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

7 ай бұрын

Thank you!

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

6 ай бұрын

Maybe I'll step it up soon 🤣🤣🤣

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 KZread tutor! Guess who is my go-to KZread tutor now🤭❤️

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!

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

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

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

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

Excellent professor!

This is brilliantly done!

Bruh, you have such a pleasant voice.

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

You are literally awesome ❤

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

Beautiful . I love your enthusiasm . I just subscribed .

@PrimeNewtons

5 ай бұрын

Thanks for subbing!

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

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

@PrimeNewtons

7 ай бұрын

Happy to hear that!

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

Very good. Thanks 🙏

Very good as usual 👍🏻

Great video, it helped me so much!

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.)

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

2 ай бұрын

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).

nice work

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

Your 'Nice' word is very nice❤

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

I am from ethiopia i always see your vidio

Thank you

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

@PrimeNewtons

7 ай бұрын

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

Thanks to find something to have good time

Can you make full concept clearing video of differentiation

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

@PrimeNewtons

5 ай бұрын

I think I'll do factorials soon

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

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

At what point does differentiation turn into tetration or vice versa

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

@Prime Newtons Your thinking is as organized as your writing

@PrimeNewtons

Ай бұрын

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

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

Great hat.

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

5 ай бұрын

I already have that

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

@first-namelast-name

7 ай бұрын

GGEZW 😎

are tetrations derivatable?

Nice equation

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

Gotta integrate this now, just to check.

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)=?

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

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

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....)

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

How seductive

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

Comment for the algorithm

So many eggs 😂

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

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

:D

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...😂

So, after need to find extremum of this :D

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

2 ай бұрын

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.

Had to click

if y=x x=1=y

Its a 11 th grade question😅

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

@bhaskarporey3768

7 ай бұрын

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

@niom9446

7 ай бұрын

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

@miscostsmusic1880

7 ай бұрын

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

this equation is a mosnter

algo

Who are here from cbse board😅

easy! dy/da = 0!

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

@PrimeNewtons

5 ай бұрын

😀😀

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