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[music]. We've seen a little bit of this linear

Â approximation business already. A long time ago, we saw this.

Â That, f of x plus h is approximately f of x.

Â Well, that much at least is really just because f is say continuous, right?

Â Nearby inputs should be sent to nearby outputs, but let's suppose that f is

Â differential, then we can say more, all right?

Â I'm going to add to this h times the derivative of f at x.

Â What is the derivative measure, right? It's the ratio of how much the output

Â changes to an input change infinitesimally.

Â But for an actual input change of h, if I take this, which is the ratio of output

Â change to input change and I multiply it by how much the input change is.

Â This quantity should tell me how much I expect the output to change when I go from

Â X to X plus H. Where does this formula actually come

Â from? Well, let's think back to the definition

Â of derivative. Right?

Â What's the definition of derivative? The derivative of F at X is the limit as h

Â approaches 0 of f of x plus h minus f of x, all over h.

Â Alright, that's the definition of derivative.

Â Now how does that help us here? Well, if I just pick some value of h, an

Â actual value of h, that's not 0 but close to 0, then I get that f prime of x should

Â be approximately f of x plus h for that small but p-, you know, non, non zero

Â value of h, minus f of x over h. So this is approximately equal, as long as

Â h is small. Now I multiply both sides of this

Â approximate quantity by h and I'll find that h times f prime of x should be

Â approximately F of X plus H minus F of X. And then I'll add F of X to both sides,

Â and what I'll get is that H times F prime of X plus F of X should be approximately F

Â of X plus H, and that's the statement that we had before, right?

Â That the value of function X plus H is approximately the value of the function of

Â X plus something I'm getting from the derivative.

Â The derivative of f at x times how much I wiggled the input by.

Â So I've got this formula, but why do I care about this formula at all?

Â Well, I care because this is really what I wanted in the first place, right?

Â I really wanted to understand how wiggling the input would effect the output, and

Â fundamentally, you know, the derivative isn't just some random limit that I cooked

Â up for you to evaluate. It's supposed to say something meaningful

Â about the function, and this is what's telling us what the derivative means,

Â right? The derivative really means something

Â about how wiggling the input affects the output.

Â There's another reason to care about this formula.

Â We can use this formula, in fact we've already used this formula to numerically

Â approximate a function that might be hard to compute otherwise.

Â Um,, let's look at a function f of x will be say, the cube root of x, and let's look

Â at the input 125 and f at 125, right? What's the cube of 125?

Â That's 5, because 5 times 5 times 5, that's 125.

Â So the cube of 125 is 5. Now let's try to wiggle the input and see

Â what happens. Let's wiggle by h, which will be three.

Â Three's not all that small, but compared to 125, three's pretty small.

Â And I'd like to know what f of a plus h is, right?

Â I want to know what the cube root of 125 plus 3, 128 is.

Â And you know that's going to be hard to know, right?

Â So I'm only going to know Approximately. But I can use the derivative here, right?

Â Because the derivative tells me how wiggling the input affects the output.

Â So as you can get the derivative of this function at the input A.

Â So let's differentiate this function. And this is X to the one third, so the

Â derivative is one third times X to the minus two thirds, right?

Â This minus 2 3rd as 1 3rd minus 1. It's a power rule.

Â Let's compute the derivative at the input 125.

Â So that means I look at 1 3rd times 125 to the negative 2 3rd power.

Â Well, 125 to the negative 3rd, the negative 2 3rds power.

Â That's 1 25th, and 1 3rd times 1 25th, that is 1 75th.

Â So this tells me what the derivative is at a 125, now how does that help?

Â Remember the other formula for this linear approximation game, that f of a plus h,

Â which is what I'm trying to compute, should be approximately f of a plus h

Â times f prime of a, and I know what all of these quantities are now.

Â F of a is 5, h is 3. And F prime at A is 1 75 and 5 plus 3

Â times 1 75 that's 5.04. Now, how close is that to the actual

Â retail value of the cube root of 128. Well, it turns out that the cube root of

Â 128 is actually about 5.0396841 and it keeps on going, which is awfully close to

Â 5.04. So our layer approximation game is working

Â really quite well. This idea that the complicated function

Â like the cube root function, with its curved graph can be approximated very

Â well, at least for short intervals by straight lines, right?

Â This idea of replacing curved object with perfectly straight lines, that's a key

Â idea of calculus. In short, curves are way too complicated

Â for us to understand. Lines, on the other hand, just straight

Â lines, are much easier for us to think about.

Â