Proof: Derivative of e^x is e^x

[@bejjwkwnsb7q777]

Thank you very much. This is the best proof I've seen so far. The other ones here on KZbin usually involve some kind of unsatisfactory implicit differentiation. Not this one!

[@LarsSandgren]

Me during the first part of the video: Isn't he just going in circles?.. Me during second part of the video: We are just going in circles.... Me during final part of the video: WE ARE SUPPOSED TO GO IN A CIRCLE!!! I don't know who made this proof but, hot diggity it is jaw dropping

[@abouillet5147]

it's kind of a redundant proof actually. you can simply set m=1/deltax (delta x->0, m->inf), then sub in and get e^x * lim m-> inf of m(e^(1/m)-1), where in the limit, e^(1/m) becomes 1, then the term in brackets becomes (1-1) and collapses the limit leaving you with e^x

[@Kzaman-yg5br]

6:50 can I apply L Hospital rules because it is now 0/0 form.

[@dinofish3262]

U can do that but you would need to take the derivative of m and ln(1+m). The proof is using first principles, which means all the information u have is the limit definition and algebra.

[@timding6241]

At 3:56 , I do not understand why the e to the power of x is "independent" of the limit and can go to the front of the whole limit/right hand side. Many videos say the same thing but never explain that. Is there a video that explains that? Thank you!

[@zx4706]

Observe that we are taking the limit of the expression (inside the limit) as h→0, meaning we only need to evaluate those terms or expressions containing the variable h. Since fhe factor e^x does not contain any variable h (hence, we call it independent), then it can be treated as a constant multiple of the expression, which can be put outside the limit expression per limit rule on constant multiples.

[@geraldvaughn8403]

How did you bring the limit inside the natural log? What rule is that?

[@phlaelo866]

The limit doesn’t affect anything outside the log so it doesn’t matter. It’s functionally the same

[@ranjansingh9972]

Natural log is monotonic. I believe that allows us to 'jump' the limit on the inside. Vaguely remember this from somewhere.

[@philw9102]

Thank you, I found this really clear. Including approaches that didn’t work was very helpful too (I’ve watched other ‘proofs’ that evaluated 0/0 as 1 with no justification).

[@saravanarajeswaran2626]

on the other side , any function say a^x , the derivative would become a^x times limit as h approaches 0 of a^h - 1/h , which gives some random irrational number (which ln a) , so we would ask , so is there any number ,so that we take derivative the limit becomes 0, yes that is e , so most likely it is one of e definition you can say from another persepective

[@oscarludwinmalaveranarcaya7080]

Best proof of this derivative, others imply knowing what Taylor series is but that's ridiculous if we're working with the definition of derivatives

[@GigachadHGDYDBHDB]

When you put m approaches 0. 1/m is undefined. So m=1/k Then k approaches infinity then use the usual definition of e and bsm

[@billtruttschel]

Why are you allowed to move the limit inside the argument of the ln function?

[@austinbradshaw5021]

cause the ln function is being treated as a coefficient I think.

[@MochiClips]

The function is continuous so if x_n converges then Lim f(x_n) = f(lim x_n) :) (Where f is any continuous function e.g. ln)

[@doyourealise]

amazing! thanks a ton, subscribed! wil be finishing all of your videos.

[@renesperb]

If the chain rule and the derivative of the log-function is known ,then there is a very fast and simple proof: set y(x) = e^x and use that d/dx(ln y(x)) =1/y(x) *y'(x) ,but by definition ln(e^x) = x ,hence we have 1/y(x) * y'(x) = 1,that is (e^x) ' = e^x.

[@MasterWuMathematics]

But what if that derivative is not known??

[@AvneeshKumar-x4o]

​@@MasterWuMathematicsthen try to find out the derivative of lnx first

[@ethanluvisia8678]

This was amazing, thank you!

[@MundiaMukelabai]

the video is well explain but I only have one question. can you show us where and why e is defined as given in the video because we just adopted it and not shown it also how e is defined by the supposed given limit

[@tzbq]

I'm pretty sure that's just the original definition of e, idk why tho

[@Xayuap]

this one was satisfactory I rather predefine e than predefine a magical exp

[@hamidmohaddes2774]

We can use lhopitalle rule

[@NumbToons]

What a beautiful video and presentation

[@soured7967]

best explanation ever, tyvvvvm

[@ksmyth999]

Thanks for the video. I think it was well presented. But I wouldn't say it is based on first principles, because you have already made an assumption about e. Is it not possible to only assume there is a function such that f''(x) = f(x) and then derive its characteristics? The function would be k^x. One would then have to determine a value for k. Or is this not possible? I suppose as with most theorems or derivations, the problem is deciding where the buck stops.

[@MasterWuMathematics]

Thank you for your comment. Very good! That’s what mathematics is all about. What I meant from first principles is I’ve formulated the result based on the definition of the what a derivative is. And I have not assumed e as such, because that’s what is is defined as as well. What you’re proposing I’m sure there’s a way but it will not be within the scope of my math channel. My goal with my channel over the next few years is to steer it toward solving mathematical problems in engineering and physics (i.e. applied mathematics). Calculus is very important to these fields which is why I’m covering it extensively.

[@JessicaShaw-ym4vc]

@@MasterWuMathematics Hiya! How did you define e in terms of m? I understand how you manipulated it, just not how you found it. Thanks!

[@dqrksun]

Nice proof sir

[@spudhead169]

Right at the start you have circular reasoning. You see, d/dx e^x = e^x IS a definition of e. Such that e is the only value of n that satisfies d/dx n^x = n^x. All definitions are equivalent, so what you are doing by taking the definition of e at the start is effectively saying: to prove d/dx e^x = e^x, we first start by asserting d/dx e^x = e^x. You see the problem? Using the limit definition of e is exactly the same as using the derivative definition of e. But more than that, YOU CANNOT PROVE A DEFINITION, if you could you wouldn't need it to be a definition in the first place. No matter what you do, you simply cannot prove it without some kind of circular reasoning. Maybe we will develop the maths tools someday that allow us to prove the value of e, but until then, we can only define it.

[@nynthes]

Hi, what program do you use to make these videos?

[@tahmidislamtasen1602]

Best proof found on internet

[@MasterWuMathematics]

Thank you very much!

[@uvuvwevwevweonyetenyevweug5884]

How about we say that e^x=1+x/1!+x^2/2!+x^3/3!+x^4/4!... Since e^x is optimal growth. And if we derive this we find 0+1+x/1!+x^2/2!+x^3/3!... And so on. Hence e^x is equal to its derivative.

[@satya7198]

Great content sir

[@MasterWuMathematics]

Thank you :)

[@katiatzo]

THANK YOU!

[@MasterWuMathematics]

You’re welcome!

[@ruzgar1372]

really good proof

[@Xayuap]

it is not the only f(x) = 0 = f'(x) also, for instance

[@MurshidIslam]

All f(x) = ce^x has that property where c is a real number. When c = 0, we get your example.

[@a_man80]

y'=y, y'/y=1 (y≠0), ln|y|=x+c |y|=e^(x+c)=e^x•e^c=Ce^x (C=e^c) y1=Ce^x , y2=-Ce^x , check y=0 0'=0 , y=Ce^x , C is a real number

[@michaeltajfel]

You need to add the ‘initial condition’ y(0) = 1

[@AscendantPerfection]

Beautiful ❤️

[@davydorynbaev]

Thank you!

[@ledinhvu-h5q]

exellent !

[@MasterWuMathematics]

Thank you!

[@ramoimas9792]

You are the man. You are him. You know where the dog is, you know victoria's secret. You can see john cena. My pronouns are he not him. Cuase I can never be you.

[@MasterWuMathematics]

I'm nothing special, honestly. But thank you!

[@stefankrimbacher7917]

Very nice.

[@Happy_Abe]

Not the only function. Also y=0

[@fsponj]

I used to think so but then I realised that any multiple of e^x is also it's own derivative (a*e^x)'=a*e^x. If we set a=0 → (0)'=0. So the derivative of f(x)=0 is just a special case of f(x)=f'(x)=a*e^x

[@Happy_Abe]

@@fsponjright!

[@timm1328]

the exponential function is defined as that unique function that is its own derivative. proof is not necessary.

[@dr.rahulgupta7573]

Sir We can use the definition of e^h (in terms of factorials of natural numbers and different powers of h .) to find the value of ( e^h --1)/h. when h tends to zero . Clearly it is 1 . Hence d/dx. e^x =e^x . proved .

[@justafunnyclown6831]

As he said when h tends zero than the value will be undefined 0/0

[@samuelprieto3873]

Coulda skipped a bunch of steps but still excellent proof ;)

[@idjles]

There is a kind of circular argument here. You’ve assumed without proof what e^irrational means. Usually a^x, where x is irrational , is defined via e^x, and then you have to prove that e^(a+b) =e^a*e^b, for all Real a,b, which you haven’t. But I like the substitution with the log - that is cool, and avoiding Taylor Series.

[@ZipplyZane]

You can define irrational exponents with limits of their ratio al approximations. Using e^x or ln(x) is just a convenience. See the Math StackExchange question: "Can you raise a number to an irrational exponent?" for more information.

[@thecarlostheory]

Very nice

[@englishfortelugupeoples]

❤