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.