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 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 TH-cam tutor! Guess who is my go-to TH-cam tutor now🤭❤️
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
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
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
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?
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 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
![เบิ้ล ปทุมราช - อ้ายกินดอง น้องลาบงัว ft. สแน็ก อัจฉรีย์ [OFFICIAL MV]](http://i.ytimg.com/vi/-OK--KaSPcI/mqdefault.jpg)
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.
thank you so much, i've been struggling with differentiation and i have a test tomorrow for it ♥️
The best maths channel I found till date, I'm so interested in learning all these
Thank you, you explain everything extremely well and make math very enjoyable!
This is brilliantly done!
I'm subscribing this channel, because you deserve for it
You are literally awesome ❤
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 TH-cam tutor! Guess who is my go-to TH-cam tutor now🤭❤️
Excellent professor!
Bruh, you have such a pleasant voice.
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!
Great video, it helped me so much!
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!
Very good as usual 👍🏻
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
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.)
hey man your voice is so cool.....i was kinnda like dancing ....
Very good. Thanks 🙏
"Why am I not multiplying? Because I don't want to"😂
This is like me also sometimes when I teach my friends and classmates
Thank you
Gotta integrate this now, just to check.
Your 'Nice' word is very nice❤
I am writing this question before watching the video : i guess you are going to use natural log (since it more convenient to derivative ) ?????
Thanks to find something to have good time
I am from ethiopia i always see your vidio
nice work
In general differentiation decrease the equation but in this case not applied
you have such a beautiful hat. from where did you get that?
@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 🤔
how do you solve x^x^x = 3 using Lambert W function
Nice equation
Great hat.
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).
Can you
make full concept clearing video of differentiation
After this I did it with 5 x’s above the original x. It barely fit in a single line 😂
That was fun! Where do you find these crazy problems? Lol
Lol. Usually, someone sends me a problem like this.
Differentiate x ^^ x (^^ means superpower like ³3 means 3^3^3)
Have you done videos on factorials? Would love to learn it from you.
I think I'll do factorials soon
Hey, just saw your video about tetration, it would be x with 4 in left top corner
GGEZW 😎
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
I like you but this is all above my head. I still gave you a Like.
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
would it be easier to say y = x*y, then ln y = y lnx, and then differentiate from there?
Wow
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)=?
At what point does differentiation turn into tetration or vice versa
are tetrations derivatable?
if y=x x=1=y
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...😂
Comment for the algorithm
How seductive
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
Sir please solve my indefinite integration:- integral of(32-x^5)^(1/5) dx
So many eggs 😂
So, after need to find extremum of this :D
Its a 11 th grade question😅
2:44 shouldn't 3^3^3 be 3^9 instead of 3^27
this equation is a mosnter
Had to click
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
: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.
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.
algo
easy! dy/da = 0!
Who are here from cbse board😅
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!