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 :)
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!
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.
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.)
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
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
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
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?
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
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
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.
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....)
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)=?
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 RUclips tutor! Guess who is my go-to RUclips tutor now🤭❤️
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
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 :)
Maybe I'll step it up soon 🤣🤣🤣
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.
I can’t believe one of my favorite channels is a math channel.
I always enjoy your approach to math problems
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.
The best maths channel I found till date, I'm so interested in learning all these
thank you so much, i've been struggling with differentiation and i have a test tomorrow for it ♥️
This is brilliantly done!
Thank you, you explain everything extremely well and make math very enjoyable!
Bruh, you have such a pleasant voice.
great teacher. really awesome
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.)
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
Beautiful . I love your enthusiasm . I just subscribed .
Thanks for subbing!
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
Your channel is amazing. I'm from Brazil and you helped me a lot. thanks
Happy to hear that!
I'm subscribing this channel, because you deserve for it
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
You are literally awesome ❤
Excellent professor!
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?
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).
I am writing this question before watching the video : i guess you are going to use natural log (since it more convenient to derivative ) ?????
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
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
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
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.
I already have that
That was fun! Where do you find these crazy problems? Lol
Lol. Usually, someone sends me a problem like this.
i just had to LAUGH at 12.20 :D MY GOD.. you need a bigger board !!!! :D
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....)
Awesome video!
Can you
make full concept clearing video of differentiation
hey man your voice is so cool.....i was kinnda like dancing ....
"Why am I not multiplying? Because I don't want to"😂
This is like me also sometimes when I teach my friends and classmates
How do you erase your board so throughly?
In general differentiation decrease the equation but in this case not applied
Very good. Thanks 🙏
how do you solve x^x^x = 3 using Lambert W function
you have such a beautiful hat. from where did you get that?
Hey, just saw your video about tetration, it would be x with 4 in left top corner
GGEZW 😎
Your 'Nice' word is very nice❤
Thank you
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)=?
would it be easier to say y = x*y, then ln y = y lnx, and then differentiate from there?
@Prime Newtons Your thinking is as organized as your writing
I hope that's a compliment because I'm still trying to organize my thinking 🤔
At what point does differentiation turn into tetration or vice versa
Differentiate x ^^ x (^^ means superpower like ³3 means 3^3^3)
I am from ethiopia i always see your vidio
Great video, it helped me so much!
Have you done videos on factorials? Would love to learn it from you.
I think I'll do factorials soon
are tetrations derivatable?
Gotta integrate this now, just to check.
Thanks to find something to have good time
I like you but this is all above my head. I still gave you a Like.
Very good as usual 👍🏻
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.
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.
Great hat.
nice work
Who are here from cbse board😅
Me
Nice equation
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 RUclips tutor! Guess who is my go-to RUclips tutor now🤭❤️
Sir please solve my indefinite integration:- integral of(32-x^5)^(1/5) dx
After this I did it with 5 x’s above the original x. It barely fit in a single line 😂
Wow
if y=x x=1=y
the step where you did x^(-1)-x^x=x^(-1-x) is wrong
He didn't....that's x^(-1)/x^x which is x^(-1-x) and it is correct.
@@bhaskarporey3768oh right I didn’t see more. But he did write x^(-1)-x^x=x^(-1-x) though
@@bhaskarporey3768he did write the division sign, but the two dots were barely visible lmao
Comment for the algorithm
How seductive
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 many eggs 😂
2:44 shouldn't 3^3^3 be 3^9 instead of 3^27
No it shouldn't. 3^(3^3)=3^(3*3*3)=3^(9*3)=3^27
Had to click
So, after need to find extremum of this :D
this equation is a mosnter
Its a 11 th grade question😅
:D
algo
easy! dy/da = 0!
No; not t, use u sub 2. LOL!!!
😀😀
is thet ⁴x??? I just watched your video 'bout tetration (8 months ago), I really enjoyed it!