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U2S5V09 Recitation video.txt
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U2S5V09 Recitation video.txt
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#
# File: content-mit-18-01-1x-captions/U2S5V09 Recitation video.txt
#
# Captions for MITx 18.01.1x module [MeVXXJv4IWQ]
#
# This file has 156 caption lines.
#
# Do not add or delete any lines. If there is text missing at the end, please add it to the last line.
#
#----------------------------------------
Hi, welcome back to recitation.
Last time in lecture, you started learning
about implicit differentiation.
And you saw some examples of how implicit differentiation can
be used to compute derivatives of functions
defined implicitly.
So let's do another example today.
So here I have a curve that's defined
by the implicit equation y cubed plus x cubed equals 3xy.
And I'd like to know what the slope
of the tangent line to that curve
is at the point (4/3, 2/3).
So before we start doing anything,
let me just make a couple of observations.
If you don't believe me that the point (4/3, 2/3)
is on this curve, you can always check by plugging the values in
and confirm that really, yes, 4/3 cubed plus 2/3 cubed is
equal to 3 times 2/3 times 4/3.
So it's how I found this point is maybe a little magical
because, as you can see, this equation is really a tough one
to solve for y.
A natural thing to want to do when
asked this question is to solve for y and get an equation for y
in terms of x, and then take the derivative using
the various differentiation rules that you've learned.
But here, this is--
I mean it's--
I'll let you in on a secret.
There is a way to do this, but it's really hard.
It's really ugly, and it's beyond the scope
of this course.
So really we're much better off treating
this as an implicit differentiation problem
than as an explicit differentiation problem.
So having said that, why don't you take a minute or two,
try and have a go at this yourself,
and then we'll come back and work through it together.
All right, so welcome back.
We were in the middle-- we were just about to start, actually,
solving this problem computing the slope of the tangent line
to the curve y cubed plus x cubed equals 3xy,
at this point (4/3, 2/3).
So the slope of the tangent line is the value y prime
of x at that point.
So we need to answer this question.
What we need to do is we need to find the derivative of y.
But as I said earlier, this is tough to do explicitly
to find y in terms of x, so we're going to use
implicit differentiation.
So we start with this equation y cubed plus x cubed equals 3xy,
and we can take a derivative with respect to x.
So some parts-- so let's start with it
in the order it's given.
So you have y cubed.
If you take a derivative of y cubed with respect to x, what
you need to use is the chain rule, because y is implicitly
a function of x.
And so y cubed is the chain rule.
It's the cubed function applied to the y function.
And this is true of implicit differentiation in general.
That the reason that we can do this, the really fundamental
reason that this works is that we have the chain rule,
and that it lets us evaluate derivatives of compositions.
So in our case, we have--
so we take a derivative of the whole thing.
I'm going to write-- this is a little sloppy notation,
but I hope you know what I mean.
We have this identity, and so we're
going to take a derivative of the whole thing.
And so the first part on the left, we
get the derivative of y cubed.
So by the chain rule, we first take
the derivative of the cube function at y,
and then multiply by the derivative of y.
So this is the derivative of y cubed,
it just gives us 3y squared.
So that's what happens when you just deal with the cubed part.
But then we need to multiply by-- in the chain rule--
by the derivative of the inside, which in this context
is dy by dx.
Plus the derivative x cubed, that's more straightforward,
nothing really complicated going on here.
We've seen this for a little while now.
It's just 3x squared equal.
So on the right now, what we have--
we don't actually have a chain rule,
we have we have a product rule situation here.
We have 3 times x times y.
So 3 is just a constant, so we could just pull it out
in front.
And so we take the derivative of xy.
So we take the derivative of the first times the second,
plus the derivative of the second times the first.
So the derivative of the first is just--
sorry, x is the first, so its derivative is 1.
So we get 3 times the second is y, plus--
so when we take the first times the derivative of the second,
which is dy by dx.
So because this is an identity, it
holds for all values of x and y, this equality
follows just by taking the derivative of both sides.
Good.
So now the thing we want is that we
want the slope of the tangent line at a particular point.
So we want to isolate dy/dx.
That's the thing we're trying to find.
So if you were only interested in dy/dx,
this is actually a linear equation in some sense.
We have dy-- a constant-- something, or--
not a constant-- something times dy/dx plus something
equals something plus something times dy/dx.
There are no squares of dy/dx is what I really mean.
So that's nice.
It makes it relatively easier to solve so we can just combine
all the terms with dy/dx.
And then let's say, we'll combine maybe then put them
over here and put everything else over there.
So over here we get dy/dx.
So we have a 3y squared minus a 3x.
And on the other side, we have a 3y minus a 3x squared.
And so-- this is times, multiplication there.
And so we want dy by dx just by itself,
so we can just divide through by 3y squared minus 3x.
So then we have dy/dx is equal to--
well, so there are a lot of 3s here.
There's a constant multiple of 3 on this side, constant multiple
of 3 on the side.
Those are going to cancel.
So this is y minus x squared over y squared minus x.
So at any point xy on this curve,
the slope of the tangent line is given by this expression here.
And we're interested in a particular point
in this problem.
We're interested in the point (4/3 comma 2/3).
So let me take that back up here.
So, at the point (4/3 comma 2/3), we have dy by dx.
So we just plug that value of y and that value
of x into this formula that we've got.
So that's 2/3 minus 4/3 squared is 16/9 over--
well, let's see.
2/3 squared is 4/9 minus 4/3.
So we have a little bit of a rational number
arithmetic here.
Maybe I'll multiply top and bottom through by 9
to get 6 minus 16 over 4 minus 12.
So this is negative 10 over negative 8, which is 5 over 4.
And if we go back to the picture that I drew,
actually looks pretty reasonable over here, right?
This slope with this tangent line
is actually a little bit bigger than 1 there.
Great.
So that's that.