So uncanny - I was just thinking the other day that I don't actually know what e is and wouldn't know how to explain it if someone asked me to. And boom, you upload the exact video I needed. Great explanation! I really love your channel for teaching me in-depth proofs and derivations that I wish I'd been taught in school. Thank you
I loved this video. Thank you for going through things step by step slowly. It was hard to get my head around e even after reading the textbook and looking at other videos. Thank you! ☺️ I learned that when f(x) = a^x, Since the derivative equation is constant rate of change, meaning change in y over change in x, considering change in x as h: f’(x)= (f(x+h) - f(x)) / h Going back to finding the derivative of f(x) = a^x For f(x+h) replace x in a^x with x+h f’(x)= limit h->0 (a^(x+h)) - (a^x) / h Split a^(x+h) into (a^x * a^h) Group a^x and bring it out of the limit = a^x limit h->0 (a^h - 1)/h limit h->0 (a^h - 1)/h is 0/0, which means we have to calculate numers close to 0 to assume what the function (a^h - 1)/h is converging to as h approaches 0. It is shown that limit h->0 (a^h - 1)/h = ln(a) As the constant a approaches a certain number, we know limit h->0 (a^h - 1)/h gets close to 1x We call that certain number e, and it is around 2.71… Since (e^h - 1)/h = 1, And (a^h - 1)/h = ln(a), We get that ln(e) = 1. Now I’m confused what “ln” exactly is… and am suspicious that the definition of ln and e go hand in hand. I assume it’s not that BECAUSE (a^h - 1)/h = ln(a), ln(e) = 1, but the other way around: ln(e) = 1, ln = loge What is log: when b^x = y (b to the x power is what? y) inverse function logb(y) = x (y is b to the power of what? x) Loge(e)=1
This morning while I was in the shower, I thought to myself, "Can I use the limit definition of a derivative to prove d/dx e^x = e^x ?" ... Along comes BPRP to help me complete that thought.
Most methodical and clear presentation on exponential functions and the base letter 'e' that I've seen. (And I went through quite a few videos until this one!) p.s. Love the enthusiasm! subscribed!
not rigorous math i guess, but it gives some intuition. lets call the number solving the following equation a lim h->0 (a^h-1)/h = 1 solve for a: lim h->0 a^h-1=h lim h->0 a^h=h+1 lim h->0 a=(1+h)^(1/h) let x=1/h then h=1/x then lim h->0 is the same as lim x->inf so a= lim x->inf (1+ 1/x)^x which is the know definiton of e. I'm however not sure if you can solve an equation involving a limit with the methods i used. Would be cool if a proper mathematician told me if my "proof" is correct and if not what the mistakes are.
It does not seem to me like you can without first defining the natural logarithm. In other words, you must first prove that lim h -> 0 (a^h - 1)/h = Ln(a). But proving this is far from obvious too. I’ve seen so-called proofs of this before, but I don’t actually know if there are any proofs or not.
@@SvenBeh Your math is actually correct, it only needs very slight refinement. what you used was lim(a/b)= lim(a)/lim(b) @VeryEvilPettingZoo will also agree, I think.
I changed the right hand side of the definition at 14:12 to say lim(h->0) 1. You can then do regular algebra to both sides to get lim(h->0) e = lim(h->0)((h+1)^(1/h)) make a new variable n=1/h e = lim(n->inf)((1+1/n)^n) presto! there's the famous compound interest formula for derivatives.
According to limit laws, we need to know that separated limits exist then we can separete them, so lim( 2^X .(2^h-1)/h) should'nt be written as 2^X . lim ((2^h-1)/h) as h goes to zero because we don't know whether lim((2^h-1)/h) is a real number. as in lim(X^2 . 1/X) as x goes to zero can't be written as lim(X^2) . lim(1/X) because lim(1/X) is not a real number.
I wouldn't plug in numbers, I'd just set the e^h-1/h limit equal to the limit of some constant c and then use some manipulation to get the limit of the constant c into the compound interest form
You could also do the taylor series of a^h since a^h= e^(ln(a)*h) since we know the derivative of e^x. [1+ln(a)*h+{(ln(a)^2)*h^2)/2}+...+]-1 /h = (ln(a)*h + ln(a)^2*h^2 +...)/h =lim(h->0) (ln(a)+ln(a)^2*h+...) =ln(a)
Is it possible to prove the value of the limit instead of just guessing that it approaches something, because how do you know that there isn't a hole somewhere when -0.001 < x < 0.001
Awesome Vid!! One question though, how do you then compute the numbers for e. I understand the concept of defying it as a special case, but how do you provide the other decimals.
If u mean to get the decimal approximation for e, then one way is to use the Taylor Series of e^x centered at 0 then plug in x=1. I.e. 1+1+1/2!+1/3!+...
You can turn that e definition into the interest one using algebra and substituting h=1/n, can't you? And then you can go from there to the reciprocal factorial series using the binomial expansion theorem! I love these derivations!
Depends on what you want to achieve. Here, we want a number so that that limit is 1. And we 'use the symbol' e for it. that's how to get e^x function. The other form is from the compound interest formula.
And now some crazy (yet amazing!) backward math inspired by this video Let's write the limit of t for t going to 1 = 1 .......of course :) Now let's rewrite it like this --> Lim 1/1/t. Now suppose that is the outcome after applying De l'Hopital...and take the integral of the numerator and the denominator! Hence, we get Lim (x+C)/(ln(x)+D) for x -->1 Wait a moment! If we plug X=1 we get 0/0 only and only if C=-1 and D=0! Hence we get Lim (x-1)/ln(x)=1 for x -->1 Now let's make a substitution where x=e^h, hence ln(x)=h and for x-->1 h-->0 Hence we get Lim (e^h-1)/h =1 for h---> 0
I love watching your videos. Please make more. Is there anyway I can make a donation? I would also like to know why you said the variable a cannot be 1. Ln(1) = 0. This would mean lim h->0 (1^h - 1)/h = 0. I know this is trivial, but there is nothing wrong. In fact the variable a can be any positive number (i.e. a > 0).
How would you prove that the remaining factor is actually the natural log of the base b? That sounds like a good additional video to offer... I think I see how to do it, but you're a very good explainer, so I think it'd be worth another video.
That's what I had in mind too... I think it would be good to get a video of it too, even if it's just a short one. He's got a knack for explaining thing well.
Zone-E Not a valid proof, since it uses the derivative of exponential functions, which use the limit definition shown in this video, so you are using the derivative to prove the derivative. That is invalid.
Challenge to you @blackpenredpen : which 2 factorials (not including 1! and 10!) do you've to take out for this to be a perfect square? PS. There are 2 possible answers.
blackpenredpen I don't know why people do that. Anyway why don't you make videos on inequalities , it would be great. Thanks for all videos you have provided
I always wondered, what was the first legit definition of e and the logarithm? Which definition has been derived from the other? Is it legitimate to talk about an "exponential fonction" defined for all real numbers without already having the exponential and the logarithm functions?
Hey love the vids, I got bored and thought of a really wacky integral for you, (sqrt(x-7))/x)times integral from -73 to pi of ((x^2)-7) (1-x)(cos^2)(×) dx. Literally came up with it off the top of my head lol it's just a part of a bigger equation I pulled out of the top of my head hope you like it
Hi my friend , l learn alot from you & other mathemations that they know their field. Keep on doing what your doing, but come to me to do some extras. Of course my sugesstion goes for anyother math- lover . What i´m talking about ? Let´s go to the field of applied math und discover new stuff that is applicable to our world . let´s be another Euler !!? Not just to repeat who he was as a great mathemation , rather than to put his legecy " e -Function " as developement for " New Field of Mathematic (NFM) " in use . Hope i hear from you soon. Mehrdad - Hamburg / Germany.
I like your videos. Great channel, but please don't make very long basic math explanations. We don't want to waste 2 minutes to explain that a+(b+c) is a+b+c
blackpenredpen that's an example but in this video I left you explaining how 2^(x+h)-2^x=2^x(2^h-1) and I went to bring something and I returned to find you just finishing it. I always watch your videos but I find myself skipping entire minutes to avoid long "useless" explanations like "pluging" or distributivity property or associativity. But in overall excellent channel and interressant subjects. Why not include some geometry subjects ? thanks.
Hi Aymen, Thanks for the comment. However, as a teacher for many years now, I do find these explanation necessary to many students (as of many of them aren't as strong in algebra or in other basics compare to others). I do want to make hard math problems/concepts accessible to as many students as possible. Thus, I would not consider these as "unless" explanations. Anyway, I do thank your feedback. Btw, I will occasionally make "speed-run" videos, where I will talk as fast as possible to solving a math problem. Just for fun. And maybe you'd like those better. Thanks again, bprp
I think it'd have been better to show that the number is the natural log of the base simply using the rules of exponents. However the video is already quite long so maybe not ;)
'twould be an unfair play, exploring a way you could find the derivate of an exponential function and at the same time, in the same story, supposing that the number of e and all rules of differentiation are already proven. would not you think so?
there is no derivative of (-2)^x because it is not continuous. you need a function to be continuous in an interval to have a derivative for this interval. Not for every point of a function must a derivative exist
Can someone please explain why e^x is e^any constant is 0? Wouldn't saying e^x is e^x mean whatever constant you plug in for x the slop is the value of that function?
Yes. Because you can rewrite c^x in base e, as e^(ln(c)*x). Then apply the chain rule, and you see that the derivative is ln(c)*e^(ln(c)*x), which reduces back to ln(c)*c^x.
+Tommero You might be able to isolate the unknown derivative & thereby solving it, just like BPRP sometimes does in DI-integration. ... You won't known until you try.
yes we know, the limit n→∞ of this expression. but… but how could you ground an exponential function on such an (at start) uncertain limit value as a basis - realising that Euler in his time had of course no calculator in, say, 16 decimals, perhaps not even a 10-log table in such kind of precision.
I like your video for the most part. However, for the later part of this video, why don't you speak out that limit[(2^h-1)/h] as h->0 is ln(2)? For the same token, limit[(3^h-1)/h] as h->0 is ln(3), limit[(e^h-1)/h] as h->0 is ln(e), equal to 1. To prove limit[(2^h-1)/h] as h->0 is ln(2), here is one proof: www.mathway.com/popular-problems/Calculus/549781
Thanks for the comment. But actually that isn't a proof because of circular reasoning. In order to know the derivative of 2^x, we need that limit first. So, we can't use L'H (which uses derivative) to show that limit..
Can someone please explain why e^x is e^any constant is 0? Wouldn't saying e^x is e^x mean whatever constant you plug in for x the slop is the value of that function?
Can someone please explain why e^x is e^any constant is 0? Wouldn't saying e^x is e^x mean whatever constant you plug in for x the slop is the value of that function?
I think you mean the derivative of e^(any constant) = 0. Not that e^(any constant) = 0. The reason the derivative with respect to x of e^(any constant) = 0, is that by definition, both the e and the constant are constants, that don't depend on x. Thus e^constant is itself a constant, that doesn't depend on x. No dependence on x, means its derivative wrt x is zero.
I am 52 years old, and thanks to you I spend my free hours learning mathematics with you. Greetings from Spain
Hm
So uncanny - I was just thinking the other day that I don't actually know what e is and wouldn't know how to explain it if someone asked me to. And boom, you upload the exact video I needed. Great explanation! I really love your channel for teaching me in-depth proofs and derivations that I wish I'd been taught in school. Thank you
Stylin on them with the supreme jacket
You have a wonderful talent!,you make math fun,a talent that few of us have,however, love your videos and work!keep it up!
an option that only depends on algebra is to change variables like this : Let n = (2^h ) - 1
then as h -- > 0, n -- > 0
and h = log_2 ( n + 1 )
Oh wow, so differentiating an exponental function implies itself.
I love how excited you are in this video! Your enthusiasm really shows!
I loved this video. Thank you for going through things step by step slowly. It was hard to get my head around e even after reading the textbook and looking at other videos. Thank you! ☺️
I learned that when f(x) = a^x,
Since the derivative equation is constant rate of change, meaning change in y over change in x, considering change in x as h:
f’(x)= (f(x+h) - f(x)) / h
Going back to finding the derivative of f(x) = a^x
For f(x+h) replace x in a^x with x+h
f’(x)= limit h->0 (a^(x+h)) - (a^x) / h
Split a^(x+h) into (a^x * a^h)
Group a^x and bring it out of the limit
= a^x limit h->0 (a^h - 1)/h
limit h->0 (a^h - 1)/h is 0/0, which means we have to calculate numers close to 0 to assume what the function (a^h - 1)/h is converging to as h approaches 0. It is shown that
limit h->0 (a^h - 1)/h = ln(a)
As the constant a approaches a certain number, we know limit h->0 (a^h - 1)/h gets close to 1x
We call that certain number e, and it is around 2.71…
Since (e^h - 1)/h = 1,
And (a^h - 1)/h = ln(a),
We get that ln(e) = 1.
Now I’m confused what “ln” exactly is… and am suspicious that the definition of ln and e go hand in hand. I assume it’s not that BECAUSE (a^h - 1)/h = ln(a), ln(e) = 1, but the other way around:
ln(e) = 1, ln = loge
What is log: when b^x = y
(b to the x power is what? y)
inverse function logb(y) = x
(y is b to the power of what? x)
Loge(e)=1
> limit h->0 (a^h - 1)/h
I think after this step you're meant to prove that it's a logarithm, rather than just number plugging.
Entiendo muy poco el inglés, pero a vos te entiendo perfectamente, sin duda las matemáticas es el idioma universal. Felicidades.
This morning while I was in the shower, I thought to myself, "Can I use the limit definition of a derivative to prove d/dx e^x = e^x ?" ... Along comes BPRP to help me complete that thought.
Most methodical and clear presentation on exponential functions and the base letter 'e' that I've seen. (And I went through quite a few videos until this one!) p.s. Love the enthusiasm!
subscribed!
Can you make a video on complex logarithms? Big fan!
like ln(i) or ln(-1)?
if so, I kinda did that in sin(Z)=2 vid already
blackpenredpen I have seen the video and I understand how ln(z)= ln(r) + iθ. I am confused about why the graph of the function in 3d is a spiral.
You are a great mathematician.. ❤
love this dude videos... and I definitely copping the merch. his videos has taken me from algebra, to Calc 2, and I won't stop now
Oohh i love e. But still wondering about another equations that explain e as well. Keep learning for me, thanks for your e-lecturing Prof. Steve ☺☺☺☺
My calc 1 course showed us this video, but I already saw it because I love your videos
a very good presentation as to why exp is a derivative of itself
keep up the good work, i love your videos
Any way to proof that the two definitions of E are the same? Doesn't seem like a completely obvious fact to me.
Joel Low super interested to see that
not rigorous math i guess, but it gives some intuition.
lets call the number solving the following equation a
lim h->0 (a^h-1)/h = 1
solve for a:
lim h->0 a^h-1=h
lim h->0 a^h=h+1
lim h->0 a=(1+h)^(1/h)
let x=1/h
then h=1/x
then lim h->0 is the same as lim x->inf
so a= lim x->inf (1+ 1/x)^x
which is the know definiton of e.
I'm however not sure if you can solve an equation involving a limit with the methods i used. Would be cool if a proper mathematician told me if my "proof" is correct and if not what the mistakes are.
It does not seem to me like you can without first defining the natural logarithm. In other words, you must first prove that lim h -> 0 (a^h - 1)/h = Ln(a). But proving this is far from obvious too. I’ve seen so-called proofs of this before, but I don’t actually know if there are any proofs or not.
@@SvenBeh Your math is actually correct, it only needs very slight refinement. what you used was
lim(a/b)= lim(a)/lim(b)
@VeryEvilPettingZoo will also agree, I think.
Love your videos keep it up
Subscribed! Thanks for making math fun for me again :)
Perfect!! Woohooo!!! Thank you!!!!!!!!
Thank you so much
Helping me study for iit. I am able to understand calculus now and that helps me a ton.
So good !!
You, my man, are one helluva teacher. Bravo!
fabulous explanation.
A wonderful approach to the "e" mysterious number-
I changed the right hand side of the definition at 14:12 to say lim(h->0) 1. You can then do regular algebra to both sides to get
lim(h->0) e = lim(h->0)((h+1)^(1/h))
make a new variable n=1/h
e = lim(n->inf)((1+1/n)^n)
presto! there's the famous compound interest formula for derivatives.
You are a living legend sir thank you very much!!!
Very elegant!
I like the fact that you smile while making your derivations 😀✌
Good on you, mate. Thanks.
According to limit laws, we need to know that separated limits exist then we can separete them, so lim( 2^X .(2^h-1)/h) should'nt be written as 2^X . lim ((2^h-1)/h) as h goes to zero because we don't know whether lim((2^h-1)/h) is a real number.
as in lim(X^2 . 1/X) as x goes to zero can't be written as lim(X^2) . lim(1/X) because lim(1/X) is not a real number.
Thanks for all you do!
Thank you for making this video, this has helped me a lot!
Intriguing!!
I wouldn't plug in numbers, I'd just set the e^h-1/h limit equal to the limit of some constant c and then use some manipulation to get the limit of the constant c into the compound interest form
You could also do the taylor series of a^h since a^h= e^(ln(a)*h) since we know the derivative of e^x.
[1+ln(a)*h+{(ln(a)^2)*h^2)/2}+...+]-1 /h = (ln(a)*h + ln(a)^2*h^2 +...)/h =lim(h->0) (ln(a)+ln(a)^2*h+...) =ln(a)
You can't do that given Taylor series were developed via derivatives. That'd be circular reasoning.
Could there be situations where it's actually easier to calculate
lim(h->0) (f(x-h)-f(x))/h, or perhaps
lim(h->0) (f(x)-f(x+h))/h
You made very simple
This one was tooo much helpful prof❤️
you left me speechless
marvellous
Is it possible to prove the value of the limit instead of just guessing that it approaches something, because how do you know that there isn't a hole somewhere when
-0.001 < x < 0.001
Awesome Vid!! One question though, how do you then compute the numbers for e. I understand the concept of defying it as a special case, but how do you provide the other decimals.
If u mean to get the decimal approximation for e, then one way is to use the Taylor Series of e^x centered at 0 then plug in x=1. I.e. 1+1+1/2!+1/3!+...
You can turn that e definition into the interest one using algebra and substituting h=1/n, can't you? And then you can go from there to the reciprocal factorial series using the binomial expansion theorem! I love these derivations!
Depends on what you want to achieve. Here, we want a number so that that limit is 1. And we 'use the symbol' e for it. that's how to get e^x function.
The other form is from the compound interest formula.
And now some crazy (yet amazing!) backward math inspired by this video
Let's write the limit of t for t going to 1 = 1 .......of course :)
Now let's rewrite it like this --> Lim 1/1/t. Now suppose that is the outcome after applying De l'Hopital...and take the integral of the numerator and the denominator!
Hence, we get Lim (x+C)/(ln(x)+D) for x -->1
Wait a moment! If we plug X=1 we get 0/0 only and only if C=-1 and D=0!
Hence we get Lim (x-1)/ln(x)=1 for x -->1
Now let's make a substitution where x=e^h, hence ln(x)=h and for x-->1 h-->0
Hence we get Lim (e^h-1)/h =1 for h---> 0
I=İntegral 1 to a 1/x dx=ln(a)-ln(1)=ln(a)
I=h->0 integral 1 to a x^(-1+h)dx=h->0 (a^h-1)/h
Thats meaning h->0 (a^h-1)/h=ln(a)
nice video
It is amazing how easy it is
I think we need to proof what "classical" and "limit" definitions of e, are equivalent
I love watching your videos. Please make more. Is there anyway I can make a donation? I would also like to know why you said the variable a cannot be 1. Ln(1) = 0. This would mean lim h->0 (1^h - 1)/h = 0. I know this is trivial, but there is nothing wrong. In fact the variable a can be any positive number (i.e. a > 0).
How would you prove that the remaining factor is actually the natural log of the base b? That sounds like a good additional video to offer... I think I see how to do it, but you're a very good explainer, so I think it'd be worth another video.
b^x = e^(x*ln(b))
d/dx(b^x) = e^(x*ln(b))*ln(b)
= b^x*ln(b)
That's what I had in mind too... I think it would be good to get a video of it too, even if it's just a short one. He's got a knack for explaining thing well.
Zone-E Not a valid proof, since it uses the derivative of exponential functions, which use the limit definition shown in this video, so you are using the derivative to prove the derivative. That is invalid.
b can also be 1, the derivative would just be 1^x * ln(1) = 1 * 0 = 0 ,which is correct.
Challenge to you @blackpenredpen : which 2 factorials (not including 1! and 10!) do you've to take out for this to be a perfect square? PS. There are 2 possible answers.
But what if we take x as Fibonacci constant? I have noticed that 2 and additional numbers are close to Fi and inverse Fi so
this is amazing!!!
Thank you!!!
i like u chinese guy
u fuck off
w0zi me too , he is the best teacher
w0zi lol what's that? Use report button
Neo i removed that. I don't know how that got to the comment
blackpenredpen I don't know why people do that. Anyway why don't you make videos on inequalities , it would be great. Thanks for all videos you have provided
Wooooooooow
2:25 how does it work that the 2^x is factorised out of the limit?
Hi really great video, just wondering does this mean we still plug any value of X into derivative function and get slope at that poont
Yeah, the first derivative will yield the instantaneous slope of the function
u can do by logarithmic differentiation can be better it will come in less than minute
Mangesh Waykar ?if that assumes what you are trying to prove?
Thanks man.
I always wondered, what was the first legit definition of e and the logarithm? Which definition has been derived from the other?
Is it legitimate to talk about an "exponential fonction" defined for all real numbers without already having the exponential and the logarithm functions?
Also, can we use the limite of (x^h -1)/h when h -> 0 as a definition of ln(x)?
Hey love the vids, I got bored and thought of a really wacky integral for you, (sqrt(x-7))/x)times integral from -73 to pi of ((x^2)-7) (1-x)(cos^2)(×) dx. Literally came up with it off the top of my head lol it's just a part of a bigger equation I pulled out of the top of my head hope you like it
Why is the differential e the same as the compound interest e?
Hi my friend , l learn alot from you & other mathemations that they know their field. Keep on doing what your doing, but come to me to do some extras. Of course my sugesstion goes for anyother math- lover . What i´m talking about ?
Let´s go to the field of applied math und discover new stuff that is applicable to our world . let´s be another Euler !!? Not just to repeat who he was as a great mathemation , rather than to put his legecy " e -Function " as developement for " New Field of Mathematic (NFM) " in use . Hope i hear from you soon. Mehrdad - Hamburg / Germany.
thank you sir a lot, you are really making maths more and more interesting and enjoyable....waiting for more exciting chapters...
I like your videos. Great channel, but please don't make very long basic math explanations.
We don't want to waste 2 minutes to explain that a+(b+c) is a+b+c
where did i do that in this vid?
blackpenredpen that's an example but in this video I left you explaining how 2^(x+h)-2^x=2^x(2^h-1) and I went to bring something and I returned to find you just finishing it. I always watch your videos but I find myself skipping entire minutes to avoid long "useless" explanations like "pluging" or distributivity property or associativity. But in overall excellent channel and interressant subjects. Why not include some geometry subjects ? thanks.
Hi Aymen,
Thanks for the comment. However, as a teacher for many years now, I do find these explanation necessary to many students (as of many of them aren't as strong in algebra or in other basics compare to others). I do want to make hard math problems/concepts accessible to as many students as possible. Thus, I would not consider these as "unless" explanations. Anyway, I do thank your feedback. Btw, I will occasionally make "speed-run" videos, where I will talk as fast as possible to solving a math problem. Just for fun. And maybe you'd like those better.
Thanks again,
bprp
I think it'd have been better to show that the number is the natural log of the base simply using the rules of exponents.
However the video is already quite long so maybe not ;)
'twould be an unfair play, exploring a way you could find the derivate of an exponential function and at the same time, in the same story, supposing that the number of e and all rules of differentiation are already proven.
would not you think so?
YAY!
あれ、eの定義ってそやったっけ（笑）
指数と対数のちょうど回転軸あたりに居るやつなんやな･･･ 納得！！
Ya gotta love the phrase "this number has to be legit"
Couldn't we _really_ work out the limit instead of approximating it with numbers?
can we demonstrate, that this number exists, and is unique and is irrational ?
My entire life would have been so much easier if you had been my maths teacher!
I hate e so much. It haunts my dreams in Differential Equations, and in every single Quantum Mechanical problem.
Absolutely fascinating.
Not as fascinating as the Playboy channel, mind you...but still fascinating.
*Nice thumbnail*
Hi Steve!
HIIIIIII
If you know Bangla language, the way his two.to.the power sounds, it means something in Bangla, which is kinda weird😅😅..
Hello, can you make a more formal aproach? That was good but i would like to see something cleaner
i thought you were gonna do the first derivative
What is derivative of (-2)^x ?
there is no derivative of (-2)^x because it is not continuous. you need a function to be continuous in an interval to have a derivative for this interval.
Not for every point of a function must a derivative exist
No real answer but in complex analysis gives dy/dx = ((-2)^x)(i pi +ln2) because ln(-2)=i pi + ln(2)
What's the lady saying in the beginning?
What you are watching right now is Teacher Cao Math
oh my
Can someone please explain why e^x is e^any constant is 0? Wouldn't
saying e^x is e^x mean whatever constant you plug in for x the slop is
the value of that function?
This isn't a rigorous proof 😕
I bet your calculator probably takes advantage of the derivative of 2^x to do the calculations. So you have circular logic here.
so the derivative of c^x is c^x times the natural log of c
Yes. Because you can rewrite c^x in base e, as e^(ln(c)*x). Then apply the chain rule, and you see that the derivative is ln(c)*e^(ln(c)*x), which reduces back to ln(c)*c^x.
THANK YOU CHINESE GUYYYYYY ALL THE LOVE
3:35 i was waiting for l'hospital rule :(
San Samman how can you use lhopital without knowing the derivative of 2^x ?
+Tommero You might be able to isolate the unknown derivative & thereby solving it, just like BPRP sometimes does in DI-integration.
... You won't known until you try.
I see why they call it the natural log
so why …? is it because of the fact that the base is as irrational as could be ?
By your English language I realise that your fathers is or was from south korea
I can die in peace now 😄😄
the other definition is:
e=(1+1/n)^n
:)
yes we know, the limit n→∞ of this expression. but… but how could you ground an exponential function on such an (at start) uncertain limit value as a basis - realising that Euler in his time had of course no calculator in, say, 16 decimals, perhaps not even a 10-log table in such kind of precision.
I like your video for the most part. However, for the later part of this video, why don't you speak out that limit[(2^h-1)/h] as h->0 is ln(2)? For the same token, limit[(3^h-1)/h] as h->0 is ln(3), limit[(e^h-1)/h] as h->0 is ln(e), equal to 1. To prove limit[(2^h-1)/h] as h->0 is ln(2), here is one proof: www.mathway.com/popular-problems/Calculus/549781
Thanks for the comment. But actually that isn't a proof because of circular reasoning.
In order to know the derivative of 2^x, we need that limit first. So, we can't use L'H (which uses derivative) to show that limit..
I agree. The proof goes in circular. Oops!
couldn't u just logarithmically differentiate these functions
You are beautiful...
calculus is so ungodly long... there has to be better methods like physics has.
Instead of putting in numbers you should solve the limit by using del'hospitals method .. would be prettier
Marko Jozic no, if he used lhopital's rule, he would face the original question (derivative of 2^h)
In order to do that, we would have to know the derivative of 2^h, but that's what we're trying to figure out in the first place.
Don't use the result in the proof of the result. That's circular logic
Can someone please explain why e^x is e^any constant is 0? Wouldn't
saying e^x is e^x mean whatever constant you plug in for x the slop is
the value of that function?
Can someone please explain why e^x is e^any constant is 0? Wouldn't
saying e^x is e^x mean whatever constant you plug in for x the slop is
the value of that function?
I think you mean the derivative of e^(any constant) = 0. Not that e^(any constant) = 0.
The reason the derivative with respect to x of e^(any constant) = 0, is that by definition, both the e and the constant are constants, that don't depend on x. Thus e^constant is itself a constant, that doesn't depend on x. No dependence on x, means its derivative wrt x is zero.