Implicit Differentiation and Linear Approximation | MIT 18.01SC Single Variable Calculus, Fall 2010

Uploaded by MIT on 10.01.2011


Welcome back to recitation.
In lecture you've been talking about implicitly defined
functions and implicit differentiation.
So one of the reasons that these are important is, or
that implicit differentiation is important, is it sometimes
you have a function to find implicitly and you
can't solve for it.
You don't have any algebraic method for computing the
function values as a formula, say.
So, for example, this function that I've written on the board
that I've called w of x is defined implicitly by the
equation that w of x plus 1 quantity times e to the w of x
is equal to x for all x.
So this function, some of its values you can guess.
Like at x equals 0, the function value is going to be
negative 1.
And the reason is that this can't ever be 0, so the only
way to get this side to be 0 is if w is negative 1 if
this term is 0.
So some of its values are easy to compute, but some of its
values aren't.
So for example, if I asked you what w of 3/2
is, it's very hard.
There's no algebraic way you can manipulate this equation
that will let you figure that out.
So in that situation you might still care about what the
function value is.
So what can you do?
Well, you can try and find a numerical approximation.
So in this problem I'd like you to try and estimate the
value w of 3/2 by using a linear approximation to the
function, to the curve--
A linear approximation of the function w of x in order to
compute this value.

So as a hint, I've given you, so you're trying to
compute w of 3/2.
As a hint I'm pointing out to you that w of 1 is 0.
If you put in x equals 0 and w of 0 equals--
sorry-- if you put in x equals 1 and w of 1 equals 0 on the
left hand side, you do indeed get 1, as you should.

So, OK.
So, good.
So that'll give you a hint about where you could base
your linear approximation.
So why don't you pause the video, take a few minutes to
work this out, come back, and we can work it out together.

All right.
Welcome back.
So hopefully you've had a chance to work on this
question a little bit.
So in order to do this linear approximation that we want,
what we need to know are we need to know a base point and
we need to know the derivative of the function
at that base point.
And those are the two pieces of data you need in order to
construct a linear approximation.
So we have a good candidate for a base point here, which
is the point 1, 0.
So this curve, whatever it looks like, it passes through
the point 1, 0.
And that's the point we're going to use for our
So we're going to use the linear approximation w of x is
approximately equal to w prime of 1 times x minus 1 plus w of
1 when x is approximately equal to 1.
So this is the linear approximation we're going to
use, and we have that w of 1 here is 0.
So this is, this is equal to w prime of one
times x minus one.
Just the w of 1 is 0.
It just goes away.
So in order to estimate w of x, and in particular w of 3/2,
what we need to know is we need to know the
derivative of w.
And to get the derivative of w, we need to use--
well, we have only one piece of information about w.
Which is we have that it's defined by
this implicit equation.
So in order to get the derivative of w we have to use
implicit differentiation.
So let's do that.
So if we implicitly differentiate this equation--
so let's start with the, the right hand side is going to be
really easy.
We're going to differentiate with respect to x.
The right hand side is going to be 1.
On the left hand side is going to be a little more
We have a product and then this piece, we're going to
have a chain rule situation.
We have e to the w of x.

So, OK.
So we're going to take an implicit
derivative and on the left--
so OK, so the product rule first. We take the derivative
of the first part, so that's just w prime of x times the
second part-- that's e to the w of x--
plus the first part-- that's w of x plus 1--
times the derivative of the second part.
So the second part is e to the w of x.
So that gives me an e to the w of x times w prime of x.
That's the chain rule.
So that's what happens when I differentiate
the left hand side.
And on the right hand side I take the derivative
of x and I get 1.
OK, good.

So now I've got this equation and I need to solve this
equation for w prime.
Because if you look up here, that's what I want.
I want a particular value of w prime.
And as always happens in implicit differentiation, from
the point of view of this w prime it's only involved in
the equation in a very simple way.
So there's it multiplied by functions of x and w of x, but
not, you know, it's just multiplied by something that
doesn't involve w prime at all.
And then here it's multiplied by something that doesn't
involve w prime at all.
So you can just collect your w prime's and divide through.
You know, it's just like solving a linear equation.
So here if we collect our w prime's, this is w prime of x
times-- looks like--
w of x plus 2 times e to the w of x.
Did I get that right?
Looks good.
OK, so that's still equal to 1.
So that means that w prime of x is just--
well, just, you know--
it's equal to 1 over w of x plus 2 times e to the w of x.
OK, so this is true for every x.
But I don't need this equation for every x.
I just need the particular value of w prime at 1.
So that's, so I'm going to take this equation, then, and
I'm just going to put in x equals 1.
So I put in x equals 1--
well, let me do it over here-- so I get w prime of 1.
And I just everywhere I had an x, I put in a 1.
So actually, in this equation the only place x appears is in
the argument of w.
So this is w of 1 plus 2 times e to the w of 1.
So in order to get w prime of 1 I need to
know what w of 1 is.
But I had that.
I had it, it was right back here.
There was, that was my hint to you.
Right, this is why we're using this point as a base point,
which is we know the value of w for this value of x.
So we take that value.
So w of 1 is 0.
So this is just 1 over--
well, 0 plus 2 is--
2, and e to the 0 is 1.
So it's just 1 over 2.
Sorry, that's written a little oddly.
We can make it 2 times 1.
So 1 over 2.

So I take that back upstairs to this
equation that I had here.
And I have that w of x is approximately equal to w prime
of 1 times x minus 1.
So w of x is approximately equal to w prime of 1--
we saw is--
1/2 times x minus 1.

And that approximation was good near our base point.
So that's good when x is near 1.

All right.
And then, so this is the linear approximation.
And I asked for the linear approximation, its value at
the particular point, x equals 3/2.
So w of 3/2 is approximately 1/2 times--
well, 3/2 minus 1 is also 1/2--

so this is a quarter.
OK, so this is our estimate for w of 3/2.
w of 3/2 is approximately 1/4.
If you wanted a better estimate you could try
iterating this process.
Now you might have a, you know--
or choosing some base point even closer if you could
figure out the value of w and x near that, near this point
that you're interested in--
So just to sum up what we did was we had this implicitly
defined function w.
We wanted to estimate its value at a point where we
couldn't compute it explicitly.
So what we did was we did our normal linear
approximation method.
So we wrote down our normal linear approximation formula.
The only thing that was slightly unusual is that we
had to use implicit differentiation.
In order to compute the derivative that appears in the
linear approximation, we implicitly differentiated.
So that happened just like normal, and then at the end we
plugged in the values that we were interested in, to
actually compute the particular value of that
So I'll end there.