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Professor: So, we're ready
to begin the fifth lecture.
I'm glad to be back.
Thank you for entertaining
my colleague, Haynes Miller.
So, today we're
going to continue
where he started, namely what
he talked about was the chain
rule, which is probably
the most powerful technique
for extending the
kinds of functions
that you can differentiate.
And we're going to use the
chain rule in some rather clever
algebraic ways today.
So the topic for
today is what's known
as implicit differentiation.
So implicit differentiation
is a technique
that allows you to differentiate
a lot of functions you didn't
even know how to find before.
And it's a technique -
let's wait for a few people
to sit down here.
Physics, huh?
Okay, more Physics.
Let's take a break.
You can get those after class.
All right, so we're talking
about implicit differentiation,
and I'm going to illustrate
it by several examples.
So this is one of the most
important and basic formulas
that we've already
covered part way.
Namely, the derivative of
x to a power is ax^(a-1).
Now, what we've got so far is
the exponents, 0, plus or minus
1, plus or minus 2, etc.
You did the positive integer
powers in the first lecture,
and then yesterday
Professor Miller
told you about the
negative powers.
So what we're going
to do right now,
today, is we're
going to consider
the exponents which are rational
numbers, ratios of integers.
So a is m/n.
m and n are integers.
All right, so that's
our goal for right now,
and we're going
to use this method
of implicit differentiation.
In particular, it's important
to realize that this
covers the case m = 1.
And those are the nth roots.
So when we take the
one over n power,
we're going to cover
that right now,
along with many other examples.
So this is our first example.
So how do we get started?
Well we just write down a
formula for the function.
The function is y = x^(m/n).
That's what we're
trying to deal with.
And now there's
really only two steps.
The first step is to take this
equation to the nth power,
so write it y^n = x^m.
Alright, so that's just the
same equation re-written.
And now, what we're
going to do is
we're going to differentiate.
So we're going to apply
d/dx to the equation.
Now why is it that we can apply
it to the second equation, not
the first equation?
So maybe I should call these
equation 1 and equation 2.
So, the point is, we can
apply it to equation 2.
Now, the reason is that we
don't know how to differentiate
x^(m/n).
That's something we
just don't know yet.
But we do know how to
differentiate integer powers.
Those are the things that
we took care of before.
So now we're in shape to be
able to do the differentiation.
So I'm going to write
it out explicitly
over here, without
carrying it out just yet.
That's d/dx of
y^n = d/dx of x^m.
And now you see
this expression here
requires us to do something we
couldn't do before yesterday.
Namely, this y is
a function of x.
So we have to apply
the chain rule here.
So this is the same as - this
is by the chain rule now -
d/dy of y^n times dy/dx.
And then, on the right hand
side, we can just carry it out.
We know the formula.
It's mx^(m-1).
Right, now this is our scheme.
And you'll see in a minute
why we win with this.
So, first of all, there
are two factors here.
One of them is unknown.
In fact, it's what
we're looking for.
But the other one is going
to be a known quantity,
because we know how
to differentiate y
to the n with respect to y.
That's the same formula,
although the letter
has been changed.
And so this is the same as -
I'll write it underneath here -
n y^(n-1) dy/dx = m x^(m-1).
Okay, now comes, if you
like, the non-calculus part
of the problem.
Remember the non-calculus
part of the problem
is always the messier
part of the problem.
So we want to figure
out this formula.
This formula, the
answer over here,
which maybe I'll
put in a box now,
has this expressed much more
simply, only in terms of x.
And what we have to do now
is just solve for dy/dx
using algebra, and then solve
all the way in terms of x.
So, first of all,
we solve for dy/dx.
So I do that by dividing the
factor on the left-hand side.
So I get here mx^(m-1)
divided by ny^(n-1).
And now I'm going to plug in--
so I'll write this as m/n.
This is x^(m-1).
Now over here I'm going to put
in for y, x^(m/n) times n-1.
So now we're almost done,
but unfortunately we
have this mess of exponents
that we have to work out.
I'm going to write
it one more time.
So I already recognize
the factor a out front.
That's not going to
be a problem for me,
and that's what I'm
aiming for here.
But now I have to encode
all of these powers,
so let's just write it.
It's m-1, and then it's
minus the quantity (n-1) m/n.
All right, so that's the law of
exponents applied to this ratio
here.
And then we'll do the arithmetic
over here on the next board.
So we have here m - 1
- (n-1) m/n = m - 1.
And if I multiply n
by this, I get -m.
And if the second factor is
minus minus, that's a plus.
And that's +m/n.
Altogether the two m's cancel.
I have here -1 + m/n.
And lo and behold that's
the same thing as a - 1,
just what we wanted.
All right, so this
equals a x^(n-1).
Okay, again just a
bunch of arithmetic.
From this point forward,
from this substitution
on, it's just the
arithmetic of exponents.
All right, so we've done
our first example here.
I want to give you a
couple more examples,
so let's just continue.
The next example I'll
keep relatively simple.
So we have example two, which is
going to be the function x^2 +
y^2 = 1.
Well, that's not
really a function.
It's a way of defining y as
a function of x implicitly.
There's the idea that I could
solve for y if I wanted to.
And indeed let's do that.
So if you solve for y here, what
happens is you get y^2 = 1 -
x^2, and y is equal to plus or
minus the square root of 1 -
x^2.
So this, if you like, is
the implicit definition.
And here is the
explicit function y,
which is a function of x.
And now just for
my own convenience,
I'm just going to take
the positive branch.
This is the function.
It's just really a
circle in disguise.
And I'm just going to take
the top part of the circle,
so we'll take that
top hump here.
All right, so that means
I'm erasing this minus sign.
I'm just taking the
positive branch, just
for my convenience.
I could do it just as well
with the negative branch.
Alright, so now I've
taken the solution,
and I can differentiate
with this.
So rather than using the
dy/dx notation over here,
I'm going to switch
notations over here,
because it's less writing.
I'm going to write y'
and change notations.
Okay, so I want to take
the derivative of this.
Well this is a somewhat
complicated function here.
It's the square root of 1 -
x^2, and the right way always
to look at functions like
this is to rewrite them using
the fractional power notation.
That's the first
step in computing
a derivative of a square root.
And then the second
step here is what?
Does somebody want to tell me?
Chain rule, right.
That's it.
So we have two things.
We start with one, and then
we do something else to it.
So whenever we do two
things to something,
we need to apply the chain rule.
So 1 - x^2, square root.
All right, so how do we do that?
Well, the first
factor I claim is
the derivative of this thing.
So this is 1/2 blah to the -1/2.
So I'm doing this kind
of by the advanced method
now, because we've
already graduated.
You already did the
chain rule last time.
So what does this mean?
This is an abbreviation for
the derivative with respect
to blah of blah ^
1/2, whatever it is.
All right, so that's the first
factor that we're going to use.
Rather than actually write
out a variable for it
and pass through
as I did previously
with this y and x
variable here, I'm
just going to skip
that step and let
you imagine it as being a
placeholder for that variable
here.
So this variable
is now parenthesis.
And then I have to multiply that
by the rate of change of what's
inside with respect to x.
And that is going to be -2x.
The derivative of
1 - x^2 is -2x.
And now again, we couldn't
have done this example two
before example one,
because we needed
to know that the power rule
worked not just for a integer
but also for a = 1/2.
We're using the case
a = 1/2 right here.
It's 1/2 times, and
this -1/2 here is a-1. -
So this is the case a = 1/2.
a-1 happens to be -1/2.
Okay, so I'm putting all
those things together.
And you know within
a week you have
to be doing this
very automatically.
So we're going to do
it at this speed now.
You want to do it even
faster, ultimately.
Yes?
Student: [INAUDIBLE]
Professor: The question is
could I have done it implicitly
without the square roots.
And the answer is yes.
That's what I'm about to do.
So this is an
illustration of what's
called the explicit solution.
So this guy is what's
called explicit.
And I want to contrast
it with the method
that we're going
to now use today.
So it involves a lot
of complications.
It involves the chain rule.
And as we'll see it can
get messier and messier.
And then there's
the implicit method,
which I claim is easier.
So let's see what happens
if you do it implicitly
The implicit method
involves, instead of writing
the function in this
relatively complicated way,
with the square root, it
involves leaving it alone.
Don't do anything to it.
In this previous case, we were
left with something which was
complicated, say x^(1/3)
or x^(1/2) or something
complicated.
We had to simplify it.
We had an equation one,
which was more complicated.
We simplified it then
differentiated it.
And so that was a simpler case.
Well here, the simplest
thing us to differentiate
is the one we started with,
because squares are practically
the easiest thing after first
powers, or maybe zeroth powers
to differentiate.
So we're leaving it alone.
This is the simplest
possible form for it,
and now we're going
to differentiate.
So what happens?
So again what's the method?
Let me remind you.
You're applying d/dx
to the equation.
So you have to differentiate
the left side of the equation,
and differentiate the
right side of the equation.
So it's this, and what you get
is 2x + 2yy' is equal to what?
0.
The derivative of 1 0.
So this is the chain rule again.
I did it a different way.
I'm trying to get you used
to many different notations
at once.
Well really just two.
Just the prime notation
and the dy/dx notation.
And this is what I get.
So now all I have to
do is solve for y'.
So that y', if I put the 2x
on the other side, is -2x,
and then divide by
2y, which is -x/y.
So let's compare our
solutions, and I'll apologize,
I'm going to have to erase
something to do that.
So let's compare
our two solutions.
I'm going to put this
underneath and simplify.
So what was our
solution over here?
It was 1/2(1-x^2)^(-1/2) (-2x).
That was what we got over here.
And that is the same thing, if I
cancel the 2's, and I change it
back to looking
like a square root,
that's the same thing as -x
divided by square root of 1 -
x^2.
So this is the formula
for the derivative
when I do it the explicit way.
And I'll just compare them,
these two expressions here.
And notice they are the same.
They're the same, because y
is equal to square root of 1 -
x^2.
Yeah?
Question?
Student: [INAUDIBLE]
Professor: The question is
why did the implicit method
not give the bottom
half of the circle?
Very good question.
The answer to that
is that it did.
I just didn't mention it.
Wait, I'll explain.
So suppose I stuck
in a minus sign here.
I would have gotten this
with the difference, so
with an extra minus sign.
But then when I compared
it to what was over there,
I would have had to have another
different minus sign over here.
So actually both places would
get an extra minus sign.
And they would still coincide.
So actually the implicit
method is a little better.
It doesn't even
notice the difference
between the branches.
It does the job on both
the top and bottom half.
Another way of saying
that is that you're
calculating the slopes here.
So let's look at this picture.
Here's a slope.
Let's just take a look
at a positive value
of x and just check the sign
to see what's happening.
If you take a positive value
of x over here, x is positive.
This denominator is positive.
The slope is negative.
You can see that
it's tilting down.
So it's okay.
Now on the bottom side,
it's going to be tilting up.
And similarly what's
happening up here
is that both x and y are
positive, and this x and this y
are positive.
And the slope is negative.
On the other hand, on the bottom
side, x is still positive,
but y is negative.
And it's tilting up because
the denominator is negative.
The numerator is positive,
and this minus sign
has a positive slope.
So it matches perfectly
in every category.
This complicated,
however, and it's easier
just to keep track of
one branch at a time,
even in advanced math.
Okay, so we only do it
one branch at a time.
Other questions?
Okay, so now I want to
give you a slightly more
complicated example here.
And indeed some
of the-- so here's
a little more
complicated example.
It's not going to be the
most complicated example,
but you know it'll
be a little tricky.
So this example, I'm going
to give you a fourth order
equation.
So y^4 + xy^2 - 2 = 0.
Now it just so
happens that there's
a trick to solving
this equation,
so actually you can do
both the explicit method
and the non-explicit method.
So the explicit method
would say okay well,
I want to solve for this.
So I'm going to use the
quadratic formula, but on y^2.
This is quadratic in y^2,
because there's a fourth power
and a second power, and the
first and third powers are
missing.
So this is y^2 is equal to -x
plus or minus the square root
of x^2 - 4(-2) divided by 2.
And so this x is the b.
This -2 is the c, and a =
1 in the quadratic formula.
And so the formula for y is plus
or minus the square root of -x
plus or minus the square
root x^2 + 8 divided by 2.
So now you can see this
problem of branches,
this happens actually
in a lot of cases,
coming up in an elaborate way.
You have two choices
for the sign here.
You have two choices
for the sign here.
Conceivably as many as four
roots for this equation,
because it's a fourth
degree equation.
It's quite a mess.
You should have to check
each branch separately.
And this really is that
level of complexity,
and in general
it's very difficult
to figure out the formulas
for quartic equations.
But fortunately we're
never going to use them.
That is, we're never going
to need those formulas.
So the implicit
method is far easier.
The implicit method
just says okay I'll
leave the equation
in its simplest form.
And now differentiate.
So when I differentiate,
I get 4y^3 y' plus -
now here I have to
apply the product rule.
So I differentiate the x
and the y^2 separately.
First I differentiate with
respect to x, so I get y^2.
Then I differentiate with
respect to the other factor,
the y^2 factor.
And I get x(2 y y').
And then the 0 gives me 0.
So minus 0 equals 0.
So there's the implicit
differentiation step.
And now I just want
to solve for y'.
So I'm going to
factor out 4y^3 + 2xy.
That's the factor on y'.
And I'm going to put the
y^2 on the other side.
-y^2 over here.
And so the formula for y' is
-y^2 divided by 4y^3 + 2xy.
So that's the formula
for the solution.
For the slope.
You have a question?
Student: [INAUDIBLE]
Professor: So the question
is for the y would
we have to put in what solved
for in the explicit equation.
And the answer is
absolutely yes.
That's exactly the point.
So this is not a complete
solution to a problem.
We started with an
implicit equation.
We differentiated.
And we got in the end,
also an implicit equation.
It doesn't tell us what
y is as a function of x.
You have to go back
to this formula
to get the formula for x.
So for example, let me
give you an example here.
So this hides a degree of
complexity of the problem.
But it's a degree of complexity
that we must live with.
So for example, at x = 1, you
can see that y = 1 solves.
That happens to be--
solves y^4 + xy^2 - 2 = 0.
That's why I picked
the 2 actually,
so it would be 1 + 1 - 2 = 0.
I just wanted to have a
convenient solution there
to pull out of my
hat at this point.
So I did that.
And so we now know
that when x = 1, y = 1.
So at (1, 1) along the curve,
the slope is equal to what?
Well, I have to plug in
here, -1^2 / (4*1^3 + 2*1*1).
That's just plugging in
that formula over there,
which turns out to be -1/6.
So I can get it.
On the other hand,
at say x = 2, we're
stuck using this formula
star here to find y.
Now, so let me just
make two points
about this, which are just
philosophical points for you
right now.
The first is, when
I promised you
at the beginning of
this class that we
were going to be
able to differentiate
any function you know, I
meant it very literally.
What I meant is if
you know the function,
we'll be able give a
formula for the derivative.
If you don't know how
to find a function,
you'll have a lot of trouble
finding the derivative.
So we didn't make any
promises that if you
can't find the
function you will be
able to find the
derivative by some magic.
That will never happen.
And however complex
the function is,
a root of a fourth
degree polynomial
can be pretty complicated
function of the coefficients,
we're stuck with this degree
of complexity in the problem.
But the big advantage
of his method, notice,
is that although we've
had to find star,
we had to find
this formula star,
and there are many other ways of
doing these things numerically,
by the way, which
we'll learn later,
so there's a good method
for doing it numerically.
Although we had to find star, we
never had to differentiate it.
We had a fast way of
getting the slope.
So we had to know
what x and y were.
But y' we got by an
algebraic formula,
in terms of the values here.
So this is very fast,
forgetting the slope,
once you know the point. yes?
Student: What's in
the parentheses?
Professor: Sorry, this is-- Well
let's see if I can manage this.
Is this the parentheses
you're talking about?
Ah, "say".
That says "say".
Well, so maybe I should
put commas around it.
But it was S A Y,
comma comma, okay?
Well here was at x = 1.
I'm just throwing out a value.
Any other value.
Actually there is one
value, my favorite value.
Well this is easy to
evaluate right? x = 0,
I can do it there.
That's maybe the only one.
The others are a nuisance.
All right, other questions?
Now we have to do
something more here.
So I claimed to you that
we could differentiate
all the functions we know.
But really we can
learn a tremendous
about functions which are
really hard to get at.
So this implicit
differentiation method
has one very, very
important application
to finding inverse functions,
or finding derivatives
of inverse functions.
So let's talk about that next.
So first, maybe we'll just
illustrate by an example.
If you have the function y
is equal to square root x,
for x positive, then
of course this idea
is that we should
simplify this equation
and we should square it so
we get this somewhat simpler
equation here.
And then we have a
notation for this.
If we call f(x) equal to
square root of x, and g(y) = x,
this is the reversal of this.
Then the formula for g(y)
is that it should be y^2.
And in general, if we start
with any old y = f(x),
and we just write down, this
is the defining relationship
for a function g, the property
that we're saying is that
g(f(x)) has got to
bring us back to x.
And we write that in a
couple of different ways.
We call g the inverse of f.
And also we call f
the inverse of g,
although I'm going to be
silent about which variable
I want to use, because people
mix them up a little bit,
as we'll be doing when we
draw some pictures of this.
So let's see.
Let's draw pictures of
both f and f inverse
on the same graph.
So first of all, I'm going
to draw the graph of f(x)
= square root of x.
That's some shape like this.
And now, in order to
understand what g(y) is,
so let's do the
analysis in general,
but then we'll draw it
in this particular case.
If you have g(y)
= x, that's really
just the same equation right?
This is the equation
g(y) = x, that's y^2 = x.
This is y = square root of x,
those are the same equations,
it's the same curve.
But suppose now that we wanted
to write down what g(x) is.
In other words, we wanted
to switch the variables,
so draw them as I said on the
same graph with the same x,
and the same y axes.
Then that would be, in effect,
trading the roles of x and y.
We have to rename every
point on the graph which
is the ordered pair (x, y), and
trade it for the opposite one.
And when you
exchange x and y, so
to do this, exchange
x and y, and when
you do that, graphically
what that looks
like is the following:
suppose you have a place here,
and this is the x
and this is the y,
then you want to trade them.
So you want the y here right?
And the x up there.
It's sort of the opposite
place over there.
And that is the place which is
directly opposite this point
across the diagonal line x = y.
So you reflect across this
or you flip across that.
You get this other shape
that looks like that.
Maybe I'll draw it with a
colored piece of chalk here.
So this guy here
is y = f^(-1)(x).
And indeed, if you
look at these graphs,
this one is the square root.
This one happens to be y = x^2.
If you take this
one, and you turn it,
you reverse the roles of
the x axis and the y axis,
and tilt it on its side.
So that's the picture of what an
inverse function is, and now I
want to show you that the method
of implicit differentiation
allows us to compute
the derivatives
of inverse functions.
So let me just
say it in general,
and then I'll carry
it out in particular.
So implicit
differentiation allows
us to find the derivative
of any inverse function,
provided we know the
derivative of the function.
So let's do that for
what is an example, which
is truly complicated and
a little subtle here.
It has a very pretty answer.
So we'll carry out
an example here,
which is the function y is
equal to the inverse tangent.
So again, for the
inverse tangent
all of the things
that we're going to do
are going to be
based on simplifying
this equation by taking
the tangent of both sides.
So, us let me remind
you by the way,
the inverse tangent is what's
also known as arctangent.
That's just another
notation for the same thing.
And what we're going
to use to describe
this function is the
equation tan y = x.
That's what happens
when you take
the tangent of this function.
This is how we're
going to figure out
what the function looks like.
So first of all,
I want to draw it,
and then we'll do
the computation.
So let's make the diagram first.
So I want to do
something which is
analogous to what I did over
here with the square root
function.
So first of all, I remind
you that the tangent function
is defined between two values
here, which are pi/2 and -pi/2.
And it starts out
at minus infinity
and curves up like this.
So that's the function tan x.
And so the one that
we have to sketch
is this one which we
get by reflecting this
across the axis.
Well not the axis, the diagonal.
This slope by the way, should
be less - a little lower here so
that we can have it
going down and up.
So let me show you
what it looks like.
On the front, it's going to
look a lot like this one.
So this one had curved
down, and so the reflection
across the diagonal curved up.
Here this is curving
up, so the reflection
is going to curve down.
It's going to look like this.
Maybe I should, sorry,
let's use a different color,
because it's
reversed from before.
I'll just call it green.
Now, the original curve
in the first quadrant
eventually had an asymptote
which was straight up.
So this one is going to have an
asymptote which is horizontal.
And that level is what?
What's the highest?
It is just pi/2.
Now similarly, the
other way, we're
going to do this:
and this bottom level
is going to be -pi/2.
So there's the picture
of this function.
It's defined for all x.
So this green guy
is y = arctan x.
And it's defined all the
way from minus infinity
to infinity.
And to use a notation that
we had from limit notation
as x goes to infinity, let's
say, x is equal to pi/2.
That's an example of one value
that's of interest in addition
to the finite values.
Okay, so now the
first ingredient
that we're going
to need, is we're
going to need the derivative
of the tangent function.
So I'm going to recall
for you, and maybe you
haven't worked this out yet, but
I hope that many of you have,
that if you take the derivative
with respect to y of tan y.
So this you do by
the quotient rule.
So this is of the
form u/v, right?
You use the quotient rule.
So I'm going to get this.
But what you get in the end is
some marvelous simplification
that comes out to cos^2 y.
1 over cosine squared.
You can recognize the cosine
squared from the fact that you
should get v^2 in
the denominator,
and somehow the numerators all
cancel and simplifies to 1.
This is also known
as secant squared y.
So that something that
if you haven't done yet,
you're going to have to
do this as an exercise.
So we need that
ingredient, and now we're
just going to
differentiate our equation.
And what do we get?
We get, again, (d/dy tan y)
times dy/dx is equal to 1.
Or, if you like, 1 / cos^2 y
times, in the other notation,
y', is equal to 1.
So I've just used the formulas
that I just wrote down there.
Now all I have to
do is solve for y'.
It's cos^2 y.
Unfortunately, this
is not the form
that we ever want to
leave these things in.
This is the same problem we
had with that ugly square root
expression, or with
any of the others.
We want to rewrite
in terms of x.
Our original question was
what is d/dx of arctan x.
Now so far we have the following
answer to that question:
it's cos^2 (arctan x).
Now this is a correct answer,
but way too complicated.
Now that doesn't
mean that if you
took a random
collection of functions,
you wouldn't end up with
something this complicated.
But these particular functions,
these beautiful circular
functions involved
with trigonometry all
have very nice formulas
associated with them.
And this simplifies
tremendously.
So one of the
skills that you need
to develop when you're
dealing with trig functions
is to simplify this.
And so let's see now that
expressions like this all
simplify.
So here we go.
There's only one
formula, one ingredient
that we need to use to
do this, and then we're
going to draw a diagram.
So the ingredient again, is the
original defining relationship
that tan y = x.
So tan y = x can be
encoded in a right triangle
in the following way: here's
the right triangle and tan
y means that y should be
represented as an angle.
And then, its
tangent is the ratio
of this vertical to
this horizontal side.
So I'm just going to pick
two values that work,
namely x and 1.
Those are the simplest ones.
So I've encoded this
equation in this picture.
And now all I have to do is
figure out what the cosine of y
is in this right triangle here.
In order to do that, I need to
figure out what the hypotenuse
is, but that's just
square root of 1 + x^2.
And now I can read off
what the cosine of y is.
So the cosine of y is 1
divided by the hypotenuse.
So it's 1 over square root,
whoops, yeah, 1 + x^2.
And so cosine squared
is just 1 / 1 + x^2.
And so our answer over here, the
preferred answer which is way
simpler than what
I wrote up there,
is that d/dx of tan inverse
x is equal to 1 over 1 + x^2.
Maybe I'll stop here
for one more question.
I have one more calculation
which I can do even
in less than a minute.
So we have a whole
minute for questions.
Yeah?
Student: [INAUDIBLE]
Professor: What happens
to the inverse tangent?
The inverse tangent--
Okay, this inverse tangent
is the same as this y here.
Those are the same thing.
So what I did was I skipped
this step here entirely.
I never wrote that down.
But the inverse
tangent was that y.
The issue was what's
a good formula
for cos y in terms of x?
So I am evaluating that, but
I'm doing it using the letter y.
So in other words, what
happened to the inverse
tangent is that I
called it y, which
is what it's been all along.
Okay, so now I'm
going to do the case
of the sine, the inverse sine.
And I'll show you
how easy this is
if I don't fuss with-- because
this one has an easy trig
identity associated with it.
So if y = sin^(-1)
x, and sin y = x,
and now watch how simple it is
when I do the differentiation.
I just differentiate.
I get (cos y) y' = 1.
And then, y', so that
implies that = 1 / cos y,
and now to rewrite
that in terms of x,
I have to just recognize that
this is the same as this,
which is the same as 1 /
square root of 1 - x^2.
So all told, the derivative
with respect to x of the arcsine
function is 1 / square
root of 1 - x^2.
So these implicit
differentiations
are very convenient.
However, I warn you
that you do have
to be careful about the range of
applicability of these things.
You have to draw a
picture like this one
to make sure you know
where this makes sense.
In other words, you have to pick
a branch for the sine function
to work that out,
and there's something
like that on your problem set.
And it's also
discussed in your text.
So we'll stop here.
