Proof of the derivative of e^x: A Step-by-Step Proof and Explanation

In this video, we will use the limit definition of the derivative, also known as the First Principle, to thoroughly prove what the derivative of e^x is. To do this, we will utilize Euler's number, and I will provide a detailed explanation so that you can easily grasp the concept. If there are any parts that you find difficult to understand or if you have any questions, please feel free to leave a comment.
If you're eager to explore more differentiation techniques for various functions, click the link below.
• Easy Differentiation w...
Stay tuned for a variety of upcoming differentiation videos. Additionally, I recommend watching the easy integration videos using the Box Rule below, as they will complement your learning experience.
• Easy Integration witho...
Also, there are various other videos available on my channel and more will be uploaded continuously, so feel free to visit and check them out.
/ @mathwithalex
www.mathwithalex.com/
#derivatives #differentiation #mathwithalex

[@MadScientyst]

This is what's called a 'rigorous proof' folks. Brilliant exposition & well done bro....a new subscriber!

[@mathwithalex]

Thank you! You've just become the illustrious 257th subscriber to my channel. Stay tuned for more mathematical adventures!

[@rudiyantoars6907]

Tks, I’m enjoyed

[@samueldeandrade8535]

Well, in the last part he shouldn't have carried e^x with that limit. Calculate the limit separately, then go back to the calculation and finish it.

[@tw5718]

Half expected the proof to be circular. Glad it wasn't. Nice

[@mathwithalex]

Thanks for tuning in. I'll bring you even better content in the future. Stay tuned!

[@Matematicand01]

A really ellegant proof! Congratulations!

[@mathwithalex]

Wow, thank you. I hope the video was helpful.

[@nektariosorfanoudakis2270]

Much better proof than the one I'd found when I was 18, where I used the definition of exp as a limit of a sequence and used Bernoulli's inequality and sandwich theorem to find the final limit and I'm not sure if I did it correctly anymore either. 😂

[@mathwithalex]

It seems like you truly have a passion for mathematics. Attempting such proofs at the age of 18 is truly impressive! I can't recall anything constructive I did at my 18, well... it's not in my memory. I genuinely appreciate your valuable insights and comments. I should definitely try the method you suggested. As a novice KZbinr, I'm well aware of my shortcomings. Comments like yours mean a lot to me and give me substantial encouragement. Thank you very much.

[@darcash1738]

Are they calling it sandwich theorem now? That’s awesome, I like that name much better than squeeze theorem

[@abbasfadhil1715]

@@mathwithalexim 16 but i didn’t discover or reinvent any mathematics just watching vids and trying to absorb as much as i can as well as doing practice problems I currently on my way to finish calc 1 yet still struggling to graph functions using calculus hope u do vid on that thx i knew a lot proofs for why the derivative of e is itself but using the definition has always its specialty

[@pauselab5569]

Another way to define e is to first prove that e^x is never 0 in real numbers. Then show that the wronskian of any 2 function that is it’s derivative is always defined and zero. So there is a unique function up to a factor such that f’=f. Call that function e^x. Then Taylor series to prove that e is a real number.

[@andunyaa]

Very Imresive

[@mathwithalex]

You're welcome. Your words mean a lot and are a great source of encouragement for a novice KZbinr like me. I sincerely request your continued interest and support in the future😄

[@squeezy8414]

Damn that's some neat writing! I always thought of this derivative being a kind of definition, like if you consider the function b^x for any base b, we know that its derivative is proportional to the function itself: Let f(x) = b^x (for well defined functions, let's say b > 0) Using the limit definition of the derivative we get: f'(x) = lim h -> 0 (b^(x+h) - b^x)/h = lim h -> 0 (b^x * b^h - b^x)/h = lim h -> 0 (b^x(b^h - 1))/h = b^x * lim h -> 0 (b^h - 1)/h = K b^x (where K is some constant of proportionality defined by the h limit above) Now to determine K for any base (remembering that we're trying to avoid circular reasoning), we don't really have a better way than just plugging in really small values of h. For example let b = 2, the corresponding limit value seems to approach ~0.693 if you let h get really small (which we now know to be ln(2) but pretend we don't know this). With b = 3 you get K ~= 1.099 - so suppose there exists a number between 2 and 3, let's call it 'e', where this K value is exactly 1 (i.e. lim h -> 0 (e^h - 1)/h = 1) and we have defined e in the following way: e is the number such that the function f(x) = e^x has derivative e^x (i.e. f(x) = f^(n)(x) for all positive integers n) and we can use numerical methods to find that e ~= 2.718. I'm not exactly sure how this relates to the definition you used but I'm 100% sure there's a connection - great video and let me know what you think!

[@mathwithalex]

Wow! This is truly amazing! It's so touching that you took the time to leave such a detailed comment. I'm deeply grateful for your thorough explanation with formulas. Defining the value of 'b' between 2 and 3, where lim h->0 (b^h-1)/h equals 1, to prove the differentiation of the exponential function is a fresh approach I've never thought about. Thanks to you, I'm learning new possibilities and methods. It would be wonderful if many others could also see this, as it's too valuable to keep to myself. Thank you so much for sharing your valuable insights with detailed explanations and formulas. Your kind explanation made it easy and quick to understand. Thank you.

[@squeezy8414]

@@mathwithalex Haha no worries, thanks for the content. I should mention that I originally got this idea from a 3blue1brown video: "What's so special about Euler's number e? | Chapter 5, Essence of calculus" so look at that if you'd like a better explanation with nice visuals :)

[@mathwithalex]

@@squeezy8414 It's wonderful to be able to exchange opinions and share knowledge like this. Please continue to show your interest and provide valuable insights in the future as well. Thank you.😁

[@averageenjoyer4404]

You can define e in various ways, the challenge lies in proving that they indeed yield the same number

[@literallyjustayoutubecomme1591]

You're right, you can define e^x precisely as the solution of a particular initial value problem.

[@darcash1738]

This was a fun proof, but I remember when I first tried to prove it, I stopped at e^x * lim h -> 0 (e^h - 1)/h. I saw that evaluating it became 0+/0+, and 0-/0- from each side. At this point, I realized these were approaching zero at the same speed, I guess you could say. Like how x^2 approaches infinity faster than x bc it has a higher order. It seemed logical to me to say that the remaining part was equal to 1 for this reason. Is there a mathematical term for this, being of the same magnitude/order I guess you might call it?

[@mathwithalex]

The points you raised seem conceptually close to L'Hôpital's rule. It appears that it could be helpful for a quick interpretation, and understanding methods for algebraic solutions would be beneficial as well. Thank you for your insights.

[@samueldeandrade8535]

But how did you figured it out that e^h-1 and h go to 0 at the same speed?

[@darcash1738]

@@samueldeandrade8535 i sorta just pictured the graph of e^x in my head, which may or may not be a bad way to think abt it. The closer you get to 0, the more it matches a linear function. Which does make a lot of sense after the fact of knowing the derivative of e^x too, since the derivative at x=0 is e^0, or 1, meaning it matches the derivative of x, which is also 1. Tho it may just be what math w/Alex replied to me about. I may have also inadvertently been thinking abt L’Hops rule without realizing it, which would definitely be cheating since we wouldn’t know the derivative yet 😂

[@samueldeandrade8535]

@@darcash1738 man, be happy that the way you think is clearly above average. And keep on studying. You have lots of potential.

[@darcash1738]

@@samueldeandrade8535 thanks man. I like to believe that there’s often a better way to think about any given topic, so as long as we keep thinking about the way we think, we can keep improving. That’s why i wanted to learn these derivative definitions. I realized I was taking for granted all these derivatives I learned without knowing why they are true, so I wanted to learn em all. Wishing you the best for your endeavors too, bro 🤝

[@gokaytaspnar1355]

I rewrite lim h -> inf (1 + x/h)^h using binomial expansion 1 + (hC1) * 1 * 1/h + (hC2) * 1 * 1/h^2 ... And found that e^x = 1 + x/1! + x^2/2! ... Without using it's derivative

[@rudiyantoars6907]

Tks

[@mathwithalex]

You are always welcome! 😁

[@BedrockBlocker]

Not bad, to note us that we used continuity of the exponential function and it's inverse here. I find it a little confusing that you use an "x" as multiplication operator, but overall a neat proof.

[@mathwithalex]

Thank you for your valuable feedback. I hadn't even thought that using 'x' for multiplication could be confusing. I will definitely keep that in mind for my next video. Please continue to share your great insights in the future as well. Thank you.

[@BedrockBlocker]

​@@mathwithalex Thank you for your genuine response, I appreciate your contributions to the math community :-)

[@mathwithalex]

Such interest and support mean a lot to a beginner KZbinr like me, giving me great strength and courage. Please continue to show your interest and support. Thank you."@@BedrockBlocker

[@winonamama8358]

silly goose

[@griffgruff1]

If one accepts the power series of e^x as being valid, then the identity is obtained by simply taking the derivative wrt x.

[@mathwithalex]

Thank you for your valuable input. Utilizing a series could be a great way to enhance understanding. Please feel free to continue sharing your insights. Thank you!

[@jamiewalker329]

You are interchanging two limiting processes, and you need to prove that the derivative of the power series, is the sum of the derivatives of the terms. You need to show the radius of convergence of e^x is infinity. So not quite as simple as you may think.

[@samueldeandrade8535]

​@@jamiewalker329 I guess what he wanted to suggest was "considering e^x is the analytical function 1+x/1!+x²/2!+... , then ...".

[@KartikPatel-nt4ff]

😊😅😅😅😅😅well information good show 😅😅

[@mathwithalex]

Thank you so much!!!

[@davannaleah]

From the series expansion of e^x, for very small x, we approximate e^x as 1+x...remember this is not the limit! as x->0. So (e^(x+h) - e^x)/h becomes e^x(e^h - 1)/h -> e^x(1+h - 1)/h -> e^x(h)/h == e^x! QED

[@mathwithalex]

Thank you for the valuable input. Please continue to share your insightful thoughts.

[@davannaleah]

Similarly for s = lim n->inf; s = (1+1/n)^n ln s = n*ln(1+1/n) For large n; ln(1+1/n) approximates as 1/n Hence ln s = s/n = 1 So s = e

[@Guillau213]

This is a circular reasoning. To get the series expansion, you need the derivative of e^x. So it is not valid in this case!

[@davannaleah]

@@Guillau213 is not a proof, just shows how similar things relate back to the same

[@nicolascamargo8339]

Wow

[@mathwithalex]

Muchas gracias. Me alegra que te haya gustado.

[@InnerView_]

The question is related to the variables h and n. Is it legal to interchange variables in the middle of a mathematical proof, and substituted? The expression was originally about h approaching zero, and then it became about n approaching zero. Is that allowed?

[@mathwithalex]

Sorry for the delay in response. It seems you are referring to the h and n at 1:48. (If not, please clarify.) I substituted e^h-1 with n, and as h approaches 0 in the limit condition, e^h tends to 1. Therefore, e^h-1 becomes 1-1, approaching 0, and consequently, n tends to 0. I hope this answers your inquiry. Feel free to reach out if you have further questions. Thank you.😀

[@InnerView_]

@@mathwithalexFrom what I understand- and I could be wrong. Substituting limit variables is not a valid method. The equivalence that was established was: e^h - 1 = n, and by that logic, the limit variable n would also be equal to e^h - 1 and not h. That is why it's frowned upon if the limit variables are changed in the middle of a rigorous mathematical proof, i.e. first principles.

[@samueldeandrade8535]

Man, what did you smoke? Hahahahaha.

[@InnerView_]

@@samueldeandrade8535 Ask god.

[@samueldeandrade8535]

@@InnerView_ hahaha. Man, there is nothing wrong with changing variables in this case. God is giving you bad sh1t to smoke.

[@nuraka3176]

Isn't e = lim n--> + infinity (1+1/n)^n ?

[@mathwithalex]

Exactly. In fact, these two are equivalent. When you substitute 1/n with x, as n approaches infinity, x approaches 0, and the expression becomes lim x->0 (1+x)^(1/x), which is also equal to e. Feel free to ask any questions, share your thoughts, or suggest any improvements anytime. Thank you

[@nuraka3176]

Oh wow, math is truly brilliant, thanks !

[@trix609]

0:18 To 0?

[@mathwithalex]

Does 'approaching zero' refer to the limit expression for Euler's number e? If so, you might find information about Euler's number in this link: www.youtube.com/@mathwithalex/community. Please check it out. Thank you.

[@samueldeandrade8535]

Yep. Usually the limit defining e is written as e = lim (1+1/n)^n , when n tends to infinity, which is equivalent to e = lim (1+n)^(1/n), when n tends to 0.

[@trix609]

@@samueldeandrade8535 oh, tx

[@pauselab5569]

it's by definition of e. not exactly a reason why that is the case, it just happens to be that number. if you wanted to find say a function whose derivative is itself, you can see that that function has to be an exponential function. then you can just define the exponential function e^x as being the one. then you try to figure out what is e numerically by using taylor series

[@mathwithalex]

Thank you for your valuable feedback. Please continue to stay engaged and share your thoughts freely. Your input is greatly appreciated. Thank you.

[@Niloctronic]

At 2:45, I didn't understand how could you move the limit operator to inside the logarithm's function. Why was that allowed ? Thanks for the video.

[@mathwithalex]

I apologize for the delayed response. When examining the properties of the limit of composite functions, if f(x) is continuous at x=b and the limit of g(x) as x approaches a equals b, then the limit of f(g(x)) as x approaches a is equal to f(lim(g(x) as x approaches a)) = f(b). In our case, f(x) = ln(x) and g(x) is (1+x)^(1/x). Thank you for watching my video and for your interest.😃

[@samueldeandrade8535]

Exactly. I think that's a mistake too. I mean, the limit of (e^h-1)/h appears when you try to prove e^x and ln(x) are continuous. So, proving it using the continuity of ln(x) is kinda circular.

[@frunsebischkek1050]

The expression (1+n)^1/n approaches e, if n -> infinity. How can you simply see, that it is a certain number and not 0? What about (10+n)^1/n or (n-10)^1/n for example? Great channel!

[@mathwithalex]

Hello, I apologize for the delayed response. Thanks for posing an interesting question. I just worked on it and posted the results on the community tab of my channel. Please take a look and share your thoughts. Thank you. www.youtube.com/@mathwithalex/community

[@samueldeandrade8535]

The expression (1+n)^1/n approaches 1, if n -> infinity. It approaches e, if n -> 0. If you want to know why, I can tell you. About (10+n)^1/n, it approaches 1, if n -> infinity, and it approaches infinity, if n -> 0 (n positive).

[@robertveith6383]

​@@samueldeandrade8535 --- Both of you, the fractional exponent, 1/n, must be inside grouping symbols.

[@samueldeandrade8535]

@@robertveith6383 no, it doesn't. Any reasonable person understands that a^1/n is supposed to mean a^(1/n) The way we wrote is fair abuse of notation.

[@avonbarksdale5821]

That limit being equal to e is not obvious to me. Even if it is the definition it could be interesting to learn more about that specific property.

[@mathwithalex]

Thanks for your valuable feedback. Regarding the proof of the Euler's number, please check out the link below. www.youtube.com/@mathwithalex/community I've summarized and posted three methods of proof on the community tab of my channel. I hope it can be helpful, and I look forward to your continued interest. Cheers!

[@newsgo1876]

Check this out. It also explains e well: kzbin.info/www/bejne/h2rTpoidechleLs

[@KingGisInDaHouse]

That’s literally where we get e from just with the infinity written as a reciprocal of zero.

[@uggupuggu]

Why didnt you let e = (1+h)^1/h as h goes to 0 and e^h =(1+h) thus the final limit = h/h which is 1

[@samueldeandrade8535]

Because that's not rigorous. It is cool. But people prefer rigorous arguments.

[@robertveith6383]

The exponent 1/h must be inside parentheses.

[@devar7502]

no f way

[@randomname9291]

This is a circular proof. The limit definition of e derives from the Taylor series of e^x, meaning it relies on the fact that the derivative of e^x is e^x. You essentially proved that if the derivative of e^x is e^x (a requirement for the limit) then the derivative of e^x is e^x (what you tried proving). The truth is there is no actual way to prove this, because that is the definition of e.

[@mathwithalex]

Thanks for leaving a good comment.

[@kirat1332]

It's not really the definition of e, There is a lot of definition of the function x->e^x -Only functions such as f(x+y) = f(x)f(y) and f(0) = 1 -The reciproc function of logarithm -Taylor series

[@randomname9291]

@@kirat1332 I never said it was the singular definition of e, I said it derives from the original definition of e.

[@samueldeandrade8535]

I think it is a circular proof. But not because of anything you said. The limit definition of e does NOT derives from Taylor series of e^x. You are completely wrong about that. The limit definition of e is the original one, from 1685. Taylor series were introduced in 1715. So you are completely wrong.

[@randomname9291]

@@samueldeandrade8535 the only proof of the limit definition of e I’ve ever seen is the Taylor series, so I’d be interested as to how someone had gotten to it through a different method if you’d be willing to share.

[@_kopcsi_]

another way is using taylor series

[@mathwithalex]

Expanding the function e^x around 0 using the Taylor series, we have e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! + ... And if we differentiate both sides, we still get the same result. Therefore, you are saying that the derivative of e^x is e^x. This is another way to prove it! Thank you for your insightful comment.

[@_kopcsi_]

​@@mathwithalex yes, and for me it is a nice derivation, because with this way we can intuitively see WHY exp(x) has such a weird Taylor-series with that factorial (and the same goes with trigonometric and hyperbolic functions which are fundamentally related to the exponential function). all the mystery boils down to the differentiation rule of polynomials. factorials usually pop up when we deal with consecutive differentiations of polynomials. another topic where this plays an important role is when we deal with the differential operator in the exponential argument which actually leads to the shift operator of functions (which is an important property in Fourier analysis, and which can be derived by using the Taylor-series of the exponential function and the binomial theorem). furthermore, when we analytically extend the domain of the factorial unary operator from integers to the real or complex numbers, we still use the rule of differentiation of polynomials with some trick.

[@mathwithalex]

@@_kopcsi_ Thank you for your detailed explanations and opinions. I appreciate the information and insights shared through comments, which allow me to learn from others' thoughts and discover new things. As a beginner's channel with much room for improvement, I kindly ask for your understanding and encouragement. Starting with the basics of calculus, I aim to gradually create videos with diverse and valuable content. Please provide me with your guidance. Thank you.

[@_kopcsi_]

@@mathwithalex you are welcome, and that's perfectly fine. I understand that this video and the channel aims mainly beginners of the topic. I just mentioned some examples where the differentiation rule of polynomials combined with the unary opartion factorial play some role. good luck with your channel.

[@mathwithalex]

@k0p1k47 Please continue to keep an eye on and share your valuable thoughts on mathematics. Have a wonderful weekend. Thank you!😁

[@jan-willemreens9010]

... also known as " FIRST PRINCIPLES " ...

[@mathwithalex]

Yes, that's correct. It's also called the First Principle. Thank you for sharing your opinion. I look forward to more valuable input in the future.

[@maxvangulik1988]

lim{h->0}((e^(x+h)-e^x)/h) =e^x•lim{h->0}((e^h-1)/h) this limit is of the form 0/0, so L'hopital's rule applies =e^x•lim{h->0}(e^h/1) =e^x ok cool

[@mathwithalex]

Thank you for your valuable feedback. When proving the derivative of e^x, it's actually more appropriate not to use L'Hôpital's Rule because it involves taking the derivative of e^h with respect to h in the numerator. However, I appreciate your suggestion. Furthermore, I'm currently working on a video that explains the proof and applications of L'Hôpital's Rule, so please stay tuned for more content. Thank you. 😀

[@maxvangulik1988]

@@mathwithalex i would consider it recursion, but only under the assumption that one knows that dh/dh=1

[@Guillau213]

This is a circular reasoning. To use the l'Hospital Rule, you need to know the derivating of e^h, which you are trying to prove. Not good in this case!

[@samueldeandrade8535]

"ok cool". Nope. Ok wrong.

[@maxvangulik1988]

@@samueldeandrade8535 it's correct, just really bad.