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[MUSIC] Up until now, we've been considering the functions that you can

Â get by starting with variables and numbers, and combining them using sums,

Â products, quotients, and differences. So we can write down, you know, functions

Â like f(x)=x2+x/(x+1^)^10+x, all of this, -1/x.

Â But there's more things in heaven and earth that are dreamt of in your rational

Â functions. For instance, can you imagine a function

Â f, which is its own derivative? I'm looking for a functions, that if I

Â differentiate it, I get back itself. Now, if you're thinking cleverly, you

Â might be able to cook up such a function very quickly.

Â What if f is just the zero function? Or if I differentiate the zero function,

Â differentiate a constant function, that's zero.

Â So this would be an example of function in its own deriviative.

Â But, that's not a very exciting example. [SOUND] So let's try to think of a

Â nonzero function, which is its own derivative.

Â How might we try to find such a function? So to make this concrete, I'm looking for

Â a function f, so if I differentiate it, I get itself and just make sure that it's

Â not the zero function. Let's have this function output one if I plug in zero.

Â Now, how could I rig this function to have the correct derivative at zero?

Â If the derivative of this function itself, the derivative of this function

Â at zero should also be one. Can you think of a function whose value

Â at zero is one and whose derivative at zero is one?

Â Yes. Here is a function, f(x)=1+x.

Â This function's value with zero is one, and this function's derivative at zero is

Â also one. But if the derivative of f is f, then the

Â derivative of the derivative of f is also the derivative of f, which is also f.

Â So, the second derivative must be f as well.

Â So, if this function is its own derivative, the second derivative of f

Â would also be equal to f. Now, specifically, at the point zero,

Â that means the second derivative of the function at the point zero would be the

Â function's value with zero which should be equal to one.

Â is this function's second derivative at zero equal to one?

Â No. If I differentiate this function twice, I

Â just get the zero function, but I can fix this at least to the point

Â zero. If I add on x^/2, now, this function's derivative at zero

Â is one and this function's second derivative at zero is one.

Â Since f is its own derivative, the third derivative of f must also be f.

Â No worries. If the thid derivative of f is also equal

Â to f, which is a consequence of the derivative of f being equal to f.

Â That means the third derivative of f at zero is equal to one, but this thing's

Â third derivative is just zero. But if I add on x^3/6, now, if I take the

Â third derivative of this function and plug in zero, I get out one.

Â The fourth derivative of f must also be f.

Â Okay, yeah. I gotta deal with the fourth derivative.

Â I'm out of space here, but no worries, I'll just get more paper.

Â Here, I've written down a function whose value at zero is one, whose derivative at

Â zero is one, whose second derivative at zero is one, whose third derivative at

Â zero is one, whose fourth derivative at zero is one.

Â And you can see, this is sort of building me closer and closer to a function which

Â is its own derivative. If I try to differentiate this function,

Â what do I get? Well, the derivative of one is zero, but the derivative of x is

Â one, and the derivative of x^2/2 is x, and the derivative of x^3/6, well, that's

Â x^2/2, and the derivative of x^4/24, well, that's x^3/6.

Â And yeah, I mean, this function isn't its own

Â derivative, but things are looking better and better.

Â But the fifth derivative of f must also be equal to f.

Â Okay, yeah. The fifth derivative.

Â I'll just add on another term, x^5/120.

Â And if you check, take the fifth derivative now of this function,

Â its value at zero is one. I've written down a function,

Â so that if I take its fifth derivative at zero,

Â I get one. The sixth derivative of f must be equal

Â to f. The sixth derivative I am out of room,

Â but here we, go. Here is a polynomial whose value first,

Â second, third, fourth, fifth and sixth derivative at the point zero are all one.

Â And you can see how this is edging us a little bit closer still to a function

Â which is its own derivative, because if I differentiate this function,

Â yeah, the one goes away, but the x gives me the one back, and the x^2/2, when I

Â differentiate that, gives me the x. X^3/6, when I differentiate that, gives

Â me x^2/2. X^4/24, when I differentiate that gives

Â me x^3/6. x^5/120, when I differentiate that, gives

Â me x^4/24. X^6/720, when I differentiate that, I've got x^5/120. And now, of

Â course, these aren't the same, but I'm doing better.

Â The seventh derivative must be equal to f.

Â To get the seventh derivative at zero to be correct, I'll add on x^7/5040.

Â The eighth derivative, I'll add on x^8/40,320.

Â The ninth derivative, I'll add on x^9/362,880.

Â Okay, okay. This is, isn't working out.

Â We're not really succeeding in writing down a function which is its own

Â derivative. Let's introduce a new friend, the number

Â e to help us. Here is how we're going to get to the

Â number e. This limit,

Â the limit of 2^h-1/h as h approaches zero is about 0.69, a little bit more.

Â On the other hand, this limit, the limit of three to the h minus one

Â over h as h approaches zero is a little bit more than one,

Â it's about 1.099. If you think of this as a function that

Â depends not on two or three, you could define a function g(x), right?

Â The limit as h approaches zero of x to the h minus one over h. In that case,

Â this first statement, the statement about the limit of two to the h minus one over

Â h, that's really saying that g(2) is a bit less than one.

Â And, this statement over here, and if you think of this as a function g, this

Â statement is really saying that g(3) is a bit more than 1.

Â Now, if you're also willing to concede that this function g is continuous, which

Â is a huge assumption to make, but let's suppose that's the case.

Â If that's the case, I've got a continuous function, let's say, and if I plug in

Â two, I get a value that's a little bit less than one,

Â and if I plug in three, I get a value that's a little bit more than one.

Â Well, by the intermediate value theorem, that would tell me there must be some

Â input so that the output is exactly one. I'm going to call that input e.

Â In other words, e is the number, so that the limit of e^h-1/h as h approaches zero

Â is equal to one, and this number is about 2.7183 blah,

Â blah, blah. Now lets consider the function f(x)=e^x.

Â So let's think about this function f(x)=e^x.

Â Now, what's the derivative of this function? Well, from the definition,

Â that's the limit as h approaches zero of f of x plus h minus f of x over h.

Â Now, in this case f is just e^x, so this is the limit as h approaches zero

Â of e to the x plus h minus e to the x over h.

Â And this is e to the x plus h minus e to the x over h,

Â so I can write this as e to the x times e to the h.

Â This is the limit then as h goes to zero of e to the x, e to the h minus e to the

Â x over h, Now I've got a common factor of e to the

Â x. So I'll pull out that common factor and

Â I've got the limit as h approaches zero of e to the x times e to the h minus one

Â over h. Now, as far as h is concerned, e to the x

Â is a constant, and this is the limit of a constant times

Â something, so I can pull that constant out.

Â This is e to the x time the limit as h goes to zero of e to the h minus one over

Â h. But I picked the number e precisely, so

Â that this limit was eqal to one. And consequently, this is e to the x

Â times one, this is just e to the x. Look,

Â I've got a function whose derivative is the same function.

Â We've done it. We've found a function which is its own

Â derivative. The derivative of e^x is e^x.

Â E^x is honestly different from this polynomials and rational functions.

Â We couldn't have produced that number e without using a limit.

Â E^x is the function that only calculus could provide us with.

Â