:47.930 --> 00:06:52.960
So I get here mx^(m-1)
divided by ny^(n-1).
00:06:56.030 --> 00:07:02.020
And now I'm going to plug in--
so I'll write this as m/n.
00:07:02.020 --> 00:07:04.540
This is x^(m-1).
00:07:04.540 --> 00:07:10.540
Now over here I'm going to put
in for y, x^(m/n) times n-1.
00:07:15.690 --> 00:07:18.700
So now we're almost done,
but unfortunately we
00:07:18.700 --> 00:07:22.070
have this mess of exponents
that we have to work out.
00:07:22.070 --> 00:07:25.060
I'm going to write
it one more time.
00:07:25.060 --> 00:07:28.584
So I already recognize
the factor a out front.
00:07:28.584 --> 00:07:30.250
That's not going to
be a problem for me,
00:07:30.250 --> 00:07:31.980
and that's what I'm
aiming for here.
00:07:31.980 --> 00:07:34.700
But now I have to encode
all of these powers,
00:07:34.700 --> 00:07:36.390
so let's just write it.
00:07:36.390 --> 00:07:41.790
It's m-1, and then it's
minus the quantity (n-1) m/n.
00:07:46.270 --> 00:07:50.250
All right, so that's the law of
exponents applied to this ratio
00:07:50.250 --> 00:07:50.750
here.
00:07:50.750 --> 00:07:58.410
And then we'll do the arithmetic
over here on the next board.
00:07:58.410 --> 00:08:08.700
So we have here m - 1
- (n-1) m/n = m - 1.
00:08:08.700 --> 00:08:12.460
And if I multiply n
by this, I get -m.
00:08:12.460 --> 00:08:15.560
And if the second factor is
minus minus, that's a plus.
00:08:15.560 --> 00:08:18.220
And that's +m/n.
00:08:18.220 --> 00:08:21.080
Altogether the two m's cancel.
00:08:21.080 --> 00:08:23.740
I have here -1 + m/n.
00:08:23.740 --> 00:08:27.492
And lo and behold that's
the same thing as a - 1,
00:08:27.492 --> 00:08:29.510
just what we wanted.
00:08:29.510 --> 00:08:31.900
All right, so this
equals a x^(n-1).
00:08:31.900 --> 00:08:39.560
Okay, again just a
bunch of arithmetic.
00:08:39.560 --> 00:08:42.530
From this point forward,
from this substitution
00:08:42.530 --> 00:08:51.180
on, it's just the
arithmetic of exponents.
00:08:51.180 --> 00:08:58.000
All right, so we've done
our first example here.
00:08:58.000 --> 00:09:00.590
I want to give you a
couple more examples,
00:09:00.590 --> 00:09:04.030
so let's just continue.
00:09:04.030 --> 00:09:08.720
The next example I'll
keep relatively simple.
00:09:08.720 --> 00:09:15.590
So we have example two, which is
going to be the function x^2 +
00:09:15.590 --> 00:09:18.060
y^2 = 1.
00:09:18.060 --> 00:09:21.200
Well, that's not
really a function.
00:09:21.200 --> 00:09:29.180
It's a way of defining y as
a function of x implicitly.
00:09:29.180 --> 00:09:34.700
There's the idea that I could
solve for y if I wanted to.
00:09:34.700 --> 00:09:36.490
And indeed let's do that.
00:09:36.490 --> 00:09:42.490
So if you solve for y here, what
happens is you get y^2 = 1 -
00:09:42.490 --> 00:09:47.380
x^2, and y is equal to plus or
minus the square root of 1 -
00:09:47.380 --> 00:09:52.940
x^2.
00:09:52.940 --> 00:09:58.200
So this, if you like, is
the implicit definition.
00:09:58.200 --> 00:10:00.910
And here is the
explicit function y,
00:10:00.910 --> 00:10:04.140
which is a function of x.
00:10:04.140 --> 00:10:06.360
And now just for
my own convenience,
00:10:06.360 --> 00:10:09.310
I'm just going to take
the positive branch.
00:10:09.310 --> 00:10:13.450
This is the function.
00:10:13.450 --> 00:10:15.900
It's just really a
circle in disguise.
00:10:15.900 --> 00:10:19.600
And I'm just going to take
the top part of the circle,
00:10:19.600 --> 00:10:24.920
so we'll take that
top hump here.
00:10:24.920 --> 00:10:27.900
All right, so that means
I'm erasing this minus sign.
00:10:27.900 --> 00:10:35.660
I'm just taking the
positive branch, just
00:10:35.660 --> 00:10:36.470
for my convenience.
00:10:36.470 --> 00:10:40.980
I could do it just as well
with the negative branch.
00:10:40.980 --> 00:10:46.000
Alright, so now I've
taken the solution,
00:10:46.000 --> 00:10:49.540
and I can differentiate
with this.
00:10:49.540 --> 00:10:53.070
So rather than using the
dy/dx notation over here,
00:10:53.070 --> 00:10:55.510
I'm going to switch
notations over here,
00:10:55.510 --> 00:10:56.620
because it's less writing.
00:10:56.620 --> 00:10:59.670
I'm going to write y'
and change notations.
00:10:59.670 --> 00:11:04.450
Okay, so I want to take
the derivative of this.
00:11:04.450 --> 00:11:11.570
Well this is a somewhat
complicated function here.
00:11:11.570 --> 00:11:15.770
It's the square root of 1 -
x^2, and the right way always
00:11:15.770 --> 00:11:21.720
to look at functions like
this is to rewrite them using
00:11:21.720 --> 00:11:26.210
the fractional power notation.
00:11:26.210 --> 00:11:28.910
That's the first
step in computing
00:11:28.910 --> 00:11:32.600
a derivative of a square root.
00:11:32.600 --> 00:11:38.030
And then the second
step here is what?
00:11:38.030 --> 00:11:40.740
Does somebody want to tell me?
00:11:40.740 --> 00:11:43.650
Chain rule, right.
00:11:43.650 --> 00:11:44.300
That's it.
00:11:44.300 --> 00:11:45.320
So we have two things.
00:11:45.320 --> 00:11:47.720
We start with one, and then
we do something else to it.
00:11:47.720 --> 00:11:50.090
So whenever we do two
things to something,
00:11:50.090 --> 00:11:52.020
we need to apply the chain rule.
00:11:52.020 --> 00:11:55.240
So 1 - x^2, square root.
00:11:55.240 --> 00:11:57.180
All right, so how do we do that?
00:11:57.180 --> 00:11:58.830
Well, the first
factor I claim is
00:11:58.830 --> 00:12:01.220
the derivative of this thing.
00:12:01.220 --> 00:12:06.500
So this is 1/2 blah to the -1/2.
00:12:06.500 --> 00:12:09.760
So I'm doing this kind
of by the advanced method
00:12:09.760 --> 00:12:11.390
now, because we've
already graduated.
00:12:11.390 --> 00:12:14.450
You already did the
chain rule last time.
00:12:14.450 --> 00:12:15.920
So what does this mean?
00:12:15.920 --> 00:12:20.940
This is an abbreviation for
the derivative with respect
00:12:20.940 --> 00:12:27.697
to blah of blah ^
1/2, whatever it is.
00:12:27.697 --> 00:12:30.280
All right, so that's the first
factor that we're going to use.
00:12:30.280 --> 00:12:34.480
Rather than actually write
out a variable for it
00:12:34.480 --> 00:12:36.890
and pass through
as I did previously
00:12:36.890 --> 00:12:39.160
with this y and x
variable here, I'm
00:12:39.160 --> 00:12:41.840
just going to skip
that step and let
00:12:41.840 --> 00:12:45.370
you imagine it as being a
placeholder for that variable
00:12:45.370 --> 00:12:45.870
here.
00:12:45.870 --> 00:12:48.960
So this variable
is now parenthesis.
00:12:48.960 --> 00:12:52.370
And then I have to multiply that
by the rate of change of what's
00:12:52.370 --> 00:12:55.050
inside with respect to x.
00:12:55.050 --> 00:12:58.580
And that is going to be -2x.
00:12:58.580 --> 00:13:02.830
The derivative of
1 - x^2 is -2x.
00:13:02.830 --> 00:13:09.030
And now again, we couldn't
have done this example two
00:13:09.030 --> 00:13:11.730
before example one,
because we needed
00:13:11.730 --> 00:13:17.470
to know that the power rule
worked not just for a integer
00:13:17.470 --> 00:13:19.690
but also for a = 1/2.
00:13:19.690 --> 00:13:22.740
We're using the case
a = 1/2 right here.
00:13:22.740 --> 00:13:29.540
It's 1/2 times, and
this -1/2 here is a-1. -
00:13:29.540 --> 00:13:33.430
So this is the case a = 1/2.
00:13:33.430 --> 00:13:39.790
a-1 happens to be -1/2.
00:13:39.790 --> 00:13:41.940
Okay, so I'm putting all
those things together.
00:13:41.940 --> 00:13:44.380
And you know within
a week you have
00:13:44.380 --> 00:13:45.970
to be doing this
very automatically.
00:13:45.970 --> 00:13:47.900
So we're going to do
it at this speed now.
00:13:47.900 --> 00:13:49.780
You want to do it even
faster, ultimately.
00:13:49.780 --> 00:13:50.280
Yes?
00:13:50.280 --> 00:13:53.630
Student: [INAUDIBLE]
00:13:53.630 --> 00:13:56.110
Professor: The question is
could I have done it implicitly
00:13:56.110 --> 00:13:58.060
without the square roots.
00:13:58.060 --> 00:13:59.440
And the answer is yes.
00:13:59.440 --> 00:14:02.040
That's what I'm about to do.
00:14:02.040 --> 00:14:04.570
So this is an
illustration of what's
00:14:04.570 --> 00:14:07.430
called the explicit solution.
00:14:07.430 --> 00:14:13.800
So this guy is what's
called explicit.
00:14:13.800 --> 00:14:17.024
And I want to contrast
it with the method
00:14:17.024 --> 00:14:18.440
that we're going
to now use today.
00:14:18.440 --> 00:14:20.410
So it involves a lot
of complications.
00:14:20.410 --> 00:14:21.701
It involves the chain rule.
00:14:21.701 --> 00:14:23.700
And as we'll see it can
get messier and messier.
00:14:23.700 --> 00:14:27.260
And then there's
the implicit method,
00:14:27.260 --> 00:14:29.830
which I claim is easier.
00:14:29.830 --> 00:14:36.170
So let's see what happens
if you do it implicitly
00:14:36.170 --> 00:14:41.010
The implicit method
involves, instead of writing
00:14:41.010 --> 00:14:43.610
the function in this
relatively complicated way,
00:14:43.610 --> 00:14:47.380
with the square root, it
involves leaving it alone.
00:14:47.380 --> 00:14:50.050
Don't do anything to it.
00:14:50.050 --> 00:14:52.820
In this previous case, we were
left with something which was
00:14:52.820 --> 00:14:56.820
complicated, say x^(1/3)
or x^(1/2) or something
00:14:56.820 --> 00:14:57.370
complicated.
00:14:57.370 --> 00:14:59.326
We had to simplify it.
00:14:59.326 --> 00:15:01.450
We had an equation one,
which was more complicated.
00:15:01.450 --> 00:15:03.970
We simplified it then
differentiated it.
00:15:03.970 --> 00:15:05.590
And so that was a simpler case.
00:15:05.590 --> 00:15:09.590
Well here, the simplest
thing us to differentiate
00:15:09.590 --> 00:15:13.470
is the one we started with,
because squares are practically
00:15:13.470 --> 00:15:16.725
the easiest thing after first
powers, or maybe zeroth powers
00:15:16.725 --> 00:15:18.830
to differentiate.
00:15:18.830 --> 00:15:19.940
So we're leaving it alone.
00:15:19.940 --> 00:15:21.689
This is the simplest
possible form for it,
00:15:21.689 --> 00:15:23.640
and now we're going
to differentiate.
00:15:23.640 --> 00:15:24.570
So what happens?
00:15:24.570 --> 00:15:26.640
So again what's the method?
00:15:26.640 --> 00:15:27.810
Let me remind you.
00:15:27.810 --> 00:15:30.210
You're applying d/dx
to the equation.
00:15:30.210 --> 00:15:33.640
So you have to differentiate
the left side of the equation,
00:15:33.640 --> 00:15:35.810
and differentiate the
right side of the equation.
00:15:35.810 --> 00:15:51.100
So it's this, and what you get
is 2x + 2yy' is equal to what?
00:15:51.100 --> 00:15:52.760
0.
00:15:52.760 --> 00:15:56.390
The derivative of 1 0.
00:15:56.390 --> 00:15:58.630
So this is the chain rule again.
00:15:58.630 --> 00:16:00.390
I did it a different way.
00:16:00.390 --> 00:16:02.690
I'm trying to get you used
to many different notations
00:16:02.690 --> 00:16:04.420
at once.
00:16:04.420 --> 00:16:05.350
Well really just two.
00:16:05.350 --> 00:16:10.600
Just the prime notation
and the dy/dx notation.
00:16:10.600 --> 00:16:14.370
And this is what I get.
00:16:14.370 --> 00:16:19.670
So now all I have to
do is solve for y'.
00:16:19.670 --> 00:16:24.270
So that y', if I put the 2x
on the other side, is -2x,
00:16:24.270 --> 00:16:27.760
and then divide by
2y, which is -x/y.
00:16:30.630 --> 00:16:34.600
So let's compare our
solutions, and I'll apologize,
00:16:34.600 --> 00:16:39.080
I'm going to have to erase
something to do that.
00:16:39.080 --> 00:16:44.480
So let's compare
our two solutions.
00:16:44.480 --> 00:16:46.460
I'm going to put this
underneath and simplify.
00:16:46.460 --> 00:16:48.880
So what was our
solution over here?
00:16:48.880 --> 00:16:51.500
It was 1/2(1-x^2)^(-1/2) (-2x).
00:16:56.720 --> 00:17:02.170
That was what we got over here.
00:17:02.170 --> 00:17:06.992
And that is the same thing, if I
cancel the 2's, and I change it
00:17:06.992 --> 00:17:08.450
back to looking
like a square root,
00:17:08.450 --> 00:17:11.536
that's the same thing as -x
divided by square root of 1 -
00:17:11.536 --> 00:17:13.960
x^2.
00:17:13.960 --> 00:17:18.380
So this is the formula
for the derivative
00:17:18.380 --> 00:17:21.340
when I do it the explicit way.
00:17:21.340 --> 00:17:29.550
And I'll just compare them,
these two expressions here.
00:17:29.550 --> 00:17:32.630
And notice they are the same.
00:17:32.630 --> 00:17:37.860
They're the same, because y
is equal to square root of 1 -
00:17:37.860 --> 00:17:40.230
x^2.
00:17:40.230 --> 00:17:40.730
Yeah?
00:17:40.730 --> 00:17:41.230
Question?
00:17:41.230 --> 00:17:46.140
Student: [INAUDIBLE]
00:17:46.140 --> 00:17:48.530
Professor: The question is
why did the implicit method
00:17:48.530 --> 00:17:50.930
not give the bottom
half of the circle?
00:17:50.930 --> 00:17:53.200
Very good question.
00:17:53.200 --> 00:17:57.000
The answer to that
is that it did.
00:17:57.000 --> 00:17:59.520
I just didn't mention it.
00:17:59.520 --> 00:18:00.890
Wait, I'll explain.
00:18:00.890 --> 00:18:05.300
So suppose I stuck
in a minus sign here.
00:18:05.300 --> 00:18:08.040
I would have gotten this
with the difference, so
00:18:08.040 --> 00:18:10.080
with an extra minus sign.
00:18:10.080 --> 00:18:12.480
But then when I compared
it to what was over there,
00:18:12.480 --> 00:18:15.620
I would have had to have another
different minus sign over here.
00:18:15.620 --> 00:18:19.310
So actually both places would
get an extra minus sign.
00:18:19.310 --> 00:18:20.630
And they would still coincide.
00:18:20.630 --> 00:18:22.760
So actually the implicit
method is a little better.
00:18:22.760 --> 00:18:23.940
It doesn't even
notice the difference
00:18:23.940 --> 00:18:24.940
between the branches.
00:18:24.940 --> 00:18:28.930
It does the job on both
the top and bottom half.
00:18:28.930 --> 00:18:31.390
Another way of saying
that is that you're
00:18:31.390 --> 00:18:33.200
calculating the slopes here.
00:18:33.200 --> 00:18:35.170
So let's look at this picture.
00:18:35.170 --> 00:18:36.740
Here's a slope.
00:18:36.740 --> 00:18:39.090
Let's just take a look
at a positive value
00:18:39.090 --> 00:18:42.770
of x and just check the sign
to see what's happening.
00:18:42.770 --> 00:18:46.982
If you take a positive value
of x over here, x is positive.
00:18:46.982 --> 00:18:48.190
This denominator is positive.
00:18:48.190 --> 00:18:49.106
The slope is negative.
00:18:49.106 --> 00:18:52.620
You can see that
it's tilting down.
00:18:52.620 --> 00:18:53.960
So it's okay.
00:18:53.960 --> 00:18:59.520
Now on the bottom side,
it's going to be tilting up.
00:18:59.520 --> 00:19:01.690
And similarly what's
happening up here
00:19:01.690 --> 00:19:05.275
is that both x and y are
positive, and this x and this y
00:19:05.275 --> 00:19:06.167
are positive.
00:19:06.167 --> 00:19:07.250
And the slope is negative.
00:19:07.250 --> 00:19:10.430
On the other hand, on the bottom
side, x is still positive,
00:19:10.430 --> 00:19:11.780
but y is negative.
00:19:11.780 --> 00:19:15.168
And it's tilting up because
the denominator is negative.
00:19:15.168 --> 00:19:17.084
The numerator is positive,
and this minus sign
00:19:17.084 --> 00:19:19.050
has a positive slope.
00:19:19.050 --> 00:19:23.330
So it matches perfectly
in every category.
00:19:23.330 --> 00:19:26.920
This complicated,
however, and it's easier
00:19:26.920 --> 00:19:30.000
just to keep track of
one branch at a time,
00:19:30.000 --> 00:19:32.850
even in advanced math.
00:19:32.850 --> 00:19:37.590
Okay, so we only do it
one branch at a time.
00:19:37.590 --> 00:19:43.970
Other questions?
00:19:43.970 --> 00:19:47.360
Okay, so now I want to
give you a slightly more
00:19:47.360 --> 00:19:49.210
complicated example here.
00:19:49.210 --> 00:19:52.690
And indeed some
of the-- so here's
00:19:52.690 --> 00:19:54.490
a little more
complicated example.
00:19:54.490 --> 00:19:56.900
It's not going to be the
most complicated example,
00:19:56.900 --> 00:20:17.980
but you know it'll
be a little tricky.
00:20:17.980 --> 00:20:22.520
So this example, I'm going
to give you a fourth order
00:20:22.520 --> 00:20:23.020
equation.
00:20:23.020 --> 00:20:31.980
So y^4 + xy^2 - 2 = 0.
00:20:31.980 --> 00:20:35.320
Now it just so
happens that there's
00:20:35.320 --> 00:20:38.700
a trick to solving
this equation,
00:20:38.700 --> 00:20:41.420
so actually you can do
both the explicit method
00:20:41.420 --> 00:20:46.210
and the non-explicit method.
00:20:46.210 --> 00:20:50.379
So the explicit method
would say okay well,
00:20:50.379 --> 00:20:51.420
I want to solve for this.
00:20:51.420 --> 00:20:55.760
So I'm going to use the
quadratic formula, but on y^2.
00:20:55.760 --> 00:20:59.230
This is quadratic in y^2,
because there's a fourth power
00:20:59.230 --> 00:21:02.891
and a second power, and the
first and third powers are
00:21:02.891 --> 00:21:03.390
missing.
00:21:03.390 --> 00:21:09.940
So this is y^2 is equal to -x
plus or minus the square root
00:21:09.940 --> 00:21:19.570
of x^2 - 4(-2) divided by 2.
00:21:19.570 --> 00:21:22.790
And so this x is the b.
00:21:22.790 --> 00:21:29.750
This -2 is the c, and a =
1 in the quadratic formula.
00:21:29.750 --> 00:21:37.200
And so the formula for y is plus
or minus the square root of -x
00:21:37.200 --> 00:21:45.070
plus or minus the square
root x^2 + 8 divided by 2.
00:21:45.070 --> 00:21:47.880
So now you can see this
problem of branches,
00:21:47.880 --> 00:21:50.540
this happens actually
in a lot of cases,
00:21:50.540 --> 00:21:53.066
coming up in an elaborate way.
00:21:53.066 --> 00:21:54.690
You have two choices
for the sign here.
00:21:54.690 --> 00:21:56.610
You have two choices
for the sign here.
00:21:56.610 --> 00:21:59.410
Conceivably as many as four
roots for this equation,
00:21:59.410 --> 00:22:02.031
because it's a fourth
degree equation.
00:22:02.031 --> 00:22:02.780
It's quite a mess.
00:22:02.780 --> 00:22:06.000
You should have to check
each branch separately.
00:22:06.000 --> 00:22:09.180
And this really is that
level of complexity,
00:22:09.180 --> 00:22:11.750
and in general
it's very difficult
00:22:11.750 --> 00:22:17.840
to figure out the formulas
for quartic equations.
00:22:17.840 --> 00:22:21.630
But fortunately we're
never going to use them.
00:22:21.630 --> 00:22:24.830
That is, we're never going
to need those formulas.
00:22:24.830 --> 00:22:31.850
So the implicit
method is far easier.
00:22:31.850 --> 00:22:35.230
The implicit method
just says okay I'll
00:22:35.230 --> 00:22:38.980
leave the equation
in its simplest form.
00:22:38.980 --> 00:22:40.520
And now differentiate.
00:22:40.520 --> 00:22:47.300
So when I differentiate,
I get 4y^3 y' plus -
00:22:47.300 --> 00:22:50.920
now here I have to
apply the product rule.
00:22:50.920 --> 00:22:56.090
So I differentiate the x
and the y^2 separately.
00:22:56.090 --> 00:22:59.720
First I differentiate with
respect to x, so I get y^2.
00:22:59.720 --> 00:23:03.220
Then I differentiate with
respect to the other factor,
00:23:03.220 --> 00:23:04.410
the y^2 factor.
00:23:04.410 --> 00:23:08.950
And I get x(2 y y').
00:23:08.950 --> 00:23:10.440
And then the 0 gives me 0.
00:23:10.440 --> 00:23:16.100
So minus 0 equals 0.
00:23:16.100 --> 00:23:21.970
So there's the implicit
differentiation step.
00:23:21.970 --> 00:23:26.260
And now I just want
to solve for y'.
00:23:26.260 --> 00:23:32.570
So I'm going to
factor out 4y^3 + 2xy.
00:23:32.570 --> 00:23:35.740
That's the factor on y'.
00:23:35.740 --> 00:23:39.780
And I'm going to put the
y^2 on the other side.
00:23:39.780 --> 00:23:43.400
-y^2 over here.
00:23:43.400 --> 00:23:55.110
And so the formula for y' is
-y^2 divided by 4y^3 + 2xy.
00:23:55.110 --> 00:24:01.420
So that's the formula
for the solution.
00:24:01.420 --> 00:24:06.947
For the slope.
00:24:06.947 --> 00:24:07.780
You have a question?
00:24:07.780 --> 00:24:16.340
Student: [INAUDIBLE]
00:24:16.340 --> 00:24:18.350
Professor: So the question
is for the y would
00:24:18.350 --> 00:24:22.140
we have to put in what solved
for in the explicit equation.
00:24:22.140 --> 00:24:24.120
And the answer is
absolutely yes.
00:24:24.120 --> 00:24:25.280
That's exactly the point.
00:24:25.280 --> 00:24:30.950
So this is not a complete
solution to a problem.
00:24:30.950 --> 00:24:32.710
We started with an
implicit equation.
00:24:32.710 --> 00:24:33.990
We differentiated.
00:24:33.990 --> 00:24:36.660
And we got in the end,
also an implicit equation.
00:24:36.660 --> 00:24:39.540
It doesn't tell us what
y is as a function of x.
00:24:39.540 --> 00:24:43.040
You have to go back
to this formula
00:24:43.040 --> 00:24:45.520
to get the formula for x.
00:24:45.520 --> 00:24:49.310
So for example, let me
give you an example here.
00:24:49.310 --> 00:24:54.660
So this hides a degree of
complexity of the problem.
00:24:54.660 --> 00:24:58.550
But it's a degree of complexity
that we must live with.
00:24:58.550 --> 00:25:10.460
So for example, at x = 1, you
can see that y = 1 solves.
00:25:10.460 --> 00:25:16.750
That happens to be--
solves y^4 + xy^2 - 2 = 0.
00:25:16.750 --> 00:25:18.400
That's why I picked
the 2 actually,
00:25:18.400 --> 00:25:21.000
so it would be 1 + 1 - 2 = 0.
00:25:21.000 --> 00:25:23.060
I just wanted to have a
convenient solution there
00:25:23.060 --> 00:25:25.630
to pull out of my
hat at this point.
00:25:25.630 --> 00:25:26.670
So I did that.
00:25:26.670 --> 00:25:30.250
And so we now know
that when x = 1, y = 1.
00:25:30.250 --> 00:25:41.740
So at (1, 1) along the curve,
the slope is equal to what?
00:25:41.740 --> 00:25:52.200
Well, I have to plug in
here, -1^2 / (4*1^3 + 2*1*1).
00:25:52.200 --> 00:25:54.290
That's just plugging in
that formula over there,
00:25:54.290 --> 00:25:59.170
which turns out to be -1/6.
00:25:59.170 --> 00:26:00.670
So I can get it.
00:26:00.670 --> 00:26:13.940
On the other hand,
at say x = 2, we're
00:26:13.940 --> 00:26:32.890
stuck using this formula
star here to find y.
00:26:32.890 --> 00:26:37.170
Now, so let me just
make two points
00:26:37.170 --> 00:26:40.020
about this, which are just
philosophical points for you
00:26:40.020 --> 00:26:42.420
right now.
00:26:42.420 --> 00:26:45.427
The first is, when
I promised you
00:26:45.427 --> 00:26:47.010
at the beginning of
this class that we
00:26:47.010 --> 00:26:48.840
were going to be
able to differentiate
00:26:48.840 --> 00:26:53.230
any function you know, I
meant it very literally.
00:26:53.230 --> 00:26:56.021
What I meant is if
you know the function,
00:26:56.021 --> 00:26:58.020
we'll be able give a
formula for the derivative.
00:26:58.020 --> 00:27:00.210
If you don't know how
to find a function,
00:27:00.210 --> 00:27:02.490
you'll have a lot of trouble
finding the derivative.
00:27:02.490 --> 00:27:05.370
So we didn't make any
promises that if you
00:27:05.370 --> 00:27:06.830
can't find the
function you will be
00:27:06.830 --> 00:27:09.300
able to find the
derivative by some magic.
00:27:09.300 --> 00:27:10.450
That will never happen.
00:27:10.450 --> 00:27:12.900
And however complex
the function is,
00:27:12.900 --> 00:27:16.130
a root of a fourth
degree polynomial
00:27:16.130 --> 00:27:20.300
can be pretty complicated
function of the coefficients,
00:27:20.300 --> 00:27:23.990
we're stuck with this degree
of complexity in the problem.
00:27:23.990 --> 00:27:27.220
But the big advantage
of his method, notice,
00:27:27.220 --> 00:27:29.340
is that although we've
had to find star,
00:27:29.340 --> 00:27:31.180
we had to find
this formula star,
00:27:31.180 --> 00:27:34.100
and there are many other ways of
doing these things numerically,
00:27:34.100 --> 00:27:36.250
by the way, which
we'll learn later,
00:27:36.250 --> 00:27:39.940
so there's a good method
for doing it numerically.
00:27:39.940 --> 00:27:43.190
Although we had to find star, we
never had to differentiate it.
00:27:43.190 --> 00:27:46.080
We had a fast way of
getting the slope.
00:27:46.080 --> 00:27:48.280
So we had to know
what x and y were.
00:27:48.280 --> 00:27:50.980
But y' we got by an
algebraic formula,
00:27:50.980 --> 00:27:54.450
in terms of the values here.
00:27:54.450 --> 00:27:57.140
So this is very fast,
forgetting the slope,
00:27:57.140 --> 00:28:02.382
once you know the point. yes?
00:28:02.382 --> 00:28:03.840
Student: What's in
the parentheses?
00:28:03.840 --> 00:28:06.790
Professor: Sorry, this is-- Well
let's see if I can manage this.
00:28:06.790 --> 00:28:16.522
Is this the parentheses
you're talking about?
00:28:16.522 --> 00:28:17.022
Ah, "say".
00:28:17.022 --> 00:28:17.310
That says "say".
00:28:17.310 --> 00:28:19.185
Well, so maybe I should
put commas around it.
00:28:19.185 --> 00:28:24.560
But it was S A Y,
comma comma, okay?
00:28:24.560 --> 00:28:28.900
Well here was at x = 1.
00:28:28.900 --> 00:28:33.100
I'm just throwing out a value.
00:28:33.100 --> 00:28:34.200
Any other value.
00:28:34.200 --> 00:28:36.360
Actually there is one
value, my favorite value.
00:28:36.360 --> 00:28:39.700
Well this is easy to
evaluate right? x = 0,
00:28:39.700 --> 00:28:42.610
I can do it there.
00:28:42.610 --> 00:28:45.400
That's maybe the only one.
00:28:45.400 --> 00:28:55.820
The others are a nuisance.
00:28:55.820 --> 00:29:03.750
All right, other questions?
00:29:03.750 --> 00:29:06.060
Now we have to do
something more here.
00:29:06.060 --> 00:29:10.470
So I claimed to you that
we could differentiate
00:29:10.470 --> 00:29:11.580
all the functions we know.
00:29:11.580 --> 00:29:13.220
But really we can
learn a tremendous
00:29:13.220 --> 00:29:17.910
about functions which are
really hard to get at.
00:29:17.910 --> 00:29:20.330
So this implicit
differentiation method
00:29:20.330 --> 00:29:30.750
has one very, very
important application
00:29:30.750 --> 00:29:38.010
to finding inverse functions,
or finding derivatives
00:29:38.010 --> 00:29:40.650
of inverse functions.
00:29:40.650 --> 00:29:51.790
So let's talk about that next.
00:29:51.790 --> 00:29:55.700
So first, maybe we'll just
illustrate by an example.
00:29:55.700 --> 00:29:59.310
If you have the function y
is equal to square root x,
00:29:59.310 --> 00:30:04.600
for x positive, then
of course this idea
00:30:04.600 --> 00:30:07.170
is that we should
simplify this equation
00:30:07.170 --> 00:30:10.350
and we should square it so
we get this somewhat simpler
00:30:10.350 --> 00:30:11.860
equation here.
00:30:11.860 --> 00:30:14.080
And then we have a
notation for this.
00:30:14.080 --> 00:30:21.690
If we call f(x) equal to
square root of x, and g(y) = x,
00:30:21.690 --> 00:30:25.340
this is the reversal of this.
00:30:25.340 --> 00:30:33.150
Then the formula for g(y)
is that it should be y^2.
00:30:33.150 --> 00:30:48.310
And in general, if we start
with any old y = f(x),
00:30:48.310 --> 00:30:52.700
and we just write down, this
is the defining relationship
00:30:52.700 --> 00:30:57.050
for a function g, the property
that we're saying is that
00:30:57.050 --> 00:31:01.040
g(f(x)) has got to
bring us back to x.
00:31:01.040 --> 00:31:04.620
And we write that in a
couple of different ways.
00:31:04.620 --> 00:31:08.260
We call g the inverse of f.
00:31:08.260 --> 00:31:13.400
And also we call f
the inverse of g,
00:31:13.400 --> 00:31:15.960
although I'm going to be
silent about which variable
00:31:15.960 --> 00:31:19.300
I want to use, because people
mix them up a little bit,
00:31:19.300 --> 00:31:31.540
as we'll be doing when we
draw some pictures of this.
00:31:31.540 --> 00:31:32.620
So let's see.
00:31:32.620 --> 00:31:42.310
Let's draw pictures of
both f and f inverse
00:31:42.310 --> 00:31:50.260
on the same graph.
00:31:50.260 --> 00:32:02.130
So first of all, I'm going
to draw the graph of f(x)
00:32:02.130 --> 00:32:06.470
= square root of x.
00:32:06.470 --> 00:32:11.260
That's some shape like this.
00:32:11.260 --> 00:32:16.390
And now, in order to
understand what g(y) is,
00:32:16.390 --> 00:32:20.030
so let's do the
analysis in general,
00:32:20.030 --> 00:32:23.420
but then we'll draw it
in this particular case.
00:32:23.420 --> 00:32:31.780
If you have g(y)
= x, that's really
00:32:31.780 --> 00:32:34.460
just the same equation right?
00:32:34.460 --> 00:32:37.440
This is the equation
g(y) = x, that's y^2 = x.
00:32:37.440 --> 00:32:40.650
This is y = square root of x,
those are the same equations,
00:32:40.650 --> 00:32:43.330
it's the same curve.
00:32:43.330 --> 00:32:49.807
But suppose now that we wanted
to write down what g(x) is.
00:32:49.807 --> 00:32:51.890
In other words, we wanted
to switch the variables,
00:32:51.890 --> 00:32:55.650
so draw them as I said on the
same graph with the same x,
00:32:55.650 --> 00:32:59.800
and the same y axes.
00:32:59.800 --> 00:33:04.340
Then that would be, in effect,
trading the roles of x and y.
00:33:04.340 --> 00:33:07.670
We have to rename every
point on the graph which
00:33:07.670 --> 00:33:12.290
is the ordered pair (x, y), and
trade it for the opposite one.
00:33:12.290 --> 00:33:15.250
And when you
exchange x and y, so
00:33:15.250 --> 00:33:23.800
to do this, exchange
x and y, and when
00:33:23.800 --> 00:33:27.030
you do that, graphically
what that looks
00:33:27.030 --> 00:33:30.440
like is the following:
suppose you have a place here,
00:33:30.440 --> 00:33:33.580
and this is the x
and this is the y,
00:33:33.580 --> 00:33:35.030
then you want to trade them.
00:33:35.030 --> 00:33:39.620
So you want the y here right?
00:33:39.620 --> 00:33:41.360
And the x up there.
00:33:41.360 --> 00:33:44.400
It's sort of the opposite
place over there.
00:33:44.400 --> 00:33:51.130
And that is the place which is
directly opposite this point
00:33:51.130 --> 00:33:55.790
across the diagonal line x = y.
00:33:55.790 --> 00:33:58.410
So you reflect across this
or you flip across that.
00:33:58.410 --> 00:34:01.120
You get this other shape
that looks like that.
00:34:01.120 --> 00:34:10.780
Maybe I'll draw it with a
colored piece of chalk here.
00:34:10.780 --> 00:34:24.090
So this guy here
is y = f^(-1)(x).
00:34:24.090 --> 00:34:26.224
And indeed, if you
look at these graphs,
00:34:26.224 --> 00:34:27.390
this one is the square root.
00:34:27.390 --> 00:34:34.787
This one happens to be y = x^2.
00:34:34.787 --> 00:34:36.370
If you take this
one, and you turn it,
00:34:36.370 --> 00:34:39.890
you reverse the roles of
the x axis and the y axis,
00:34:39.890 --> 00:34:43.620
and tilt it on its side.
00:34:43.620 --> 00:34:51.150
So that's the picture of what an
inverse function is, and now I
00:34:51.150 --> 00:34:56.070
want to show you that the method
of implicit differentiation
00:34:56.070 --> 00:34:59.890
allows us to compute
the derivatives
00:34:59.890 --> 00:35:03.250
of inverse functions.
00:35:03.250 --> 00:35:05.100
So let me just
say it in general,
00:35:05.100 --> 00:35:07.310
and then I'll carry
it out in particular.
00:35:07.310 --> 00:35:16.390
So implicit
differentiation allows
00:35:16.390 --> 00:35:32.880
us to find the derivative
of any inverse function,
00:35:32.880 --> 00:35:53.380
provided we know the
derivative of the function.
00:35:53.380 --> 00:35:58.360
So let's do that for
what is an example, which
00:35:58.360 --> 00:36:02.510
is truly complicated and
a little subtle here.
00:36:02.510 --> 00:36:04.770
It has a very pretty answer.
00:36:04.770 --> 00:36:09.700
So we'll carry out
an example here,
00:36:09.700 --> 00:36:19.660
which is the function y is
equal to the inverse tangent.
00:36:19.660 --> 00:36:25.900
So again, for the
inverse tangent
00:36:25.900 --> 00:36:30.020
all of the things
that we're going to do
00:36:30.020 --> 00:36:32.360
are going to be
based on simplifying
00:36:32.360 --> 00:36:36.250
this equation by taking
the tangent of both sides.
00:36:36.250 --> 00:36:38.260
So, us let me remind
you by the way,
00:36:38.260 --> 00:36:41.780
the inverse tangent is what's
also known as arctangent.
00:36:41.780 --> 00:36:45.210
That's just another
notation for the same thing.
00:36:45.210 --> 00:36:49.770
And what we're going
to use to describe
00:36:49.770 --> 00:36:55.565
this function is the
equation tan y = x.
00:36:55.565 --> 00:36:56.940
That's what happens
when you take
00:36:56.940 --> 00:36:59.110
the tangent of this function.
00:36:59.110 --> 00:37:01.690
This is how we're
going to figure out
00:37:01.690 --> 00:37:19.650
what the function looks like.
00:37:19.650 --> 00:37:23.370
So first of all,
I want to draw it,
00:37:23.370 --> 00:37:26.610
and then we'll do
the computation.
00:37:26.610 --> 00:37:32.120
So let's make the diagram first.
00:37:32.120 --> 00:37:33.630
So I want to do
something which is
00:37:33.630 --> 00:37:35.879
analogous to what I did over
here with the square root
00:37:35.879 --> 00:37:38.060
function.
00:37:38.060 --> 00:37:43.740
So first of all, I remind
you that the tangent function
00:37:43.740 --> 00:37:52.850
is defined between two values
here, which are pi/2 and -pi/2.
00:37:52.850 --> 00:37:55.010
And it starts out
at minus infinity
00:37:55.010 --> 00:37:58.560
and curves up like this.
00:37:58.560 --> 00:38:08.050
So that's the function tan x.
00:38:08.050 --> 00:38:11.350
And so the one that
we have to sketch
00:38:11.350 --> 00:38:14.870
is this one which we
get by reflecting this
00:38:14.870 --> 00:38:21.780
across the axis.
00:38:21.780 --> 00:38:25.150
Well not the axis, the diagonal.
00:38:25.150 --> 00:38:33.080
This slope by the way, should
be less - a little lower here so
00:38:33.080 --> 00:38:37.580
that we can have it
going down and up.
00:38:37.580 --> 00:38:42.000
So let me show you
what it looks like.
00:38:42.000 --> 00:38:44.830
On the front, it's going to
look a lot like this one.
00:38:44.830 --> 00:38:50.620
So this one had curved
down, and so the reflection
00:38:50.620 --> 00:38:52.480
across the diagonal curved up.
00:38:52.480 --> 00:38:54.320
Here this is curving
up, so the reflection
00:38:54.320 --> 00:38:56.070
is going to curve down.
00:38:56.070 --> 00:38:58.760
It's going to look like this.
00:38:58.760 --> 00:39:02.360
Maybe I should, sorry,
let's use a different color,
00:39:02.360 --> 00:39:04.450
because it's
reversed from before.
00:39:04.450 --> 00:39:10.420
I'll just call it green.
00:39:10.420 --> 00:39:15.630
Now, the original curve
in the first quadrant
00:39:15.630 --> 00:39:17.990
eventually had an asymptote
which was straight up.
00:39:17.990 --> 00:39:24.090
So this one is going to have an
asymptote which is horizontal.
00:39:24.090 --> 00:39:27.260
And that level is what?
00:39:27.260 --> 00:39:29.950
What's the highest?
00:39:29.950 --> 00:39:30.780
It is just pi/2.
00:39:33.470 --> 00:39:35.910
Now similarly, the
other way, we're
00:39:35.910 --> 00:39:40.040
going to do this:
and this bottom level
00:39:40.040 --> 00:39:42.940
is going to be -pi/2.
00:39:42.940 --> 00:39:47.480
So there's the picture
of this function.
00:39:47.480 --> 00:39:50.300
It's defined for all x.
00:39:50.300 --> 00:39:57.530
So this green guy
is y = arctan x.
00:39:57.530 --> 00:39:59.530
And it's defined all the
way from minus infinity
00:39:59.530 --> 00:40:05.290
to infinity.
00:40:05.290 --> 00:40:11.410
And to use a notation that
we had from limit notation
00:40:11.410 --> 00:40:21.290
as x goes to infinity, let's
say, x is equal to pi/2.
00:40:21.290 --> 00:40:24.823
That's an example of one value
that's of interest in addition
00:40:24.823 --> 00:40:28.250
to the finite values.
00:40:28.250 --> 00:40:31.430
Okay, so now the
first ingredient
00:40:31.430 --> 00:40:34.580
that we're going
to need, is we're
00:40:34.580 --> 00:40:37.190
going to need the derivative
of the tangent function.
00:40:37.190 --> 00:40:40.060
So I'm going to recall
for you, and maybe you
00:40:40.060 --> 00:40:43.180
haven't worked this out yet, but
I hope that many of you have,
00:40:43.180 --> 00:40:48.790
that if you take the derivative
with respect to y of tan y.
00:40:48.790 --> 00:40:55.500
So this you do by
the quotient rule.
00:40:55.500 --> 00:40:59.150
So this is of the
form u/v, right?
00:40:59.150 --> 00:41:00.630
You use the quotient rule.
00:41:00.630 --> 00:41:06.090
So I'm going to get this.
00:41:06.090 --> 00:41:09.560
But what you get in the end is
some marvelous simplification
00:41:09.560 --> 00:41:12.580
that comes out to cos^2 y.
00:41:12.580 --> 00:41:14.740
1 over cosine squared.
00:41:14.740 --> 00:41:17.460
You can recognize the cosine
squared from the fact that you
00:41:17.460 --> 00:41:19.240
should get v^2 in
the denominator,
00:41:19.240 --> 00:41:26.630
and somehow the numerators all
cancel and simplifies to 1.
00:41:26.630 --> 00:41:32.560
This is also known
as secant squared y.
00:41:32.560 --> 00:41:38.010
So that something that
if you haven't done yet,
00:41:38.010 --> 00:41:48.450
you're going to have to
do this as an exercise.
00:41:48.450 --> 00:41:50.220
So we need that
ingredient, and now we're
00:41:50.220 --> 00:41:59.180
just going to
differentiate our equation.
00:41:59.180 --> 00:42:00.540
And what do we get?
00:42:00.540 --> 00:42:15.460
We get, again, (d/dy tan y)
times dy/dx is equal to 1.
00:42:15.460 --> 00:42:22.930
Or, if you like, 1 / cos^2 y
times, in the other notation,
00:42:22.930 --> 00:42:30.090
y', is equal to 1.
00:42:30.090 --> 00:42:35.740
So I've just used the formulas
that I just wrote down there.
00:42:35.740 --> 00:42:37.860
Now all I have to
do is solve for y'.
00:42:37.860 --> 00:42:44.060
It's cos^2 y.
00:42:44.060 --> 00:42:47.040
Unfortunately, this
is not the form
00:42:47.040 --> 00:42:49.610
that we ever want to
leave these things in.
00:42:49.610 --> 00:42:52.180
This is the same problem we
had with that ugly square root
00:42:52.180 --> 00:42:54.500
expression, or with
any of the others.
00:42:54.500 --> 00:42:58.070
We want to rewrite
in terms of x.
00:42:58.070 --> 00:43:05.280
Our original question was
what is d/dx of arctan x.
00:43:05.280 --> 00:43:08.260
Now so far we have the following
answer to that question:
00:43:08.260 --> 00:43:15.210
it's cos^2 (arctan x).
00:43:15.210 --> 00:43:31.780
Now this is a correct answer,
but way too complicated.
00:43:31.780 --> 00:43:33.180
Now that doesn't
mean that if you
00:43:33.180 --> 00:43:35.100
took a random
collection of functions,
00:43:35.100 --> 00:43:37.560
you wouldn't end up with
something this complicated.
00:43:37.560 --> 00:43:41.000
But these particular functions,
these beautiful circular
00:43:41.000 --> 00:43:42.900
functions involved
with trigonometry all
00:43:42.900 --> 00:43:45.610
have very nice formulas
associated with them.
00:43:45.610 --> 00:43:48.740
And this simplifies
tremendously.
00:43:48.740 --> 00:43:50.360
So one of the
skills that you need
00:43:50.360 --> 00:43:54.660
to develop when you're
dealing with trig functions
00:43:54.660 --> 00:43:56.690
is to simplify this.
00:43:56.690 --> 00:44:03.620
And so let's see now that
expressions like this all
00:44:03.620 --> 00:44:07.950
simplify.
00:44:07.950 --> 00:44:10.400
So here we go.
00:44:10.400 --> 00:44:12.264
There's only one
formula, one ingredient
00:44:12.264 --> 00:44:14.180
that we need to use to
do this, and then we're
00:44:14.180 --> 00:44:15.460
going to draw a diagram.
00:44:15.460 --> 00:44:18.530
So the ingredient again, is the
original defining relationship
00:44:18.530 --> 00:44:22.150
that tan y = x.
00:44:22.150 --> 00:44:27.780
So tan y = x can be
encoded in a right triangle
00:44:27.780 --> 00:44:34.520
in the following way: here's
the right triangle and tan
00:44:34.520 --> 00:44:38.590
y means that y should be
represented as an angle.
00:44:38.590 --> 00:44:40.820
And then, its
tangent is the ratio
00:44:40.820 --> 00:44:43.500
of this vertical to
this horizontal side.
00:44:43.500 --> 00:44:46.250
So I'm just going to pick
two values that work,
00:44:46.250 --> 00:44:48.990
namely x and 1.
00:44:48.990 --> 00:44:51.930
Those are the simplest ones.
00:44:51.930 --> 00:44:57.670
So I've encoded this
equation in this picture.
00:44:57.670 --> 00:45:01.560
And now all I have to do is
figure out what the cosine of y
00:45:01.560 --> 00:45:03.540
is in this right triangle here.
00:45:03.540 --> 00:45:06.081
In order to do that, I need to
figure out what the hypotenuse
00:45:06.081 --> 00:45:13.280
is, but that's just
square root of 1 + x^2.
00:45:13.280 --> 00:45:18.580
And now I can read off
what the cosine of y is.
00:45:18.580 --> 00:45:23.560
So the cosine of y is 1
divided by the hypotenuse.
00:45:23.560 --> 00:45:32.480
So it's 1 over square root,
whoops, yeah, 1 + x^2.
00:45:32.480 --> 00:45:39.900
And so cosine squared
is just 1 / 1 + x^2.
00:45:39.900 --> 00:45:43.050
And so our answer over here, the
preferred answer which is way
00:45:43.050 --> 00:45:45.670
simpler than what
I wrote up there,
00:45:45.670 --> 00:46:04.690
is that d/dx of tan inverse
x is equal to 1 over 1 + x^2.
00:46:04.690 --> 00:46:06.770
Maybe I'll stop here
for one more question.
00:46:06.770 --> 00:46:10.240
I have one more calculation
which I can do even
00:46:10.240 --> 00:46:11.500
in less than a minute.
00:46:11.500 --> 00:46:16.360
So we have a whole
minute for questions.
00:46:16.360 --> 00:46:20.480
Yeah?
00:46:20.480 --> 00:46:26.620
Student: [INAUDIBLE]
00:46:26.620 --> 00:46:34.210
Professor: What happens
to the inverse tangent?
00:46:34.210 --> 00:46:41.530
The inverse tangent--
Okay, this inverse tangent
00:46:41.530 --> 00:46:44.030
is the same as this y here.
00:46:44.030 --> 00:46:46.140
Those are the same thing.
00:46:46.140 --> 00:46:50.480
So what I did was I skipped
this step here entirely.
00:46:50.480 --> 00:46:52.070
I never wrote that down.
00:46:52.070 --> 00:46:54.560
But the inverse
tangent was that y.
00:46:54.560 --> 00:46:56.690
The issue was what's
a good formula
00:46:56.690 --> 00:47:01.080
for cos y in terms of x?
00:47:01.080 --> 00:47:04.510
So I am evaluating that, but
I'm doing it using the letter y.
00:47:04.510 --> 00:47:06.600
So in other words, what
happened to the inverse
00:47:06.600 --> 00:47:10.350
tangent is that I
called it y, which
00:47:10.350 --> 00:47:15.340
is what it's been all along.
00:47:15.340 --> 00:47:17.450
Okay, so now I'm
going to do the case
00:47:17.450 --> 00:47:20.400
of the sine, the inverse sine.
00:47:20.400 --> 00:47:22.890
And I'll show you
how easy this is
00:47:22.890 --> 00:47:27.420
if I don't fuss with-- because
this one has an easy trig
00:47:27.420 --> 00:47:29.760
identity associated with it.
00:47:29.760 --> 00:47:37.870
So if y = sin^(-1)
x, and sin y = x,
00:47:37.870 --> 00:47:40.970
and now watch how simple it is
when I do the differentiation.
00:47:40.970 --> 00:47:42.440
I just differentiate.
00:47:42.440 --> 00:47:50.780
I get (cos y) y' = 1.
00:47:50.780 --> 00:48:00.580
And then, y', so that
implies that = 1 / cos y,
00:48:00.580 --> 00:48:03.080
and now to rewrite
that in terms of x,
00:48:03.080 --> 00:48:10.550
I have to just recognize that
this is the same as this,
00:48:10.550 --> 00:48:14.910
which is the same as 1 /
square root of 1 - x^2.
00:48:14.910 --> 00:48:19.600
So all told, the derivative
with respect to x of the arcsine
00:48:19.600 --> 00:48:30.420
function is 1 / square
root of 1 - x^2.
00:48:30.420 --> 00:48:32.540
So these implicit
differentiations
00:48:32.540 --> 00:48:34.700
are very convenient.
00:48:34.700 --> 00:48:38.190
However, I warn you
that you do have
00:48:38.190 --> 00:48:42.670
to be careful about the range of
applicability of these things.
00:48:42.670 --> 00:48:44.550
You have to draw a
picture like this one
00:48:44.550 --> 00:48:47.210
to make sure you know
where this makes sense.
00:48:47.210 --> 00:48:50.380
In other words, you have to pick
a branch for the sine function
00:48:50.380 --> 00:48:52.050
to work that out,
and there's something
00:48:52.050 --> 00:48:53.330
like that on your problem set.
00:48:53.330 --> 00:48:56.030
And it's also
discussed in your text.
00:48:56.030 --> 00:48:58.070
So we'll stop here.