I have graduated 3 months ago, at the start of the calculus class 2 years ago i hated calculus but here i am, loving calculus and enjoying every second of your awesome videos.
How do we know ln(x) is a logarithm? I once had a professor “define” ln(x) as a function whose derivative is 1/x. He then proceeded to show the ln(x) is indeed a logarithm, and it has the base e. I’d like to see this again. It was very inspiring, but I have forgotten how it was done.
Just watched it again as there were a few things I wasn’t sure of. I really liked it when he explained one trick to use was because the natural log is a continuous function, and the limit of a continuous function is a continuous function of the limit, you can move the limit inside the parentheses to simplify things. Cool stuff.
Another way : exp(ln(x)) = x Derivative of both sides : ln(x)' * exp(ln(x)) = 1 Replace exp(ln(x)) by x and divide the whole equation by it : ln(x)' = 1/x
Great video man! I feel like you've made me so much smarter; this time I was actually able to see ahead a little bit, that the argument of ln would end up being e^1/x (this was around when you brought the derivative into the u world)
This is just the application of the first principle definition of the derivative. You know how to do limits and should be well versed in algebra manipulation. It's not a big leap to do this problem. This is the sort of exercise a student should do away from school.
I've always been told that the derivative rule for f'(x) of ln(x) has always been 1/x but I've never understood how that was proven. Thank you for the explanation.
There's usually one of these proofs for it somewhere in the textbook. Since the teacher probably sees proving them as reinventing the wheel, and not necessary to understand the subject, they probably just skip showing why these derivative rules work.
You are my new favorite high school math teacher. In my AP calculus class, we were never taught how to derive this. Only taught to memorize that d/dx ln(x) = 1/x
I have always wanted a more detailed explanation of this result. This is the best I’ve seen on the subject. Considering things like Euler’s identity and the quantum wave equation and other uses of the exponential function, it seems to me it’s the most useful of all the special functions.
blackpenredpen could you solve the non elementary integral of x^x. You did the (easier) derivative so please do the difficult integral or let Payem do it
Ahsoka Tano How is he suppose to solve it if it is non-elementary? Do you understand what solving an integral is? And do you understand what non-elementary is?
Zach Cate if an integral is non-elementary, by definition, that means you cannot solve it. It will be defined by a special function. For example, the fresnel integral
Okay for anyone that is confused this is a matter of pedanticism. "Solving an integral" technically refers to definite integrals. The original comment probably just wants the indefinite integral and is using the word "solve" to mean "to do" as in ordinary english. Again, all a matter of mathematical vs normal language.
12:05 when I saw this, I was like... OMG I just realized what the hell I've been watching for the past 12 minutes... I was more intrigued by what he was able to do in terms of modifying the formulae, but then noticed he brought it down to 1/x, I love this guy.
Wonderful videos. It is a long time ago that I studied complex variables, differential and integral calculus and algebra. So it is great fun watching this guy do with ease what most of us struggled with when learning the basic elements of these important mathematical techniques. I can generally follow him right to the end once I see where he headed. The mathematical manipulations seem to be firmly rooted in my brain. The algorithms he applies for problem solving are much less so.
The most impressive thing about these videos is not the math, it's his ability to write with 2 or 3 markers in the same hand while holding them all at the same time. And that his writing is still legible while he does it. I can barely read my own handwriting when i write with just 1 pencil
To differentiate ln(x) I use this trick: 1 = 1 1 = d/dx x 1 = d/dx [e^(ln(x))] 1 = e^(ln(x)) * d/dx(ln(x)) d/dx(ln(x)) = 1 / [e^(ln(x))] d/dx(ln(x)) = 1/x This also works for all inverse functions like arcsin(x), arcos(x) & arctan(x).
I have a fourth proof: If we differentiate e^ln x, instead of resulting in x, we use the chen lu, where u = ln(x). That results in e^(ln x) * du/dx. However, if we use the power rule, it results in 1. Therefore, x * du/dx = 1. We solve for du/dx = 1/x.
This reminds me of when I was a COBOL programmer, we would have discussions about whether you could have positive zero and negative zero. This was because the sign of a number was contained in the units digit. So, when comparing numbers it was important to take this into account. But I would say to my colleagues that zero was neither positive nor negative, it was separate from other numbers.
Hey , I recently started reading Thomas calculus and found that lnx was actually defined as definite integral of 1/t from 1 to x. So i think a proof is not needed stating the definition is enough. Anyway hats off to the great content
How it is defined, really depends on who you ask. Historically, natural log was discovered before the number e, and it was defined as this integral. But in modern times, we usually define it as the inverse of e^x, and define e^x as the special case of the exponential where it is its own derivative. The modern definition is much more useful, to learn what logs are for the first time. These two definitions are internally consistent, but you need to start with one to prove the other.
such a long proof but very well thought out. I was definitely doing a shorter proof for my test (luckily, not sure if I could survive writing this for my test.. lol). Dloga(x)=1/x*ln(a) D(log(e^x))=1/xlne=1/xloge(e)=1/x*1=1/x but of course mine is already making assumptions (that derivative of loga(x)=1/x*ln(a)) instead of figuring it out with definition of e. Great work, definitely I learned something.
The proof of the derivative of e^x uses the exact same ln lim n->0 so it’s better to derive with limits and not other operations that rely on said limits because it can end up begging the question
6:30 Hum. Ah. But when we say limit(h->0) that implies in any direction right ? As we work with real numbers we can have h < 0 or h > 0 and both directional limits (or whatever the proper name for that is) must give the same result for the derivative to exist. But when we substitute for h->infinity, we only check the side h>0, right ? So shouldn't we also substitute u=-1/t and verify that we have the same result ? Or else prove that the derivative must exists in which case only one side is enough to get the value.
Very well explained as usual, only one thing: I ask my students to avoid “canceling ln with e”; I want them to say that log of a power with the same base is the exponent.
@@stewartzayat7526 Since that is the definition of a log, I prefer my students to repeat it as frequently as possible: it's the best way to capture it completely. It's part of my campaign against voodoo maths: you know, strange things like quantities that change their sign while flying over the equal sign and all that. An easy way to forget that there are equivalence principles behind that, and no flying stuff. Also: linguistics is a central part of our learning processes: our first impact with new stuff is via a language, so it makes lot of difference, imho.
Logarithm is the inverse function of exponentiation and viceversa, that's why log_a(a^x) = x and a^(log_a(x)) = x. I would prefer to say this rather than what you have written here (that includes the equivalent of what you have said for explaining that a^(log_a(x)) = x).
@@diegocabrales and of course you would be right, but my 36 years of experience teaching maths make me say that your students would benefit less from that explanation.
I think it could have been made a bit more clear at 3:29 that the 1/h exponent is supposed to be evaluated for (1+h/x) before the log is taken. (But I still got the point.)
I just wanted to say, that for some reason, LOGb(X)=ln(X)/ln(b) has always been my favorite relationship in "Logarithmic Functions" and THANKS for the bonus at the end!!!
"when we are taking the limit of a continious function, it's the continious function of the limit" ?? Can somebody explain how this works and why we can simply move the limit around like this?
Back when I learned this we defined the logarithm function in terms of the integral 1/x dx, then proved that this function had the properties expected of a logarithm.
But bprp, there is a problem! lim x-> 0 [(1+u)^(1/u)]^(1/x) isn't exactly a continious function. The limit is equal to e, then the function would be e^(1/x), whose domain doesn't include 0. So technically it is just continious only for the values of x that you are interested in, as in the original function 1/x, the domain also excludes 0, so there is no good in trying to find out the derivative in that point. Just a technical detail but it's important isn't it?
The domain does not need to include 0 for it to be continuous in the neighborhood of x = 0. Remember, we are evaluating the limit of the value of the function as x APPROACHES 0, NOT the value of the function AT x = 0. The function is continuous for all nonzero values of x. Therefore, the limit and the function commute and "can be interchanged".
I once tried to find a good thermodynamic definition of 'heat'. Every thermo text I have, and there are more than a handful, defines heat its own way. The most honest says that heat is what we measure with a calorimeter (Sort of like defining temperature as the thing we measure with a thermometer.). And what is a calorimeter? Why it's an instrument we use to measure heat of course. In the same spirit we can prove that the ln of x= the integral of 1/x. ( ln x = exp(integral(1/x dx) ). To do that, we must show that exp(integral(1/x dx) ) = x. Working out the first few terms of the Taylor expansion for the integral exponent will show that this equality is true. (I verified this with Mathematica. ) Since the definition of ln x is that it is the inverse function of the exponential: ln( exp(x) ) = x, we can substitute what we just proved for x in the definition: ln (exp (integral(1/x dx) )= integral(1/x dx). From the fundamental theorem of calculus (another inverse operation) we know that (ln (x) )' = d/dx (integral (1/x dx) = 1/x. QED I read Mat Hunt's answer and I think he and I are thinking the same thing. I just wanted to spell it out in full, except for that Taylor's series bit.
Was it proved when this was discovered that the limit of a continuous function is said function of the limit? I know in the beginning, calculus was not super rigorous. And if not... how did they figure out the derivative of ln(x)?
The prood might not be but that's actually quite an intuitive statement. And if I find it intuitive, someone like napier or leibniz definitely also did...
In my old Ellis and Gullick calc text, it defines ln x as the integral (anti-derivative) of the function 1/x (because you can't integrate 1/x with the power rule, ln x is "invented" to be the solution). That means basically, by definition, d ln x/dx = 1/x. From that everything else flows, including the definition of e that you get from the limit. It's really beautiful, and it all follows from that definition. So your approach of substituting e into the limit, from the Ellis and Gullick approach, is just tautology. If you define ln x like this, and e to have this relationship with ln x, then you get back to the definition of ln x.
But another approach is to first define the exponential function as f'=f and f(0)=1. ln is then the reciprocal function... So, it depends on where you start from.
@@xnqmap Yep. But ultimately, it turns into a matter of "proving" the definition. I agree it's a fun exercise, especially in the way he's done it, but I don't see it as really mathematically meaningful.
This approach is an elegant application of the Fundamental Theorem of Calculus (also used in Thomas). As you say, everything flows from this-including a definition for exponentiation that extends from rational exponents to real exponents. These proofs for the derivative of ln(x) should start off with a definition for the natural logarithm. When you use ln(e^x)=x, you need to explain what e^x means for irrational x. The beauty of using FTOC to define ln is that ln(x) is defined for all real x>0. Once it is shown that ln is a one-to-one function, exp is defined as the inverse of ln. Now exp(x) is defined for all real x, exp(ln(x))=x, and ln(exp(x))=x. Along the way, the IVT is use to define e as the number for which ln(e)=1, and the familiar properties of ln are derived. Once it is shown that e^r=exp(r) for all rational r, the definition of e^x is extended to all real x by defining e^x=exp(x). From this the derivative of e^x can be found by differentiating both sides of ln(e^x)=x. Exponentiation for other bases can now be extended from rational exponents to real exponents: b^x = e^(x*ln(b)). Obviously I’ve omitted details, but it is beautiful application of tools from calculus.
Idea - Solution to limit 1/x where x->0. Assume all derivatives have a defined initial value. dy_dx = 1/x. As x goes to zero there must exist a valid y(0) value. Just choose a limit such that it fits slope. Call it the perpendicular limit for 1/x. Or it is a freeze limit (black holes) when expressed in an equation. Depending on how fast you approach 0 by the 0.9 value you freeze the y values. import numpy as np # 1/x is a random number generator? when expressed in equation 0 = yx**2 - x x = np.random.rand(1000) y = np.random.rand(1000) while True: err = x**2*y - x x = 0.9*x + 0.001*err y = y + 0.001*err print(np.mean(err**2))
f(x)=lin(x) let linx=y e^linx=e^y x=e^y dx=e^ydy=xdy dy=dx/x divide the two sides by dx we get : dy/dx =1/x but y=linx so d(linx)/dx=1/x Is it is ok ? Thank you very much
if you draw an arrow so that a constant becomes an exponent of a logarithm, don't point the arrow outside of the logarithm, since that just proves the commutativaty of multiplication
lim n->inf (1)^n=1 lim n->inf (1+Ln(k)/n)^n = k That's what I was saying, sorry if it wasn't clear. It's just that he says "one to the infinity power" and that's one, every time, but "one plus an infinitesimal bit", that's another story
Ignacio Except you are wrong. Saying that he was supposed to say "one plus one infinitesimal bit" is ridiculously ignorant. There is no such a thing as "an infinitesimal bit". There are no infinitesimal numbers. The real numbers contain no infinitesimals. 1^♾ is universally well-known and well-defined as an indeterminate form of limits. This is what BPRP was referring to.
Just because 1 raised to any power is 1, infinity is not a number. 1 raised to infinity power is indeterminate. We have to use l'Hopital's rule to do more work.
You can do this in two ways. You can use the integral definition of log(c) and use the fundamental theorem of calculus or you can note that log(x) is the inverse function of exp(x), and just use the expression for differentiating the inverse function.
You forgot one. The inverse function theorem: if f(x) and g(x) are inverse functions of each other then g'(x) = 1/f'(g(x)). Therefore, d ln(x)/dx = 1/e^[ ln(x)] = 1/x. ◼
Here's a mind blowing question: Is Log_2_(3) bigger/smaller/equal than/to Log_3_(5)? That's a so called "coffin" problem, in other words that's one of the questions that russian teachers were use to ask to a russian jew in order to prevent him/her to enter the math faculty (no kidding!)
got this pretty quicky, hope this is correct: log_2(3)=x and log_3(5)=y thus 2^x=3 and 3^y=5 setting these equal we get: 2^x=3/5*3^y now take the log2 again: x=log_2(3/5)+y*log_2(3) we have a function f(y) = x set up new function f*(y)=f(y)-y note that if this is positive, the x value is greater than the y value and vice versa f*(y)=log_2(3/5)+y*(log_2(3)-1) =log_2(3/5)+y*log_2(3/2) the first part is positive since the inside is greater than 1 the second part is positive because y is positive and the inside of the log is also greater than 1. Since 0
Thanks i like you so much, maths is magic ♥️. I try to find this by focus on the definition of a function wich is derivating if this limite was not infinity and i encounter a lot of problème by not knowing this definition of e and also "the limit of a continuous function is the function of the limit. Thanks a lot ♥️ Sorry i dont speak english very well but i learned more and more each days
I have graduated 3 months ago, at the start of the calculus class 2 years ago i hated calculus but here i am, loving calculus and enjoying every second of your awesome videos.
Alex Ramyeon thank you!!!!!!
Honestly speaking calculus is more fun than GTA and all other video games.
@@blackpenredpen my way docs.google.com/document/d/e/2PACX-1vQ0SB1cs5gR0S17zmIhfFuQqmhGsw8_jn_QoL1n6AjI26wsu2bOPIxzCrw1D0SK-fCca1FUR-xAQ-gI/pub
@@tintinfan007 you're probably speaking about some random mobile gta spinoff
I love the way he teaches
You know its about to get real when he starts using the blue pen (-:
yup, that's right!
Somewhat relevant xkcd: xkcd.com/849/
He has also used a purple pen a few times.
😂😂😂😂😂😂😂😂
Multiple Colour Pen
8:25 my brain to me after a test
Underrated
Lol
Lol
lol
lol
8:11
Talking to your GF
under rated comment
Perfect
Lol
ayyy
Very good lmao
10:43 I really love the satisfaction I get when my mind snaps and know how the demostration continues before the video. Great video!
Clicked at around 7:50 for me, so satisfying
was about to cmment that same thing! It's such a great feeling
How do we know ln(x) is a logarithm? I once had a professor “define” ln(x) as a function whose derivative is 1/x. He then proceeded to show the ln(x) is indeed a logarithm, and it has the base e. I’d like to see this again. It was very inspiring, but I have forgotten how it was done.
Are you asking how to prove the properties based on that definition? If so, I have a video here th-cam.com/video/4D-M5qB_l6k/w-d-xo.html
@hobo doc id be happy to receive those pages on scrubster@gmx.de
I was taught that ln(x) is by definition, log base e of x. The term ln itself means the natural logarithm.
@@lewisbotterill4948 look
@@cellcomsanggau424 ?
Another proof using parametric equation:
x = e^t
y = t
dx/dt = e^t
dy/dt = 1
(dy/dt)/(dx/dt) = dy/dx = 1/e^t = 1/x
Wonder proof buddy! Three different proofs: Limits, implicit differentiation and parametric equations.
Wonderful proof buddy! Three different proofs: Using parametric equations, limits and implicit differentiation.
Just brilliant. Congratulations, I’m going to teach this one tomorrow
nicely done
For any log , 1/x .ln a, a its the base of the log , If a = e, the derivative is 1/x
I did a general proof
Just watched it again as there were a few things I wasn’t sure of. I really liked it when he explained one trick to use was because the natural log is a continuous function, and the limit of a continuous function is a continuous function of the limit, you can move the limit inside the parentheses to simplify things. Cool stuff.
I loved calculus in college and now i am 71 and watch these videos to fill the gaps in my understanding and keep dementia away
Another way :
exp(ln(x)) = x
Derivative of both sides :
ln(x)' * exp(ln(x)) = 1
Replace exp(ln(x)) by x and divide the whole equation by it :
ln(x)' = 1/x
wow!
it's literally in the video
Great video man! I feel like you've made me so much smarter; this time I was actually able to see ahead a little bit, that the argument of ln would end up being e^1/x (this was around when you brought the derivative into the u world)
Its a shame we dont get teached this stuff in school but are just supposed to remember f'(x)=1/x of F(x)=ln(x)
I remember my school teaching us a variation of the 2nd method, namely
y = ln x
=> e^y = x
Therefore dx/dy = e^y
dy/dx = 1/(dx/dy) = 1/e^y = 1/x
FF same situation(((
Of f(x)=ln(x). Capital f often implies integration. Especially because integrated function is defined by it F'(x)=f(x) then you are ok.
they told us:
e^ln(x)=x
diferentiating gives:
e^ln(x)*d/dx(ln(x))=1
d/dx(ln(x))=1/e^ln(x)=1/x
This is just the application of the first principle definition of the derivative. You know how to do limits and should be well versed in algebra manipulation. It's not a big leap to do this problem. This is the sort of exercise a student should do away from school.
I really enjoyed your last few videos, and I am glad you're back to uploading more videos containing your explanations
I've always been told that the derivative rule for f'(x) of ln(x) has always been 1/x but I've never understood how that was proven. Thank you for the explanation.
There's usually one of these proofs for it somewhere in the textbook. Since the teacher probably sees proving them as reinventing the wheel, and not necessary to understand the subject, they probably just skip showing why these derivative rules work.
You are my new favorite high school math teacher. In my AP calculus class, we were never taught how to derive this. Only taught to memorize that d/dx ln(x) = 1/x
I have always wanted a more detailed explanation of this result. This is the best I’ve seen on the subject. Considering things like Euler’s identity and the quantum wave equation and other uses of the exponential function, it seems to me it’s the most useful of all the special functions.
dayuumm now that's impressive, finding the derivative of ln(x) using the standard definition of a derivative
Elegant proofs for the derivative of ln(x). I like the intelligent and creative ways you used to develop and establish your proofs. Thanks.
The second lim going into the continues function was so eye opening and satisfying
blackpenredpen could you solve the non elementary integral of x^x. You did the (easier) derivative so please do the difficult integral or let Payem do it
Ahsoka Tano How is he suppose to solve it if it is non-elementary? Do you understand what solving an integral is? And do you understand what non-elementary is?
Angel Mendez-Rivera ima be real with you that made no sense
Zach Cate if an integral is non-elementary, by definition, that means you cannot solve it. It will be defined by a special function. For example, the fresnel integral
Okay for anyone that is confused this is a matter of pedanticism. "Solving an integral" technically refers to definite integrals. The original comment probably just wants the indefinite integral and is using the word "solve" to mean "to do" as in ordinary english.
Again, all a matter of mathematical vs normal language.
Smart moves and thank you. To avoid confusion in approach 1 instead of twice using u I will use U and then w.
12:05 when I saw this, I was like... OMG I just realized what the hell I've been watching for the past 12 minutes... I was more intrigued by what he was able to do in terms of modifying the formulae, but then noticed he brought it down to 1/x, I love this guy.
This was so intertwining I was guessing what to do and when he showed what to do it made sense feels amazing
How easily he changes markers is amazing to watch
This mad lad really just used the limit definition. Can we get this guy a medal?
Wonderful videos. It is a long time ago that I studied complex variables, differential and integral calculus and algebra. So it is great fun watching this guy do with ease what most of us struggled with when learning the basic elements of these important mathematical techniques. I can generally follow him right to the end once I see where he headed. The mathematical manipulations seem to be firmly rooted in my brain. The algorithms he applies for problem solving are much less so.
Dear friend, you are not only genius but you a great guru (teacher). My regards - Sudarshan🙏
The most impressive thing about these videos is not the math, it's his ability to write with 2 or 3 markers in the same hand while holding them all at the same time. And that his writing is still legible while he does it. I can barely read my own handwriting when i write with just 1 pencil
And hold a microphone in the other hand. Might as well start juggling at that point.
Finally, since the basketball secret has been revealed I can find some sleep!
Marian P. Gajda in fact, in was in that previous video as well, just allllll the way at the end.
I live close to where you recorded that basketball video! I was pleasantly surprised when I saw that
Derek Anderson
Are you serious???
How did u even recognize that place!!!
@@blackpenredpen COINCIDENCE?
@@blackpenredpen it is possible, but very unlikely
To differentiate ln(x) I use this trick:
1 = 1
1 = d/dx x
1 = d/dx [e^(ln(x))]
1 = e^(ln(x)) * d/dx(ln(x))
d/dx(ln(x)) = 1 / [e^(ln(x))]
d/dx(ln(x)) = 1/x
This also works for all inverse functions like arcsin(x), arcos(x) & arctan(x).
Thank YOU so much for sharing your beautiful smile and passion!! It makes me so much more excited to learn and genuinely happy :))
I have a fourth proof: If we differentiate e^ln x, instead of resulting in x, we use the chen lu, where u = ln(x). That results in e^(ln x) * du/dx. However, if we use the power rule, it results in 1. Therefore, x * du/dx = 1. We solve for du/dx = 1/x.
This reminds me of when I was a COBOL programmer, we would have discussions about whether you could have positive zero and negative zero. This was because the sign of a number was contained in the units digit. So, when comparing numbers it was important to take this into account. But I would say to my colleagues that zero was neither positive nor negative, it was separate from other numbers.
Hey , I recently started reading Thomas calculus and found that lnx was actually defined as definite integral of 1/t from 1 to x. So i think a proof is not needed stating the definition is enough. Anyway hats off to the great content
How it is defined, really depends on who you ask. Historically, natural log was discovered before the number e, and it was defined as this integral. But in modern times, we usually define it as the inverse of e^x, and define e^x as the special case of the exponential where it is its own derivative. The modern definition is much more useful, to learn what logs are for the first time.
These two definitions are internally consistent, but you need to start with one to prove the other.
Oh my god you are incredible! I learned a thing or two because of you! Loved it ❤️
جميل ورائع ومميز ما يقوم به هذا الشاب،،، فعلا عقليه فذه،، 🌹🌹🌹
I love these type of people on TH-cam
such a long proof but very well thought out. I was definitely doing a shorter proof for my test (luckily, not sure if I could survive writing this for my test.. lol).
Dloga(x)=1/x*ln(a)
D(log(e^x))=1/xlne=1/xloge(e)=1/x*1=1/x
but of course mine is already making assumptions (that derivative of loga(x)=1/x*ln(a)) instead of figuring it out with definition of e. Great work, definitely I learned something.
Wait, doesn't the power 1/h belong inside the brackets?
A most elegant solution to d/dx Ln(x)...I didn't imagine it would take 3 substitutions.
16:43 Yeah! It even works for e: ln(e)=1 so you get back the 1/x :)
Just found your channel. Thanks for creating this content and keep up the good work.
Dang, when he finally pulled out the e term, I got super excited. Nice job!
y=lnx
e^y=x
(e^y)’=x’
(e^y)*y’=1
y’=1/(e^y)
(lnx)’=1/x (substitution of the given parts)
The proof of the derivative of e^x uses the exact same ln lim n->0 so it’s better to derive with limits and not other operations that rely on said limits because it can end up begging the question
@@AhmedKhan-qk3xi idk what u are saying but ig ur right
@@AhmedKhan-qk3xi I think it is just way shorter
6:30 Hum. Ah. But when we say limit(h->0) that implies in any direction right ?
As we work with real numbers we can have h < 0 or h > 0 and both directional limits (or whatever the proper name for that is) must give the same result for the derivative to exist.
But when we substitute for h->infinity, we only check the side h>0, right ?
So shouldn't we also substitute u=-1/t and verify that we have the same result ?
Or else prove that the derivative must exists in which case only one side is enough to get the value.
I wonder if there is a numerical analysis class he teaches. This guy is a good teacher.
Well done guy!
You sort it out!
Keep it up!
Go always deep n in every detail to enlightening.
Again you've done it!
Which was proven first, the derivative of e^x or that of ln x?
So clear explanation, Greatest Math Teacher in the WORLD, Thank You Sir!.
Very elegant description of this important derivative
That moment when a channel about two colors of pen PULLS OUT A THIRD COLOR
THANK YOU!!! All other videos I found only explained how you used the derivative not actually showing proof on why it’s 1/x
Very well explained as usual, only one thing: I ask my students to avoid “canceling ln with e”; I want them to say that log of a power with the same base is the exponent.
I think that's just a linguistic thing. As long as your students know _why_ it works, I don't think it matters what they say
@@stewartzayat7526 Since that is the definition of a log, I prefer my students to repeat it as frequently as possible: it's the best way to capture it completely. It's part of my campaign against voodoo maths: you know, strange things like quantities that change their sign while flying over the equal sign and all that. An easy way to forget that there are equivalence principles behind that, and no flying stuff.
Also: linguistics is a central part of our learning processes: our first impact with new stuff is via a language, so it makes lot of difference, imho.
Logarithm is the inverse function of exponentiation and viceversa, that's why log_a(a^x) = x and a^(log_a(x)) = x. I would prefer to say this rather than what you have written here (that includes the equivalent of what you have said for explaining that a^(log_a(x)) = x).
@@diegocabrales and of course you would be right, but my 36 years of experience teaching maths make me say that your students would benefit less from that explanation.
I think it could have been made a bit more clear at 3:29 that the 1/h exponent is supposed to be evaluated for (1+h/x) before the log is taken. (But I still got the point.)
I just wanted to say, that for some reason, LOGb(X)=ln(X)/ln(b) has always been my favorite relationship in "Logarithmic Functions" and THANKS for the bonus at the end!!!
X = b ^ logb(x), then take logd of both sides and bring the exponent down. Then solve for logb(x).
Wasn't this video already posted once?? I remember seeing it
Yes but it got deleted for some reason.
Pranav B yes. But I forgot to consider cases in that previous vid. And also this is a better pf
I think it's a mix of 2 videos
Oh btw, I did add a bonus part at the end lol
th-cam.com/video/XYYdl9A__mo/w-d-xo.html
I like all the math problem and solutions 👍👍
"when we are taking the limit of a continious function, it's the continious function of the limit" ?? Can somebody explain how this works and why we can simply move the limit around like this?
Back when I learned this we defined the logarithm function in terms of the integral 1/x dx, then proved that this function had the properties expected of a logarithm.
CIERTO BRO YO TAMBIEN LO APRENDI AL REVES,QUE EL LOGARITMO SE DEFINE JUSTO POR LA INTEGRAL DE 1/X
Proving this was actually a question on one of our calc exams
Love the second proof of lnx's derivative
But bprp, there is a problem! lim x-> 0 [(1+u)^(1/u)]^(1/x) isn't exactly a continious function. The limit is equal to e, then the function would be e^(1/x), whose domain doesn't include 0. So technically it is just continious only for the values of x that you are interested in, as in the original function 1/x, the domain also excludes 0, so there is no good in trying to find out the derivative in that point. Just a technical detail but it's important isn't it?
Kiritsu it's limit u->0, not x... Remember, when u do the substitution, u change the limit too
The domain does not need to include 0 for it to be continuous in the neighborhood of x = 0. Remember, we are evaluating the limit of the value of the function as x APPROACHES 0, NOT the value of the function AT x = 0. The function is continuous for all nonzero values of x. Therefore, the limit and the function commute and "can be interchanged".
I like to use the inverse function rule I came up with: d/dx f⁻¹(x) = 1/f'(f⁻¹(x)) which outputs 1/x.
Awesome explanation
i think to explain why you can shoot the limit into a continous function you would need into to analysis.
I once tried to find a good thermodynamic definition of 'heat'. Every thermo text I have, and there are more than a handful, defines heat its own way. The most honest says that heat is what we measure with a calorimeter (Sort of like defining temperature as the thing we measure with a thermometer.). And what is a calorimeter? Why it's an instrument we use to measure heat of course.
In the same spirit we can prove that the ln of x= the integral of 1/x. ( ln x = exp(integral(1/x dx) ). To do that, we must show that exp(integral(1/x dx) ) = x.
Working out the first few terms of the Taylor expansion for the integral exponent will show that this equality is true. (I verified this with Mathematica. ) Since the definition of ln x is that it is the inverse function of the exponential: ln( exp(x) ) = x,
we can substitute what we just proved for x in the definition: ln (exp (integral(1/x dx) )= integral(1/x dx). From the fundamental theorem of calculus (another inverse operation) we know that
(ln (x) )' = d/dx (integral (1/x dx) = 1/x. QED
I read Mat Hunt's answer and I think he and I are thinking the same thing. I just wanted to spell it out in full, except for that Taylor's series bit.
the first definition are amazing
I appreciate the derivative 1/x now
There is a rule for the derivative of the inverse funtion; we could use that
Oh wow that makes it really simple
This rule is exactly what you get if you apply his 2nd method to a general function and only in the end you plug in the specific function log(x)
I love u-sub when doing algebra and calculus. SO useful.
thanks brother for clear information ❤🤗🤗😎😎
Lim x->0[ln(1+x)/x] = 1
A standard limit can also be used
Was it proved when this was discovered that the limit of a continuous function is said function of the limit? I know in the beginning, calculus was not super rigorous. And if not... how did they figure out the derivative of ln(x)?
The prood might not be but that's actually quite an intuitive statement. And if I find it intuitive, someone like napier or leibniz definitely also did...
they did it by definition, and the definition already existed before
@@me_hanics you need the fact the function is continuous for that
I don't like that he says ln and e "cancel out." He's always so thorough and meticulous with everything he does, and then he does that.
they do tho?
The closer he gets to 1/x, the bigger his smile becomes. :D
In my old Ellis and Gullick calc text, it defines ln x as the integral (anti-derivative) of the function 1/x (because you can't integrate 1/x with the power rule, ln x is "invented" to be the solution). That means basically, by definition, d ln x/dx = 1/x. From that everything else flows, including the definition of e that you get from the limit. It's really beautiful, and it all follows from that definition. So your approach of substituting e into the limit, from the Ellis and Gullick approach, is just tautology. If you define ln x like this, and e to have this relationship with ln x, then you get back to the definition of ln x.
But another approach is to first define the exponential function as f'=f and f(0)=1. ln is then the reciprocal function... So, it depends on where you start from.
@@xnqmap Yep. But ultimately, it turns into a matter of "proving" the definition.
I agree it's a fun exercise, especially in the way he's done it, but I don't see it as really mathematically meaningful.
This approach is an elegant application of the Fundamental Theorem of Calculus (also used in Thomas). As you say, everything flows from this-including a definition for exponentiation that extends from rational exponents to real exponents. These proofs for the derivative of ln(x) should start off with a definition for the natural logarithm. When you use ln(e^x)=x, you need to explain what e^x means for irrational x. The beauty of using FTOC to define ln is that ln(x) is defined for all real x>0. Once it is shown that ln is a one-to-one function, exp is defined as the inverse of ln. Now exp(x) is defined for all real x, exp(ln(x))=x, and ln(exp(x))=x. Along the way, the IVT is use to define e as the number for which ln(e)=1, and the familiar properties of ln are derived. Once it is shown that e^r=exp(r) for all rational r, the definition of e^x is extended to all real x by defining e^x=exp(x). From this the derivative of e^x can be found by differentiating both sides of ln(e^x)=x. Exponentiation for other bases can now be extended from rational exponents to real exponents: b^x = e^(x*ln(b)). Obviously I’ve omitted details, but it is beautiful application of tools from calculus.
1/ln(b) can also be written as logb(e) so you don't have it under a fraction 👍
(the derivative of a general logb(x) function would be (1/x) * logb(e))
Nice piece of Mathematics.
Idea - Solution to limit 1/x where x->0. Assume all derivatives have a defined initial value. dy_dx = 1/x. As x goes to zero there must exist a valid y(0) value. Just choose a limit such that it fits slope. Call it the perpendicular limit for 1/x. Or it is a freeze limit (black holes) when expressed in an equation. Depending on how fast you approach 0 by the 0.9 value you freeze the y values.
import numpy as np
# 1/x is a random number generator? when expressed in equation 0 = yx**2 - x
x = np.random.rand(1000)
y = np.random.rand(1000)
while True:
err = x**2*y - x
x = 0.9*x + 0.001*err
y = y + 0.001*err
print(np.mean(err**2))
f(x)=lin(x)
let linx=y
e^linx=e^y
x=e^y
dx=e^ydy=xdy
dy=dx/x
divide the two sides by dx we get :
dy/dx =1/x but y=linx
so d(linx)/dx=1/x
Is it is ok ?
Thank you very much
if you draw an arrow so that a constant becomes an exponent of a logarithm, don't point the arrow outside of the logarithm, since that just proves the commutativaty of multiplication
1 to the infinity power is 1, 1 plus an infinitesimal value, all to the infinity power, could be anything.
Except it converges to e
lim n->inf (1)^n=1
lim n->inf (1+Ln(k)/n)^n = k
That's what I was saying, sorry if it wasn't clear. It's just that he says "one to the infinity power" and that's one, every time, but "one plus an infinitesimal bit", that's another story
Ignacio but since when does 1/0=infinity? am i being stupid or something but i swear it isn’t defined
Ignacio Except you are wrong. Saying that he was supposed to say "one plus one infinitesimal bit" is ridiculously ignorant. There is no such a thing as "an infinitesimal bit". There are no infinitesimal numbers. The real numbers contain no infinitesimals. 1^♾ is universally well-known and well-defined as an indeterminate form of limits. This is what BPRP was referring to.
Just because 1 raised to any power is 1, infinity is not a number. 1 raised to infinity power is indeterminate. We have to use l'Hopital's rule to do more work.
You can do this in two ways. You can use the integral definition of log(c) and use the fundamental theorem of calculus or you can note that log(x) is the inverse function of exp(x), and just use the expression for differentiating the inverse function.
Could you do integral of 1/x from -1 to 1? It looks physically 0 but ended up undefined, why?
youre my inspiration
Excellent proof process! 中国人来TH-cam看中国人用英语讲课了
Let’s see how good your math skills are when you put down your ball of power
You forgot one. The inverse function theorem: if f(x) and g(x) are inverse functions of each other then g'(x) = 1/f'(g(x)). Therefore, d ln(x)/dx = 1/e^[ ln(x)] = 1/x. ◼
well done professor
There’s a very easy one: f(x) = ln x, e^f(x) = x => f’(x) *e^f(x) = 1 => f’(x) * x = 1
I watch your videos for inspiration and help as I just started year 7
What about this : y=lnx . x=e^y. dx/dy = e^y. Take reciprocal on both sides. dy/dx=1/e^y = 1/x
Absolutely beautiful. Great explanations! Thank you.
YOU ARE PHENOMENON!!!!!
Thanks bro. Awesomeness
Here's a mind blowing question:
Is Log_2_(3) bigger/smaller/equal than/to Log_3_(5)?
That's a so called "coffin" problem, in other words that's one of the questions that russian teachers were use to ask to a russian jew in order to prevent him/her to enter the math faculty (no kidding!)
Zonnymaka wow!!! Super interesting problem. I will have to think hard on it
How is it going? :)
Tip: don't waste you time (as i did) comparing the 2 logs...you'll never get out alive of that!
Zonnymaka my initial try was to set up a function but no luck. I will have more time tmr or so to think about it.
got this pretty quicky, hope this is correct:
log_2(3)=x and log_3(5)=y
thus 2^x=3 and 3^y=5
setting these equal we get:
2^x=3/5*3^y
now take the log2 again:
x=log_2(3/5)+y*log_2(3)
we have a function f(y) = x
set up new function f*(y)=f(y)-y
note that if this is positive, the x value is greater than the y value and vice versa
f*(y)=log_2(3/5)+y*(log_2(3)-1)
=log_2(3/5)+y*log_2(3/2)
the first part is positive since the inside is greater than 1
the second part is positive because y is positive and the inside of the log is also greater than 1.
Since 0
just looked it up on wolframAlpha, seems to be correct
I love you, plain and simple.
Thanks i like you so much, maths is magic ♥️. I try to find this by focus on the definition of a function wich is derivating if this limite was not infinity and i encounter a lot of problème by not knowing this definition of e and also "the limit of a continuous function is the function of the limit. Thanks a lot ♥️
Sorry i dont speak english very well but i learned more and more each days
great teacher
Thank you so much sir.