Proofs of derivatives of ln(x) and e^x | Taking derivatives | Differential Calculus | Khan Academy
Вставка
- Опубліковано 8 лют 2025
- Courses on Khan Academy are always 100% free. Start practicing-and saving your progress-now: www.khanacadem...
Doing both proofs in the same video to clarify any misconceptions that the original proof was "circular".
Watch the next lesson: www.khanacadem...
Missed the previous lesson?
www.khanacadem...
Differential calculus on Khan Academy: Limit introduction, squeeze theorem, and epsilon-delta definition of limits.
About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content.
For free. For everyone. Forever. #YouCanLearnAnything
Subscribe to KhanAcademy’s Differential Calculus channel:
/ channel
Subscribe to KhanAcademy: www.youtube.co...
in this episode: Sal gets sassy
This is so funny
i guess im asking the wrong place but does any of you know a method to log back into an instagram account??
I somehow forgot the account password. I would appreciate any tips you can give me.
@Ariel Fernando Instablaster :)
😅😅😅😅😅😅😮
@@landonraylan99963:42 😮😮😅😮😮😅😅😅😅😅😅😅
I rarely logged into youtube just to comment a compliment before, because if that vid deserves compliments it shouldve received a lot of them
But then, this vid of you, sir, is brilliante.
I've seen a proof that defines the natural logarithm as the antiderivative of 1/x and then proves that it obeys all of the laws of logarithms. Then they define e as the value of x that makes ln x = 1. This is pretty cool as well. I love all of your videos.
Sal I'm just amazed by the fact that not only you 're a genius but you are also modest and kind!!! Your first proofs were completely clear and not at all circular but you still felt the need to do another video to clear things up (although they were completely clear in the first place!!)!!
YOU ROCK!!!
To those that ask why delta x / x can equal 1 / n , this is possible because ln(x) is defined for the domain x > 0. As such delta x / x is an expression of all numbers greater than 0. If you do the same for n, i.e. say that n > 0, 1 / n is also an expression for all positive numbers greater than 0. As such the two statements are equivalent, but it's important to remember your domains, that both x and n are > 0.
As a fellow maths tutor I do really like the second part of your proof.. deriving the Derivative of Exp(x) in two different ways.. I don';t remember seeing that before
I must admit I like your proof of Ln(x) alot better this time, This one was much clearer. You did seem a tad bit defensive, however. ;)
This is a truly awesome collection of videos. Keep up the GREAT work! You really should win some sort of UA-cam award man!
y = e^x
ln y = ln(e^x) =x
1/y * (y') = 1, by the chain rule
y' = y
that is to say d/dx( e^x) = e^x
of course then you'd need to prove d/dx ln x
= 1/x
that is
Braden Ripple this is circular logic, you're assuming properties of e^x (and ln), and by assuming those you have already assumed all of their properties because you use the definition of e. I mean you pointed it out yourself but still
@@ApplepieFTW how is that circular? I dont get it.
I really like his way of proving many of the concepts. Many teachers fail to deliver their understanding of a concept to the students, but he seems to be a great at clarifying everything with the way of his proof.
Incredible is all I have to say. Never knew math is so beautiful.
@iamincorruptible
When we say n-->∞ we really mean really really large numbers (you can't really plug in ∞). Try this expression
(1+ 1/n)^n for n=1 million and you 'll see that you are going to get 2.71something (which is something aproaching e), then try it for example with n=1 billion and you 'll get something that has even more digits of e.
Hope this helps.
Thank you so much this is the best video I have found on this
Excellent explanation. This is invaluable!!
Thanks so much!
So beautifully explained
You can arrive at the proof of the derivative of the ln(x) by first stating that ln(x) is the inverse function of e^x, then just use the rule for taking a derivative of an inverse function.
d/dx of inverse function where f(x) is original function { e^x } and g(x) { ln(x) } is the inverse is as follows:
g'(x)= 1 / f '(g(x))
it ends up working out a lot easier.
If you start with the derivative of e^x being itself then you can prove 1/x is the derivative of ln x by doing the proof in reverse.
d/dx(e^[lnx]) = d/dx(x) = 1
d/dx(lnx)*e^lnx = 1
d/dx(lnx)*x = 1
d/dx(lnx) = 1/x
Around the 6 minutes mark what allows you to take out Ln to get the definition right for e by itself? I’m struggling following all that is going on but. I would like to understand more about how all these rules work
60% of this video: explaining the proof
40% of this video: explaining that IT IS NOT A CIRCULAR PROOF, and that at NO POINT DURING THE PROOF did he ever mention d/dx( e^x) = e^x
Many thanks for the fantastic mini-lecture!
[e^(x+deltax)-e^x]/deltax = [e^x]*[e^(deltax)-1]/deltax
as deltax approachs zero the limit of [e^x]*[e^(deltax)-1]/deltax approaches e^x
Another approach could be to start by assuming there exists a function a^x such that d/dx a^x=a^x, then prove that such a function can only exist if a=e.
i think you "clearified" it perfectly :)
This proof uses the fact that ln x is continuous when interchanging Lim and ln.
Not sure if you'd want to mention this or not, but a lot of text-books use h instead of delta-x. I'm not sure if there's a difference, but I know that h is really reserved for a very, very small value while delta-x simply means change in x.
h is used as difference for the gradient in order to simplify the whole expression, just like the delta x, so there's no real difference; besides, changing variable name doesn't effect anything at all, so it's chill (changing a to b won't change the meaning of a, as long as it's variable and not some defined constant)
There is no difference in h and delta x due to the limit being delta x approaches 0
If you the absolute best calculus book as far as proofs go, use the calculus book by M Spivak. This book is so unique and refreshing .
Also , there are many different ways to define lnx or e to the power of x.For example , some books define Ins as the area under the curve of positive part of 1 over x from zero to infinity .Also for e to the power of x,you can define it as the function such that the function and it derivative are the same. What would be interesting is a source that goes in a circle starting from one definition goes to another and another and finally come back to what was started with.
Thanks Sal!
@anthonybb2008 Instead of deltax, which I can not write, I will use h.
x is some number external to the limit; therefore, x/h could just as well be 4/h, and because as long as x is not zero, this is not zero.
For any n you may pick you can pick nonzero h so that x/h = n, and for any nonzero h you can pick n such that x/h = n. Assuming h is nonzero is okay because the value of the function at h = 0 does not affect the limit in any way.
Thanks for clarifying.
Well that a great way to prove the derivative of e^x and ln x,. Another way you can do it simply by using fundamental definitions as usual and then use series expansion of e^x and ln x to simplify the work.
Sal on offensive! Go go Sal!
Lol
The way he did it was a less obvious way, but it is a very quick and easy method once you've seen it (I'd never seen it proved like that). The obvious way arguable isn't a proof because it is boldly uses the definition of e.. lim(h->0)((a^(x+h)-a^x)/h) = a^xlim(h->0)((a^h+1)/h). Now, e is DEFINED such that lim(h->0)((e^h-1)/h)=1, and the result follows
This is really amazing stuff. But I'm sure that this is not how this derivative was originally derived. Does anyone know who, how and when these derivaitves were discovered?
it is still circular, but it is not your fault!!
The circle come from the begining when you work with natural logarithm when we know that basis is "e". To make the proof complete you must work with logarithm of any basis, but you chose "e" basis. First you must work with logarithm in general, then you can go forward.
This proof is satisfying to me, cause I understand the underlying logic. But to make it perfect you must define derivative of log - then - ln - then other stuff....
This is beautiful! I really Love math
What property allows Sal to just pull out the natural log out of the limit at 6:30? Is there a proof or something because I don't get how that is equal.
The natural log has nothing to do with the limit. You need to understand what a limit is to get it: think about it. The limit is evaluated based on whatever variable youre talking about. Therefore, anything not dependent on the variable in uestion, you can pull out and its the same. Try it on anything
SAL! u r awsome! u dont blame those who blame u! thats just SO KIND! :)
thanks! helped a lot!
congratulation, for the new tools you have and making the videos HD, your Videos are helpful in term of improving my math skills, thank you for your time.
With all respect those critics about circular do have some point but not totally voiced correctly: You need to prove what is the derivative w.r.t. X of F(X):, and not just what you did. i.e., derive the expression for the derivative of ln(F(X)), is that indeed (1/F(X)) times F'(X)? if so then is fine and rigorous. Thanks, great videos.
Real mental gymnastics.
Keep it up!!!
@metalmaniac767
e is not a variable (like x or y or t or whatever) it's just a constant (like any other number, 1, 2, pi, root 2, whatever)
so e^3 is also just another number that you could work out if you want and the derivative of a constant is just 0.
for example, if instead of e you used 2, it would be like saying the derivative of 2^3 should be 3(2^2), which is really saying the derivative of 8 is 12, which doesn't make sense since there aren't any variables. (hope i'm understandable >.>)
@ssfshock1 Of course he is. I'm not a teacher, I was referring to my professors' lectures.
Good initiative. However, I believe that the audience could handle a the material at a greater complexity. These proofs took some longwinding turns, all not necessary.
eg e^x proof
y=e^x
ln(y)=x
y'/y=1
y'=y=e^x
The way I see it, the accusation of circularity comes from the definition of e. I have seen a proof that shows e=lim n-->inf(1+1/n)^n, but assumes we know d[e^x]/dx=e^x. But it's fine to call lim n-->inf(1+1/n)^n the definition of e because e is what you get from plugging in bigger and bigger numbers for n.
Why would you need a proof for e= lim n->inf (1 + 1/n)^n? Isn't that the earliest definition? That's exactly what e IS. Everything else s derived off of it.
You can proove that e is equal to the limit, as n aproches infinity, of (1 + 1/n)^n by just applying Newton's binomial theorem, which is appliable to any real n, not just natural numbers
Why did you substitute delta x as 1/n. For example you could substitute 1/n^2 but that would give a different result.
if you substitute n^2, that does not change anything when it comes to equating n->inf to delta x -> 0 since it does not matter weather infitinity is square or infinity to the power of 1, x/infiinity ( however large) will always approach 0. On the other hand, the proof here is based on definition of e as a limit. using n^2 instead of n will complicated your algebra but wont make a different result.
Awesome i just got into this in college.
Dude hi? you there? Or even alive? Hello?
great video!
This is great, I understand all the maths, but one problem I'm having: Using our previous rules of the derivative, how come e^3 does not equal 3(e^2)? From this is seems that the derivative of e^x is not equal to e^x. However, Sal clearly just proved that it IS, so what have I got wrong??
Please explain: Why did you substitute {delta(x)}/x for 1/n ? Why not some other variable (i.e. u)?
I would give so much to be as smart as Khan.
i like the way you proof
I thought his explanation was very clear.
Perhaps those of you who dont get it, need to stop, rewind the video and listen more carefully. Also if you dont know the Log properties, or lower math, then you need to understand THAT before you can understand this.
I think there is a huge mistake somewhere in this prove:
we just derived that dln(x)/dx = 1/x; setting x to e^x and substituting gives:
dln(e^x)/dx = 1/e^x
dxln(e)/dx = 1/e^x
dx/dx = 1/e^x
1 = 1/e^x
e^x = 1,
which certainly isn't true because x could be set to be anything - 1, for example: e^2!=1 - and can't have one distinct solution.
Please, correct me if I'm wrong -been really cracking my head about this
No. Sal's proof is correct. Yours isn't. You broke it by your substitution because you "captured the variable" by using the same symbol to mean two different things. To avoid that, let me use another symbol in your derivation so you could see this better:
(d/dx)[ ln(x) ] = 1/x
Now suppose that `x` is itself a function of some other variable, let's call it `u`, and the relation between them is:
x = e^u
NOW you can do your substitution as follows:
(d/dx)[ ln(e^u) ] = 1/e^u
and use the law of exponents to simplify the left side:
(d/dx)[ u·ln(e) ] = 1/e^u
(d/dx)[ u·1 ] = 1/e^u
but now the problem with your derivation becomes apparent: you have
du/dx = 1/e^u
but you cannot do much with it, because on the left you have `du/dx` instead of `dx/dx` as it was in your erroneous "proof". That is, `u` is itself a function of `x` (the inverse function of `x = e^u`), and you need to do the chain rule first to translate this expression to the new variables (which you always have to do after substitution - something you didn't do in your "proof" because due to confused symbols you didn't know it was necessary).
Since
x = e^u
we can solve it for `u` by taking the natural logarithm of both sides:
ln x = ln(e^u)
ln x = u · ln(e)
ln x = u
and after substituting back to your equation you get:
(d/dx)u = 1/e^u
(d/dx)[ln x] = 1/e^u
And now, if you differentiate on the left, you get:
1/x = 1/e^u
so `x = e^u`. But this is something we already know :P It's our substitution. So this is not the way to go.
Instead, we want to know what is the derivative of `e^u`, that is, of the function `x` with respect to its independent variable `u`, or `dx/du`. NOT `du/dx`.
But this is simple! Just take this one:
du/dx = 1/e^u
and flip it around, and you'll get:
dx/du = e^u
which is indeed the correct derivative of the function `x = e^u` with respect to `u` :)
Which indeed proves that:
(dx/du) = (d/du)[e^u] = e^u
QED
Sci Twi yeah, I already realised I forgot to substitute de^x into dx (this is the exact reason I don't get as high a number of marks as I should in exams), but thanks for such a detailed explanation ;)
Excellent!
@iamincorruptible the thing is that a number that is that 1/n approaches 0 slower than the exponent n pulls the expression up to a higher value. Dunno if you get what i mean.
Since d/dx[lnx] = 1/x was proven independent of d/dx[e^x] = e^x the proof is not circular.
Great video
Good stuff, Thanks!
@iamincorruptible sorry i made a mistake. he also made a small typo mistake its e^n = lim n-->∞ (1 + !/n)^n these is not equal to e
Um... Hi, I have a question and please don't insult my intel if the question is lucid.
In the begining you were useing a limit as dX ->0 (....).
Then you went to another lim but you still used (....). Shouldn't you include, "as long as X is not equal to 0." IDK I just learned about limits derivitives and logs so I'm probably just missing something.
He is not being defesnsive, he was just being humble..
Brilliant.
For the first proof, why can't you just say that limit of ((x+dx)/x)=x/x because dx approaches 0? then you get that the whole thing is 1*1/x which works much easier.
THANK YOU!!!
at 3:10 a substitution is made .
how did he decide that he should do that?
and why did he choose 1/n?
in lesson 28 he choose △x/x=n,but n->0,in this lesson n->oo,each of these can done this.
They can substitute because they're equivalent at the limit he is taking.
2:41 is incorrect. Pause it and you will see. Ln[(1+delta/x)^(1/delta)]. You didn't remember your log properties regarding exponents. Example, ln(x)² is not the same at as 2ln(x).
Yup. And I don't see how it's supposed to open up if done correctly or if I'm just too tired to do it. I solved it using infinite series representation of e^h (replaced delta(x) with h).
Limit(h->0) (e^(x+h) - e^x)/h
Take e^(x+h) apart...
=Limit(h->0) (e^x * e^h - e^x)/h
Take out the common factor e^x...
=Limit(h->0) (e^x (e^h - 1))/h
Replace e^h with it's infinite series representation...
=Limit(h->0) (e^x (1 + h + h^2/2! + h^3/3! + h^4/4! ... - 1))/h
Clean 1 - 1 inside the brackets...
=Limit(h->0) (e^x (h + h^2/2! + h^3/3! + h^4/4! ...))/h
Bring the 1/h inside the brackets...
=Limit(h->0) (e^x (1 + h/2! + h^2/3! + h^3/4! ...)
Let the h approach zero...
=e^x (1)
=e^x
He ends up reversing the same incorrect process so it ends up working out.
Isn't that just using the logarithm power rule: logb(x^y) = y ∙ logb(x)??
I may be wrong, but to me ln(x)^2 is the same as 2 * ln(x) -- (you can simply test with any value for x)
ln(x²) = 2*ln(x), but ln(x)² is not. It's very different thing to put the exponent inside the parentheses than outside. I.g. ln(e)=1. Now ln(e)² = 1² is very different thing than 2*ln(e) = 2*1 = ln(e^2).
Omnia in numeris can you help with my question at the top of the comments please
6:22 Still a problem with this proof here. How can we just switch the limit and the natural log functions? They both do separate things to their arguments, so how can we simply switch them around justifiably? In my mind, this is similar to taking |x^3| and making it (|x|)^3. Can someone please explain?
Kevin Berry simply, define the increment delta x as x divided by some arbitrary natural number n. Now saying that this increment tends to zero is equivalent to 1/n tending to zero, which in turn is equivalent to n tending to infinity. You can substitute the limit at delta x tending to zero with a limit at n tending to infinity. Then, of course, you have to substitute all the variables in the expression, whose limit you are trying to find. If you look for example at some numerical methods texts you will see, that indeed delta x is a spacing of the grid, which is defined by the domain divided by a number of sections of that domain. The more sections you divide the domain into, the more accuracy you get. The drawback in numerical computing is that by increasing the number of divisions your computational effort increases. Here you don't have to worry about it ;)
Hope this helps.
Kevin Berry The following is an intuitive response to your question of why the Limit and Log are interchangeable, as at approx 6.22 in the video.
If we first agree that the definition of the limit can be changed from "as n ==> infinity" to "as (1 + 1/n)^n ==> e". Then, in the interest of brevity, I will reword the expression "(1+1/n)^n" as "X", and rephrase the limit to be "as X ==> e"
In doing so, the expressions as written in the video would be:
(a) Lim X ==> e, of Ln [ X ]
(b) Ln [ Lim as X ==> e, of X ]
Evaluating each of these expressions at (or virtually at) the limit, they both end with the same result:
(a) = Ln [ e ]
(b) = Ln [ e ]
Kevin Berry The limit can "move through" the natural log function here because the limit of the input ((1 + 1/n)^n) exists and because the ln function is continuous at the value of that limit (which is e).
In general, if lim x->a g(x)=b and f(x) is continuous at b then lim x->a f(g(x)) = f(lim x->a g(x)) = f(b)
The neatural log has nothing to do with the limit? Anything that has no dependency on the variable in uestion can be factored out.
And, yes, you can: lim x->n (x^3) = (lim x->n (x))^3.
The cube is nothing to do with the variable, so you can take it out of the limit. Absolutely
general proof, then proving it for e or ln or whatever is immediate. However, I do not believe that ii is necessary
5:41 I don't fully understand why 1/x is independent. Isn't n defined using x?
thanks!
Thanks Sal
Well… almost. You have to prove that the limit exists first so that any sequence will have the same limit before you transition to n.
You can use the quotient rule to figured that out.
very good video.. not circular at all
If I understand correctly and I don’t think I do, in the last part Sal explains that the derivative of natural log of e to the x equals the derivative of e to the x... which makes no sense. Why does he use the first to prove the former?
thanks. this was clearly explained!
mistake @ 4:00
where you say "we take the limit as its denominator approaches ZERO, we're gonna make delta x go to zero" umm.. i think u mean "its denominator approaches infinity*" as this would make delta x go to zero. dont quote me though, i might be wrong since i suck at calculus :S
what software and tablet do you use?
Maths is alive.
THIS IS THE AWESOMEST VIDEO ON UA-cam!
Sal, I think we have a problem in your proof. At 5:40 you say 1/x has nothing to do with n, it's a constant term. I don't think so because you've defined n in terms of x. In fact n = 1/( X(DeltaX) ).
Now, we can use the fact that the limit of a product = the product of the limits but this doesn't give what we want.
Lim( 1/X) * Lim(Ln(e)) as n --> inf, for both limits
= Lim( 1/X) * 1, as n --> inf. However, 1/X = 1/(nDeltaX)
= Lim ( 1/(nDeltaX) ), as n --> inf which = 0 .
This approach appears to suggest that d/dx ( lnx ) = 0 which is not what we want.
Please let me know if I've made an error in my analysis and many thanks for your great work. - Brendan McCann
No, you can treat it as a constant term. Doesnt matter if you define n in terms of x, actually.
3:11 Wait, why can you just set that equal to 1/n, does that mean I can set it equal to anything I want, like 2, e, pi, doesn't make sense to me.
+Zack Walker (JapaneseLearning101) - he isn't setting it to equal anything, as your example of 2, e, pi suggests. He is merely making an algebraic substitution. He is yet to find the value of n - n is just some variable.
The actual "value" of n is x/delta(x) - so as x changes, so does n. Hence n is just some variable in terms of two other variables.
I hope this explains it :) feel free to contact me
Yes, you can pretty much substitute it with "anything", as long as it is in any way helpful to you in simplifying the formula. Some "anythings" will do the job, other "anythings" will make the formula more cumbersome.
But it's not just some careless substitution: when you substitute, you need to change anything in the formula that depends on that substitution.
Think of it like switching to a different coordinate system in which instead of the `x` axis you have a `w` axis with different scale and perhaps a different direction too. So in order for this to work, you need to convert all `x` coordinates into `w` coordinates in your formula, which may simplify it or may not, depending on how much your new coordinate system "lays along with the graph" and how much it uses its symmetry to help you.
Here's a simple example not involving calculus, just plain old algebra:
Let's say you have a function:
y = x²/(1-x)
and you want to graph it. You could substitute some values for `x` to see what `y`s will come out, but this is too much work, and you can miss some "kinks" of the graph.
If only it had something simpler, like `t`, in the denominator instead of that weird `1-x`... :q
So you have an idea: How about substituting `1 - x = t`?
But when you substitute, you have to substitute it everywhere the "old coordinate" (x) appears. So you solve your substitution for `x` to see how it depends on your "new coordinate" (t): `x = 1 - t`.
So you substitute `1 - t` for every `x` in your old expression:
y = x²/(1-x)
y = (1-t)²/t
and after expanding the numerator:
y = (1² - 2·1·t + t²)/t
and simplifying:
y = (1 - 2·t + t²)/t
you can split the fraction to get:
y = 1/t - 2·t/t + t²/t
y = 1/t - 2 + t
or in a somewhat more understandable form:
y = (t - 2) + 1/t
and this is easy to draw and reason about it! :)
It's just a linear function `y = t - 2`, from which a standard hyperbola `y = 1/t` has been subtracted :)
So the line `y = t - 2` is its horizontal asymptote (well, now it will be diagonal), and it has another (vertical) asymptote at `t = 0`, as usual.
You can graph this easily on your grid paper :)
The graph would look exactly the same as your original function `y = x²/(1-x)`, but it uses a different coordinate for the horizontal axis: it uses `t` instead of `x`. So you need to switch back to your original coordinate now. But this is easy :) We know that `x = 1 - t`, so we can simply write another number for `x` under each tick mark for `t` calculating it from that formula. E.g. the vertical asymptote in the "new coordinate" (t) is at `t = 0`, but when you substitute this for `t` in your formula for `x`, you'll see that `x = 1 - t = 1 - 0 = 1`, so the vertical asymptote is at `x = 1` in the "old coordinate" (x). Also the direction of the axis is flipped, so you need to account for that.
Fun fact: There is a clever substitution which can make this graph even more symmetric, reducing it simply to `u = 1/w`. Can you find a coordinate system (or a substitution of variables) for which this happens? :>
@@scitwi9164 brother thank you. You are amazing your explanation is very good. You made my day.
No, because 2, e, pi, are constants. You can set it eual to 1/n, because n is a new variable that he created just for that.
You can substitute any VARIABLE you want, but setting it eual to some random value is obviously wrong Bruh
How did you originally come up with this equation.
Does anyone knows what tools is he using to write in this video?
...
It looks so much neater now!
If you wrote this derivation on paper there would be far less distraction and the viewer could even read it with no trouble.
People should not gain immunity from critism simply because they are held in high regard.
What does delta x mean? I understand that its change in x, but how would you allgebraically rewrite it? you can't write x_f - x_i. Is it just a dummy variable?
Get two points of a graphic: a point X and a point a. Δx we define as the difference between x and a
@ 6:40
Can the limit "definition" of e be proved without assuming d/dx ln x = 1/x?
Sorry to nitpick, but you can't use the Taylor-series method to generate the series, because in doing so you need the derivative of e^x, which what you're trying to prove. You have to come up with the series some other way that doesn't need derivation of e^x to generate it.
There are other ways of generating the power series which don't use derivatives explicitly. Instead they use polynomial long division and the binomial formula. It just "happens" that the coefficients in the power series agree with the subsequent derivatives.
Im confused. e is literally defined as this. This is e. when you say "e", this is what you are talking about. Why would you need a proof? This is like asking for the proof that pi = circumference / diameter. There is none, thats just what the constant is.
Well if that isn't explained clearly I don't know what is... just for the record the last video explained it perfectly clearly as I remember... 2 seperate ones tho i think...
This is exhausting.
I got it eventually.
The only problem I see is that you are assuming that the limit definition of e is a good one, i.e. that the limit really exists.
Sal, I am not arguing the proof on derivative of ln x, but help me here please. When I graph the function of ln x, I only see a graph that is rising. X is undefined in the negative numbers. So, how can 1/x be the derivative when the slope of ln x is always positive?
We could have just used the derivative of inverse function formula to find the derivative of e^x after we found the derivative for ln(x).
After further consideration, delta x can be a negative number. We do not usually use negative values for delta x when proving derivatives, as it complicates the calculations immensely. So, point in fact, the expression delta / x can be any number except 0, and 1 / n can be any number except 0. We do not have to impose any other restrictions of the domains of the two expressions and they are therefore still equivalent.
how is delta x/x = 1/n?
He assumed a new variable n such that ∆x/x=1/n
He just made a new variable
So u used 2 ways of solving d/dx (ln e^x) to solve d/dx e^x?
Yes. He proved d/dx lnx = 1/x
d/dx ln (e^x) = d/dx x × ln(e) = d/dx x × 1 = 1
Bcs d/dx x = 1 and 1 × 1 = 1
Now there is the chain rule
[f(g(x))]' = f'(g(x)) × g'(x)
d/dx ln(e^x) = (1 / e^x) × d/dx e^x
As d/dx ln(e^x) = d/dx ln(e^x)
This is logical I know
(1/e^x) × d/dx e^x = 1 | × e^x
(e^x/e^x) × d / dx e^x = e^x
1 × d/dx e^x = e^x
d/dx e^x = e^x
@J0EH3AD Yes, but that's assumed before the proof. What you're saying is like saying we have to go all the way back to the beginning and prove all of the axioms needed for this proof to work.
I agree that you need to define what the definition of "log" and "ln" is, but I disagree with what you are saying about needing to make an initial general proof. There's no reason as to why you need the general proof in the first place. However, I can see where you're coming from because if you do have the
I don't fully get the substitution of Δx/x = 1/n... I mean, was it because it looked like we were going to end up with the definition of e? My math level is quite low, so I don't get how we could get from Δx/x to 1/n if it wasn't because we already knew the answer.
You can substitute anything into anything. If you want to substitute dx/x = 1/n, you can if you want ---- If you end up at the correct answer, you did it right. So what if you KNEW you were going to get it?
Thats silly. You can do anything you want if its logically coniistent
I think you should consider one sided limits when you substitute to get limit given by Bernoulli
What makes the substitution legal?
??? Substitution is always legal as long as its logically consistent. What would stop you from doing it?