Why is the derivative of e^x equal to e^x?
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- Опубліковано 9 тра 2024
- What's the number e and why is the derivative of e^x = e^x? Take a course from Brilliant to learn more about calculus 👉 brilliant.org/blackpenredpen/ (20% off with this link)
In this video, we will learn the derivatives of exponential functions and we will see how we can define the number e. Calculus 1, AP calculus, calculus A-level lesson.
For more math for fun videos like this, be sure to subscribe 👉 bit.ly/3o2fMNo
0:00 We will talk about why the derivative of e to the x is e to the x
0:30 Derivative of 2^x by the definition of derivative
3:43 Defining the number e
8:07 Differentiate b^x
9:52 Check out Brilliant
10:51 Bonus: derivative of ln(x)
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Summary: by definition!
yes sir
@蛋已经破仍爱三角函数微积分
🦒: I think mathematician
@蛋已经破仍爱三角函数微积分 Ash Ketchum, if he was a mathematician.
@蛋已经破仍爱三角函数微积分 🦒
Love from India 🇮🇳🇮🇳❤️❤️❤️
As a math major, I really enjoy how you explain these things. A lot of these videos are not trivial at all
Thanks. I appreciate your nice comment.
Even when I specialized in theoretical (Pure) math as a 2nd major back in '14, I still come to these videos to 'refresh' myself on some concepts that while the solution is trivial, it uses a lot of previous fundamental math to show how it works out. For all pure math majors; if you survive the gauntlet, the abstract concepts never leave your head 😅
@@blackpenredpen Yo hablo español pero igual logro entender lo que explicas, muy buen video para entender el significado porque la derivada de e^x es e^x
@@YTMini_alonso porque las matemáticas son un idioma internacional
@@IoT_ beautifully said
Fun fact, based on this definition you can actually rearrange the equation to arrive at the definition of e based on compound interest:
lim(h-->0)((e^h - 1)/h) = 1
Let h_0 be a small number such that
(e^h_0 - 1)/h_0 = 1
e^h_0 - 1 = h_0
e^h_0 = 1 + h_0
e = (1 + h_0)^(1/h_0)
Convert back to limit
e = lim(h-->0)(1+h)^(1/h)
i.e. compound interest recursion formula with infinitesimally small interest rate for an infinitely large number of compounding periods.
Damn I was just finished typing this up in my comment when I saw your comment 👌
Oh that's actually really cool
When you write "let h_0 be a small number such that: (e^(h_0)-1)/h_0=1"
How can you tell that such h_0 exist? And in particular if you rearrange the equation one may get e^(h_0)=h_0+1, and if I'm not mistaken there exist an inequality which tells that e^x>=x+1, whit equality at x=0
@@Evan-ne5bu I suppose it is equivalent of using the property that limit of f(g(x)) is equal to f(limit of g(x)) where f=(e^h - 1)/h and g=h. So h_0 is just a simpler way of writing lim(h-->0)(h).
Also, didn't know about that inequality, nice! It is true for x>0 if you graph it. In this case, the equation e^(h_0)=1+h_0 is true if we are talking about limits since h_0 is a small number approaching 0
@@Evan-ne5bu Also fun fact #2, using base 'a' in the exponential instead of base 'e' you can show that lim(x-->0)((a^x - 1)/x) = ln(a). If anyone knows a way to directly approach this limit let me know, I've been curious haha
It's really important to understand that it's not that the derivate of e^x is just e^x, but that's WHY the number e exists. Videos like yours really inspire me to share my own videos as well!
Nope.
The reason e exists is because when multiples I think it was 1.0001 the values of whole numbers were interesting to Euler and he wanted to calculate what the base was
@@Asiago9 what?
As a veteran calculus teacher, I enjoy learning something from almost every BPRP video. Thank you!
I am glad to hear this. Thank you.
As a veteran, one can conclude this is circular reasoning which meant euler's number has NOTHING to do with reality. Which meant the actual value of pi is completely different. It is also ignorant reasoning to presuppose the perimeter of circumscribed polygon is always longer than the circumference of the unit circle at different number of sides. There must be ways to prove that and yet nobody even attempted it. Clown world this is.
@@lukiepoole9254 this dude thinks that he's more knowledgeable about math than the collective efforts of the smartest mathematicians over thousands of years
@@lukiepoole9254 what
@@Maple_MK Why are there not a single absolutely undeniably irrefutable rigorous proof for circumference of circle? The pi based on circumference of circle isn't 3.14159265; It is literally 4sqrt(1/golden ratio).
I love how this shows the process of discovering the derivatives and defining things along the way; it makes math much more interesting when you see how these things might’ve been originally figured out.
Perfectly executed! I love how you managed to connect everything so wonderfully and I could see the beauty in it.
For a few minutes I thought you were intentionally breaking up the white board into golden ratio rectangles!
That’s a great idea! I should do that one day.
Or by writing the value of it
Like:1.618........................
😅
how is this comment 19 hours ago? u dont seem to be a member....
@@blackpenredpen How the comment was done 11 hours ago?🤔
@@avishek_paul The video premiered a day ago which allowed people to comment early, but was only uploaded recently. I know, it's weird.
Oh! I wanted to know about this for a long time. Thank you so much bprp for this tutorial!
I'm preparing for math exams in may and this is the kind of content I need right now, I'll pray for more blackpenredpen content suggested for me by the youtube algorithm 🙏
e is actually e-rational
Thank you! This video helped me realize the relationship to define ln(a) as the limit of (a^h-1)/h as h goes to 0.
very clear explanation, I was confused about the meaning of e for a long time!
Very good.
That was a good and needed (for me) return to the definitions.
Haven't visited your channel in ages. Aside from the mathematics which is always entertaining, you deserve an Excellence In Bearding Award (EIBA).
Finally after 2 short videos one long video
Really innovative and cool 😎
I really thought you were gonna say d/dx(2^x)=e^x and 0.2 is even
Please start making integral battles as you used to do before, those are really fun.
But before defining e in such a way, you need to prove that there exists a unique number with the described property and I can't say that that's quite easier then proving that the derivative of e^x is e^x using some other definition of e.
That was also my reaction when he did it. The existence and uniqueness of such a number are not guaranteed.
You could perhaps build a fonction that for all input a outputs the limit of (a^h-1)/h as h tends to 0, then prove it's continuous and strictly increasing, and finally apply the intermediate value theorem. That could work but that sounds more complicated than using an other definition.
Yeah it's a flawed but intuitive way of defining it.
I guess you could look at the function f(t) = d/dx (t^x) and see that f is continuous, and apply the Intermediate Value Theorem to t=2 and t=3
Excellent question! Good discussion, would like to see what more people say.
Let’s take a hypothetical number
(N^h - 1)
________. h-->0
h
Where this equals 1.
How do we know if it doesn’t settle in a value smaller than 1? Maybe it tends to a value like .85327 or something?
I’m assuming people can just say
If we plug in 2 for N, it’s lower than 1
If we plug in 3 for N, it’s more than 1
-> Therefore it’s somewhere between the two.
But plugging in number isn’t that satisfying, it’s not really a proof :/
What we can do is compare the outputs as the inputs get bigger
(a^h-1)/h Vs ((2a)^h-1)/h. The input is twice as big, but what happens to the output?
Both have a common denominator so let’s take that out.
Now what is bigger?
(a^h-1) or ((2a^h)-1)
Clearly ((2a^h)-1) is always gonna be larger than the previous value i.e (a^h - 1) as “a” gets larger.
Example:
(5^a)-1 < [(2*5)^a]-1
But the question is, how much? There’s probably a way that proves it isn’t asymptotic and extends to infinity,
I’ll try and think of a proof, but this is an excellent discussion and a reminder to always question our intuition. Good job guys!
@@skylardeslypere9909 The problem with that is that this relies on t^x being well-defined, which itself relies on the exponential function and the logarithmic function also being well-defined a priori.
We can use the definition given here to prove that e = 1/0! + 1/1! + 1/2! +1/3! + ... then we can prove that the sum converges hence proving the existence of e and it's uniqueness
Finally someone who explained it properly. Really helpful video.
for the exponential function b^x, the derivative by first principles becomes (b^x) lim h→0 ((b^h) - 1)/h = (b^x) lim h→0 ((b^h) - (b^0))/h, which is the instantaneous gradient at x = 0, and b = e is the only case where the gradient at x = 0 is 1
idk, I think it's a fun visualization of what that weird limit is actually representing
That's not a visualization.
This reminded me of your 2017 video "what is e, and the derivative of exponential functions"
It's great to come back to basics and try to find different definitions for them, in my case when we got acquianted to e^x we were told to see it as the inverse of ln(x) which is by definition the integral from 1 to x of dt/t, from this we was able to derive the exponential properties (derivative, limit, ..)
btw, can you make some videos discussing saquences of functions and series of functions (different types of convergence, use of every type, how to prove practically each type, ...) and maybe improper integrals as well because it looks fun ^^ (when it exists, how should we manipulate it) along with parametric integrals like integral from 0 to infinity of f(x,t) dt ( continuity, differantiability)
Thanks, you really explained it well!
This has to be the best explanation I've seen for this!
Why?
man i am addicted to your videos .It contains many information
what playlist in your channel should i take to review for AP Calculus AB and BC? Tks for your videos
Just finished a level maths exams but even though I’m not doing maths at uni, I watch these vids every day, they’re so interesting
Excellent video! Thank you for sharing
Something really cool you should have pointed out is that the limit as h goes to 0 of (x^h-1)/h could be a way of approximating ln x
Finally a logical conclusion. Excellent explanation 🙏
You re brillant ! You re far away the best maths teacher i ever seen thank you very much
i saw an ig reel by you some time ago and now I'm learning.. lol thanks! the pokeball really ties it all together tbh
3B1B made also video about this in his Calculus Series.
This is how I learned this at school...e was the value of a such that d/dx(a^x) = a^x. Very good and clear explanation from BPRP
Hello Blackpenredpen ....i recently studied about the agrand plane and polar representation of complex numbers ......can you please make a video on it ....
You can actually evaluate the limit of (2^h-1)/h by using the variable substitution u=2^h-1, turning it into the limit as u->0 u/log2(u+1) = 1/((1/u)log2(u+1)) =1/log2((1+u)^(1/u)).
Change of variables to t=1/u gives lim t->inf 1/log2((1+1/t)^t). This includes the limit definition of e which can be shown to exist by the monotone sequence theorem.
Then you have 1/log2(e) =log2(2)/log2(e) = ln(2) by the change of base formula.
I found out about this from Khan Academy.
Justo así es como se hace, él lo hice de una manera muy estilo ingeniero.
Jajaja
I think when I learned it at school (in the math power lessons at grammar school with a teacher who had a math PhD), we started by pointing out that f(x)=1/x must have an antiderivative F (I'm avoiding to use the terms ln, e etc.). This cannot be calculated by using the (x^n)' = n*x^(n-1) formula because it would involve a division by 0, respectively it doesn't work for n=0. We chose to calculate the one with the constant leading to F(1)=0 and then made deductions about its inverse function G. We showed that G'=G, that G is of the form G(x) = Z^x by proving that for every x1 and x2, G(x1+x2)= G(x1)*G(x2). Finally we calculated G(1) = Z^1 = Z (which ie 2.71.....) and voila, we were done.
So interesting. It's beautiful how you explained the deep truths behind these common derivatives. I understand the reason so much better now
That's the basics of derivatives. If you didn't know that, you never really learned derivatives.
The basics of derivatives is slope of a curve, not e. Proofs for the derivative of e^x that are not circular are also notoriously hard to find.
Theorem: There exists one unique function verifying f'=f and f(0)=1.
We define this to be the exponential function exp(x).
Therefore [exp(x)]'=exp(x) by definition
That's a bit of a cop out reasoning though.
@@xinpingdonohoe3978 sure but it's probably the easiest way to define the exponential function. We work with this one in real analysis. Even for the bprp case you need to prove unicity which he doesn't.
@@SimsHacks it's the easiest but it's hardly doing much. You're just listing criteria, naming it e^x and claiming it's correct by definition.
I do admit that sometimes definitions can be useful, like proving the integral from 0 to ∞ of e^(-x^2) - (√x)e^(-x) dx = 0, but there has to be some work going into it otherwise it's almost arbitrary.
I’d be suspicious of any mathematician who accepted this without understanding the motivation, or necessity, for e.
@@SingaporeSkaterSam Wow, you are seriously arrogant. Do you think people with Ph.D's in mathematics do not know the history behind e? I am baffled by this comments section. No respect for mathematicians in here, I guess.
Love to see some of the basic (but very important stuff). Many calculus students will find this extremely helpful. I always hated the definition of ln(x) as an integral of 1/x. It's WEIRD and backwards and comes out of nowhere.
Historically, that is the way we defined ln(x), since this property of natural log was discovered before the number e.
To learn it more intuitively, I think it makes more sense to define e as the value of base b, such that d/dx b^x = b^x. Then, use this as the starting point to prove the historical ways that ln(x) and e as a limit, were previously defined.
thanks sir, nice explanation.
Funny enough if you arrange the limit (e^h-1)/h =1 as h->0, you get e=(1+h)^(1/h) as h->0.
e=(1+h)^(1/h) as h->0 is a generally used interest formula, that is continuous growth.
Can you send the link for that derivative table in the background
super informational video thank you
Very nice and helpful concept of video sir
Maybe the best explanation of the derivation of e on youtube. The only thing missing is that you should've algebraically derived the formula for e from the difference quotient. That way viewers can see exactly how the formula for e is algebraically derived from trying to isolate an exponent base whose derivative is identical to that exponential function.
I think one intuitive way to define e is by defining e^x generally:
What we want is that e^0=1, and that the derivative is the same as the original function.
Lets set f_0 (x)=1 to achieve f(0)=1. Now to get closer to f(x)=f ' (x), we set f_1 (x)=1+x, and f_2 (x)=1+x+x^2 /2 and so on. In the limit, we get the formula that is the way to define e^x where c is a matrix or a complex number. Now with enough analytical skills, you can prove that this function f behaves like the following: f(x)=(f(1))^x. So if we set f(1)=e, then f(x)=e^x. This even works in the limit version of the derivative: ( f(x+h)-f(x) )/h=(e^x e^h -e^x)/h=(e^x (1+h+ O(h^2))-e^x)/h=(e^x +x e^x -e^x)/h=e^x h/h=e^x
Now to prove that this definition of e is the same as the others (for example ((n+1)/n)^n for n = infinity) is not simple, but not noteworthy hard.
hey bro, exelent videos, could you explain the main idea behind laplace transform, i mean the fundamentals
Alternative way to find e:
1. Graph f(x)=2^x
2. Stretch horizontally by scale factor f’(0)
3. g(x)=2^[x/f’(0)]=[2^(1/f’{0})]^x
4. Deduce e=2^[1/f^(0)]
What do you think about hyperreal numbers and nonstandard calculus?
wow! what a superb video. crazy to think there are nearly half a million or so people in the same situation: their teachers just gave them some generic compound interest problem to find how exponential growth works (if lucky), told e is 2.718……… and then told that e^x differentiates to itself, very few show the why.
Best video on the subject :)
Nice teaching sir 👍🏻
Best video so far!
I love watching your videos. The Werefrog do miss the "black pen red pen yay" at the start of the video, though.
9:26 wow thats a new way of doing a^x that I didn't know of. Mind blown 🤯
Really nice treatment!
i am think if we need to show the uniqueness of e when define as the number st lim (e^h-1/h)=1, i.e. e is the unique solution to lim (x^h-1)/h=1
I haven't heard you say "let's do some math for fun!" In a minute. I neeeeeed it!
Respected BPRP I have a problem that I have been trying a long time but did reach the dsolution, would you please solve this geometry problem :
In an isosceles triangle ABC, AB=AC. Side AB is produce to D such that AD = BC. If angle ABC is of 40 degrees, find angle ADC.
Answer to Q1 & Q2:
.
.
.
.
.
Q1: What is log base b of x?
Let's name that value a, then a is log base b of x b^a=x by definition of the logarithm.
Let's write that with base e instead (i.e. b= e^(ln b) ) : x=b^a=(e^ln(b))^a=e^(a*ln(b))
Take the natural logarithm of this: ln(x)=ln(b^a)=a*ln(b) a=ln(x)/ln(b)
Q2: What is the derivative of log base b of x?
Just differentiate the answer of Q1: d/dx(a)=d/dx(ln(x)/ln(b))=(1/x)/ln(b)=1/(x*ln(b))
You could also find e^x sum definition and d/dx the sum aka taking the derivative of each fraction
We cant really use that to prove the derivative of e^x, because we need the derivative the get the taylor series expansion, which would result in circular reasoning.
@@Ninja20704 Ye the taylor Series expansion is also e^x’s definition in summation form right? Cause then when you take the derivative it ends up the same, since infinity-1 is simply infinity
@@Ninja20704 in fact, you can define exp as the sum of the serie without saying that it s a Taylor serie
@@Ninja20704 i was bored so i also did it with d/dx of sin x using it’s sum form/Taylor series and ended up with the sum form of cos x
@@Ninja20704 nope. Taylor series is a way to find the expansion but you don't need it to prove that sum from 0 to infinity of (x^n)/n! converges. We can just define e^x as that series
I like your videos, but, I wish in this one you could have found a way to differentiate b^x without calculator or the ln(b)*e^x rule.
Blackpenredpen the way to evaluate the limit of 2^h -1/h surely would be, by doing a substitution where u = 2^h - 1
Therefore evaluating lim u-> 0 of u/log2(1+u), then reicprocating and and doing 1 over allows you to take 1/u to be the coefficient of the logarithim to use as an exponential. 😊
Summary: same reason Pikachu is Pikachu
Simple and lovely.
Really cool video!
My preferred definition of e^x is it's the solution to the differential equation y'(x)=y(x) with the initial condition y(0)=1. I like this definition because it's simplest sounding definition, you get the differentiability of e^x for free and I think it's an easy starting point to show the other 3 common definitions on wikipedia are equivalent without making circular arguments. It also makes defining the inverse, ln(x), in terms of the integral of 1/x relatively straight forward. All you have to do is start with y^-1(y(x))=x, differentiate both sides, use chain rule and solve for dy^-1/dx. You can also easily use the initial condition to deduce y^-1(1)=0 or ln(1)=0.
Using seperation of variables, we can define y=exp(x) to satisfy the integral equation from 1 to y of dt/t=x. Dividing the interval [1, x] into n equal parts and making successive linear approximations you can deduce y(x) is approximately (1+x/n)^n. You can then make this exact by taking the limit as n approaches infinity, which is the compound interest definition of exp(x). Assuming you can interchange the limit and derivative it's easy to check this limit satisfies the differential equation. After that you can apply the binomial theorem and evaluate the limit as n approaches infinity to get the power series representation of exp(x).
To prove that exp(x) is in fact an exponential function, you can use the power series representation to show that exp(x+y)=exp(x)exp(y), a property that is only applies to exponential functions, assuming the function is continuous and maybe meets a few other criteria. You could then define the limit of (1+1/n)^n as n approaches infinity to be e and use that as your base. As you can probably tell it takes a significant amount of work to rigorously justify the e^x notation using my preferred definition and BPRP's definition is arguably more intuitive. However some other people in the comments pointed out some problems BPRP's definition, so I think I prefer this one.
I'm just doing the same thing, wondering why derivative of e^x is just the same function, so I use the definition of Derivative but in little different from the video
d/dx (e^x) = lim a->0 [e^(x+a) - e^x]/a
d/dx (e^x) = lim a->0 [e^x*e^a - e^x]/a
d/dx (e^x) = lim a->0 e^x*[e^a - 1]/a
d/dx (e^x) = e^x*lim a->0 [e^a - 1]/a
Subtitute [e^a - 1] = t then a = ln(t+1) and when *a* approach 0, *t* is also approach 0
d/dx (e^x) = e^x*lim t->0 t/ln(t+1)
ln(t+1) is similar with t when t is approach 0 so lim t->0 t/ln(t+1) = 1
d/dx (e^x) = e^x*1
d/dx (e^x) = e^x
That's it.
You can also define ln(x) as the limit of (x^h-1)/h as h -> 0
❤️...love u sir...love from India...I like all your video...You always motivates me to learn something new in maths
Thank you so much ❤
Who likes the video of bprp and his way of teaching me by myself he is super math man👏👏👏
Can you make a Video about variation and deffrantiation deference
From this definition of e we can directly get another one:
We need x^h -1 to be h soo x^h is 1+h and one way is to simply put the exponent in the definition of e to be 1/h and the base 1+h soo we get limit as h goes to 0 of (1+h)^(1/h)
How do you prove that lim(h-->0)((b^h - 1) / h ) converges ? For b=2 or for any other non trivial b ?
I mean, I can clearly see that it does if I use my calculator but how do you prove it ?
In France, e is defined as exp(1) and exp(x) is defined as exp(x)'=exp(x). No other explanation on what is e.
This is fine, and probably the best definition there is, since it is not hopelessly circular like almost every other definition there is, all while being simple.
Ça dépend peut etre des profs mais moi on m'avait bien défini e telle que e= lim(1+1/x)^x avec x-->+∞. Pour ce qui est de la fonction exponentielle, on nous la définit comme l'unique fonction f telle que f=f' et f(0)=1. On note cette fonction exp
d/dx[0]=0
Therefore e^x=0 by that definition
@@ablasphemite1940 There is the additional requirement that exp(0) = 1. This immediately excludes exp(x) = 0 from being possible. The equation dy/dx = y has exactly one solution satisfying y(0) = 1. This solution is exp. Then, we define e := exp(1). Well, in theory, this comes about more naturally as [exp^(-1)][lim (1 + ε)^(1/ε) (ε -> 0)] = 1, but since e = lim (1 + ε)^(1/ε) (ε -> 0) and the above equation implies lim (1 + ε)^(1/ε) (ε -> 0) = exp(1), we have e = exp(1). Anyway, every property of the exponential function can be derived extremely easily from this definition. Deriving the Taylor series is literally trivial. The functional equation can be derived from the Taylor series using the binomial theorem and the Cauchy product theorem. Also, exp(z) = lim (1 + z·ε)^(1/ε) (ε -> 0) can be proven using Tannery's theorem, hence further justifying the connections. The properties of the natural logarithm can also be derived very easily from this.
What are you referring to as the domain and codomain in order to say that the inverse of f(x)=e^x is ln(x) ?
2:15 Here's how the ancient people probably did it:
By using u substitution and letting 2^h=u, we can change the limit as h approaches 0 to the limit as u approaches 1
We can say that h is log base 2 of u from this statement.
From there we can take the derivative with respect to u. By using the definition of the derivative, we can find the generalized derivative for any function log base n of x, using logarithm properties and the definition of e as a limit. From there we can find that d/dx log n of x= 1/(x*ln(n)). using that derivative we can plug in ln(2) as the solution of that limit and get 2^x*ln(2). Sorry if the text formatting on this is confusing, i could post a video about it
Father of Calculus
How are you able to just plug in b for x at 8:35? What allows for that?
Can you derivate e^x if the definition of e is e = lim (1 + 1/n)^n as n goes to infinity?
There is for me a problem in this proof :
At the beginning, you try to calculate the derivative of 2^x but at this point if you don't have the exponential function you cannot (I can be wrong) define what is 2^x and moreover you cannot compute 2^0.0001 for your limit.
I edit :
In fact, you can define 2^(p/q) with p and q two integers with a q-th root therefore you can define 2^x on Q. Finally It's likely that you can use the density of Q to define 2^x for all real values but again, it feels complicated to me.
2^x=( 2*2*2.....*2*2*2) " X" no. Of times ......
@@kabsantoor3251 No, that is incorrect. You cannot define 2^x as being the power series you mentioned, because that power series is in turn defined in terms of ln(2), which you approximated as 0.693, but ln(2) is itself defined in terms 2^x. You have thus given a circular definition, which is to say, not a definition at all.
@@valentinenprepa1379 You make a fair argument, though: 2^x is not typically regarded as well-defined by peer-reviewed publications by mathematicians for arbitrary real x. Of course, there are different constructions for which using the notation 2^x to denote a well-defined expression in certain specifiv contexts is very practical, but this notation would rarely be used otherwise. As for the exponential function, it is just defined as a power series, most of the time, and then 2^x is just taken to be as an abbreviation for exp[ln(2)·x], although in practice, such abbreviations are actually rarely used.
@@angelmendez-rivera351 @Angel Mendez-Rivera No circular reasoning. See if I define 2^x as a shorthand for lim (1+ x/n)^k*n for some(yet not defined) k and n-->infinity and expand the binomial via binomial theorem, it's not hard to show that the derivative of 2^x is this number k times 2^x. We fix k by returning to the original binomial and looking at the value of the infinite power series for k=1. This way we're led to the number e. Clearly e is the number such that e^x (important to understand in this context as the infinite polynomial rather than e multiplied a certain number of times) is its own derivative. Changing to any other number say 2 changes k as defined by the binomial - that's the proportionality constant between the rate of change and the function itself.
We can even evaluate that limit by using the expansion
2^x = e^(xln2)
So e^(xln2) = 1+ xln2 +....
We plug that in
And get lim h->0 1+hln2...-1/h
Lim h->0 hln2+.../h
We get ln2 as limit as all other terms would give zero
We can also prove it by using the infinite series of sum from n = 0 to infinity of x^n/n!
When we take the derivative of the whole series, then we notice that the derivative is equal to the original series
Q.E.D
When I taught calculus to high schoolers I used the unit on implicit/logarithm differentiation to prove that e^x is it's own derivative.
ya but that blows their minds, most high schoolers I know cant handle implicit differentiation.
@John Spence I didn't find that to be true. It's just an extension of the chain rule, and critical for related rates. I taught differential Calc from the perspective that it's just Algebra with limits. The chain rule is easily compared to solving an equation using "opposite operations". Most of my students seemed to get that.
@@spirome28 ya I’ve since changed my position and you are correct. The only problem I find with it is it feels like a cheat of the long form proof methods.
I see what you're saying and I respect that. Keep in mind my comment was coming from my experiences working with teenagers who generally didn't plan on majoring in theoretical math at university. relating it back to Alg2 gave me the most success with the most students. personally, I barely remember "Advanced Calc" from when I was in college where we just went through Calc 1 to Diff Eq and proved the all the theorems. That's where my hatred for delta-epsilon bands comes from lol.
@@spirome28 lol
Why you not write log base 10 like lg? And why you not using arctg, arcsin, arccos?
As illustrative as that was, the only way one really knows to start at 2^x is either using calculator to determine e is approximately 2.718 OR using the lim n approaches infinity (1+1/n)^n definition of e which really more of a result that follows from some not so obvious calculus
When I was in fifth grade, I had started to really love math. It was my passion, but then my parents found out I was getting good at math so they kinda forced me to study more. This made me not feel as interested in math again, and I didn't listen to them, so I didn't study. But then, 1 year later, I decided that I wanna do math again. This is gonna be the start of my journey, hopefully my parents don't make me stop again...
You’re in 6th grade? Start at algebra
Let y=e^x
Taking log both the sides,
Ln(y)=xln(e)
Ln(y)=x
Differentiating both the sides with respect to x,
1/y(dy/dx)=1
Dy/dx=y
But y=e^x
Hence dy/dx= e^x
Ingenious like usual. BTW, your beard is epic! :-D
You could set y=eˣ so that ln y=x. Differentiating both sides with respect to x yields y'/y=1 or y'=y=eˣ.
This is how I was taught about e, and logarithms didn't come in til later. I'm in NSW, australia.
great video!
This is so cool !
can someone explain why this is a proove? Like he used the fact that d/dx(e^x)=e^x to "proove" it at 9:03. Or am I wrong
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From the point of view of complex variables every derivative of a real function could be computed just by evaluating the function in a complex number thus no sense to be called DIFFERENTIAL calculus no need to make differences instead plugging complex numbers. What is the opposite of evaluating a function? In order to redefine integration by the way Could it be posiible?
Finally! Someone explained this in a way that didn't seem like circular logic. I asked my professor about this when I first took calculus my freshman year of college. The answer I got never explained how we knew that d/dx(e^x) = e^x. It was frustrating because I felt like I was supposed to take it on faith. Your first demonstration opened the door to understanding why all the e and ln related rules exist. Thank you.
your teacher did not answer since this eq diff is the one defining the exponential function
The existence of e is strong evidence that we live in a computer simulation where someone was just too lazy to come up with different constants so they just plugged one universal constant in for a whole bunch of different uses.
Lmao