[MUSIC] The last thing we've covered is the definition of differentiability and it's quite tricky, right? Of course, I've just said to you that your life just got sufficiently hard here, sufficiently more complicated. So let us just see what it actually means. Well, first of all, there is a quite fine line between differentiability and derivatives. And we've actually have looked at this at our square root from absolute value of x multiplied by y case. You do remember that it's tangent plane. I'm going to just write by capital here is zed equals to 0, right? So the thing we're going to see here that it is differentiable in case our effort towards xy function, our square root from absolute value is actually can be properly approximated by just simple 0, right? And it's kind of [INAUDIBLE] in here. So let us check this thoroughly. So what we're going to check here? We're going to check that square root of x multiplied by y. It's actually little, or infinitesimal towards the distance between the point xy and point 0,0, right, xy and 0,0 here. So by the Pythagoras Theorem, it's square root from x squared plus y squared, right? Okay, so what we're going to do to check whether or not it's true. As x and y approaches 0 simultaneously, so we need to check the definition of the small annotation, which is that the relation between left part and the right part is infinitesimal. So in order to do this, we need to calculate the limit where x and y books approaches zero. And we are looking at square root from x multiplied by y and devised by square root from x squared plus y squared. And I'm going to we just use the common square root here and write this down. And at this very point, you all know the answer because we've actually spoke about this limit whilst we were actually defining the limit of multi variate functions in the previous videos. Because if we can see the just the case of ax equal to y equals to some parameter t, then we clearly see that this function turns into just square root out of one half. Because well, in nominator is it is just t squared and in the denominator, it's just squared times 2 so it approaches one-half. Well, square root of one-half and not zero [INAUDIBLE]. This relation does not hold. I'm going to go in right in quality sign here and our function square root of absolute value of x multiplied by y is not differentiable at the point 0,0. In other words, the plans that we've actually come up with a candidate to tangent plane is not a perfect approximation here. And let us see just a picture here. It's quite expectable, right? Because here we see at 0,0 point, some peak, right, this kind of [INAUDIBLE]. And as you'll remember in a single word cases, this peak, this sharp edge is the first thing that for sign that you are going to have no just a perfect linear approximation here. So it's basically [INAUDIBLE] in a multi variate case, but is your, remember the construction, this surface picture is quite hard. So it's nice that we can do it just analytically without any extensive computer ports in help. Here is another tricky example here. Assume that we are looking at this [INAUDIBLE] defined function. It's defined at every point as multiplication of polynomial function and sine function. And the very zero point, it is just as [INAUDIBLE]. Firstly, let us think about its partial derivatives. Well, in order to do so firstly, we need to substitute our function y at our function, which is your, which results in two x squared multiplied by sine of x [INAUDIBLE] minus 2. If x is not a 0, right, and a 0 if x is 0, you will need to remember that if it was a definition, it's going to be most cases the same [INAUDIBLE] definition. And well, life doesn't get easy at this way point but it gets easier later because as we've stated that's the case where we're going to need to revisit our definition of the derivative. And to do so, we need to just write our definition as a limits, right? As x approaches 0, we need to find the limit of the change of the function, which is x squared multiplied by sine of x [INAUDIBLE] minus 1. And the value of the function at the point of differentiation, which is 0 here, right? And the change of our argument which is x minus 0, which is 0. And as a result, we get as x approaches 0, function x because we can just, Divide by x in the denominator and denominator as a result. We get x multiplied by sine function of power minus one, and it's, there are limits we've actually seen earlier because the x functions is infinitesimal. And point zero and sine function is buying bounded. Thus we get 0 as a result. So our partial derivative here equals to 0, the same stunt for the partial derivative towards y. So the function we were actually looking at has the same candidate as partial derivative of and as the same candidate as tangent planes here, which is that just equals to the plane 0, right? Let us reiterate this, okay. So does it actually means that we are going to have a nightmarish example as in the previous video again and function is non-differentiable? Well, let's see. Well, actually decided that we need is in the previous case to find the limit of the function, minus it's a tangent plane, which is 0. It's quite, [INAUDIBLE] divided by the square root of x squared plus y squared and here is where it's become quite nice because you can just divide the denominator and numerator by the denominator. And as a result, the power one is in x squared plus y squared turns into a square root in the denominator. But it is still infinitesimal function and sign is still well, [INAUDIBLE] approach is doesn't approach anything but it is bounded. But bounded multiplied by infinitesimal is still infinitesimal and thus we get our 0 is the result. So just let us take the fact. These function ugly, well, not the prettiest one actually is differentiable. It's our 0,0 point but more pretty square root is not differentiable. That's life, sorry. Okay, so that was another example. And once again, I'm just going to stretch this out is that we're going to be extremely cautious with our definition of differentiability and our calculation of partial derivatives from now. [SOUND]
