Differentiate x^x^x^x

In this video, I showed how to differentiate x^x^x^x

[@akiya9216]

The questions you do are normally quite easy for me and would be pretty boring to go through, but watching you do them is really enjoyable. Very very very fun, you are good at teaching :)

[@PrimeNewtons]

Maybe I'll step it up soon 🤣🤣🤣

[@ADWYETYATRIPATHYBEI]

It's been 6 years since I opened a Math book... and this vid just brought back memories of school and college days!!! You deserve way more subscribers

[@PrimeNewtons]

Thank you!

[@et427gamer9]

I am in algebra two so this all goes over my head but I still enjoy it significantly! Cant wait to get to higher level math like this. I can tell you love the subject and that love will transfer to your students. Keep it up!

[@cherryisripe3165]

Everything seems so simple with your explanations and pedagogy. Thank you so much.

[@nanasung2701]

thank you so much, i've been struggling with differentiation and i have a test tomorrow for it ♥️

[@sunil.shegaonkar1]

I have problem with the last term at 16:16, it does not account for X^x, this term is not in y' but factored out. This is a multiplier of the greater bracket. Factoring increases stack of 2 to 3, but greater bracket has x^x which has No term in y'. Rest is wonderful, I had no idea how to find derivative of 3 stacked function

[@ThenSaidHeUntoThem]

This is brilliantly done!

[@saiprasadpadhy6832]

The best maths channel I found till date, I'm so interested in learning all these

[@rishichava355]

Thank you, you explain everything extremely well and make math very enjoyable!

[@HJ28_398]

I saw you were doing the same thing over and over so I wanted to see some properties of the pattern. I hope I write this correctly, but the derivative of x tetrated to A is sum{Q} {ln(x)^Q * prod{W} {x^^(A+2-W}} / x Where Q is between and equal to 0 and A (the order of the tetration) W is between and equal to 1 and Q+2 (↓↓and also formatting because I don't really know how to express the big functions in text↓↓) sum{Q} {function of Q} is the sum of all the functions with varying Q values, Q is used only for that one set of operations, as with prod{W} {function of W} is the product of the outputs of the function with varying W values. An example using the variables from the video: d/dx (x^x^x^x) = y(u(t(x(1(ln(x)^3)+ln(x)^2)+ln(x)^1)+ln(x)^0)/x a very nice and algorithmic process! Maths is so interesting.

[@dreanq]

Always pleasant to see you resolve thesse questions

[@christianlibertarian5488]

I can’t believe one of my favorite channels is a math channel.

[@pk2712]

Beautiful . I love your enthusiasm . I just subscribed .

[@PrimeNewtons]

Thanks for subbing!

[@dannap8831]

great teacher. really awesome

[@justpassingbyy]

Bruh, you have such a pleasant voice.

[@gokubaianassauro4533]

Your channel is amazing. I'm from Brazil and you helped me a lot. thanks

[@PrimeNewtons]

Happy to hear that!

[@treborg777]

I always enjoy your approach to math problems

[@ananthianandan553]

I'm subscribing this channel, because you deserve for it

[@arbenkellici3808]

You are amazing proffesor You might be an excellent Hollywood actor as well I dont know how many subscribers you have, but beleive me, you deserve a lot more

[@lukaskamin755]

So cool, I'm from Ukraine and we learn maths with slightly different approaches, though, of course, math is the same, no doubt. I definitely enjoy your videos, with such a hillarious attitude, and perfect clear language (being non-native English speaker, I can totally understand everything

[@mansourativo9658]

"Why am I not multiplying? Because I don't want to"😂 This is like me also sometimes when I teach my friends and classmates

[@ananthianandan553]

You are literally awesome ❤

[@kathieharine5982]

Excellent professor!

[@Calcprof]

If y = x^x^x^x^......, then y = x^y, and differentiate implicitly. and solve. This gives y' in terms of x, y, and log x. You have to be a little careful of boundary values, but I think you can handle these. BTW: y can be easily expressed in terms of the Lambert W function y = - W(-log[x])/log[x]. Since W'[x] = W[x]/(x (1 + W[x])), this can be used to calculate y' and express it entirely in terms of x, log[x] and W[x]. (You have to be a little careful of which branch of the solutions of z = x e^x you have, but all of this can be sorted out.)

[@amritpatel3794]

To make the operation clear, use brackets, like (x)^(x)^(x)^(x) is different than ((((x)^x)^x)x) which is equal to (x)^3x

[@roseonmars-06]

hey man your voice is so cool.....i was kinnda like dancing ....

[@Deadpool-rw1pk]

I am writing this question before watching the video : i guess you are going to use natural log (since it more convenient to derivative ) ?????

[@wavingbuddy3535]

i tried this myself and got the same answer but i wrote mine as: x^( x^x^x + x^x + x) * ( ln(x)^3 + ln(x)^2 + ln(x)/x + 1/(x^(x+1))) very satisfying video as usual, love your charisma when you're going through the steps

[@dennisroote9145]

GREAT CHANNEL!! - now lets see the integral..

[@CRnk153]

Hey, just saw your video about tetration, it would be x with 4 in left top corner

[@first-namelast-name]

GGEZW 😎

[@josephparrish7625]

That was fun! Where do you find these crazy problems? Lol

[@PrimeNewtons]

Lol. Usually, someone sends me a problem like this.

[@yetismacker7053]

Awesome video!

[@littlegrass320]

how do you solve x^x^x = 3 using Lambert W function

[@Notking444.]

Can you make full concept clearing video of differentiation

[@souverain1er]

@Prime Newtons Your thinking is as organized as your writing

[@PrimeNewtons]

I hope that's a compliment because I'm still trying to organize my thinking 🤔

[@PaiboonKityadhiguna]

Thank you very much for opening my eyes, Professor! so much appreciate in your way to explain and solve that question quite easily. Well if I could ask you about what is the differentiate of X^X^X^2x, what is it should be then?

[@jumpman8282]

Hi! The easiest way (in my opinion) to tackle this type of problem is to start by differentiating 𝑦 = 𝑥^(2𝑥). Taking the natural log of both sides, we get ln 𝑦 = 2𝑥 ln 𝑥. Implicit differentiation (chain rule on the left-hand side and product rule on the right) then gives us 1 ∕ 𝑦⋅𝑑𝑦 ∕𝑑𝑥 = 2⋅ln(𝑥) + 2x⋅1 ∕ 𝑥 = 2(ln(𝑥) + 1) (Note that we don't have to worry about 𝑥 = 0 in the denominator since 𝑥^(2𝑥) is not defined for 𝑥 = 0 anyway, so 𝑥 ∕ 𝑥 = 1) ⇒ 𝑑𝑦 ∕ 𝑑𝑥 = 𝑦⋅2(ln(𝑥) + 1) = 𝑥^(2𝑥)⋅2(ln(𝑥) + 1). So, 𝑑 ∕ 𝑑𝑥[𝑥^(2𝑥)] = 𝑥^(2𝑥)⋅2(ln(𝑥) + 1). - - - Now we can differentiate 𝑦 = 𝑥^𝑥^(2𝑥), using the exact same method. Take the natural log of both sides: ln 𝑦 = 𝑥^(2𝑥) ln 𝑥. Implicit differentiation: 1 ∕ 𝑦⋅𝑑𝑦 ∕ 𝑑𝑥 = 𝑑 ∕ 𝑑𝑥[𝑥^(2𝑥)]⋅ln(𝑥) + 𝑥^(2𝑥)⋅1 ∕ 𝑥 = 𝑥^(2𝑥)⋅2(ln(𝑥) + 1)⋅ln(𝑥) + 𝑥^(2𝑥)⋅1 ∕ 𝑥 = 𝑥^(2𝑥)⋅(2(ln²(𝑥) + ln 𝑥) + 1 ∕ 𝑥). ⇒ 𝑑𝑦 ∕ 𝑑𝑥 = 𝑦⋅𝑥^(2𝑥)⋅(2(ln²(𝑥) + ln 𝑥) + 1 ∕ 𝑥) = 𝑥^𝑥^(2𝑥)⋅𝑥^(2𝑥)⋅(2(ln²(𝑥) + ln 𝑥) + 1 ∕ 𝑥). So, 𝑑 ∕ 𝑑𝑥[𝑥^𝑥^(2𝑥)] = 𝑥^𝑥^(2𝑥)⋅𝑥^(2𝑥)⋅(2(ln²(𝑥) + ln 𝑥) + 1 ∕ 𝑥). - - - Finally, we can differentiate 𝑦 = 𝑥^𝑥^𝑥^(2𝑥). ln 𝑦 = x^𝑥^(2𝑥) ln 𝑥. 1 ∕ 𝑦⋅𝑑𝑦 ∕ 𝑑𝑥 = 𝑥^𝑥^(2𝑥)⋅𝑥^(2𝑥)⋅(2(ln²(𝑥) + ln 𝑥) + 1 ∕ 𝑥)⋅ln(𝑥) + 𝑥^𝑥^(2𝑥)⋅1 ∕ 𝑥 = 𝑥^𝑥^(2𝑥)⋅(𝑥^(2𝑥)⋅(2(ln³(𝑥) + ln²𝑥) + ln(𝑥) ∕ 𝑥) + 1 ∕ 𝑥) ⇒ 𝑑𝑦 ∕ 𝑑𝑥 = 𝑥^𝑥^𝑥^(2𝑥)⋅𝑥^𝑥^(2𝑥)⋅(𝑥^(2𝑥)⋅(2(ln³(𝑥) + ln²𝑥) + ln(𝑥) ∕ 𝑥) + 1 ∕ 𝑥). So, in the end we have 𝑑 ∕ 𝑑𝑥[𝑥^𝑥^𝑥^(2𝑥)] = 𝑥^𝑥^𝑥^(2𝑥)⋅𝑥^𝑥^(2𝑥)⋅(𝑥^(2𝑥)⋅(2(ln³(𝑥) + ln²𝑥) + ln(𝑥) ∕ 𝑥) + 1 ∕ 𝑥).

[@maeveoconnor821]

Great video, it helped me so much!

[@luca_151]

would it be easier to say y = x*y, then ln y = y lnx, and then differentiate from there?

[@surendrakverma555]

Very good. Thanks 🙏

[@childrenofkoris]

i just had to LAUGH at 12.20 :D MY GOD.. you need a bigger board !!!! :D

[@gopikayala6551]

In general differentiation decrease the equation but in this case not applied

[@СулейманВеликолепный-ц4й]

Help me please🙏 I've a arcsin(1/3) and I need to find that, but I need an exact value. I mean I needn't a number like 0,3472.....I need an expression. For example: Arcsin(1/4(√5-1))=π/10 Arcsin(1/2)=π/6 Arcsin(1/3)=??? Arcsin(1/3)=?

[@devcoachingclasses1]

Your 'Nice' word is very nice❤

[@treborg777]

How do you erase your board so throughly?

[@gghelis]

Gotta integrate this now, just to check.

[@Midkipper]

Have you done videos on factorials? Would love to learn it from you.

[@PrimeNewtons]

I think I'll do factorials soon

[@sam123-h5r]

I am from ethiopia i always see your vidio

[@jamiepianist]

First video I watched in 2025 :D

[@OmChouhan-ps6sk]

you have such a beautiful hat. from where did you get that?

[@superiorjr154]

At what point does differentiation turn into tetration or vice versa

[@hiitstan]

Differentiate x ^^ x (^^ means superpower like ³3 means 3^3^3)

[@aaditya8283]

Sir can you plz bring a video pf proper explanation of why e^x differentiation and integration is e^x always bcz your explanation are easy to understand😊😊 love you from India.

[@PrimeNewtons]

I already have that

[@weisanpang7173]

The last term in y' is x^x^x • 1/x, but you changed it to (1/x)/x^x, shouldnt the denomitor be x^x^x ?

[@MazinBugshan]

Very good as usual 👍🏻

[@DSN.001]

are tetrations derivatable?

[@nengimotejaphet2565]

I usually feel shy watching your videos ’cause you’re so flirty😌😹 but this one! I can’t even lie you did this explanation better than organic chemistry tutor, and he was my go-to KZbin tutor! Guess who is my go-to KZbin tutor now🤭❤️

[@kavvame]

Thanks to find something to have good time

[@flowingafterglow629]

You what would have been a really cool way to end that video would be to evaluate the derivative at some point (not x = 0 or 1, though). Something like, 2. This is the slope of the line x^x^x^x at x = 2..... It's a beautiful expression, but it's fun to remember a use of the derivative.... (maybe in the next video set y' = 0 and find critical points....)

[@jensberling2341]

Is there a mistake? You said 2^2^2=2^4. Then you said; 3^3^3=3^27. That must be a misprint. Pls, responds, Dr. Newton.

[@jumpman8282]

A power tower is evaluated top down. However, some calculators interpret 3^3^3 as (3^3)^3 instead of 3^(3^3), so you've got to be careful.

[@darcash1738]

After this I did it with 5 x’s above the original x. It barely fit in a single line 😂

[@陈兆堂]

Good ❤

[@mehmetdurna3115]

Nice equation

[@AS-ix3qd]

nice work

[@dellaih_studies]

Who are here from cbse board😅

[@IIT2033IAS]

Me

[@jamesburrelljr.8561]

I like you but this is all above my head. I still gave you a Like.

[@pritamsur1926]

Sir please solve my indefinite integration:- integral of(32-x^5)^(1/5) dx

[@roddos]

Great hat.

[@leoniii1247]

Woah this is way harder than I thought... I thought the answer was x^x^x^x * x^3 * ln(x) and I got no idea why the video is that long...😂

[@meow11119]

Thank you

[@niom9446]

the step where you did x^(-1)-x^x=x^(-1-x) is wrong

[@bhaskarporey3768]

He didn't....that's x^(-1)/x^x which is x^(-1-x) and it is correct.

[@niom9446]

⁠@@bhaskarporey3768oh right I didn’t see more. But he did write x^(-1)-x^x=x^(-1-x) though

[@miscostsmusic1880]

@@bhaskarporey3768he did write the division sign, but the two dots were barely visible lmao

[@theupson]

i know of no finesse for the actual labor of the problem, but the whole construction is more legible, maybe, if you start with y = s^t^u^v s=t=u=v=x and use multivariate chain rule. if that's out of bounds, switch the first three "x" for e^logx. y=exp(logx * exp (logx * exp (x*logx))) and the disassembly via chain rule and the product rule subtasks flows pretty naturally

[@lirich0]

Comment for the algorithm

[@luggis7574]

So many eggs 😂

[@FollowJesusCnfistofNazareth]

How seductive

[@Alisdead]

So, after need to find extremum of this :D

[@annxu8219]

if y=x x=1=y

[@stew880]

2:44 shouldn't 3^3^3 be 3^9 instead of 3^27

[@mrdeath1803]

No it shouldn't. 3^(3^3)=3^(3*3*3)=3^(9*3)=3^27

[@segunfolorunso5255]

Wow

[@dellaih_studies]

Its a 11 th grade question😅

[@prateek1.9]

this equation is a mosnter

[@mikedubovs1574]

Had to click

[@beaverbuoy3011]

:D

[@lec_hd]

algo

[@ZAPT-sg5ws]

[@owoLight]

easy! dy/da = 0!

[@christopherguerra7236]

No; not t, use u sub 2. LOL!!!

[@PrimeNewtons]

😀😀

[@pabs356]

is thet ⁴x??? I just watched your video 'bout tetration (8 months ago), I really enjoyed it!