So since we have actually come up with
way derivative for multivariate case,
we should think about
second partial derivatives as we did in
the single variate case when we thought about
the derivative and then thought
about the speed or the speed of change, right?
So, let us draw another funny scheme here.
So first of all we have our function here, right.
That's our function level.
Then we do differentiate it by X and Y respectively.
Thus we get partial derivatives of
the gradient whatsoever right here, right?
So that's nice and expectable and we do know it,
and then comes the turn
of the second partial derivatives.
As always we think about
second partial derivative as
a derivative of a derivative.
But now we can differentiate either by X or Y.
Thus instead of one second derivative,
we get well one,
two, three,four partial derivatives here.
Well towards X two time times towards X and Y,
Y and X and Y two times.
Well, the notation we actually do see here
because well it's just tries
to stress out the order and
the set of differentiation that
we've actually come up with here.
In other words what we do,
we just differentiate the function one
time and then do it another differentiation.
So we compute partial derivative towards X and
then we compute the derivative of it's
derivative towards Y, right?
So and to finish it,
we are going to speak may be on notation,
and it's just an operator notation,
and that's expectable because
when we been talking
about the derivative for single-variable functions,
we've talked about the operator of
taking first derivative d towards dx, right?
So here it's still the same,
assumes that we have our first
partial derivative towards X for example,
it is a procedure when we
actually get as an input function F,
and as a result we return our partial derivatives.
So just by putting F behind
or just the symbols in this notation,
we get F which is an argument for this procedure
and then basically the symbol for
this procedure it's just an instruction what to do.
So in order to calculate the second partial derivative,
we need to apply
this operator several attempts, two times actually.
Firstly we will apply the rate of differentiation by
X for example and then differentiation by Y.
Okay? As you may see
the order of differentiation here is tricky because,
if you just multiply this fraction and fraction of
differentiation by Y and partial derivative for X, right?
You get in the denominator order
partial Y partial X, right?
But as for the notation
we discussed earlier will just simply use indexes,
we've written the last differentiation last one right?
So it's tricky about
right now we are going to discuss the problem of order
of differentiation itself and it's
not going to be such a mass later on.
But now let us consider some examples.
First let's take a look at
the function X power to Y-squared.
Once again I'm going to just reminds you that if
we have such multistage power,
then we are going to move from top to the bottom, right?
So something like this.
So we need to firstly compute XY squared and
then put X in the power of Y-squared.
Okay? So let's start with basic,
let's start with gradient.
Okay?
Partial derivative towards X
partial derivatives towards Y.
So towards X, it is just simply a polynomial function.
Thus we get AX powers of Y squared minus one,
multiplied by Y squared, right?
We need to write it down,
and still what's X.
What's Y it's just an exponential function which with
a nice exponential form here.
So we're going to write our logarithm X multiplied by
X about Y squared and then we
should remember that it is
a composition of two functions which
is A in power of X and X squared, right?
So we need to multiply everything by 2Y.
Nice, so then we can proceed with for example,
second derivative towards X two times,
and in order to do this we need to
differentiate the result of
our first partial derivative by X,
Y squared is a constant towards X so it remains,
and then we get in other case of polynomial function.
So we need just to move the power as
a multiplier behind and decrease the power by one.
So we get as a result XY
squared and multiply by Y-squared minus one,
multiplied by X power Y squared minus two.
So, let us assume
another one which is derivative towards X and
Y. I'm going to just
write that we need to differentiate the same towards Y,
and thus we get the derivative of the multiplication,
which is two Y's multiplied by
X powered Y-squared minus one,
plus Y squared multiplied by logarithm of X by
the same X powered Y squared minus one,
multiplied by two Y's.
Awesome, well and maybe for the last one,
I'm now going to compute the derivative
towards Y and then towards X,
just because as we will see later there is
a distance to coincide here.
So what is for the case
of derivative towards Y two times,
we going to need to differentiate
logarithm X multiplied by X powered Y
squared multiplied by two Ys towards Y,
which is once again logarithm X is a constant,
then we get two as a constant and then we
get the derivative of the product,
and okay, let's just write it down.
It's boring. So and
they're not going to waste any precious time here.
Okay, what was the process of computing
all the possible paths partial derivatives
for our given function,
that was quite well demanding.
Okay, so we can see next video.