So now, we've got our a bit complicated definition here. Let's see what it actually means for our some given functions. For example, let us revisit the definition once again. I'm going to stress something out for you. Firstly, we all need to understand that basically we can enroll in two possible manners here. We can either substitute or fix all the variables except the variable of differentiation for example, we need to substitute y with b and then find the derivative by the definition or any other technique. Here comes the difference. One can differentiate our function at every point and then substitute our variable x with a point we're actual interested in, or just find a derivative at this very point. It's not a big difference or every thing will be true. So let us see what it actually means. For example, let us take on quite simple extra function here, square root from absolute value of x product y. Assume that we're in need of finding partial derivatives at zero point. So what we're going to do? We are going to consider function which is a restriction of powers of four y equals to zero. So as a result, we get function F to root x and zero which is square root from x multiplied by zero which is simply zero, for all possible axis right? So the derivative of zero it's pretty obvious that it is zero and as a result it is zero if we substitute x with zero. Same thing applies obviously for the derivative. Partial derivative to [inaudible]. Why is so we've actually find both partial derivative at a given point right? That was easy. So maybe assume that we need to find the derivative at some arbitrary point right? In order to do so first of all, let us look at this nice surface here. That's kind of what we've extracted right, because we've worked at our partial derivatives are equal to zero. So we get horizontal tangent lines of four hour zero zero point. So as for arbitrary point, let us cut them as a more complicated but actually more nice function here, more smooth function. Here x squared plus y squared multiply it by sum of product x and y and we are going to calculate our derivative at pretty much any point. Here is what I'm going to do, I'm going to just assume that we're going to substitute. For example y with a b just on the [inaudible]. So what we're gonna do we're gonna write our function that x squared plus b squared multiplied by sine of bx and right here we are going to just find the derivative combined x. We're looking at the derivative of the project and thus we are going to roll as usual. We need to calculate the derivative of the first one which is x times two and multiply it by the second, and then add vice-versa. The first one, the first function x squared plus b squared multiplied by the derivative of sine of bx which is essentially a composition of two functions of sine with cosine as a derivative and bx which is quite easy. The derivative of it's equals to b. So we find our partial derivative is on the sine will have to do is substitute our X with points a and del give us our answer. Sometimes people usually omit the step of substituting y with b and to x plus y. Because obviously we are talking about the values of this partial derivative at any point. At point wherever a and b can be [inaudible] from the real plane. So we just can't skip our substitution with a and b and just roll with the derivative in terms of x and y. That's always true, but I'm just stressing you out but that you always think about fixating your variables in the first place. So now we know how to calculate our derivatives with substitution and that was quite easy.