We will show that the derivative of ln(x), namely the natural logarithmic function, is 1/x. We will use the definition of the derivative and also implicit differentiation. Subscribe to @bprpcalculusbasics for new calculus tutorials
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.
Are you asking how to prove the properties based on that definition? If so, I have a video here ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-4D-M5qB_l6k.html
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 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.
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.
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)
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
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
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.
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.
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 can also use the formula for inverse derivatives. This is how I did it: Let g(x) = the inverse of f(x) g’(x) = 1/(f’(g(x)) Let f(x) = e^x Therefore f’(x) = e^x & g(x) = lnx g’(x) = 1/(e^lnx) g’(x) = 1/x Therefore the derivative of lnx is 1/x. To prove the formula I used, you can let g(x) = inverse of f(x) So, x = f(g(x)) Differentiating both sides, you get: 1 = f’(g(x))*g’(x) g’(x) = 1/(f’(g(x))
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.
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.
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
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.
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!!!
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.
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.
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.
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
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.
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).
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.
I like your channel, its content and overall disposition (subscribed long back). I would like to mentiom that a lot of times you seem to show some very simple or basic algebraic manupilation in great detail as if you are showing that to some beginner who is not very bright. I request you to be consistent in knowledge density (idk how to describe thos) throughout the vdo and spread your time evenly on the topics and ponder over stuff which really calls for it. In this vdo, time spent on definition of e with t to u is probably 6 times than necessary, my personal feeling. Lastly, as I like this channel, I complained. If I was indifferent, I wouldn't have cared.
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.
At the third step you could have broken the fraction into 1+h/x and then divided and multiplied the base h with x instead of using it is as power then you would have got 1/x lim h->0 ln(1+h/x) divided by h/x and then by using the limit you could’ve simply got 1/x
