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Professor: So, we're ready
to begin the fifth lecture.

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I'm glad to be back.
 

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Thank you for entertaining my
colleague, Haynes Miller.

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So, today we're going to
continue where he started,

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namely what he talked about
was the chain rule, which is

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probably the most powerful
technique for extending the

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kinds of functions that
you can differentiate.

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And we're going to use the
chain rule in some rather

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clever algebraic ways today.
 

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So the topic for today is
what's known as implicit

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differentiation.
 

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So implicit differentiation is
a technique that allows you

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to differentiate a lot of
functions you didn't even

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know how to find before.
 

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And it's a technique - let's
wait for a few people

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to sit down here.
 

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Physics, huh?
 

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Ok, more Physics.
 

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Let's take a break.
 

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You can get those after class.
 

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All right, so we're talking
about implicit differentiation,

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and I'm going to illustrate
it by several examples.

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So this is one of the most
important and basic formulas

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that we've already
covered part way.

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Namely, the derivative of x
to a power is a x ^ a - 1.

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Now, what we've got so far is
the exponents, 0 plus or minus

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1 plus or minus 2, etc.
 

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You did the positive integer
powers in the first lecture,

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and then yesterday Professor
Miller told you about

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the negative powers.
 

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So what we're going to do right
now, today, is we're going to

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consider the exponents which
are rational numbers,

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ratios of integers.
 

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So a = m / n. m and
n are integers.

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All right, so that's our goal
for right now, and we're

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going to use this method of
implicit differentiation.

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In particular, it's important
to realize that this

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covers the case m = 1.
 

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And those are the nth roots.
 

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So when we take the one over n
power, we're going to cover

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that right now, along with many
other examples. so this

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is our first example.
 

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So how do we get started?
 

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Well we just write down a
formula for the function.

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The function is y = x^ m / n.
 

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That's what we're
trying to deal with.

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And now there's really
only two steps.

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The first step is to take this
equation to the nth power,

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so write it y ^ n = x ^m.
 

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Alright, so that's just the
same equation re-written.

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And now, what we're going
to do is we're going

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to differentiate.
 

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So we're going to apply
d / dx to the equation.

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Now why is that we can apply
it to the second equation,

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not the first equation?
 

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So maybe I should call these
equation 1 and equation 2.

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So, the point is, we can
apply it to equation 2.

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Now, the reason is that
we don't know how to

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differentiate x^ m / n.
 

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That's something we
just don't know yet.

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But we do know how to
differentiate integer powers.

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Those are the things that
we took care of before.

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So now we're in shape to be
able to do the differentiation.

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So I'm going to write it out
explicitly over here, without

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carrying it out just yet.
 

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That's d / dx of y^
n = d / dx of x ^m.

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And now you see this expression
here requires us to do

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something we couldn't
do before yesterday.

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Namely, this y is
a function of x.

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So we have to apply
the chain rule here.

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So this is the same as - this
is by the chain rule now

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- (d/dy y ^ n) dy / dx.
 

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And then, on the right hand
side, we can just carry it out.

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We know the formula.
 

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It's mx ^ m - 1.
 

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Right, now this is our scheme.
 

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And you'll see in a minute
why we win with this.

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So, first of all, there
are two factors here.

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One of them is unknown.
 

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In fact, it's what
we're looking for.

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But the other one is going to
be a known quantity, because we

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know how to differentiate y
to the n with respect to y.

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That's the same formula,
although the letter

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has been changed.
 

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And so this is the same as -
I'll write it underneath here

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- (n y^ n - 1) dy
/ dx = m x^m - 1.

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Okay, now comes, if you
like, the non-calculus

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part of the problem.
 

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Remember the non-calculus part
of the problem is always the

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messier part of the problem.
 

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So we want to figure
out this formula.

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This formula, the answer over
here, which maybe I'll put in

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a box now, has this
expressed much more simply,

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only in terms of x.
 

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And what we have to do now is
just solve for dy / dx using

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algebra, and then solve all
the way in terms of x.

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00:06:39 --> 00:06:44
So, first of all, we
solve for dy/ dx.

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So I do that by dividing the
factor on the left hand side.

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00:06:47 --> 00:06:56
So I get here m x ^m
- 1 / n y ^ n - 1.

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00:06:56 --> 00:07:02
And now I'm going to plug in --
so I'll write this as m / n.

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00:07:02 --> 00:07:04
This is x^ m - 1.
 

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00:07:04 --> 00:07:15
Now over here I'm going to put
in for y, (x ^ m / n)( n - 1).

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So now we're almost done, but
unfortunately we have this

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mess of exponents that
we have to work out.

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I'm going to write
it one more time.

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So I already recognize
the factor a out front.

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00:07:28 --> 00:07:30
That's not going to be a
problem for me, and that's

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what I'm aiming for here.
 

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But now I have to encode
all of these powers, so

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let's just write it.
 

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It's m - 1, and then it's minus
the quantity (n - 1) t m / n.

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All right, so that's the
laws exponents applied

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to this ratio here.
 

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And then we'll do the
arithmetic over here

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on the next board.
 

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So we have here m - 1 -
(n - 1) m / n = m - 1.

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And if I multiply n
by this, I get - m.

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And if the second factor is
minus minus, that's a plus.

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And that's + m /n.
 

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Altogether the two m's cancel.
 

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I have here - 1 + m / n.
 

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And lo and behold that's
the same thing as a -

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1, just what we wanted.
 

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All right, so this
equals a x^n - 1.

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00:08:31 --> 00:08:39
Okay, again just a
bunch of arithmetic.

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From this point forward, from
this substitution on, it's just

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the arithmetic of exponents.
 

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All right, so we've done
our first example here.

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I want to give you a
couple more examples,

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00:09:00 --> 00:09:04
so let's just continue.
 

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00:09:04 --> 00:09:08
The next example I'll
keep relatively simple.

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00:09:08 --> 00:09:14
So we have example two,
which is going to be the

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00:09:14 --> 00:09:18
function x^2 + y^2 = 1.
 

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00:09:18 --> 00:09:21
Well, that's not
really a function.

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00:09:21 --> 00:09:29
It's a way of defining y as
a function of x implicitly.

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00:09:29 --> 00:09:34
There's the idea that I could
solve for y if I wanted to.

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00:09:34 --> 00:09:36
And indeed let's do that.
 

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00:09:36 --> 00:09:42
So if you solve for y here,
what happens is you get y^2 = 1

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00:09:42 --> 00:09:52
- x ^2, and y = plus or minus
the square root of 1 - x ^2.

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00:09:52 --> 00:09:58
So this, if you like, is
the implicit definition.

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00:09:58 --> 00:10:00
And here is the explicit
function y, which

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00:10:00 --> 00:10:04
is a function of x.
 

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00:10:04 --> 00:10:06
And now just for my own
convenience, I'm just going

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00:10:06 --> 00:10:09
to take the positive branch.
 

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00:10:09 --> 00:10:13
This is the function.
 

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00:10:13 --> 00:10:15
It's just really a
circle in disguise.

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And I'm just going to
take the top part of

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00:10:18 --> 00:10:20
the circle, so we'll
 

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00:10:20 --> 00:10:24
take that top hump here.
 

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00:10:24 --> 00:10:27
All right, so that means I'm
erasing this minus sign.

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00:10:27 --> 00:10:35
I'm just taking the
positive branch, just

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00:10:35 --> 00:10:36
for my convenience.
 

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00:10:36 --> 00:10:40
I could do it just as well
with the negative branch.

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00:10:40 --> 00:10:47
Alright, so now I've taken
the solution, and I can

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00:10:47 --> 00:10:49
differentiate with this.
 

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00:10:49 --> 00:10:53
So rather than using the dy /
dx notation over here, I'm

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00:10:53 --> 00:10:55
going to switch notations over
here, because it's

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00:10:55 --> 00:10:56
less writing.
 

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00:10:56 --> 00:10:59
I'm going to write y '
and change notations.

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00:10:59 --> 00:11:04
Okay, so I want to take
the derivative of this.

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00:11:04 --> 00:11:11
Well this is a somewhat
complicated function here.

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00:11:11 --> 00:11:15
It's the square root of 1 -
x^2, and the right way always

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00:11:15 --> 00:11:22
to look at functions like this
is to rewrite them using the

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fractional power notation.
 

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00:11:26 --> 00:11:29
That's the first step in
computing a derivative

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00:11:29 --> 00:11:32
of a square root.
 

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00:11:32 --> 00:11:38
And then the second
step here is what?

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00:11:38 --> 00:11:40
Does somebody want to tell me?
 

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00:11:40 --> 00:11:43
Chain rule, right.
 

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That's it.
 

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00:11:44 --> 00:11:45
So we have two things.
 

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We start with one, and then
we do something else to it.

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00:11:47 --> 00:11:50
So whenever we do two things
to something, we need to

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00:11:50 --> 00:11:52
apply the chain rule.
 

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00:11:52 --> 00:11:55
So 1 - x^2, square root.
 

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00:11:55 --> 00:11:57
All right, so how
do we do that?

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00:11:57 --> 00:12:00
Well, the first factor I
claim is the derivative

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00:12:00 --> 00:12:01
of this thing.
 

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00:12:01 --> 00:12:06
So this is 1/2 ^ - 1/2.
 

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00:12:06 --> 00:12:09
So I'm doing this kind of
by the advanced method

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00:12:09 --> 00:12:11
now, because we've
already graduated.

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00:12:11 --> 00:12:14
You already did the
chain rule last time.

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00:12:14 --> 00:12:15
So what does this mean?
 

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00:12:15 --> 00:12:22
This is an abbreviation for the
derivative with respect to blah

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00:12:22 --> 00:12:28
of blah ^ 1/2, whatever it is.
 

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00:12:28 --> 00:12:30
All right, so that's the first
factor that we're going to use.

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00:12:30 --> 00:12:34
Rather than actually write out
a variable for it and pass

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00:12:34 --> 00:12:38
through as I did previously
with this y and x variable

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00:12:38 --> 00:12:42
here, I'm just going to skip
that step and let you imagine

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00:12:42 --> 00:12:45
it as being a placeholder
folder that variable here.

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00:12:45 --> 00:12:48
So this variable is
now parenthesis.

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00:12:48 --> 00:12:51
And then I have to multiply
that by the rate of

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00:12:51 --> 00:12:55
change of what's inside
with respect to x.

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00:12:55 --> 00:12:58
And that is going to be - 2x.
 

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00:12:58 --> 00:13:02
The derivative of
1 - x^2 = - 2x.

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00:13:02 --> 00:13:09
And now again, we couldn't have
done this example two before

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00:13:09 --> 00:13:14
example one, because we needed
to know that the power rule

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00:13:14 --> 00:13:18
worked not just for a integer
but also for a = 1/2.

207
00:13:19 --> 00:13:22
We're using the case
a = 1/2 right here.

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00:13:22 --> 00:13:29
It's 1/2 times, and this
- 1/2 here is a - 1.

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00:13:29 --> 00:13:32
So this is the case a = 1/2.
 

210
00:13:33 --> 00:13:39
a - 1 happens to be -1/2.
 

211
00:13:39 --> 00:13:41
Okay, so I'm putting all
those things together.

212
00:13:41 --> 00:13:44
And you know within a week
you have to be doing

213
00:13:44 --> 00:13:45
this very automatically.
 

214
00:13:45 --> 00:13:47
So we're going to do
it at this speed now.

215
00:13:47 --> 00:13:49
You want to do it even
faster, ultimately.

216
00:13:49 --> 00:13:50
Yes?
 

217
00:13:50 --> 00:13:53
Student: [INAUDIBLE]
 

218
00:13:53 --> 00:13:56
Professor: The question is
could I have done it implicitly

219
00:13:56 --> 00:13:58
without the square roots.
 

220
00:13:58 --> 00:13:59
And the answer is yes.
 

221
00:13:59 --> 00:14:02
That's what I'm about to do.
 

222
00:14:02 --> 00:14:05
So this is an illustration
of what's called the

223
00:14:05 --> 00:14:07
explicit solution.
 

224
00:14:07 --> 00:14:13
So this guy is what's
called explicit.

225
00:14:13 --> 00:14:17
And I want to contrast it
with the method that we're

226
00:14:17 --> 00:14:18
going to now use today.
 

227
00:14:18 --> 00:14:20
So it involves a lot
of complications.

228
00:14:20 --> 00:14:21
It involves the chain rule.
 

229
00:14:21 --> 00:14:23
And as we'll see it can
get messier and messier.

230
00:14:23 --> 00:14:27
And then there's the
implicit method, which

231
00:14:27 --> 00:14:29
I claim is easier.
 

232
00:14:29 --> 00:14:36
So let's see what happens if
you do it implicitly The

233
00:14:36 --> 00:14:41
implicit method involves,
instead of writing the function

234
00:14:41 --> 00:14:45
in this relatively complicated
way, with the square root, it

235
00:14:45 --> 00:14:47
involves leaving it alone.
 

236
00:14:47 --> 00:14:50
Don't do anything to it.
 

237
00:14:50 --> 00:14:52
In this previous case, we were
left with something which was

238
00:14:52 --> 00:14:57
complicated, say x ^ 1/3 or x ^
1/2 or something complicated.

239
00:14:57 --> 00:14:59
We had to simplify it.
 

240
00:14:59 --> 00:15:01
We had an equation one,
which was more complicated.

241
00:15:01 --> 00:15:03
We simplified it then
differentiated it.

242
00:15:03 --> 00:15:05
And so that was a simpler case.
 

243
00:15:05 --> 00:15:10
Well here, the simplest thing
us to differentiate is the one

244
00:15:10 --> 00:15:13
we started with, because
squares are practically the

245
00:15:13 --> 00:15:16
easiest thing after first
powers, or maybe zeroeth

246
00:15:16 --> 00:15:18
powers to differentiate.
 

247
00:15:18 --> 00:15:19
So we're leaving it alone.
 

248
00:15:19 --> 00:15:21
This is the simplest possible
form for it, and now we're

249
00:15:21 --> 00:15:23
going to differentiate.
 

250
00:15:23 --> 00:15:24
So what happens?
 

251
00:15:24 --> 00:15:26
So again what's the method?
 

252
00:15:26 --> 00:15:27
Let me remind you.
 

253
00:15:27 --> 00:15:30
You're applying d /
dx to the equation.

254
00:15:30 --> 00:15:33
So you have to differentiate
the left side of the equation,

255
00:15:33 --> 00:15:35
and differentiate the right
side of the equation.

256
00:15:35 --> 00:15:51
So it's this, and what you get
is 2 x + 2 y y ' = to what?

257
00:15:51 --> 00:15:52
0.
 

258
00:15:52 --> 00:15:56
The derivative of 1 = 0.
 

259
00:15:56 --> 00:15:58
So this is the
chain rule again.

260
00:15:58 --> 00:16:00
I did it a different way.
 

261
00:16:00 --> 00:16:02
I'm trying to get you
used to many different

262
00:16:02 --> 00:16:04
notations at once.
 

263
00:16:04 --> 00:16:05
Well really just two.
 

264
00:16:05 --> 00:16:10
Just the prime notation
and the dy / dx notation.

265
00:16:10 --> 00:16:14
And this is what I get.
 

266
00:16:14 --> 00:16:19
So now all I have to
do is solve for y '.

267
00:16:19 --> 00:16:23
So that y ', if I put the 2
x on the other side is - 2

268
00:16:23 --> 00:16:30
x, and then divide by
2y, which is -x /y.

269
00:16:30 --> 00:16:34
So let's compare our solutions,
and I'll apologize, I'm

270
00:16:34 --> 00:16:39
going to have to erase
something to do that.

271
00:16:39 --> 00:16:44
So let's compare
our two solutions.

272
00:16:44 --> 00:16:46
I'm going to put this
underneath and simplify.

273
00:16:46 --> 00:16:48
So what was our
solution over here?

274
00:16:48 --> 00:16:56
It was 1/2 (1 - x
^2) ^ -1/2 ( -2 x).

275
00:16:56 --> 00:17:02
That was what we got over here.
 

276
00:17:02 --> 00:17:07
And that is the same thing, if
I cancel the 2's, and I change

277
00:17:07 --> 00:17:08
it back to looking like a
square root, that's the same

278
00:17:08 --> 00:17:13
thing as - x / square
root of 1 - x ^2.

279
00:17:13 --> 00:17:18
So this is the formula for
the derivative when I

280
00:17:18 --> 00:17:21
do it the explicit way.
 

281
00:17:21 --> 00:17:29
And I'll just compare them,
these two expressions here.

282
00:17:29 --> 00:17:32
And notice they are the same.
 

283
00:17:32 --> 00:17:40
They're the same, because y
= square root of 1 - x^2.

284
00:17:40 --> 00:17:40
Yeah?
 

285
00:17:40 --> 00:17:41
Question?
 

286
00:17:41 --> 00:17:46
Student: [INAUDIBLE]
 

287
00:17:46 --> 00:17:48
Professor: The question is why
did the implicit method not

288
00:17:48 --> 00:17:50
give the bottom half
of the circle?

289
00:17:50 --> 00:17:53
Very good question.
 

290
00:17:53 --> 00:17:57
The answer to that
is that it did.

291
00:17:57 --> 00:17:59
I just didn't mention it.
 

292
00:17:59 --> 00:18:00
Wait, I'll explain.
 

293
00:18:00 --> 00:18:05
So suppose I stuck in
a minus sign here.

294
00:18:05 --> 00:18:08
I would have gotten this with
the difference, so with

295
00:18:08 --> 00:18:10
an extra minus sign.
 

296
00:18:10 --> 00:18:12
But then when I compared it to
what was over there, I would

297
00:18:12 --> 00:18:15
have had to have another
different minus sign over here.

298
00:18:15 --> 00:18:19
So actually both places would
get an extra minus sign.

299
00:18:19 --> 00:18:20
And they would still coincide.
 

300
00:18:20 --> 00:18:22
So actually the implicit
method is a little better.

301
00:18:22 --> 00:18:23
It doesn't even notice
the difference

302
00:18:23 --> 00:18:24
between the branches.
 

303
00:18:24 --> 00:18:28
It does the job on both
the top and bottom half.

304
00:18:28 --> 00:18:32
Another way of saying that
is that you're calculating

305
00:18:32 --> 00:18:33
the slopes here.
 

306
00:18:33 --> 00:18:35
So let's look at this picture.
 

307
00:18:35 --> 00:18:36
Here's a slope.
 

308
00:18:36 --> 00:18:40
Let's just take a look at a
positive value of x and just

309
00:18:40 --> 00:18:42
check the sign to see
what's happening.

310
00:18:42 --> 00:18:47
If you take a positive value of
x over here, x is positive.

311
00:18:47 --> 00:18:48
This denominator is positive.
 

312
00:18:48 --> 00:18:49
The slope is negative.
 

313
00:18:49 --> 00:18:52
You can see that
it's tilting down.

314
00:18:52 --> 00:18:53
So it's ok.
 

315
00:18:53 --> 00:18:59
Now on the bottom side, it's
going to be tilting up.

316
00:18:59 --> 00:19:02
And similarly what's happening
up here is that both x and y

317
00:19:02 --> 00:19:06
are positive, and this x
and this y are positive.

318
00:19:06 --> 00:19:07
And the slope is negative.
 

319
00:19:07 --> 00:19:09
On the other hand, on the
bottom side, x is still

320
00:19:09 --> 00:19:11
positive, but y is negative.
 

321
00:19:11 --> 00:19:15
And it's tilting up because
the denominator is negative.

322
00:19:15 --> 00:19:17
The numerator is positive,
and this minus sign

323
00:19:17 --> 00:19:19
has a positive slope.
 

324
00:19:19 --> 00:19:23
So it matches perfectly
in every category.

325
00:19:23 --> 00:19:27
This complicated, however, and
it's easier just to keep track

326
00:19:27 --> 00:19:32
of one branch at a time,
even in advanced math.

327
00:19:32 --> 00:19:37
Okay, so we only do it
one branch at a time.

328
00:19:37 --> 00:19:43
Other questions?
 

329
00:19:43 --> 00:19:47
Okay, so now I want to
give you a slightly more

330
00:19:47 --> 00:19:49
complicated example here.
 

331
00:19:49 --> 00:19:53
And indeed some of the,
so here's a little more

332
00:19:53 --> 00:19:54
complicated example.
 

333
00:19:54 --> 00:19:57
It's not going to be the most
complicated example, but you

334
00:19:57 --> 00:20:17
know it'll be a little tricky.
 

335
00:20:17 --> 00:20:22
So this example, I'm going
to give you a fourth

336
00:20:22 --> 00:20:23
order equation.
 

337
00:20:23 --> 00:20:31
So y ^ 4 + x y ^2 - 2 = 0.
 

338
00:20:31 --> 00:20:37
Now it just so happens that
there's a trick to solving this

339
00:20:37 --> 00:20:41
equation, so actually you can
do both the explicit method

340
00:20:41 --> 00:20:46
and the non-explicit method.
 

341
00:20:46 --> 00:20:50
So the explicit method would
say okay well, I want

342
00:20:50 --> 00:20:51
to solve for this.
 

343
00:20:51 --> 00:20:55
So I'm going to use the
quadratic formula, but on y ^2.

344
00:20:55 --> 00:20:59
This is quadratic in y ^2,
because there's a fourth power

345
00:20:59 --> 00:21:01
and a second power, and
the first and third

346
00:21:01 --> 00:21:03
powers are missing.
 

347
00:21:03 --> 00:21:09
So this is y ^2 = - x plus
or minus the square root

348
00:21:09 --> 00:21:19
of x ^2 - 4 (-2 ) / 2.
 

349
00:21:19 --> 00:21:22
And so this x is the b.
 

350
00:21:22 --> 00:21:29
This -2 is the c, and a = 1
in the quadratic formula.

351
00:21:29 --> 00:21:36
And so the formula for y is
plus or minus the square root

352
00:21:36 --> 00:21:45
of - x plus or minus the
square root x ^2 + 8 /2.

353
00:21:45 --> 00:21:48
So now you can see this problem
of branches, this happens

354
00:21:48 --> 00:21:53
actually in a lot of cases,
coming up in an elaborate way.

355
00:21:53 --> 00:21:54
You have two choices
for the sign here.

356
00:21:54 --> 00:21:56
You have two choices
for the sign here.

357
00:21:56 --> 00:21:59
Conceivably as many as four
roots for this equation,

358
00:21:59 --> 00:22:02
because it's a fourth
degree equation.

359
00:22:02 --> 00:22:02
It's quite a mess.
 

360
00:22:02 --> 00:22:06
You should have to check
each branch separately.

361
00:22:06 --> 00:22:10
And this really is that level
of complexity, and in general

362
00:22:10 --> 00:22:16
it's very difficult to figure
out the formulas for

363
00:22:16 --> 00:22:17
quartic equations.
 

364
00:22:17 --> 00:22:21
But fortunately we're
never going to use them.

365
00:22:21 --> 00:22:24
That is, we're never going
to need those formulas.

366
00:22:24 --> 00:22:31
So the implicit method
is far easier.

367
00:22:31 --> 00:22:36
The implicit method just says
okay I'll leave the equation

368
00:22:36 --> 00:22:38
in its simplest form.
 

369
00:22:38 --> 00:22:40
And now differentiate.
 

370
00:22:40 --> 00:22:48
So when I differentiate, I get
4 y ^3 y ' +... now here I have

371
00:22:48 --> 00:22:50
to apply the product rule.
 

372
00:22:50 --> 00:22:56
So I differentiate the x
and the y ^2 separately.

373
00:22:56 --> 00:22:59
First I differentiate with
respect to x, so I get y ^2.

374
00:22:59 --> 00:23:02
Then I differentiate with
respect to the other

375
00:23:02 --> 00:23:04
factor, the y ^2 factors.
 

376
00:23:04 --> 00:23:08
And I get x (2 y y ').
 

377
00:23:08 --> 00:23:10
And then the 0 gives me 0.
 

378
00:23:10 --> 00:23:16
So - 0 = 0.
 

379
00:23:16 --> 00:23:21
So there's the implicit
differentiation step.

380
00:23:21 --> 00:23:26
And now I just want
to solve for y '.

381
00:23:26 --> 00:23:32
So I'm going to factor
out 4 y ^3 + 2 x y.

382
00:23:32 --> 00:23:35
That's the factor on y '.
 

383
00:23:35 --> 00:23:39
And I'm going to put the
y ^2 on the other side.

384
00:23:39 --> 00:23:43
- y ^2 over here.
 

385
00:23:43 --> 00:23:55
And so the formula for y '
is - y ^2 / 4 y ^3 + 2 x y.

386
00:23:55 --> 00:24:01
So that's the formula
for the solution.

387
00:24:01 --> 00:24:07
For the slope.
 

388
00:24:07 --> 00:24:07
You have a question?
 

389
00:24:07 --> 00:24:16
Student: [INAUDIBLE]
 

390
00:24:16 --> 00:24:19
Professor: So the question is
for the y would we have to

391
00:24:19 --> 00:24:22
put in what solved for in
the explicit equation.

392
00:24:22 --> 00:24:24
And the answer is
absolutely yes.

393
00:24:24 --> 00:24:25
That's exactly the point.
 

394
00:24:25 --> 00:24:30
So this is not a complete
solution to a problem.

395
00:24:30 --> 00:24:32
We started with an
implicit equation.

396
00:24:32 --> 00:24:33
We differentiated.
 

397
00:24:33 --> 00:24:36
And we got in the end, also
an implicit equation.

398
00:24:36 --> 00:24:39
It doesn't tell us what y
is as a function of x.

399
00:24:39 --> 00:24:43
You have to go back to
this formula to get

400
00:24:43 --> 00:24:45
the formula for x.
 

401
00:24:45 --> 00:24:49
So for example, let me
give you an example here.

402
00:24:49 --> 00:24:54
So this hides a degree of
complexity of the problem.

403
00:24:54 --> 00:24:58
But it's a degree of complexity
that we must live with.

404
00:24:58 --> 00:25:10
So for example, at x = 1, you
can see that y = 1 solves.

405
00:25:10 --> 00:25:16
That happens to solve y
^ 4 + x y ^2 - 2= 0.

406
00:25:16 --> 00:25:18
That's why I picked the
2 actually, so it would

407
00:25:18 --> 00:25:21
be 1 + 1 - 2 = 0.
 

408
00:25:21 --> 00:25:22
I just wanted to have a
convenient solution

409
00:25:22 --> 00:25:25
there to pull out of my
hat at this point.

410
00:25:25 --> 00:25:26
So I did that.
 

411
00:25:26 --> 00:25:29
And so we now know
that when x = 1, y

412
00:25:29 --> 00:25:30
= 1.
 

413
00:25:30 --> 00:25:41
So at (1, 1) along the curve,
the slope is equal to what?

414
00:25:41 --> 00:25:52
Well, I have to plug in here,
- 1 ^2 / 4 * 1^3 + 2 * 1 * 1.

415
00:25:52 --> 00:25:54
That's just plugging in that
formula over there, which

416
00:25:54 --> 00:25:59
turns out to be - 1/6.
 

417
00:25:59 --> 00:26:00
So I can get it.
 

418
00:26:00 --> 00:26:21
On the other hand, at say x =
2, we're stuck using this

419
00:26:21 --> 00:26:32
formula star here to find y.
 

420
00:26:32 --> 00:26:38
Now, so let me just make two
points about this, which are

421
00:26:38 --> 00:26:42
just philosophical points
for you right now.

422
00:26:42 --> 00:26:46
The first is, when I promised
you at the beginning of this

423
00:26:46 --> 00:26:48
class that we were going to
be able to differentiate

424
00:26:48 --> 00:26:53
any function you know, I
meant it very literally.

425
00:26:53 --> 00:26:56
What I meant is if you know the
function, we'll be able give

426
00:26:56 --> 00:26:58
a formula for the derivative.
 

427
00:26:58 --> 00:27:00
If you don't know how to find a
function, you'll have a lot of

428
00:27:00 --> 00:27:02
trouble finding the derivative.
 

429
00:27:02 --> 00:27:06
So we didn't make any promises
that if you can't find the

430
00:27:06 --> 00:27:07
function you will be able to
find the derivative

431
00:27:07 --> 00:27:09
by some magic.
 

432
00:27:09 --> 00:27:10
That will never happen.
 

433
00:27:10 --> 00:27:14
And however complex the
function is, a root of a fourth

434
00:27:14 --> 00:27:18
degree polynomial can be pretty
complicated function of the

435
00:27:18 --> 00:27:22
coefficients, we're stuck with
this degree of complexity

436
00:27:22 --> 00:27:23
in the problem.
 

437
00:27:23 --> 00:27:27
But the big advantage of his
method, notice, is that

438
00:27:27 --> 00:27:29
although we've had to find
star, we had to find this

439
00:27:29 --> 00:27:33
formula star, and there are
many other ways of doing these

440
00:27:33 --> 00:27:36
things numerically, by the way,
which we'll learn later, so

441
00:27:36 --> 00:27:39
there's a good method for
doing it numerically.

442
00:27:39 --> 00:27:41
Although we had to find
star, we never had

443
00:27:41 --> 00:27:43
to differentiate it.
 

444
00:27:43 --> 00:27:46
We had a fast way of
getting the slope.

445
00:27:46 --> 00:27:48
So we had to know
what x and y were.

446
00:27:48 --> 00:27:51
But y ' we got by an
algebraic formula, in

447
00:27:51 --> 00:27:54
terms of the values here.
 

448
00:27:54 --> 00:27:57
So this is very fast,
forgetting the slope, once

449
00:27:57 --> 00:28:02
you know the point. yes?
 

450
00:28:02 --> 00:28:03
Student: What's in
the parentheses?

451
00:28:03 --> 00:28:05
Professor: Sorry, this is...
 

452
00:28:05 --> 00:28:06
Well let's see if I
can manage this.

453
00:28:06 --> 00:28:16
Is this the parentheses
you're talking about?

454
00:28:16 --> 00:28:18
Well, so maybe I should
put commas around it.

455
00:28:18 --> 00:28:24
But it was "s a y",
comma comma, okay?

456
00:28:24 --> 00:28:28
Well here was at x = 1.
 

457
00:28:28 --> 00:28:33
I'm just throwing out a value.
 

458
00:28:33 --> 00:28:34
Any other value.
 

459
00:28:34 --> 00:28:36
Actually there is one
value, my favorite value.

460
00:28:36 --> 00:28:39
Well this is easy to
evaluate right? x =

461
00:28:39 --> 00:28:42
0, I can do it there.
 

462
00:28:42 --> 00:28:45
That's maybe the only one.
 

463
00:28:45 --> 00:28:55
The others are a nuisance.
 

464
00:28:55 --> 00:29:03
All right, other questions?
 

465
00:29:03 --> 00:29:06
Now we have to do
something more here.

466
00:29:06 --> 00:29:10
So I claimed to you that we
could differentiate all

467
00:29:10 --> 00:29:11
the functions we know.
 

468
00:29:11 --> 00:29:14
But really we can learn a
tremendous about functions

469
00:29:14 --> 00:29:17
which are really
hard to get at.

470
00:29:17 --> 00:29:20
So this implicit
differentiation method has one

471
00:29:20 --> 00:29:32
very, very important
application to finding inverse

472
00:29:32 --> 00:29:38
functions, or finding
derivatives of

473
00:29:38 --> 00:29:40
inverse functions.
 

474
00:29:40 --> 00:29:51
So let's talk about that next.
 

475
00:29:51 --> 00:29:55
So first, maybe we'll just
illustrate by an example.

476
00:29:55 --> 00:30:01
If you have the function y = to
square root x, for x positive,

477
00:30:01 --> 00:30:06
then of course this idea is
that we should simplify this

478
00:30:06 --> 00:30:09
equation and we should square
it so we get this somewhat

479
00:30:09 --> 00:30:11
simpler equation here.
 

480
00:30:11 --> 00:30:14
And then we have a
notation for this.

481
00:30:14 --> 00:30:21
If we call f (x) = the square
root of x, and g ( y) = x,

482
00:30:21 --> 00:30:25
this is the reversal of this.
 

483
00:30:25 --> 00:30:33
Then the formula for g(y)
is that it should be y ^2.

484
00:30:33 --> 00:30:48
And in general, if we start
with any old y = f(x), and we

485
00:30:48 --> 00:30:52
just write down, this is the
defining relationship for a

486
00:30:52 --> 00:30:58
function g, the property that
we're saying is that g ( f(x))

487
00:30:58 --> 00:31:01
has got to bring us back to x.
 

488
00:31:01 --> 00:31:04
And we write that in a
couple of different ways.

489
00:31:04 --> 00:31:08
We call g the inverse of f.
 

490
00:31:08 --> 00:31:14
And also we call f the inverse
of g, although I'm going to be

491
00:31:14 --> 00:31:17
silent about which variable I
want to use, because people mix

492
00:31:17 --> 00:31:21
them up a little bit, as
we'll be doing when we draw

493
00:31:21 --> 00:31:31
some pictures of this.
 

494
00:31:31 --> 00:31:32
So let's see.
 

495
00:31:32 --> 00:31:42
Let's draw pictures of
both f and f inverse

496
00:31:42 --> 00:31:50
on the same graph.
 

497
00:31:50 --> 00:32:01
So first of all, I'm going
to draw the graph of

498
00:32:01 --> 00:32:06
f(x) = square root of x.
 

499
00:32:06 --> 00:32:11
That's some shape like this.
 

500
00:32:11 --> 00:32:18
And now, in order to understand
what g ( y) is, so let's do the

501
00:32:18 --> 00:32:21
analysis in general, but then
we'll draw it in this

502
00:32:21 --> 00:32:23
particular case.
 

503
00:32:23 --> 00:32:32
If you have g (y) = x,
that's really just the

504
00:32:32 --> 00:32:34
same equation right?
 

505
00:32:34 --> 00:32:37
This is the equation g (y)
= x, that's y ^2 = x.

506
00:32:37 --> 00:32:40
This is y = square root of x,
those are the same equations,

507
00:32:40 --> 00:32:43
it's the same curve.
 

508
00:32:43 --> 00:32:50
But suppose now that we wanted
to write down what g( x) is.

509
00:32:50 --> 00:32:52
In other words, we wanted to
switch the variables, so draw

510
00:32:52 --> 00:32:55
them as I said on the same
graph with the same x,

511
00:32:55 --> 00:32:59
and the same y axes.
 

512
00:32:59 --> 00:33:04
Then that would be, in effect,
trading the roles of x and y.

513
00:33:04 --> 00:33:08
We have to rename every point
on the graph which is of the

514
00:33:08 --> 00:33:12
ordered pair (x, y), and trade
it for the opposite one.

515
00:33:12 --> 00:33:20
And when you exchange x and y,
so to do this, exchange x and

516
00:33:20 --> 00:33:27
y, and when you do that,
graphically what that looks

517
00:33:27 --> 00:33:31
like is the following: suppose
you have a place here, and this

518
00:33:31 --> 00:33:35
is the x and this is the y,
then you want to trade them.

519
00:33:35 --> 00:33:39
So you want the y here right?
 

520
00:33:39 --> 00:33:41
And the x up there.
 

521
00:33:41 --> 00:33:44
It's sort of the opposite
place over there.

522
00:33:44 --> 00:33:51
And that is the place which is
directly opposite this point

523
00:33:51 --> 00:33:55
across the diagonal line x = y.
 

524
00:33:55 --> 00:33:58
So you reflect across this
or you flip across that.

525
00:33:58 --> 00:34:01
You get this other shape
that looks like that.

526
00:34:01 --> 00:34:10
Maybe I'll draw it with a
colored piece of chalk here.

527
00:34:10 --> 00:34:24
So this guy here
is y = f^(-1)(x).

528
00:34:24 --> 00:34:26
And indeed, if you look at
these graphs, this one

529
00:34:26 --> 00:34:27
is the square roots.
 

530
00:34:27 --> 00:34:35
This one happens to be y = x^2.
 

531
00:34:35 --> 00:34:37
If you take this one, and you
turn it, you reverse the roles

532
00:34:37 --> 00:34:43
of the x axis and the y axis,
and tilt it on its side.

533
00:34:43 --> 00:34:51
So that's the picture of what
an inverse function is, and now

534
00:34:51 --> 00:34:54
I want to show you that the
method of implicit

535
00:34:54 --> 00:34:59
differentiation allows us to
compute the derivatives of

536
00:34:59 --> 00:35:03
inverse functions.
 

537
00:35:03 --> 00:35:05
So let me just say it in
general, and then I'll carry

538
00:35:05 --> 00:35:07
it out in particular.
 

539
00:35:07 --> 00:35:22
So implicit differentiation
allows us to find the

540
00:35:22 --> 00:35:39
derivative of any inverse
function, provided we know the

541
00:35:39 --> 00:35:53
derivative of the function.
 

542
00:35:53 --> 00:35:59
So let's do that for what
is an example, which is

543
00:35:59 --> 00:36:02
truly complicated and
a little subtle here.

544
00:36:02 --> 00:36:04
It has a very pretty answer.
 

545
00:36:04 --> 00:36:10
So we'll carry out an
example here, which is the

546
00:36:10 --> 00:36:19
function y = tan^(-1).
 

547
00:36:19 --> 00:36:29
So again, for the inverse
tangent all of the things that

548
00:36:29 --> 00:36:32
we're going to do are going to
be based on simplifying this

549
00:36:32 --> 00:36:36
equation by taking the
tangent of both sides.

550
00:36:36 --> 00:36:39
So, us let me remind you by the
way, the inverse tangent is

551
00:36:39 --> 00:36:41
what's also known
as arctangent.

552
00:36:41 --> 00:36:45
That's just another notation
for the same thing.

553
00:36:45 --> 00:36:51
And we're going to use to
describe this function is

554
00:36:51 --> 00:36:55
the equation tan y = x.
 

555
00:36:55 --> 00:36:58
That's what happens when
you take the tangent

556
00:36:58 --> 00:36:59
of this function.
 

557
00:36:59 --> 00:37:01
This is how we're going
to figure out what the

558
00:37:01 --> 00:37:19
function looks like.
 

559
00:37:19 --> 00:37:23
So first of all, I want to
draw it, and then we'll

560
00:37:23 --> 00:37:26
do the computation.
 

561
00:37:26 --> 00:37:32
So let's make the
diagram first.

562
00:37:32 --> 00:37:34
So I want to do something which
is analogous to what I did over

563
00:37:34 --> 00:37:38
here with the square
root function.

564
00:37:38 --> 00:37:43
So first of all, I remind you
that the tangent function is

565
00:37:43 --> 00:37:52
defined find two values here,
which are pi / 2 and - pi/2.

566
00:37:52 --> 00:37:55
And it starts out at minus
infinity and curves

567
00:37:55 --> 00:37:58
up like this.
 

568
00:37:58 --> 00:38:08
So that's the function tan x.
 

569
00:38:08 --> 00:38:12
And so the one that we have to
sketch is this one which we get

570
00:38:12 --> 00:38:21
by reflecting this
across the axis.

571
00:38:21 --> 00:38:25
Well not the axis,
the diagonal.

572
00:38:25 --> 00:38:32
This slope by the way, should
be less - a little lower here

573
00:38:32 --> 00:38:37
so that we can have it
going down and up.

574
00:38:37 --> 00:38:42
So let me show you
what it looks like.

575
00:38:42 --> 00:38:44
On the front, it's going to
look a lot like this one.

576
00:38:44 --> 00:38:50
So this one had curved down,
and so the reflection across

577
00:38:50 --> 00:38:52
the diagonal curved up.
 

578
00:38:52 --> 00:38:54
Here this is curving up,
so the reflection is

579
00:38:54 --> 00:38:56
going to curve down.
 

580
00:38:56 --> 00:38:58
It's going to look like this.
 

581
00:38:58 --> 00:39:02
Maybe I should, sorry, let's
use a different color, because

582
00:39:02 --> 00:39:04
it's reversed from before.
 

583
00:39:04 --> 00:39:10
I'll just call it green.
 

584
00:39:10 --> 00:39:16
Now, the original curve in the
first quadrant eventually had

585
00:39:16 --> 00:39:17
an asymptote which
was straight up.

586
00:39:17 --> 00:39:24
So this one is going to have an
asymptote which is horizontal.

587
00:39:24 --> 00:39:27
And that level is what?
 

588
00:39:27 --> 00:39:29
What's the highest?
 

589
00:39:29 --> 00:39:33
It is just pi / 2.
 

590
00:39:33 --> 00:39:37
Now similarly, the other way,
we're going to do this: and

591
00:39:37 --> 00:39:42
this bottom level is
going to be - pi/2.

592
00:39:42 --> 00:39:47
So there's the picture
of this function.

593
00:39:47 --> 00:39:50
It's defined for all x.
 

594
00:39:50 --> 00:39:57
So this green guy
is y = arctan x.

595
00:39:57 --> 00:39:59
And it's defined all
the way from minus

596
00:39:59 --> 00:40:05
infinity to infinity.
 

597
00:40:05 --> 00:40:11
And to use a notation that we
had from limit notation as x

598
00:40:11 --> 00:40:21
goes to infinity, let's
say, arctan x = pi/2.

599
00:40:21 --> 00:40:24
That's an example of one value
that's of interest in addition

600
00:40:24 --> 00:40:28
to the finite values.
 

601
00:40:28 --> 00:40:31
Okay, so now the first
ingredient that we're going to

602
00:40:31 --> 00:40:35
need, is we're going to need
the derivative of the

603
00:40:35 --> 00:40:37
tangent function.
 

604
00:40:37 --> 00:40:40
So I'm going to recall for you,
and maybe you haven't worked

605
00:40:40 --> 00:40:43
this out yet, but I hope that
many of you have, that if you

606
00:40:43 --> 00:40:48
take the derivative with
respect to y of tan y .

607
00:40:48 --> 00:40:55
So this you do by
the quotient rule.

608
00:40:55 --> 00:40:59
So this is of the
form u / v, right?

609
00:40:59 --> 00:41:00
You use the quotient rule.
 

610
00:41:00 --> 00:41:06
So I'm going to get this.
 

611
00:41:06 --> 00:41:09
But what you get in the end is
some marvelous simplification

612
00:41:09 --> 00:41:12
that comes out to cos ^2y.
 

613
00:41:12 --> 00:41:14
1 / cos squared.
 

614
00:41:14 --> 00:41:17
You can recognize the cosine
squared from the fact that

615
00:41:17 --> 00:41:19
you should get v ^2 in the
denominator, and somehow the

616
00:41:19 --> 00:41:26
numerators all cancel
and simplifies to 1.

617
00:41:26 --> 00:41:32
This is also known as sec^2y.
 

618
00:41:32 --> 00:41:38
So that something that if you
haven't done yet, you're going

619
00:41:38 --> 00:41:48
to have to do this
as an exercise.

620
00:41:48 --> 00:41:50
So we need that ingredient,
and now we're just going to

621
00:41:50 --> 00:41:59
differentiate our equation.
 

622
00:41:59 --> 00:42:00
And what do we get?
 

623
00:42:00 --> 00:42:15
We get, again, (d /dy
tan y ) dy / dx = 1.

624
00:42:15 --> 00:42:22
Or, if you like, 1 / cos
^2 y times in the other

625
00:42:22 --> 00:42:30
notation, y' = 1.
 

626
00:42:30 --> 00:42:35
So I've just used the formulas
that I just wrote down there.

627
00:42:35 --> 00:42:37
Now all I have to do
is solve for y '.

628
00:42:37 --> 00:42:44
It's cos ^2y.
 

629
00:42:44 --> 00:42:47
Unfortunately, this is not the
form that we ever want to

630
00:42:47 --> 00:42:49
leave these things in.
 

631
00:42:49 --> 00:42:52
This is the same problem we had
with that ugly square root

632
00:42:52 --> 00:42:54
expression, or with
any of the others.

633
00:42:54 --> 00:42:58
We want to rewrite
in terms of x.

634
00:42:58 --> 00:43:05
Our original question was
what is d / dx of arctan x.

635
00:43:05 --> 00:43:07
Now so far we have the
following answer to that

636
00:43:07 --> 00:43:15
question: it's cos
^2 ( arctan x).

637
00:43:15 --> 00:43:31
Now this is a correct answer,
but way too complicated.

638
00:43:31 --> 00:43:34
Now that doesn't mean that if
you took a random collection of

639
00:43:34 --> 00:43:36
functions, you wouldn't end
up with something

640
00:43:36 --> 00:43:37
this complicated.
 

641
00:43:37 --> 00:43:41
But these particular functions,
these beautiful circular

642
00:43:41 --> 00:43:43
functions involved with
trigonometry all have very nice

643
00:43:43 --> 00:43:45
formulas associated with them.
 

644
00:43:45 --> 00:43:48
And this simplifies
tremendously.

645
00:43:48 --> 00:43:52
So one of the skills that you
need to develop when you're

646
00:43:52 --> 00:43:56
dealing with trig functions
is to simplify this.

647
00:43:56 --> 00:44:02
And so let's see now
that expressions like

648
00:44:02 --> 00:44:07
this all simplify.
 

649
00:44:07 --> 00:44:10
So here we go.
 

650
00:44:10 --> 00:44:13
There's only one formula, one
ingredient that we need to use

651
00:44:13 --> 00:44:15
to do this, and then we're
going to draw a diagram.

652
00:44:15 --> 00:44:17
So the ingredient again,
is the original defining

653
00:44:17 --> 00:44:22
relationship that tan y = x.
 

654
00:44:22 --> 00:44:28
So tan y = x can be encoded in
a right triangle in the

655
00:44:28 --> 00:44:35
following way: here's the right
triangle and tan y means that y

656
00:44:35 --> 00:44:38
should be represented
as an angle.

657
00:44:38 --> 00:44:41
And then, its tangent is
the ratio of this vertical

658
00:44:41 --> 00:44:43
to this horizontal side.
 

659
00:44:43 --> 00:44:46
So I'm just going to pick
two values that work,

660
00:44:46 --> 00:44:48
namely x and 1.
 

661
00:44:48 --> 00:44:51
Those are the simplest ones.
 

662
00:44:51 --> 00:44:57
So I've encoded this
equation in this picture.

663
00:44:57 --> 00:45:01
And now all I have to do is
figure out what the cos y is

664
00:45:01 --> 00:45:03
in this right trying here.
 

665
00:45:03 --> 00:45:05
In order to do that, I need to
figure out what the hypotenuse

666
00:45:05 --> 00:45:13
is, but that's just
square root of 1 + x ^2.

667
00:45:13 --> 00:45:18
And now I can read off
what the cos y is.

668
00:45:18 --> 00:45:23
So the cos y is one divided
by the hypotenuse.

669
00:45:23 --> 00:45:32
So it's 1 / square root,
whoops, yeah, 1 + x^2.

670
00:45:32 --> 00:45:39
And so cos ^2 is
just 1 / 1 + x^2.

671
00:45:39 --> 00:45:42
And so our answer over here,
the preferred answer which is

672
00:45:42 --> 00:45:48
way simpler than what I wrote
up there, is that d/dx of

673
00:45:48 --> 00:46:04
arctan x = 1 / 1 + x^2.
 

674
00:46:04 --> 00:46:06
Maybe I'll stop here
for one more question.

675
00:46:06 --> 00:46:10
I have one more calculation
which I can do even in

676
00:46:10 --> 00:46:11
less than a minute.
 

677
00:46:11 --> 00:46:16
So we have a whole
minute for questions.

678
00:46:16 --> 00:46:20
Yeah?
 

679
00:46:20 --> 00:46:26
Student: [INAUDIBLE]
 

680
00:46:26 --> 00:46:34
Professor: What happens
to the inverse tangent?

681
00:46:34 --> 00:46:39
The inverse tangent this...
 

682
00:46:39 --> 00:46:44
Ok this inverse tangent is
the same as this y here.

683
00:46:44 --> 00:46:46
Those are the same thing.
 

684
00:46:46 --> 00:46:50
So what I did was I skipped
this step here entirely.

685
00:46:50 --> 00:46:52
I never wrote that down.
 

686
00:46:52 --> 00:46:54
But the inverse
tangent was that y.

687
00:46:54 --> 00:46:57
The issue was what's a
good formula for cos

688
00:46:57 --> 00:47:01
y in terms of x?
 

689
00:47:01 --> 00:47:03
So I am evaluating
that, but I'm doing it

690
00:47:03 --> 00:47:04
using the letter y.
 

691
00:47:04 --> 00:47:08
So in other words, what
happened to the inverse tangent

692
00:47:08 --> 00:47:15
is that I called it y, which is
what it's been all along.

693
00:47:15 --> 00:47:17
Okay, so now I'm going
to do the case of the

694
00:47:17 --> 00:47:20
sine, the inverse sine.
 

695
00:47:20 --> 00:47:25
And I'll show you how easy this
is if I don't fuss with...

696
00:47:25 --> 00:47:28
because this one has
an easy trig identity

697
00:47:28 --> 00:47:29
associated with it.
 

698
00:47:29 --> 00:47:39
So if y = arcsin x, and sin y =
x, and now watch how simple it

699
00:47:39 --> 00:47:40
is when I do the
differentiation.

700
00:47:40 --> 00:47:42
I just differentiate.
 

701
00:47:42 --> 00:47:50
I get (cos y) y ' = 1.
 

702
00:47:50 --> 00:48:00
And then, y ', so that implies
that y ' = 1 / cos y, and now

703
00:48:00 --> 00:48:05
to rewrite that in terms of x,
I have to just recognize that

704
00:48:05 --> 00:48:12
this is the same as this,
which is the same as 1 /

705
00:48:12 --> 00:48:14
square root of 1 - x^2.
 

706
00:48:14 --> 00:48:18
So all told, the derivative
with respect to x of the

707
00:48:18 --> 00:48:30
arcsine function is 1 /
square root of 1 - x^2.

708
00:48:30 --> 00:48:33
So these implicit
differentiations are

709
00:48:33 --> 00:48:34
very convenient.
 

710
00:48:34 --> 00:48:39
However, I warn you that you do
have to be careful about the

711
00:48:39 --> 00:48:42
range of applicability
of these things.

712
00:48:42 --> 00:48:45
You have to draw a picture like
this one to make sure you

713
00:48:45 --> 00:48:47
know where this makes sense.
 

714
00:48:47 --> 00:48:49
In other words, you have to
pick a branch for the sine

715
00:48:49 --> 00:48:52
function to work that out,
and there's something like

716
00:48:52 --> 00:48:53
that on your problem set.
 

717
00:48:53 --> 00:48:56
And it's also discussed
in your text.

718
00:48:56 --> 00:48:57
So we'll stop here.
 

719
00:48:57 --> 00:49:00