The Easiest, Yet Hardest Derivative.

[@joelklein3501]

9:49 When you substituted 1/t with s, you said that t-->0 implies s-->inf. But actually, it can be either +-inf. Setting t to be strictly possitive will indeed imply that s-->inf, but that would only be a one side limit. We should similarly calculate for s-->-inf, and see that since both one sided limits equal 1, then the entire limit equals 1

[@glenmatthes8839]

I vaguely remember my high school calculus teacher proving this exactly as you did. We knew nothing about Taylor series expansions, but we all new the definition of e as the limit of (1 + 1/x)^x as x approaches infinity.

[@steve112285]

In my opinion, exp(x) is defined as the solution to f'(x)=f(x) with f(0)=1, so there's nothing to prove. From this definition, you can easily derive the power series then the homomorphism properties so that it becomes e^x.

[@juandavidrodriguezcastillo9190]

yeah, but the video way is cooler

[@danielc.martin]

And when you try to find the integral of 1/x you saying that it is log beacuse of definition xd

[@steve112285]

@@danielc.martin Well that's actually a consequence of the definition for exp(x). If you define log(x) as the inverse function of exp(x), then since f(f^-1(x))=x, by chain rule we have f'(f^-1(x))*(f^-1)'(x)=1, so (f^-1)'(x)=1/f'(f^-1(x))=1/f(f^-1(x))=1/x.

[@mahmoudelitreby3427]

I think exp(x) is conventionally defined from its tailor series expansion

[@SimsHacks]

It's not truly a definition, it's more like a theorem that says that there exists a unique function with such properties. To prove existence, you still use most probably the power series formula lol.

[@mkqhwg]

the meme at the beginning lmaooo

[@aniritri8635]

You can use the mean value theorem at 5:30 since the quotient is differentiable on [0, x]

[@silimoon]

I like to look at the series of exp(x) and derive every summand by using the power rule instead. The first term becomes 0 and the following terms become exp(x) again.

[@wulli_]

you would have to justify the ability of differentiating termwise, since that generally only works for finite sums.

[@Errenium]

​@@wulli_is that a concern if we're specifying functions which are analytic at x, though?

[@joda7697]

I mean, the number e is kind of defined as the specific base for which the differential of an exponential funcion equals itself. That's what e _is,_ the right number such that: d(e^x)/dx = e^x holds as an equality, instead of as a proportionality only.

[@cptn_n3m012]

Or it could be defined by the power series... or the limit of (1+1/n)^n... there are many ways to define exponential

[@angelmendez-rivera351]

Well, no, not really. The exponential function itself cannot be defined without an analytic property establishing its well-definedness. Therefore, defining exp necessarily precedes any definition of e. e is just defined as exp(1), but exp itself is not defined in reference to any constants.

[@joda7697]

@@angelmendez-rivera351 The exponential function can't be defined that way for _arbitrary_ sets on which exponentiation by integer coefficients is defined, no. But you can simply apply the limit definition of the differential on specifically the Reals to an exponential function from the Reals to the (positive) Reals. Doing that, and using exponentiation rules, you can separate out the variable (for ease of reading called x) part, so what remains is: Differential of a^x with respect to x = a^x * limit{ (a^h - 1) / h} as h goes to 0 where a is a positive real number. Now we try to find the value of a such that the leftover limit evaluates to 1, and the number fulfilling this condition is e, Euler's constant.

[@angelmendez-rivera351]

@@joda7697 Your definition is circular, as it already assumes the operation is well-defined to create the definition itself, which proves my point.

[@AMB666]

Papa flammy, this one was so fucking satisfying!!

[@PapaFlammy69]

[@bettyaugustine6691]

5:24 I guess you could make it much easier using l'hospital rule

[@evanbarkman5786]

I think the first time I saw a proof of the derivative of e^x was when I figured it out myself (although I'm not sure if I was being particularly rigorous), I did the same thing you did until I got to the Lim_x->0 of (e^x - 1)/x, but then I just turned e^x to (1 +x/t)^t before expanding the product of (1+x/t)^t, the leading one cancels out the -1, and the next term simplifies to x, and everything else goes to 0 as you take the limits.

[@justinpark939]

0:01: in light of this, I hope everyone here has a great day :) (unironically of course)

[@SimsHacks]

Preface of Papa Rudin: "This is undoubtely the most important function in mathematics."

[@drekkerscythe4723]

once I saw the thumbnail I already thought e^x

[@zirkq]

trying to get e in there is so sneaky

[@mazetlionel]

🤣🤣 so beautiful !! 🤣🤣 you are the deadpan master of finding the hardest way to solve problems.

[@pontusl1421]

I think the easiest way to prove it is to just use l'hospitals rule on the differential quotient

[@prodbytukoo]

the problem with L'Hopital's rule is that you are assuming the conclusion as true when you are differentiating, you can't use it.

[@damyankorena]

The OP implies f'/f=1 So (ln(f))'=1 ln(f)=x+C f=C*e^x

[@Dakers11]

Excellent !! Very satisfying, ich will singen'.

[@nuklearboysymbiote]

Isn't it only the case for t going to 0 from positive that 1/t tends to positive infinity? Does this hold in the negative case?

[@joelklein3501]

It should hold, but you need to calculate it seperately. It's probably just a slightly different version of the possitive s limit he calculated

[@physicus372]

I'd love to know where you got your chalk board from!

[@mr.inhuman7932]

Yes, Papa Flammy, I did enjoy the Video!

[@RithwikAgarwal-r3c]

Did this in my head using L'Hopital's Rule

[@lukasmoudry9973]

It is kinda clear you have to use some of the other definitions, no? (Provided you don't use the limit equals one as your definition). If you do not use any definition, well then you have to define what is e^x... so you would use another definition anyway

[@tarikbada6968]

Oh my god so nice

[@iamtufa]

luv this guy

[@darcash1738]

e^x or 0 I'm thinking, unless im fogetting something.

[@aimsmathmatrix]

Can you do something about operators? Maybe some functional calculus?

[@joda7697]

The operator exp(d/dx) could be fun to explore. I played around with that one, within the radius of convergence, for some fucking reason, it shifts the input of any function by one. So, like, where convergence works yadda yadda, the following happens: exp(d/dx) f(x) = f(x+1) [where you use the taylor series definition for exp(), and treat the integer powers of d/dx as repeated usage of the differential operator]

[@theunicornbay4286]

​​@@joda7697 You should try the generalized shift functional operator. Essentially, its a generalization of the shift operator in the form exp(k(x)d/dx), and its closed form evaluation of a function r(x) is: exp(k(x)d/dx) r(x) = r(u^{-1}(u(x)+1)), where u^{-1}(x) is the compositional inverse of u(x), and u'(x) = 1/k(x). Tho I honestly doubt papa flammy will cover it, I dont think its his area of interest anyway lol

[@oni8337]

​@@joda7697Pretty sure Jens has already talked about the shift theorem

[@EmileCluyse]

Very nice video!

[@parithiilamaaran.h9829]

Hallo Mach mehr Videos in Physik. Danke

[@oanceatudor4443]

Jaa

[@oni8337]

ikr

[@jacobgoldman5780]

If t goes to 0 isnt that a problem because 1/t could also go to negative infinity if you approach it from the left.

[@douglasmagowan2709]

Do you need to consider s goes to - infinity in addition to s going to infinity? I know that s going to negative infinity still evaluates to e, but, is it a necessary step?

[@edmundwoolliams1240]

But why were you able to use e^(x+y) = e^x * e^y, without first proving it from one of teh definitions?

[@hillabwonS]

Googlogy videos when

[@tgx3529]

Didn't lnx arise historically as the value of the integrated area under the function 1/y on the interval (1; x)??

[@angelmendez-rivera351]

I think this approach is somewhat unnatural, with regards to the definitions and the properties being assumed as consequences of the definitions. The starting point of the proof is ad hoc, and while this does not detract from the validity of proof, it is not particularly helpful from a pedagogical standpoint, in my assessment. The most natural way to introduce the functions exp and log are, to my judgment, by using the Riemann integral of the reciprocal function. Since the reciprocal function is continuous everywhere on its domain, it is continuous everywhere on [1, t], and since it is continuous on a closed interval, it is Riemann integrable there (this would not follow if the function were merely continuous λ-almost everywhere, and in this case, boundedness would be an additional condition required, which the function satisfies anyway). Since the Riemann integral on [1, t] exists, it is a fact that there exists *some* function g such that g(t) is the Riemann integral on [1, t] of the reciprocal function. We do not know what g is a priori (remember: in this context, we are pretending to introduce the functions to those who do not already have definitions). However, simply by how the definition of g was motivated, which was straightforward, finding many other properties becomes easy. Proving g(xy) = g(x) + g(y) (for x < 0, y < 0) is a fairly standard exercise, and it follows that g is an invertible function. Using the inverse function theorem, it follows from D(exp) = exp and exp(0) = 1. All other properties of exp you would care about come from here.

[@Happy_Abe]

@4:25 doesn’t this depend on if you’re approaching 0 from the left or right? If from the left then s approaches negative infinity

[@Grassmpl]

Why does the limit defining e exist and is positive?

[@VaradMahashabde]

Differential equation from first principles?

[@diogeneslaertius3365]

Awesome video, what a comeback after those primitive arithmetic videos!

[@tys3292]

f’(your mom) = a wonderful person

[@BedrockBlocker]

Neat!

[@nogs4077]

why cant f(x) = 0 be a solution to f(x) = f'(x)?

[@Bruh-bk6yo]

It's a particular one, not general.

[@HomieSeal]

I’ve been trying to find a bunch of solutions to the differential equations f(x) = g(x) * f’(x)! (Not factorial) So far, the ones I know of are: f(x) = e^x, g(x) = 1 f(x) = sec(x), g(x) = tan(x) f(x) = csc(x), g(x) = -cot(x) Does anyone know more solutions?

[@coucoucoucou8321]

Take any differentiable nonconstant function u f(x)=exp(u(x)) g(x)=1/u'(x) Wherever u'(x)≠0 the differential equation holds

[@steve112285]

Also, f(x)=0.

[@ThsHunt]

Just use hospitals rule for the limits

[@gianfronzioilventilatore4940]

bruh...

[@anarchodave7997]

what about f(x)=0 ?

[@bavariabal]

Hello papa can you make some videos on topology ploz 😀😀👌💀💀💀🪡🔝🔘

[@acho8387]

cursed video

[@PapaFlammy69]

?

[@Darth_Niki4]

videoed curse

[@banrtv]

what is this intro?

[@avinashbabut.n4123]

8:14 My mind yells: That L'Hopital's Rule....

[@Kiyotaka141]

I was thinking the same thing looks way simpler and it’s 0/0 so seems way easier

[@MrGabriel73]

@@Kiyotaka141 to use l'hopital you would need to know first that d/dx e^x = e^x, which is precisely the result we are trying to prove

[@Kiyotaka141]

@@MrGabriel73haha i’m so stupid I took de^x/dx so for granted I was about to use it while proving it. Would be circular reasoning ofc ignore what I said previously😂

[@mwolfe99]

You could probably use squeeze theorem too.

[@domc3743]

Please never say x-x_0 = x holy fuck

[@kasramoumeni6835]

This guy's kind of scam Believe me

[@atallcosts999]

What about just using l’hospitals rule to evaluate the limit directly?

[@Gazebr]

So you want to find the derivative through the limit using that same derivative? Or am I missing your point

[@indescribablecardinal6571]

President of México AMLO and President of Argentina Milei sends you blessings.

[@no-bk4zx]

The proof is kinda gay ngl. I wish it was something more fundamental than just using the definition itself.I wanted it to be something cooler but ig e^x is already cool enough. Great video though ❤