Derivative of Exponential Function (e^x) From First Principles

[@AlongCameAirxD]

This is the only video I found that solved this problem and that limit without using the circular logic of L'Hopital's rule. You are the only one that showed that limit to truly be 1. Earned a like from me.

[@lukewarm7465]

Same here

[@agytjax]

@@lukewarm7465 He could have avoided the complicated route of proving using natural log (ln). Here is the proof : We have e^x.Lim(h->0){(e^h - 1)/h} --- (1) We know that e=Lim(h->0){(1+h)^(1/h)} Substituting the value of 'e' in (1) above, we get : => e^x.Lim(h->0){([1+h]^(h*1/h) - 1)/h} => e^x.Lim(h->0){([1+h]^(1) - 1)/h} => e^x.Lim(h->0){(1+h-1)/h} => e^x.Lim(h->0){(h)/h} => e^x.Lim(h->0){1} => e^x

[@sphakamisozondi]

Sane here. This is a satisfying explaination to this problem

[@boguslawszostak1784]

You don't have this problem if you DEFINE ln(x) as the integral from 1 to x of 1/u du, and the function e^x as its inverse function.

[@spudhead169]

Unfortunately it's still kind of circular. d/dx e^x = e^x is itself a definition of e, in that e is the only value of n that satisfies d/dx n^x = n^x . You cannot prove a definition, if you could you wouldn't need it to be a definition. Trying to prove it will always result in a circular argument with a derivative because such a proof requires the use of the definition of e in some form, and since all definitions are equivalent, using the definition of e is equivalent to using the derivative definition and thus circular.

[@nadineabusaleh9401]

The way he looks at maths as it is magic and charm gave we a really beautiful vibes , i have never seen a teacher that is calm and has this clarity before . I hope he continues .

[@emmanuelmasemola1014]

Very sincere, very clear, I wish we were together during my university days, these are the kind of channels that deserve subscription, you don't need to tell us to subscribe , we have fallen inlove with your content.

[@herbertsusmann986]

It has literally been 50 years now since I learned this stuff so many of the details I've forgotten (like how to derive things like this from first principles). This is an elegant way to do that for e^x!

[@Einstein.Albert.official]

man you have a beautyful handwrigting

[@AcryllixGD]

Honestly one of the best videos ive ever watched! Im an a level student in the uk learning about calculus and this video made it so clear as to why this was the case! Really good video!

[@PrimeNewtons]

Thank you.

[@xebby9]

The BEST explanation I've watched about this derivative

[@sphakamisozondi]

I have never, not even in my maths books I used at university, have someone explained why, _lim_ _h -->0_ *{exp(h) - 1}/h =1* Well done sir. Much love from South Africa

[@hiderr6805]

You may want to substitute (e^h - 1) with (1/n) instead of n. This way you would get easily to the most commonly known definition of e, that is lim n->inf (1+1/n)^n instead of (1+n)^(1/n)

[@PrimeNewtons]

Thank you. I will investigate that option

[@robread-jones3698]

We all know there is something inherently beautiful in mathematics, but that explanation with its cool, calm, clear and entertaining delivery really emphasized that point. It was a joy to watch. A video has to be something particularly special to get both a like and a subscribe out of a grumpy old git like me. Job done here. I'm looking forward to watching more of your videos.

[@Katutowavicle]

I wanna thank not just for the great explanation but the positive energy you put in the video

[@MrDipanmehta]

This is the most critical video - unlike so many other dealing with this topic. However, this doesn't comes on top of youtube search try adding some keywords or description mentioning "exponential function". This is amaziing video thank you.

[@PrimeNewtons]

Thank you for the suggestion, I have made some changes.

[@GicaKontraglobalismului]

Great! I have always calculated the derivative of the exponential using the derivative of its inverse, that is, of the logarithm, and always thought the direct calculation impossible. In Romanian Language "to learn" is said "a invata" which is formed of words "in" and "viata" which mean "in" and "life" ; in other words, the Romanian the word for "to learn" actually means "to be alive" which is exactly what you said in the end. Your mind already thinks Romanian! I also appreciate your style, the blackboard, the chalk, and last but not least your calligraphy!

[@PrimeNewtons]

Wow! This is inspirational! Thank you for your comment.

[@labibbidabibbadum]

Liked, subscribed, coming back. I’m helping my teenage son who is just starting with calculus. This kind of clarity in teaching is wonderful.

[@PrimeNewtons]

Thanks for the feedback

[@debjanimukherjee502]

Reminiscing my college days with you and enjoying my retired life ❤

[@mihaipuiu6231]

Sir,you are a good teacher.Why? Because your writing is very nice, you work on a clean table, but very important...your proof is very clear and you explain like MICHAEL PENN. Thanks, SIR.

[@obadamh7030]

I finally found someone to clear it up simply, I really owe you

[@MrWildcathendrix]

I've just studied this demonstration in my Math 1 book for my first year of Computer Science Engineering university course, it's exactly the same as you write, but the way that you explain it makes math much more fun!

[@RobertKashila]

The guy is a logarithmic genius 👌

[@renesperb]

A nice and clear presentation,and , in contrast to many other videos of this type , a good handwriting , making it easy to read.

[@donald_w]

You are an incredible teacher! Thank you for explaining this so well and not overlooking the small details 😊

[@mohamedsaith4532]

Wow!!!!!! How amazing explanation 👏🏻👏🏻👏🏻👏🏻👏🏻

[@agytjax]

From 4:35 onwards, you could have avoided the complicated route of proving using natural log (ln). Here is the proof : We have e^x.Lim(h->0){(e^h - 1)/h} --- (1) We know that e=Lim(h->0){(1+h)^(1/h)} Substituting the value of 'e' in (1) above, we get : => e^x.Lim(h->0){([1+h]^(h*1/h) - 1)/h} => e^x.Lim(h->0){([1+h]^(1) - 1)/h} => e^x.Lim(h->0){(1+h-1)/h} => e^x.Lim(h->0){(h)/h} => e^x.Lim(h->0){1} => e^x Q.E.D

[@clemensvorbauer1183]

no, you are not allowed to take the limit h->0 twice…

[@SanePerson1]

A nice side result from this extremely nice demonstration is hidden in the penultimate line. I'm so accustomed to taking the derivative of e^x, that I forget what constant I should use when taking the derivative of a^x. But the entire derivation you've given doesn't change for that case, EXCEPT that in the middle panel, one should use the base-a log instead of the natural log (ln). so you get (d/dx)a^x = [1/log(e)]a^x, where the log is the base-a log. In particular, this recovers the conversion factor for base-10 log and natural log: 1/log(e) ≈ 2.303.

[@catnip2906]

Dear Sir. Thanks for the clarity. I was blind but now I see.

[@goldCrystalhaze]

I saw an explanation of the derivative of a^x in a lecture, which I never actually understood and I was going to search for a better explanation these days. Your video came by chance and it is fantastic! Thank you so much! I have subscribed to your channel.

[@No-cg9kj]

e and ln love to sneak their way into everything lol. If you haven't got to calc 2 yet be prepared to see them a lot haha.

[@averagehooligan620]

Mindblown. Been searching for this.

[@kopisusu3781]

this really cleared things up for me. thankyou very much!

[@kingbeauregard]

Oooh, I like your style! You're really clear, and your enthusiasm is infectious. Subscribed!

[@petersamantharadisich6095]

it might be easier to simply start with that definition of e and expand in a power series e^h = lim_n (1+h+h^2/2+...+h^n/n!) then subtract 1 e^h - 1 = lim_n (h+h^2/2+...h^n/n!) then divide by h (e^h - 1)/h= lim_n (h+h^2/2+...h^n/n!)/h = lim_n (1+h+h^2/2+...h^(n-1)/n!) then take limit with respect to h (limit is 1 and doesn't depend on n), then with respect to n (still 1)

[@PrimeNewtons]

I agree. I wanted to stay within knowledge from precalculus and highlight that manipulation I showed.

[@znhait]

This is circular reasoning. You just gave the Maclaurin series for e^x, which is obtained by finding the derivative of e^x. If someone is finding derivatives from first principle, the assumption is that no result that includes the derivative--other than the definition itself--can be used.

[@petersamantharadisich6095]

@@znhait I don't think it is, as you can derive the power series by expanding the definition used in the video... e^x=lim_n(1+x/n)^n = lim_n{sum_j (x/n)^j × n!/[j!(n-j)!]} = lim_n{sum_j x^j/j! × [n/n][(n-1)/n]...[(n-j+1)/n]} = lim_n{sum_j x^j/j! × [1][1-1/n]...[1-(j-1)/n]} = lim_n {sum_j x^j/j!} × lim_n{[1][1-1/n]...[1-(j-1)/n]} = lim_n {sum_j x^j/j!} The above does not refer to derivatives of e^x, there is no need to use circular reasoning to get the result this way.

[@Ray1tx]

Wonderful explanation!

[@aram5642]

Greatest blackboard and chalk I have seen of all math videos here. The lighting would benefit from some angle or diffusor though ;)

[@PrimeNewtons]

Thank you. I am still trying to find the pest lighting conditions for videos. I hope the newer videos are better lit in your opinion.

[@souverain1er]

Great explanation. Love it. Never learnt this in calculus

[@tomvitale3555]

We've been saved from destruction and made the world a better place to live! 😁 Excellent description!

[@invisiblelemur]

Beautiful. Thank you for getting me as excited about this as you are!!

[@idolgin776]

It was really cool when the exponential definition of e popped out. Never seen such manipulation before!

[@anonymous-ui7il]

I am binging on your videos, it has helped me a lot with calculus.

[@PrimeNewtons]

I'm glad you find them helpful. Thanks for the feedback.

[@kiturundee9077]

Beautiful video. Love the energy 😀

[@felipecanogiraldo2499]

Very greatful of this explenation, great teacher, great video, great smile haha. Keep it on like that. Greetings from colombia !

[@Rob1066-]

Great pure math explanations!

[@tafadzwachigumbu4276]

This is a very good presentation. Thank you sir.

[@PrimeNewtons]

Thanks

[@sakangbenjamin]

Always on point sir God bless for your impactation

[@hiderr6805]

What an amazing video! Thank you so much! So cool, the only source I found using only elementary methods...

[@PrimeNewtons]

Thank you

[@kemumawhitney5439]

Your classes are enjoyable

[@muwememwanza3815]

Great video just learned something new

[@okeuwechue9238]

Great explanation. Thnx. An alternative explanation would also be expressing the natural exponential function as a Taylor series expansion and then differentiating each individual term to show that the resultant expression is the same as the original series

[@PrimeNewtons]

That would not be from first principle, though.

[@okeuwechue9238]

True :-)@@PrimeNewtons

[@randalltucker9343]

Very nicely done, sir! Great video!

[@tcmxiyw]

I think your explanations are beautiful, but when you say something like “the limit of the function is the function of the limit”, please justify it by saying “because the function is continuous”. It is interesting to note that the derivative of f(x)=e^x at any point can be found once we know f’(0).

[@icafe36485]

Hi Master, I enjoy your teaching method💐🌹👏

[@PrimeNewtons]

Thank you 😊

[@punditgi]

My first principle is to always watch Prime Newtons! 🥰

[@masoudhabibi700]

Thank for one more time.... master

[@binhql]

Great! You've just made by day :D Appreciate it a lot.

[@PrimeNewtons]

Happy to help

[@theeligator8728]

thank you sooo much i love your positivity! keep going +1 follower gonna recommend to my peers.

[@tfdtfdtfd]

Excellent video avoiding the unelegant definition of e as the "eh-1" limit......we should probably add a few technical details about the existence of limits when you split into products, ratios, swap order of ln and lim.....these generally hold true here due to continuity

[@PrimeNewtons]

Thanks for the extra notes. Appreciated!

[@muzza881]

I like to use the series expansion of exp(x) for this. Then exp(x+h) = 1+(x+h)+(x+h)^2/2! +........... Exp(x+h)-exp(h) term by term gives 1-1 + x+h-x + 0.5(x^2+2hx+h^2-x^2) + ........ Subtracting, dividing by h, and taking the limit gives us back exp(x). I suppose you can claim that the definition of the Taylor series already used derivatives to all orders of exp(x).

[@barthennin6088]

It appears to be a circular argument. ln(e^x)=x and e^ln(x)=x

[@PrimeNewtons]

Lol. That's what a lawyer would say. In mathematics, they are called inverse functions.

[@nemo5619]

Fabulous video, hats off!

[@Aferz]

Thank you SOOOOOO MUCH! You made me understand it and now I feel so good and so YEEEEEES YEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEES

[@PrimeNewtons]

❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️

[@patelharsh5133]

Thanks sir

[@shcottam]

Dude, this is pretty sick

[@Bob-sq7ev]

Thank you sir this helped me a lot ❤❤❤

[@K-drama-LegendKing]

thanks for not using circular logic this makes so much sense the video is amazing, i would love to check your other videos although i know the the derivatives but the way you explained this one im excited to see the other derivatives

[@user-dp9yn7zf4l]

Amazing video, first time learn the derivative of e^x this well! I have a question, when we taking the reciprocal, do we need to show that the denominator is not zero (at about 7:45)?

[@12388696]

Well done

[@馬瑞基]

It is so helpful

[@ΛαζαροςΙωαννιδης-φ5υ]

Bravo. Perfect.

[@atri5280]

ॐनमःशिवाय 🙏

[@ukidding]

you have v nice writing

[@PrimeNewtons]

Thanks a lot 😊

[@komalshah1535]

Fantastic sir!

[@JulesMoyaert_photo]

Nice demo!

[@the_n_ecromancer]

"you see that? That makes life a lot easier"😂😂😂

[@sergiolucas38]

Great video, you're very didactic and your letters are quite pretty as well, thank you :)

[@PrimeNewtons]

Thank you

[@TofaraRungano]

Culculas simplified ❤

[@nievalesterloydp.7399]

Ang angas

[@HelloBillyyu]

Hi I have just watched the video. Great work! Many thanks. Can I explore another approach here? We could find the Maclaurin series of e^h and it is be 1 + h + h^2 / 2 … then (e^h - 1) / h = 1 + h^3/2 + … then the limit of this is 1 if h goes to 0. This method is generally applicable to many ‘nasty’ limit calculations. Happy to chat. Cheers!

[@PrimeNewtons]

Yes, that's an option. However, this video was to highlight first principles.

[@paulwood3460]

Excellent proof. 👏👏👏👍👍 just one criticism..before starting the the proof of derivative e^x just state the fact that (lim n->0 (1+n)^1/n) = e Mathematics is simply wonderful.

[@robertveith6383]

The exponent, 1/n, must be inside grouping symbols: (1 + n)^(1/n)

[@paulwood3460]

@@robertveith6383 surely you mean (1+n)^1/n 😀

[@znhait]

This is excellent work. I'm wondering if there isn't an easier way...or just a more obvious to come up with this limit. Otherwise, the definition of e could have been used a lot earlier.

[@PrimeNewtons]

Great point!

[@Pauladam2216]

Nice

[@reddottgamer3047]

Finnally earth become a livable place.

[@ThenSaidHeUntoThem]

You are hilarious 😂

[@Rayglobster]

This is perfect

[@geraldomelo2751]

The l'hopital theorem can also be applied.

[@durjoysaha2896]

That helps a lot❤

[@_cran]

I know you mostly do calc but can you make a video about fourier series/transform-inverse transform and a video about laplace transform-inverse transform? It'd be pretty educating I think since I just know the logic of it's graph I know it's formula but I don't understand how or why it works to just integrate something with e^-ikx then re-integrate it with e^ikx shapes the function in a different way

[@mazenzidieh]

Thanks alot

[@CarolineSikamoi-rh7iv]

Encourage though small writings make them more clear atleast

[@СергейКовалев-т1д6м]

👍👍

[@shivankargupta6675]

😍😍😍😍😍

[@ThenSaidHeUntoThem]

Coool!

[@yigitrefikguzelses291]

When we are dealing with lim n->0 1/(ln(1+n)^(1/n) isn't this expression undefined because we have 1/n in the expression. I will be so happy if you can help me

[@PrimeNewtons]

n is not 0 yet. We are just approaching 0. So the function is not undefined and you should not plug in zero into the function because then it will be erratic. Try sketching that in desmos and see what happens as you approach zero. 😉

[@yigitrefikguzelses291]

@@PrimeNewtons yeah its say approxirametly 2.718 so it's e but aren't we getting different result while approching from negative infinity,(by the way thank you for your respond)

[@kaboflotv6455]

What about y=sin(×+1) from first differentiation??kindly asking

[@PrimeNewtons]

I have to a video for sin x. Use the same idea.

[@PrimeNewtons]

Same exact process. You'll get cos(x+1)

[@JessicaShaw-ym4vc]

Hi! where does the definition of e in terms of n come from? thank you. your video was great :)

[@dr.rahulgupta7573]

Sir. Can we use the definition of e^h to simplify ( e^h --1)/h and then take the limit ?

[@PrimeNewtons]

Certainly. That would be highly recommended.

[@dr.rahulgupta7573]

@@PrimeNewtons Yes Sir .

[@fardowsakhalif6669]

M.Allah

[@fardowsakhalif6669]

Shugran

[@hypothesisnyc916]

Your teaching style is great but it bugs me that you're using implication symbols as though they are equal signs. For students to know the difference between "equals" and "implies" makes a big difference in their understanding of proofs.

[@PrimeNewtons]

I completely agree. I promise to never do that again. Could you suggest a replacement for doing my transition without using the implication symbol.

[@GiftMlinde]

Eeeee sir ❤❤❤❤

[@anestismoutafidis4575]

It stays as it is= e^x, except for x we put numbers ( N○ - C)

[@aftabahmed7018]

will smith teaches maths

[@petechen794]

It's not difficult. You may use the definition of derivative to do it. You may also do it by using ln .

[@sochegeorge7962]

Can someone please explain the move at 9:32

[@sochegeorge7962]

Should have added if n = 23, (1/23)*ln(24) is NOT equal to Ln(24)^(1/23)

[@justinnwachukwu2054]

Write your number on the board. I appreciated your teaching style

[@PrimeNewtons]

My guy, I no fit write my number for blackboard na! Wetin be dat?