BAM!!!
Mr. Tarrou.
In this video we are going to work through
two examples of finding the second derivative
implicitly.
Or in other words, we are going to find the
second derivative in terms of x and y.
So...
And we are doing it implicitly because of
course with some of these problems it is a
lot less work.
So let's start off with our circle with a
center on the origin and a radius of square
root of 6, or just x^2+y^2=6.
And we get for working towards our first derivative
2x plus 2y... and it is not just really 2y,
right?
Because we are finding the derivative with
respect to x.
So we have to follow the chain rule.
So we have 2y, I am going to bring that exponent
by one, and then we need to multiply by the
derivative of the inside function and that
is going to be dy/dx.
The derivative of a constant is equal to zero.
Ok.
We want to get that dy/dx, or that first derivative
notation alone.
We are going to subtract both sides by 2x.
We are going to 
divide both sides by 2y.
And we get the 2's cancelling out getting
-x/y. Ok.
Well now we have our first derivative implicitly
and it is -x/y.
Our second derivative 
is to find out now....
To find that we are going to follow the Quotient
Rule, excuse me.
So we are going to take the denominator which
is y times the derivative of the numerator.
The derivative of -x with respect to x is
-1.
Now minus the numerator times the derivative
of the denominator with respect to x.
This is going to be dy/dx.
This over the denominator squared.
Ok.
Well there...
Excuse me.
There is my second derivative but it is not
in terms of x and y, it is in terms of x,
y, and dy/dx.
So we need to get rid of that dy/dx and I
know what it is equal to... dy/dx is -x/y.
So I can do this substitution and take that
out, do some simplification, and when I am
done I will have a second derivative in terms
of x and y only.
So we have -y... negative times negative is
positive.
I am seeing another negative here that I am
bring in, but let's just cleaning that up.
So now plus x times -x/y over y^2.
Ok.
Well I just let those two negative cancel
out to be positive, but now I am multiplying
by a negative again.
So we have -y, positive times negative is
-x^2/y all over y squared.
Now we can clean this up a little bit so I
am going to go through these steps.
We want to find a common denominator.
I like to...
At this point I can take the denominator,
flip it up, and change that division to multiplication.
But I don't like to do that until I have one
fraction in the numerator and one in the denominator.
So I am going to take this y/1 and find a
common denominator by multiplying the numerator
and denominator by y.
So that is going to give us -y^2-x^2 over
our common denominator of y which is what
is allowing us to put this together.
I am starting to run out of room.
This over the denominator of y^2.
Ok.
So now what I have is, I have a fraction over
a fraction.
Which I kind of now pushed myself down to
the bottom of the chalk board.
But a division of y^2, it is really the division
of y^2/1.
So I am going to change that division to multiplication
by taking that denominator and flipping it
up and changing that division to multiplication.
So this is going to become -y^2-x^2 over y,
now this is where I am changing that division
sign to multiplication, I am changing this
one fraction divided by another fraction...
flipping it up... and changing it to multiply
by 1/y^2.
So that means, if we clean this up a little
bit, now...
Well I will show all the steps.
-y^2-x^2 over y^3 is our second derivative
with respect to x. Ok.
If you happen to see a substitution that,
especially with the original function, it
is going to make things a little bit simpler
then by all means do so.
Maybe your teacher is going to require that
and maybe not.
If I take these two top terms and factor out
a -1 and get y^2+x^2 over y^3.
See that y^2+x^2, that is actually defined
in my original function and it shows up pretty
easily that I can take that x^2+y^2 or y^2+x^2...
addition is commutative of course... that
is equal to 6.
So I can take this out, go back to my original
function, and say that my second derivative
can be simplifies even more to say that it
is -6/y^3.
Now if you don't see that substitution with
the original function... that substitution
is going to help simplify a lot, then this
is the second derivative in terms of x and
y and it is a valid form... but we had that
substitution that was pretty easily spotted
so there is our final answer.
Let's take a look at one more example that
I am going to allow you to attempt on your
own.
It is going to be a bit more complicated than
this one.
Then I will just kind of guide you through
the steps once we finish.
For our second and last example we have xy^2+x
is equal to 5.
Ok, so let's get the first step done together.
We have to use the Product Rule here, so we
have xy^2.
That product rule is going to give us...
Well, let's see.
x times the derivative of y^2 and again we
are doing this with respect to x.
This is going to 2y then involve the chain
rule.
Now I can write 2y and then multiply by the
derivative of the inside function dy/dx, but
just to simply the notation through this problem
because it is going to be kind of long i am
going to use the short hand version of this
which is to simply write y'.
So we are going to have 2yy'... that is my
dy/dx... plus ok that was first times the
derivative of the second now the second which
is y^2 times the derivative of the first.
The derivative of x with respect to x is dx/dx
which is going to be 1.
And that is the end of my product rule.
Now plus again the derivative of x with respect
to x is 1.
This is a constant so that is going to come
out to be zero.
We are going to solve for y', or solve for
dy/dx.
So what do we have to do here?
Well we are going to move the 1 over with
subtraction and the y^2 over with subtraction.
i am going to get 2xyy' is equal to... let's
put the y^2 first... -y^2-1.
Then we are going to divide both sides by
2xy and turn the air condition off so that
you don't have to hear that background noise.
That is better.
Dividing both sides by 2xy and we get y' or
dy/dx is equal to...
Now I am going to reveal the next part of
this when we work out the second derivative
for this from my notes.
And again like I said reveal it in small steps.
I just looked at my notes just to make them
match exactly and I actually have this written
in my notes as -1-y^2 so I am going to turn
the around.
-1-y^2 over 2xy.
Ok, so that is my first derivative in terms
of x and y.
Now we are going to find the second derivative.
We are going to have to be really careful
with our work.
There is a lot of Algebra in this.
That is actually part of the reason I want
to reveal this in stages.
I don't want to talk and think and write something
and make a small mistake that no one in this
room is going to help me catch.
We are going to have to use of course the
quotient rule and our denominator is 2xy.
So we are actually going in there as well
have to remember when it is appropriate to
use the product rule in our work.
But we have been doing a lot of derivatives.
Again, it is just sometimes hard for me to
talk and write and think at the same time,
and then I do something silly and then I have
to correct the video.
So you should be able to pause the video and
work through this.
Practice your derivative skills.
I will explain it of course when we are done
revealing the answer in small steps so you
can pause this at any point through the process
if you are trying to do this on your own and
you get stuck.
Alright let's look at this work step by step.
We have... well... using the quotient rule
right.
So it is the denominator times the derivative
of the numerator, so we have 2x times... that
is a constant so it is zero... take the 2
down and we get -2y times y' following the
chain rule... minus the numerator -1-y^2 times
the derivative of the denominator which is
2xy.
That is going to involve the product rule.
So we have the first times the derivative
of the second plus the second factor times
the derivative of the first.
Now we have a lot of multiplication and some
distributive of these binomials together.
So that is here.
We are going to take this minus sign...
Be careful of those parenthesis as you are
working these out.
There is a lot going on here.
So when you complete that distributive property
make sure that is still wrapped up in parenthesis
to remind you to take the negative that is
out front and multiply it through those parenthesis.
Combining like terms.
I see that I have coefficients that are all
even 4, 2, 2, 2, and 2 and then we have a
coefficient of 4 in the denominator so I went
ahead and cancelled.
I added some like terms and then I reduced
the coefficients by dividing them by 2.
Ok.
Now I knew that I was going to start running
out of space so I cleaned this up after I
cancelled out a 2 from all of these coefficients.
But as you look through here we have xy and
then y', and xy', and xy.
So we have a second derivative here, but it
is not in terms of x and y that I have drawn
lines all over.
So I went back to my first derivative and
started doing a substitution so I could that
requirement of having my second derivative
in terms of x and y.
So y' comes out and -1-y^2 over 2xy goes in
a couple of places.
We have an x in the numerator and denominator.
We have 2 y's up here and just doing some
basic algebra cancellation.
And when i got done with that cancellation,...
I kind of burned up some extra work here.
Maybe you may or may not choose to have done.
You could have found common denominators here,
but instead I multiplied the numerator and
denominator by 2y to cancel out this denominator
of 2 and the denominator 2y.
I just felt like that was how I wanted to
do that to simplify the algebra in the numerator
of this fraction.
So the 2y and 2, well the 2's cancel out so
y times -y is -y^2.
2y going to this second term is going to completely
cancel out so we have -1-y^2.
Then 2y times y is 2y^2 and finally 2y^4.
We have the -y^2 to multiply through those
parenthesis.
Negative times negative is positive... and
so on.
Combining like terms and bam!
Ok.
We finally got our second derivative.
I didn't come up with a real easy example
here to come up with our second derivative
implicitly.
But we have 3y^4+2y^2-1 over 4x^2y^3.
I am Mr. Tarrou.
BAM!
Go Do Your Homework:)
