So in the previous tutorial, I showed you that unlike a linear function, quadratic function, or more complex function than linear one, there is no point to talk about the rate of change, when we change X by only one unit. Because in each and every step that we take, this one unit, the rate of change will be different. Therefore, let's get rid of one unit as a measure. And let's just introduce ourselves with a new measure, which we'll call, delta_X. Delta_X will represent a change in X, rather than a change of X in of one unit. And if we want to introduce our self, what will happen? If we want to familiarize with what happened to Y as a result of this change in X, we'll have to understand how this function reacts to those changes. First, I remind you, what we were interested in? We were interested in the rate of change, how much Y will change, as a result of change in X? So in other words, we were interested in this formulation where we see delta_Y, change in Y, as a result of change in X, or in other words the rate of change. Let's take this as an example. In order to understand that, we'll have to take an initial point. Any initial point that you want on the graph. Let's call this X0. When we introduce an addition to this X0, the additional B, the delta_X, the change in X0 that we have. So the second point that we're interested in will be X0 plus delta_X. So this will be our X1, and this will be our X2. Coming back here, what we're looking for basically, is what will happen to Y when we take X equals 2. X2 minus Y when X is equal to X1, the original X1, over X2 minus X1. Or in other words, delta_X. Like I just said, if we want to use a quadratic function, let's try to convert this into the values. So whenever Y is equal to X1, the value of one will be X0_squared, because the value of X1 is actually our original point. So, whenever X is equal to X2, what will happen to this original point, is that we'll have to take X0 plus delta_X and square it, because this is the value of the second point that we have. It will be easy to detect the changes here if we open up the brackets. So, a simple rule of quadratics will show us that this is X0_squared plus 2X0 times delta_X plus delta_X_squared. If we substitute this all back here, what we will see is the following thing. We will see that instead of X equals X2, this is the expression that we're substituting first. So it will be basically, X0_squared plus 2X0 delta_X plus delta_X_squared. When we substitute the value of Y, when X is equal X1, we simply have minus X0_squared. We divide all that by the value of X2, which is this value, which is X0 plus delta_X minus the value X1, which is this value minus X0. We can continue that, and see that some things are disappearing from the equation. This and that cancel each other out, as well as this and that, cancel each other. And we are left with is, in the numerator, with 2X0 delta_X delta_X_squared. In the denominator, we're left with delta_X. Now, the following step will be to get rid of delta_X. Factorizing that, we'll be able to see that this and that, and this, disappears. And we're left with 2X0 minus delta_X over one, or simply 2X0 minus delta_X. Now, to actually translate what we have done into words, we said that the rate of change in Y as a result of change in X, will be 2X0 minus delta_X. Actually, this should be plus, yes. Plus delta_X. 2X0 plus delta_X. 2X0 plus delta_X. So, the rate of change depends on the original point where we start our calculation with, plus the bit that is the change from the original point, until the next point of X. So this is the rate of change. Now, what we're interested when we talk about derivatives, we're interested in the changes of delta_X, when delta_X is very, very, small. It's almost zero. I can now see that this rule, this assumption is bringing us to the general rule of derivatives. And I will show you why. Because we're not interested in this point because this point was arbitrarily chosen. Therefore, it can be any point X for this sake. And if we're talking about a very small value of delta_X, or very small change of delta in X, then this expression disappears completely from our equation. And what we're left with is 2X. And if you notice, and if you're familiar a little bit with derivatives, you will see the derivative of this function is actually 2X. So in other words, derivatives is actually representing a rate of change in Y, when the change in X is very, very, small. Which means that you can use this rule in order to find a derivative of any single function, regardless how hard, or how complex this function is. The only thing you need to do is to structure the rate of change of Y, and changing Y, and change and X, in the following manner. And assume that the change that we're talking about here is very, very, tiny. If we take another example, let's pick another example. Let's pick a linear function. A linear function of the following form, 3X plus seven. And as before, what we're interested in is delta_Y of a delta_X, or the rate of change in Y, as a result of change in X. What we do, we assume there is initial point X0, and there's point of X0 plus delta_X. And we plug it back here. So what we have, is we have 3X0 plus delta_X plus seven, minus 3X0 plus 7, over X0 plus delta_X minus X0. Doing the same exercise by opening our brackets. We're getting to the point where we have 3X0 plus 3_delta_X plus 7 minus 3X0 plus seven, over delta_X. This is minus of course, seven and seven cancels out, 3x and 3x0 cancels out, and we're left with 3_delta_X, over delta_X. Delta_X and delta_X cancel out, and we're left with three. And again, this will be exactly the derivative of this function.