CHAPTER 02.01 PRIOR KNOWLEDGE FOR NUMERICAL DIFFERENTIATION

In this segment we're going to talk about differentiation of continuous functions, what kind of prior knowledge you need to have so that you can conduct numerical differentiation.

So if you look at this particular figure here, this is a function of f of x being plotted as a function of x, and let's suppose we call this distance here between p and Q on the x-axis to be Delta X. So what you are basically going to do is that let's suppose I want to draw a secant line from point to point Q right here so I'm interested in finding out what the value of the derivative of the function is at Point p, and what I will do is I'll draw a secant line just to approximate it so that's how we learned in your differentiation class it will be this divided by Delta X and so that's the slope of the of of the secant line so because that's the rise and that's the run, and what we learned in our numerical differentiation class was that as Q approaches P then Delta X is approaching zero and that becomes the exact definition of the derivative of the function, that as Q approaches P the secant line becomes the tangent line at P so that's how we Define our first derivative of a continuous function. Let's go and review through an example.

So let's suppose I am being asked that hey can you find out the first derivative of the function f prime of x for this particular function you want to find f prime of 3, and we also asked to find f prime of 3 if you draw a secant line with Delta x equal to .5, so this is just to bring everything together so that you have some recall of what you learned in your numerical differentiation class, so let's go and get started.

So if you look at this that you have f of x = 7 x² and if I want to find f prime of X then we know that 7 * the derivative of x² is 2 * X and that gives us 14 x and if I want to find the value f prime of 3 that becomes 14 * 3 and that becomes 42, so that's the answer to the first one what I'm going to do is I'm also going to show you how to derive the value of f prime of x to be 14x if we start from the very definition it's just to recall and also so this is going to very much help you to figure out hey what it means to numerically find the derivative of a function.

So if you remember we said that f prime of X is equal to the limit of Delta X approaching zero f of x + Delta x - F of x / Delta X right? so if our f of x is = 7 x²

what we're going  to do is we're going to say hey f prime of x is limit of Delta X approaching zero, and we're going to write down the function at X Plus Delta X which will be this right? we'll write the function at X which is 7 x² and divided by Delta X, and now it's all about simplifying it so we'll use the A + B whole square formula here we get 7 we get x² + Delta x whole squared + 2x Delta X - 7 x² / Delta X and if we simplify it some further we get this, and from here you can see the 7 x square cancels with this one so if I take the limit of Delta X approaching 0 here this is 7 Delta x² divided by Delta X that's 7 * Delta X then this is 14x Delta X divided by Delta X that gives 14x and as you can see that as Delta X approaches zero this quantity is going to approach zero and this quantity is going to approach 14x and that's how we get the 14x problem, and that's how we got that the f prime of 3 is 14 * 3 which is 42. So this is solving for the value of the derivative of the function from the basics so you can see that how this formula works it's not magic that we find f prime x to be 14x here.

So in Part B what we asked to do is to figure out that hey if we are trying to find the value of the derivative of the function by seeing what is the slope of the secant Line, This is X = 3 so, not drawn to scale here let's suppose this is x equal to 3.5 so we are looking at a run of 0.5 that's Delta X,

and then you can see the the rise will be this amount right here, that will be the value of the function this is the value of the function f of x plus Delta X and this is the value of the function of X, so the difference will be of course f of x plus Delta x minus F of x. So if you look at the secant line the derivative of the function will be approximately equal to F of X + Delta x - F of X / Delta X rise over run, where this is the slope of the secant line, now we know that X is 3 Delta X is 0.5 we subtract three function of 3/ Delta X which is 0.5, so it is the difference between the function at 3.5 and 3 divided by 0.5 which is going to give us the slope of this secant line right here so we're trying to find the slope of this secant line. So we get 7 * 3.5 squared - 7 * 3 squared / 0.5 which in fact turns out to be 45.5 once you do the calculations, and this one is very close to 42 which is the exact value, and you can see that if you decrease the value of delta X it is also going to improve the value of the derivative of the function so I should have call this to be f prime of 3 here, and that's what we're getting so this F Prime of 3 is not the exact derivative of the function it is an approximate one but it is the slope of this secant line, and that is the end of this segment.